问:
Given linear operator $A: R^3 \to R^3$, $\forall x \in R^3: (Ax, x)=0$, prove that A is skew-symmetric operator.
Given linear operator $A: R^3 \to R^3$, $\forall x \in R^3: (Ax, x)=0$, prove that A is skew-symmetric operator.
I came up with a solution like: if $(Ax, x) = 0 \implies (x, A^{T}x)=(A^{T}x, x)=0 \implies (Ax, x) + (A^{T}x, x)= 0 = ((A+A^{T})x,x)=0 \ \forall x$ then $A+A^{T}=0$, but I'm not sure that last implication is correct.

答:
You're going to need to make more direct use of $(A x, x) = 0$. Specifically, consider that the $(i,j)^{\text{th}}$ element of $A$ (call it $a_{i, j}$) can be found by $$a_{i, j} = (A e_j, e_i)$$

We can conclude that $A$ is skew symmetric if we can show $a_{i, j} = - a_{j, i}$ for all $i, j$. To take advantage of the hypothesis, let's try using $x = e_i + e_j$:
$$
0 = (Ax, x) = (A(e_i + e_j), e_i + e_j) = (Ae_i, e_i) + (Ae_i, e_j) + (Ae_j, e_i) + (Ae_j, e_j)
$$

Of course, we already know that $(Ae_i, e_i) = (Ae_j, e_j) = 0$ by hypothesis. Thus we have
$$
(Ae_i, e_j) + (Ae_j, e_i) = a_{j, i} + a_{i, j} = 0
$$

which is what we wanted to show.


问:
Variational calculus with pointwise constraint involving derivative
I am looking for the Euler-Langrange equations for the following problem:

Find the extremal points of the functional
$$ \mathcal{F}(u)=\int_\Omega F(x,u(x),D(u)(x)) dx $$
under the the pointwise constraint $G(D(u))=0$.
Here, $u:\Omega\subset\mathbb{R}^n\to\mathbb{R}^m$ and $F:(\Omega,\mathbb{R}^m,\mathbb{R}^{m\times n})\to\mathbb{R}$. Furthermore $D(u)$ denotes the Jacobian of $u$.

Motivation:
I am looking for a trajectory $u(t,p)=(x(t,p),y(t,p))$ under the constraint $|u_p|=1$.

Known related results: I know that for pointwise constraints of the form $G(u)=0$, the above can be achieved by introducing Langrange multipliers and simply finding the extremal points of $F(x,u(x),D(u)(x))+\Lambda(x)\cdot G(u(x))$, however I am unsure if the results can be extended to the case involving the derivatives of $u$.

I would appreciate your help or references to e.g. a textbook. Thank you very much in advance. :-)

答:
Observe that after a Taylor expansion, and choosing $v(x)\in C_c^{\infty}(\Omega)$,
$$\lim_{\varepsilon\to0}\frac{1}{\varepsilon}\left(\int_{\Omega}\Lambda(x)G(D(u(x)+\varepsilon v(x))dx\int_{\Omega}\Lambda(x)G(D(u(x))dx\right)
=-\int _{\Omega}D(\lambda(x)u(x))G'(D(u(x)))v(x)dx
$$
means
$$\frac{\delta\mathcal{F}}{\delta u}=\mathrm{Usual\ stuff}-D(\Lambda(x)u(x))G'(D(u(x)).
$$

ContentLicense: CC BY-SA 4.0


问:
Distribute stones into buckets with limits
You have $s$ different stones and $b$ different buckets. How many ways $w(s, b)$ can you distribute the stones into the buckets if no bucket is allowed to contain more than $L$ stones?

Clarification: The stones are all different from each other. So, for example, if you have 3 stones, 2 buckets, and at most 2 stones per bucket, there are 6 ways to distribute the stones: (A, BC), (AB, C), (AC, B), (B, AC), (BC, A), (C, AB).

I noticed that you can express the answer for $b$ buckets in terms of the answers for $b - 1$ buckets:

$$
w(s, b) = \sum_{i=0}^{\min(s, L)} {s \choose i} w(s - i, b - 1)
$$

Also,
$$
w(s, 1) = \left\{ \begin{array}{rl}
1 & \mbox{if }s <= L\\
0 & \mbox{otherwise}
\end{array} \right.
$$

Based on those observations I came up with this Python function:

```<code>def countWays(numStones, numBuckets, L):
    # Precompute C(n, r) table
    C = [[1]]
    for n in xrange(numStones):
        C.append([1] + [C[-1][r] + C[-1][r+1] for r in xrange(n)] + [1])
    # Loop through number of buckets
    ways = [1] * (L + 1) + [0] * (numStones - L)
    for b in xrange(2, numBuckets + 1):
        ways = [sum([C[s][i] * ways[s - i] for i in xrange(min(s, L) + 1)])
                for s in xrange(numStones + 1)]
    return ways[numStones]
</code>```

Some sample output:

```<code>>>> countWays(1, 2, 1)
2
>>> countWays(3, 2, 2)
6
>>> countWays(3, 3, 2)
24
>>> countWays(10, 10, 5)
9985309740L
>>> countWays(100, 100, 5)
94759885461565034681687026066885761311439165668901259822785622795797870660611943632316455279091491854154551818337317753294675330799396303953809575897498333285450197379975598951694414085442292940800000L
</code>```

Is there a simpler solution, or closed form? Some basic principle of combinatorics I'm missing?

答:
[svenkatr posted much the same thing while I was typing this up, but I think I've gone one step farther]

A standard way to do this is to see that it's the same as the number of solutions to $$x_1+x_2+\dots+x_b=s,\qquad 0\le x_i\le L$$ Then note that this is the coefficient of $x^s$ in $$(1+x+x^2+\dots+x^L)^b$$ Rewrite as $$(1-x^{L+1})^b(1-x)^{-b}$$ Use the binomial Theorem to expand both terms, and pick out the coefficient of $x^s$.

ContentLicense: CC BY-SA 3.0


问:
Summation with factorial terms (involving Laguerre polynomials)
As part of an exercise including gamma functions and Laguerre polynomials, I need to show that for a Laguerre polynomial $L_n(x)$,
$$\int\limits_0^\infty L_n(x)x^ke^{-x}dx = 0  \textrm{ with } n \in \mathbb{N} \textrm{ and }k \in \{0,1,\dots,n-1\}$$
I've managed to narrow it down to proving that
$$\sum\limits_{i=0}^n\frac{(-1)^{n-i}(k+i)!}{(i!)^2(n-i)!}  = 0\textrm{ with } n \in \mathbb{N} \textrm{ and }k \in \{0,1,\dots,n-1\}$$
But this is where I'm stuck. I used Maple and WolframAlpha to turn this sum into
$$\frac{k!(-1)^n\Gamma(n-k)}{\Gamma(-k)\Gamma(n+1)^2}$$
Which is $0$ since $\Gamma(-k)$ would be infinity, due to the Gamma function having simple poles in the negative integers. However, neither Maple or Wolfram can tell me how I convert the sum into this fraction of gamma functions. Am I missing something obvious or are there any tricks to get there?</p>

答:
With the definition you have given,
\begin{align}\int_0^{\infty}L_n(x)x^ke^{-x}dx&=\frac{(-1)^n n!}{2\pi i}\oint_{\Sigma}\frac{\Gamma(t-n)}{\Gamma^2(t+1)}\left(\int_0^{\infty}x^{t+k}e^{-x}dx\right)dt=\\
&=\frac{(-1)^n n!}{2\pi i}\oint_{\Sigma}\frac{\Gamma(t-n)\Gamma(t+k+1)}{\Gamma^2(t+1)}dt
\end{align}
What are the singular points of the integrand?

$\Gamma(t-n)$ has simple poles at $t=n,n-1,n-2,\ldots$. However, almost all of these poles, namely $t=-1,-2,\ldots$ are compensated by the poles of one of the $\Gamma(t+1)$ in the denominator.

Also, $\Gamma(t+k+1)$ has simple poles at $-k-1,-k-2,\ldots$ but all of them are compensated by the poles of the second $\Gamma(t+1)$.

Thus we have simple poles at $t=n,n-1,\ldots,0$ - precisely those inside of $\Sigma$. But now, instead of shrinking the integration contour, let us expand it to infinity. On the circle $|t|=R$, we have (think why)
$$\frac{\Gamma(t-n)\Gamma(t+k+1)}{\Gamma^2(t+1)}=O\left(R^{k-n-1}\right)\qquad \text{as}\;R\rightarrow\infty.$$
Therefore, the integral over this circle will be (at most) $O\left(R^{k-n}\right)$, but this goes to zero as $R\rightarrow\infty$ for $k<n$.


问:
Difference between exclamation mark (!) and point (.) in MAGMA
In this MAGMA handbook, I found the following code:

```<code>K := FiniteField(2, 160); // finite field of size 2^160
E := EllipticCurve([K!1, 0, 0, 0, K.1]);
</code>```

There, I noted the notations K!1 and K.1, and I do not really understand what the difference is between these two. It seems like two ways of conversion of $1$ into the finite field $K$. But then it would be pretty useless to use both notations if they supposedly do the same. Could someone explain this to me?

答:
The ! denotes coercion, for which more information can be found here: http://magma.maths.usyd.edu.au/magma/handbook/text/9

On the other hand, K.1 picks out the first generator of K.

Consider for instance the code:

F<x> := FiniteField(3,2); \\ F.1; \\ F!1;

With output:

```<code>x; \\ 1
</code>```

So they are very much not the same (but may happen to overlap in simple cases)!

Body:
The ! denotes coercion, for which more information can be found here: http://magma.maths.usyd.edu.au/magma/handbook/text/9

On the other hand, K.1 picks out the first generator of K.

Consider for instance the code:

F<x> := FiniteField(3,2); \\ F.1; \\ F!1;

With output:

x; \\ 1

So they are very much not the same (but may happen to overlap in simple cases)!


问:
Is there a category in which, between any two objects, there is a unique morphism?
I am interested in knowing information about such a category (if it is well-defined).

Does there exist a category $\mathcal{C}$, in which there is a unique morphism between any two objects in it?

答:
Given a continuous flow f on a locally compact Hausdorff space X, call a
subset S of X an f-invariant set if S is the union of orbits of the flow f. Call S f-isolated if, with respect to the partial order defined by
inclusion, S is the largest f-invariant set contained in some compact
neighborhood N of S; e.g., the critical elements of a Morse-Smale vector
field on a manifold are isolated relative to the flow generated
by the vector field. The Conley index of an f-isolated invariant set is a
1-connected groupoid (i.e., there is exactly one morphism from A to B for
any ordered pair (A,B) of objects of the groupoid) that is a small
subcategory of the homotopy category of pointed compact Hausdorff spaces.
Conley calls such groupoids connected simple systems. The Conley index is
particularly useful in finding bifurcation orbits, e.g., heteroclinic
orbits between two critical elements. See

H.L. Kurland, The Morse index of an isolated invariant set is a
connected simple system, J.~Differential Equations 42 (1981), 234--259.

C.C. Conley, Isolated Invariant Sets and the Morse Index, CBMS
Conference Proceedings, Amer. Math. Soc., Providence, R.I., 1978.


问:
Quotient of two monotone function
Let $(f_n)$ and $(g_n)$ be two positive and monotone increasing sequences of real functions.

Is $\frac{f_n}{f_n+g_n}$ monotone? Does it converge?

My thoughts

$\frac{f_n}{f_n+g_n}=1-\frac{g_n}{f_n+g_n}=1+\frac{(-1)g_n}{f_n+g_n}$. As the numerator $(-1)g_n$ is monotone decreasing and the denominator monotone increasing, the quotient should be monotone decreasing.

If there is $N$ such that $\forall n>N$, the quotient is in some bounded subset (it is eventually bounded), then it converges.

Did I go in the right direction?

答:
In my opinion, this is easier to think about in this form:
$$
\frac{f_n}{f_n + g_n} = \frac{1}{1 + \frac{g_n}{f_n}}
$$

Now the question is, what behavior can $\frac{g_n}{f_n}$ have? Of course, as you might imagine, it can do just about anything. If $g_n$ is faster in growth than $f_n$, it tends to $\infty$. If $f_n$ grows faster than $g_n$, it tends to $0$. Or you could make it tend to any positive value you like. Or you can make it oscillate:$$
f_n = 1, 2, 5, 6, 13, 14, \ldots
$$
$$
g_n = 1, 4, 5, 12, 13, 28, \ldots
$$
$$
\frac{g_n}{f_n} = 1, 2, 1, 2, \dots
$$

So no, generally, the quotient is not monotone and does not converge.

Title: Quotient of two monotone function


答: Hint: $\frac {f_n} {f_n+g_n}=\frac1  {1+\frac {g_n} {f_n}}$. So this is montone iff  $\frac {g_n} {f_n}$ is monotone. It is easy to construct examples where the ratio of two positive increasing functions is not monotone. (Look at constant functions).  Also, the limit may not exist. I hope you can work out some exmaples.

Title: Quotient of two monotone function

Body:
Let $(f_n)$ and $(g_n)$ be two positive and monotone increasing sequences of real functions.

Is $\frac{f_n}{f_n+g_n}$ monotone? Does it converge?

My thoughts

$\frac{f_n}{f_n+g_n}=1-\frac{g_n}{f_n+g_n}=1+\frac{(-1)g_n}{f_n+g_n}$. As the numerator $(-1)g_n$ is monotone decreasing and the denominator monotone increasing, the quotient should be monotone decreasing.

If there is $N$ such that $\forall n>N$, the quotient is in some bounded subset (it is eventually bounded), then it converges.

Did I go in the right direction?


问:
Interior, Boundary, Exterior points of subset of metric space of natural numbers
Consider the metric space of natural numbers with usual metric. And consider the set A= {1,2,3,4}.

Then what are the interior, exterior and boundary points of A?

I think the set of interior points is equal to A={1,2,3,4} itself.

Because as the whole space is of natural numbers, i can find an open ball centered at any one point of A of radius 1/2 (say) which is contained in A as the ball contains only one point that is the center.

The exterior should be the interior of complement of the set A. so here exterior of A is actually N-{1,2,3,4}

Am i right with these two things?

-Please explain the case of boundary points. I think they should be {1,2,3,4}. But still i want an explaination.

答:
You are right about the interior and the exterior of $A$.

The boundary of $A$ consists of those $n\in\Bbb N$ such that every neighborhood contains at least an element of $A$ and at least one element of $\Bbb N\setminus A$. Therefore the boundary is empty, because:

if $n\in A$, the open ball centered at $n$ with radius $1$ is $\{n\}$, and therefore it contains no element of $\Bbb N\setminus A$;

if $n\notin A$, the open ball centered at $n$ with radius $1$ is $\{n\}$, and therefore it contains no element of $A$.


问:
How to find the number of pages which has one method of drawing?
The problem is as follows:

Jenny bought a manga drawing book to her daughter so she could learn
the technique for her further animation studies. The book has 720
pages so it can be colored with pencils.

However she noticed the following:

On the pages whose numbering is even and that exactly divides the
total of pages, her daughter used the stitching technique.

On the pages whose numbering is a multiple of 3 and which divides
exactly to the total number of pages, her daugther has used the
outline technique.

On those pages whose numbering is a multiple of 5 and divisor of 720
she used the shading technique.

Using the information found by Jenny's. Find the number of pages which
her daughter has used only one painting technique.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{4}\\
2.&\textrm{5}\\
3.&\textrm{7}\\
4.&\textrm{6}\\
\end{array}$

The official solution from my book states the following:

Please note that I'm requiring some help here because the concept it is not very clear.

The approach that my book uses is:

$\operatorname{factor}(720)=2^4\cdot 3^2 \cdot 5$

Then it goes to count all the divisors which are multiples of

$2$, $3$, $5$, $6$ and $30$.

I don't know why those specific numbers, my best guess was that those are resulting from multiplying all the combinations from 2, 3 and 5.

Interestingly the qunatity of divisors of $720$ is $30$. But does this has anything to do with the prior selection of those numbers?

Anyways it goes as follows: (The notation $QD_{n}$ denotes the quantity of divisors of that number multiples of $n$).

$\operatorname{QD}_{2}(720)=4\times 3 \times 2 = 24$

This comes from the factorization: $720= 2\times(2^3\cdot 3^2 \cdot 5)$

$\operatorname{QD}_{3}(720)=(4+1)(1+1)(1+1)=20$

This comes from the factorization: $720= 3\times(2^4\cdot 3^1 \cdot 5)$

$\operatorname{QD}_{5}(720)=(4+1)(2+1)=15$

This comes from the factorization: $720= 5\times(2^4\cdot 3^2)$

$\operatorname{QD}_{6}(720)=(3+1)(1+1)(1+1)=16$

This comes from the factorization: $720= 6\times(2^3\cdot 3^1 \cdot 5)$

$\operatorname{QD}_{10}(720)=(3+1)(2+1)=12$

This comes from the factorization: $720= 10\times(2^3\cdot 3^2)$

$\operatorname{QD}_{30}(720)=(3+1)(1+1)=8$

This comes from the factorization: $720= 30\times(2^3\cdot 3^1)$

Well after doing all of that my book concludes that the number of pages on which it was used only one technique is:

$4+2+1=7$

But I don't know where does it come those numbers?. They seem to be the same powers of the prime factorization from $720$.

But it lacks from a proper explanation why should those numbers be the total of pages on which only one drawing technique was used?

Perhaps does it exist a way to reverse engineering what was intended to be done here or something like that?.

I'm trying to get the idea but I'm lost, hence I need for asistance. Can someone help me with a step by step explanation on which sort of method using divisors could be used to find those pages where it was used only one technique?.

So far the only idea which I have is that such number of pages must be let's say multiple of $2$, but not multiple of $3$ or $5$. Is this right?. Please help me here as I'm lost.

答:
Every divisor of $720$ is of the form $2^r3^s5^t$ for some $r,s,t$. We can say more: $0\le r\le4$, $0\le s\le2$, $0\le t\le1$.

Now the ones that are multiples of $2$ but of neither $3$ nor $5$ are just the ones of the form $2^r$, $1\le r\le 4$, and there are four of them.

The ones that are multiples of $3$ but of neither $2$ nor $5$ are just the ones of the form $3^s$, $1\le s\le 2$, and there are two of them.

The ones that are multiples of $5$ but of neither $2$ nor $3$ are just the ones of the form $5^t$, $1\le t\le 1$, and there is just one of them.

So the total number is $4+2+1=7$.

And it's exactly as you said – these numbers are coming from the exponents $4,2,1$ in the factorization $720=2^43^25^1$.


问:
comparing angle in measurement for concave mirror
From the above picture we can prove easily $ CF = XF$ because CFX is a isosceles triangle. Now if we move the point X  more near near P, it will give the same results(I mean we will get again $ CF = XF$) but by drawing the figure on paper, I have measured that XF= CF are not equal always. What is wrong here?

答:
The geometry in the diagram is true no matter what the figure of the mirror is; that is, $\triangle XFC$ is isosceles. What is not true is that both $C$ and $F$ are independent of $X$. In what follows, $\kappa$ is the curvature of the curve at $P$.

If we wish to fix $F$ independent of $X$, then the curve needs to be a parabola. However, in that case, $C$ is dependent on $X$.

$\hspace{2cm}$

If we wish to fix $C$ independent of $X$, then the curve needs to be a circle. However, in that case, $F$ is dependent on $X$.

$\hspace{2cm}$

Note that as $r\to0$, $PC\to2PF$.


问:
Gradient and Hessian of a matrix
For $g(y) = f(D^{\frac{1}{2}}y)$ where $D^{\frac{1}{2}}$ is a matrix to the power half and $ x = D^{\frac{1}{2}}y$

Then $\nabla g(y) = \nabla f(D^{\frac{1}{2}}y) = D^{\frac{1}{2}} \nabla f(D^{\frac{1}{2}}y) = D^{\frac{1}{2}} \nabla f(x)$

What I struggle with is the form of the Hessian given in Amir Beck's Introduction to non-linear optimisation on page 63. Why is the $D^{\frac{1}{2}}$ mutliplied on the RHS in the Hessian?

$\nabla^2 g(y) = D^{\frac{1}{2}} \nabla^2 f(D^{\frac{1}{2}}y) D^{\frac{1}{2}} = D^{\frac{1}{2}} \nabla^2 f(x) D^{\frac{1}{2}}$

答:
Suppose you've calculated the gradient and Hessian of a function in terms of the variable $x$

$$\eqalign{
 \phi = f(x),\,\,\,\,\,
 p = \frac{\partial f}{\partial x},\,\,\,\,\,\,
 A = \frac{\partial p}{\partial x}
}$$
Then you learn that $x$ is not independent, but actually depends on another variable $(x = Sy).\,\,$ Note that the matrix $S$ does not need to be invertible. It might even be rectangular.

Recalculate everything in terms of the new variable
$$\eqalign{
 \phi = g(y),\,\,\,\,\,
 q = \frac{\partial g}{\partial y},\,\,\,\,\,\,
 B = \frac{\partial q}{\partial y}
}$$
using differentials
$$\eqalign{
&d\phi = p^Tdx = p^T(S\,dy) = (S^Tp)^Tdy = q^Tdy \quad
&\therefore\;\; &q = \frac{\partial g}{\partial y} = S^Tp \\
&dq = S^T\,dp = S^T(A\,dx) = S^TA(S\,dy) = B\,dy \quad
&\therefore\;\; &B =\frac{\partial q}{\partial y} = S^TAS \\
}$$
Obviously, in this case we have $\,S=S^T=D^{\frac 12}$

Title: Gradient and Hessian of a matrix


问:
The probability of getting an odd sum

Mu turn:

The number of even numbers =$3$ , the number of odd numbers = $5$

So the possibilities to get an odd sum are

EEO , EOE , OEE , OOO
Where E means even , O means odd then the probability is
$\frac{3}{8} \times \frac{2}{7}\times \frac{5}{6})\times 3 + \frac{5}{8} \times \frac{4}{7} \times \frac{3}{6} = \frac{25}{56}$

Is this correct ? I

答:
Comment:

A simulation of a million such draws of three (without
replacement) from the bag should give an answer correct
to about three decimal places. Your answer is $25/56 =
0.4464286.$ The simulated answer is $0.4459 \pm 0.0010.$
So your answer looks good.

```<code>set.seed(2011)
bag = c(1,1,2,4,4,5,5,5)
jona = replicate(10^6, sum(sample(bag, 3)))
mean((jona/2) != round(jona/2))
[1] 0.445945     # aprx P(Odd)
2*sd((jona/2) != round(jona/2))/1000
[1] 0.0009941394 # aprx 95% margin of sim. error
25/56
[1] 0.4464286    # exact
</code>```

Title: The probability of getting an odd sum

Body:
Comment:

A simulation of a million such draws of three (without
replacement) from the bag should give an answer correct
to about three decimal places. Your answer is $25/56 =
0.4464286.$ The simulated answer is $0.4459 \pm 0.0010.$
So your answer looks good.

set.seed(2011)
bag = c(1,1,2,4,4,5,5,5)
jona = replicate(10^6, sum(sample(bag, 3)))
mean((jona/2) != round(jona/2))
[1] 0.445945     # aprx P(Odd)
2*sd((jona/2) != round(jona/2))/1000
[1] 0.0009941394 # aprx 95% margin of sim. error
25/56
[1] 0.4464286    # exact


问:
See if this functional equation works for some function and write down the domain and co-domain
$f(x+y)=[f(x)]^2+y$

Prove whether or not this functional equation works for some function and, if it does, specify the domain(s) and co-domain(s).

P.S.: I've invented this problem from scratch just a couple of minutes ago. I hope to see some interesting results and proofs.

EDIT: I was thinking this while trying to solve this equation. Assume the function is constant. This means that i can write $f$ as $f(a)=c$. Therefore i can rewrite the equation as $c=c^2+y$ and so $y=c-c^2$. But that is a contradiction, since I've written a variable ($y$) in terms of a constant ($c$). As a consequence of this, let's try $y=0$: $f(x) = [f(x)]^2$. The only solutions to this are $f(x)=1$ or $f(x)=0$, but both are impossible, since we just proved that the function cannot be constant. If my reasoning is correct, this should imply that the function does not allow $y=0$ in its domain.

答:
The functional equation [for reference] is: $$f(x+y)=f(x)^2+y\ . \tag{1} $$
I'll be assuming both the domain and codomain are subsets of the reals, to be determined by $(1).$
If in this we put $y=0$ and keep $x,$ we get to
$$f(x)=f(x)^2. \tag{2}$$
This immediately implies the range of $f$ is a subset of $\{0,1\}.$

Next going back to $(1)$ we now put $x=0$ and keep $y,$ and get
$$f(y)=f(0)^2+y\ . \tag{3}$$
At this point we know as noted that $f(0)=0$ [''case A''] or else $f(0)=1$ [''case B''].

In case A $(3)$ becomes $f(y)=y.$ But since then $y$ is in the range of $f$ it is $0$ or $1$ and we now know in case A the domain must be a subset of $\{0,1\}.$

In case B $(3)$ becomes $f(y)=1+y.$ Here from $1+y \in \{0,1\}$ we see in case B the domain is a subset of $\{-1,0\}.$

Already things are quite restricted. But we can go one more step by returning to $(1)$ and this time putting $y=x$ keeping both variable to get to
$$ f(2x)=f(x)^2+x. \tag{3}$$
Going back to cases A,B we now see that in case A, $x$ cannot be $1$ else $2$ is in domain of $f.$ And in case B, $x$ cannot be $-1$ else $-2$ ends up in the domain of $f.$ The upshot is that in either case A or B the domain of $f$ is the single number $0.$ Finally the range of $f$ is in either case only $0$ by putting $x=0$ into $(3).$ I think the codomain could be just $0$ or could be $\{0,1\}.$


问:
How to find the probability of drawing colored marbles without replacement?
There are $2$ red marbles, $3$ white marbles and $5$ black marbles in a bag.
What is the probability of drawing $1$ red marble, $2$ white marbles and $3$ black marbles, if $6$ marbles are drawn without replacement?

Answer:  $$\frac{\binom{2}{1}\binom{3}{2}\binom{5}{3}}{\binom{10}{6}}=\frac{2}{7}$$

My question is why is the numerator of the answer $\binom{2}{1}\binom{3}{2}\binom{5}{3}$ instead of $1$? Below is my answer:

Total number of ways to choose $6$ marbles from $10$ marbles is $\binom{10}{6}$, so combinations such as

$BBBBBW$, $RWWBBB$, $WWWRBB$, ...

So the probabilty I calculated is $\frac{1}{\binom{10}{6}}$ because $RWWBBB$ is one of the possible combinations out of the $\binom{10}{6}$ total combinations. However this is incorrect, and is supposed to be $\frac{\binom{2}{1}\binom{3}{2}\binom{5}{3}}{\binom{10}{6}}$. I do not understand why the numerator is the product combinations because it does not make sense to multiply combinations of nondistinct objects. For example,

there are $5$ black marbles $BBBBB$ and so the number of ways to select $3$ black marbles is $1$, therefore it does not make sense to me that it is written as $\binom{5}{3}$ as above. To me this would make sense if it were asking to arrange $3$ letters at a time from $ABCDE$ where the order does not matter.

Can anyone explain why the numerator is the product of combinations instead of $1*1*1=1$?

My observation:

I noticed that using subscripts are necessary when there are fewer distinct objects than the total number of objects. See below:

The combination formula is $\binom{n}{r}$ where $n$ is total number of objects and $r$ is number of selected objects.

Let $k$ be the number of distinct objects where

if $k<n$, then subscripts are needed

if $k=n$, then subscripts are not needed

Case $1$. $k<n$

Suppose I want to select $r=2$ marbles from a bag with $1$ white marble and $2$ black marbles, so

$k=2$ and $n=3$, so $k<n$

So $\binom{n}{r}=3$ with combinations: {$W_1,B_1$}, {$W_1,B_2$}, {$B_1,B_2$}

Case $2$. $k=n$

Suppose I want to select $r=2$ marbles from a bag with $1$ white marble, $1$ black marble and $1$ red marble, so

$k=3$ and $n=3$, so $k=n$

So $\binom{n}{r}=3$ with combinations: {$W,B$}, {$W,R$}, {$B,R$}

答:
$\binom{10}6$ is the count for selecting 6 items from a set of 10 distinct items.  We don't care about the order in which these are selected, just which individual marbles are selected.

So too should we not be concerned with order of the favored event. We count it in the same sort of way --- treating the marbles as distinct items although grouped by colours.

Thus we evaluate the probability for obtaining: 1 from 2 red, 2 from 3 white, and 3 from 5 blue, when selecting any 6 from all 10 marbles as:$$\dfrac{\dbinom 21\dbinom 32\dbinom 53}{\dbinom{2+3+5}{1+2+3}}$$


问:
Sampling Distribution: Difference of Normal Random Variables
This is a self-study question. I am trying to complete this textbook problem.

Suppose that $X_{1}, X_{2}, \dots, X_{m}$ and $Y_{1}, Y_{2}, \dots, Y_{n}$ are independent random samples with $X_{i}$ and $Y_{i}$ being normally distributed with means $\mu_{1}$ and $\mu_{2}$ and variances $\sigma^2_{1}$ and $\sigma^2_{2}$, respectively. The difference between the sample means, $\bar{X} - \bar{Y}$, is then a linear combination of $m+n$ normally distributed random variables is also normally distributed. If $\sigma^2_{1} = 2, \sigma^2_{2} = 2.5$, and $m=n$, find the sample sizes so that $(\bar{X} - \bar{Y})$ will be within 1 unit of $(\mu_{1} - \mu_{2})$ with probability $0.95$.

I started by saying that $P(|\bar{X} - \bar{Y} - (\mu_{1} - \mu_{2})| \le 1) = 0.95$. I think I have to standardize the variables, but I am unsure on how to do that since I am working with two random variables at once.

The answer is $n = 17.29 \rightarrow 18$, but I am unsure how to get here.

答:
Let's look at the pieces

$$\begin{split}X&\sim N(\mu_1,\sigma_1^2)\\
Y&\sim N(\mu_2,\sigma_2^2)\end{split}$$

Then what the book means by $\bar X-\bar Y$ is a sum of m+n independent normal random variables is that $\bar X=\frac{\sum X}{m}=\sum {\frac{X}{m}}$ and likewise for $\bar Y$ with

$$\begin{split}\frac {X}{m}&\sim N\left(\frac{\mu_1}{m},\frac{\sigma_1^2}{m^2}\right)\\
\frac{Y}{n}&\sim N\left(\frac{\mu_2}{n},\frac{\sigma_2^2}{n^2}\right)\end{split}$$

As you know the sum of k independent normal random variables $X_1,...,X_k$ with means $\mu_1,...,\mu_k$ and variances $\sigma_1^2,...,\sigma_k^2$ is normally distributed with mean $\sum \mu$ and variance $\sum \sigma^2$

Also note that

$$-\frac{Y}{n}\sim N\left(-\frac{\mu_2}{n},\frac{\sigma_2^2}{n^2}\right)$$

Hence

$$\begin{split}\bar X-\bar Y&=\frac{X_1}{m}+...+\frac{X_m}{m}+\frac{Y_1}{n}+...+\frac{Y_n}{n}\\
&\sim N\left(m\cdot \frac{\mu_1}{m}-n\cdot \frac{\mu_2}{n}, m\cdot \frac{\sigma_1^2}{m^2}+n\cdot \frac{\sigma_2^2}{n^2}\right)\\
&= N\left(\mu_1-\mu_2, \frac{\sigma_1^2}{m}+\frac{\sigma_2^2}{n}\right)\end{split}$$

We seek

$$P(|\bar X-\bar Y-(\mu_1-\mu_2)|\le 1)\ge.95$$

Standardize via division by the standard error with m=n here

$$\begin{split}P\left(\left|\frac{\bar X-\bar Y-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^2+\sigma_2^2}{n}}}\right|\le \frac 1{\sqrt{\frac{\sigma_1^2+\sigma_2^2}{n}}}\right)&\ge.95\\
P(|Z|\le\sqrt{\frac{n}{\sigma_1^2+\sigma_2^2}})&\ge.95\end{split}$$

Using R we find that the critical value is

```<code>> qnorm(.975)
[1] 1.959964
</code>```

Thus

$$\begin{split}\sqrt{\frac{n}{4.5}}&\ge1.96\\
n&\ge17.29\end{split}$$

Conservatively we pick n=18

Body:
Let's look at the pieces

$$\begin{split}X&\sim N(\mu_1,\sigma_1^2)\\
Y&\sim N(\mu_2,\sigma_2^2)\end{split}$$

Then what the book means by $\bar X-\bar Y$ is a sum of m+n independent normal random variables is that $\bar X=\frac{\sum X}{m}=\sum {\frac{X}{m}}$ and likewise for $\bar Y$ with

$$\begin{split}\frac {X}{m}&\sim N\left(\frac{\mu_1}{m},\frac{\sigma_1^2}{m^2}\right)\\
\frac{Y}{n}&\sim N\left(\frac{\mu_2}{n},\frac{\sigma_2^2}{n^2}\right)\end{split}$$

As you know the sum of k independent normal random variables $X_1,...,X_k$ with means $\mu_1,...,\mu_k$ and variances $\sigma_1^2,...,\sigma_k^2$ is normally distributed with mean $\sum \mu$ and variance $\sum \sigma^2$

Also note that

$$-\frac{Y}{n}\sim N\left(-\frac{\mu_2}{n},\frac{\sigma_2^2}{n^2}\right)$$

Hence

$$\begin{split}\bar X-\bar Y&=\frac{X_1}{m}+...+\frac{X_m}{m}+\frac{Y_1}{n}+...+\frac{Y_n}{n}\\
&\sim N\left(m\cdot \frac{\mu_1}{m}-n\cdot \frac{\mu_2}{n}, m\cdot \frac{\sigma_1^2}{m^2}+n\cdot \frac{\sigma_2^2}{n^2}\right)\\
&= N\left(\mu_1-\mu_2, \frac{\sigma_1^2}{m}+\frac{\sigma_2^2}{n}\right)\end{split}$$

We seek

$$P(|\bar X-\bar Y-(\mu_1-\mu_2)|\le 1)\ge.95$$

Standardize via division by the standard error with m=n here

$$\begin{split}P\left(\left|\frac{\bar X-\bar Y-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^2+\sigma_2^2}{n}}}\right|\le \frac 1{\sqrt{\frac{\sigma_1^2+\sigma_2^2}{n}}}\right)&\ge.95\\
P(|Z|\le\sqrt{\frac{n}{\sigma_1^2+\sigma_2^2}})&\ge.95\end{split}$$

Using R we find that the critical value is

> qnorm(.975)
[1] 1.959964

Thus

$$\begin{split}\sqrt{\frac{n}{4.5}}&\ge1.96\\
n&\ge17.29\end{split}$$

Conservatively we pick n=18


问:
In a matrix ring, no zero divisors may have an inverse
In a general ring with 1, a right (left) zero divisor cannot have a right (left) inverse. In a matrix ring over a field, a stronger condition is satisfied: a (right or left) zero divisor cannot have a (right or left) inverse, with the proof I know invoking the rank plus nullity theorem (which makes sense, because infinite dimensional linear transformations can have right-inverses and be right-zero divisors). Is this more or less a peculiarity of matrices, or am I missing a more general property that they (and perhaps other spaces) possess?</p>

答:
I'm glad to see that Dedekind finite rings have been raised as the right concept to look at, but I think a little more can be done to help you see where the condition lies with respect to other conditions you might know.

Here is a diagram reproduced from a portion of my notes that gives a view (at least a partial one):

There are a lot of strange names up there, but between Lam's two books (First course on noncommutative rings and Lectures on modules and rings) you would be able to reproduce this table. In addition to those books, there is an additional paper by Lam that talks about stable range and its immediate relatives: A crash course on...

Of course, matrix rings over division rings lie in the "semisimple" box. In addition to the Dedekind finite box, I think you should also know about "stably finite". A ring $R$ is said to be stably finite if all of the $n\times n$ matrix rings over $R$ are Dedekind finite.

Since Noetherian and perfect rings have already been mentioned, I'd like to contribute by pointing out two "new" examples given here. Since there are two-sided self-injective rings which aren't Noetherian and aren't semiperfect, and unit regular rings which aren't Noetherian and aren't semiperfect, nor injective on both sides, I think both of those classes of rings are "new" to this post.

Finally, there is one thing to add to this diagram. I don't normally keep a box for "commutative ring," but if it were in this picture, it would have an arrow pointing to "Stably finite."

There is one more condition which might interest you. A ring in which every element which isn't a left zero divisor is a unit is called a right cohopfian ring. This means essentially that if multiplication by an element on the left makes an injective map $R\to R$, then it is an isomorphism. This relates closely to an alternative definition of Dedekind finite: If multiplication on the left by $a$ is surjective then it is an isomorphism. This is saying that $R$ is "hopfian". It turns out that for a ring, this condition is left-right symmetric, but the cohopfian version isn't :)

I've got a bit of info on them in this solution: https://math.stackexchange.com/a/135051/29335 . One-sided cohofian rings are also Dedekind finite. I would have to double check which things on the graph point to "cohopfian" :) They're a bit of an oddball...


答:
Consider a simpler problem: your bag has $99$ blue marbles and $1$ red marble in it. You draw a marble from the bag. What is the probability that it's blue?

If you want to answer $\frac{99}{100}$ rather than $\frac12$, you want to treat the blue marbles as distinguishable. This has nothing to do with whether you can tell the marbles apart by looking.

Rather, the problem is that we can only find probabilities by counting outcomes if we are sampling uniformly. In order to be drawing one of the $100$ marbles uniformly, the $100$ marbles should all be considered different outcomes.

Similarly, in the actual question you're dealing with, we can only answer the question by counting outcomes if all $10$ marbles are distinguishable: because only in that case are there $\binom{10}{6}$ different outcomes, each of which is equally likely. So you shouldn't be thinking of your outcomes as $BBBBBW, RWWBBB, \dots$. Rather, we should say:

There are $10$ marbles in the bag: marbles $R_1, R_2$ are red, marbles $W_1, W_2, W_3$ are white, and marbles $B_1, B_2, B_3, B_4, B_5$ are black.

We draw a uniformly random subset of $6$ of the marbles, such as $\{B_1, B_2, B_3, B_4, B_5, W_1\}$ or $\{R_2, W_1, W_3, B_2, B_4, B_5\}$.

In this case, there are in fact $\binom21 \binom32 \binom53$ outcomes that match the description "$1$ red marble, $2$ white marbles, and $3$ black marbles". There's $\{R_1, W_1, W_2, B_1, B_2, B_3\}$ and $\{R_1, W_1, W_2, B_1, B_2, B_4\}$, and so on.

These are exactly all outcomes of the form $\{R_a, W_b, W_c, B_d, B_e, B_f\}$ where $\{a\} \subset \{1,2\}$, $\{b,c\} \subset \{1,2,3\}$, and $\{d,e,f\} \subset \{1,2,3,4,5\}$. From this description, we can see why there's $\binom21 \binom32 \binom53$ of them: there's $\binom21$ ways to pick $\{a\}$, $\binom32$ ways to pick $\{b,c\}$, and $\binom53$ ways to pick $\{d,e,f\}$.

If instead our outcomes only had descriptions like "there are $3$ black marbles and $3$ white marbles" without distinguishing the marbles of each color, then our outcomes wouldn't all be equally likely. There are actually only $7$ different outcomes in this case: $$WWWBBB, WWBBBB, WBBBBB,\\ RWWWBB, RWWBBB, RWBBBB, RBBBBB.$$ But their probabilities are different, so this model doesn't lend itself to computation very well.

Title: How to find the probability of drawing colored marbles without replacement?

Body:
There are $2$ red marbles, $3$ white marbles and $5$ black marbles in a bag.
What is the probability of drawing $1$ red marble, $2$ white marbles and $3$ black marbles, if $6$ marbles are drawn without replacement?

Answer:  $$\frac{\binom{2}{1}\binom{3}{2}\binom{5}{3}}{\binom{10}{6}}=\frac{2}{7}$$

My question is why is the numerator of the answer $\binom{2}{1}\binom{3}{2}\binom{5}{3}$ instead of $1$? Below is my answer:

Total number of ways to choose $6$ marbles from $10$ marbles is $\binom{10}{6}$, so combinations such as

$BBBBBW$, $RWWBBB$, $WWWRBB$, ...

So the probabilty I calculated is $\frac{1}{\binom{10}{6}}$ because $RWWBBB$ is one of the possible combinations out of the $\binom{10}{6}$ total combinations. However this is incorrect, and is supposed to be $\frac{\binom{2}{1}\binom{3}{2}\binom{5}{3}}{\binom{10}{6}}$. I do not understand why the numerator is the product combinations because it does not make sense to multiply combinations of nondistinct objects. For example,

there are $5$ black marbles $BBBBB$ and so the number of ways to select $3$ black marbles is $1$, therefore it does not make sense to me that it is written as $\binom{5}{3}$ as above. To me this would make sense if it were asking to arrange $3$ letters at a time from $ABCDE$ where the order does not matter.

Can anyone explain why the numerator is the product of combinations instead of $1*1*1=1$?

My observation:

I noticed that using subscripts are necessary when there are fewer distinct objects than the total number of objects. See below:

The combination formula is $\binom{n}{r}$ where $n$ is total number of objects and $r$ is number of selected objects.

Let $k$ be the number of distinct objects where

if $k<n$, then subscripts are needed

if $k=n$, then subscripts are not needed

Case $1$. $k<n$

Suppose I want to select $r=2$ marbles from a bag with $1$ white marble and $2$ black marbles, so

$k=2$ and $n=3$, so $k<n$

So $\binom{n}{r}=3$ with combinations: {$W_1,B_1$}, {$W_1,B_2$}, {$B_1,B_2$}

Case $2$. $k=n$

Suppose I want to select $r=2$ marbles from a bag with $1$ white marble, $1$ black marble and $1$ red marble, so

$k=3$ and $n=3$, so $k=n$

So $\binom{n}{r}=3$ with combinations: {$W,B$}, {$W,R$}, {$B,R$}


问:
Constructing a point D on AB such that PD is perpendicular to AB
I need to write the steps of construction from this exercise.

Construct a line segment $AB$, and a point $P$, not on $AB$. Use the theorem on the previous page or one of its alternative statements to construct a point $D$ on $AB$ such that $PD ⊥ AB$. Write down your steps of construction, and explain how you used the theorem to do the construction.

This is the previous theorem: In an isosceles triangle, the median to the base is the perpendicular bisector (of the base) and also the angle bisector for the angle opposite to the base.

So far I have learned how to create an equilateral triangle with side $AB$ by using the intersections of circle $(A,AB)$ and circle $(B,BA)$.

I am unsure how to start this. I think I need to somehow create a isosceles triangle that would be triangle $PAB$ but not sure how to construct it?

答:
Draw a circle centered on $P$ that passes through $A$ (that is, radius $|PA|$)${}^*$. Extend line $AB$ if necessary to have two points of intersection with the circle; $A$ and a new point $F$. These are equidistant from $P$ and so $\triangle APF$ is isosceles, with $AF$ the base.

Now bisect the line $AF$ (you know how to do this?) at $D$ and by the theorem $PD$ is the perpendicular bisector of $\triangle APF$ and so $D$ is the point as required.

$*\;$ For practical purposes, if $\angle PAB$ appears to be close to a right angle, draw the initial $P$-centered circle through $B$ instead and proceed accordingly.


问:
Is this a Hyperbola or Ellipse inscribed or escribed around triangle?
The vertices of a triangle are $(5, 3)$ and $(5, -5)$. Determine the equation of the locus of the third vertex if the perimeter of the triangle is $20$ units.

So I assume $5$ units to be the major segment where $b$ is the minor segment, and $3$ units is the distance to foci.

By Pythagoras, I got:

$a^2=b^2+c^2$

$b= 4$ units

But still, I can't get the right locus of points given with the choices. So I was thinking if the locus of points escribing or inscribing the triangle?

答:
Your figure is not correct because the points $(5,3)$ and $(5,-5)$ are the foci of the ellipse. What you have is the center of the ellipse given by the midpoint $(5,-1)$. This said you that the ellipse is "vertical" instead of "horizontal".

You have the equation which is the definition itself of the ellipse as a locus
$$\sqrt{(x-5)^2+(y-3)^2}+\sqrt{(x-5)^2+(y+5)^2}=20-8=12$$ Simplifying you get
$$9(x-5)^2+5(y+1)^2=180$$
An ellipse "vertical" as you can verify.


问:
Effros Borel structure
Let $X$ be a Polish space and denote by $\mathcal{F}(X)$ the set of all closed subsets of $X$. The Effros Borel structure on $\mathcal{F}(X)$ is the smallest $\sigma$-algebra on $\mathcal{F}(X)$ containing the family $\big\{ \{ F \in \mathcal{F}(X) \, ; \, F \cap U \neq \emptyset \} \, ; \ U \subseteq X \text{ is open} \big\}$.

Now, given a closed set $A \subseteq X$, is it necessary true that the set $\{ F \in \mathcal{F}(X) \, ; \, F \cap A \neq \emptyset \}$ is Borel (i.e. belongs to the Effros Borel structure)? I strongly hope the answer is yes (since it would help me solve some other problem), but none of my attempts to prove it was successful.

I will appreciate any help.

答:
Unfortunately the answer is negative.

Let $H=\ell^2$ be the separable infinite dimensional Hilbert space and let $S_1\subseteq H$ be the sphere of radius $1$, that is $S_1=\{x\in H\mid \|x\|=1\}$. Note that $S_1(H)$ is closed in $H$.

As shown by Christensen (see 27.6 in Kechris Classical Descriptive Set Theory) the set $\{F\in\mathcal F(H)\mid F\cap S_1(H)\neq\varnothing\}$ is $\mathbf{\Sigma}^1_1$-complete, which implies that it cannot be Borel.

Note however that if $A$ is a compact set, then $\{F\in\mathcal F(X)\mid F\cap A\neq \varnothing\}$ is indeed Borel, because $$\{F\in\mathcal F(X)\mid F\cap A\neq \varnothing\}=\bigcap_{n\in\Bbb N}\{F\in\mathcal F(X)\mid F\cap B_{1/n}(A)\neq\varnothing\},$$
where $B_r(A)=\{x\in X\mid d(x,A)<r\}$. The left to right inclusion is obvious, while if $F$ belong to the right hand side, then easily $d(F,A)=0$, and the distance between a closed set and a compact one must be realized by some point.

Title: Effros Borel structure


问:
Upper bound on the number of ways to completely fill boxes.
I have $x$ indistinguishable items and $n$ boxes with sizes $y_1,\dots,y_n$. I am interested in the number of ways of allocating those $x$ items into the boxes such that all boxes must be either fully filled (i.e., the number of items in each box equals its size) or empty. Also, I have the condition $x=o(n)$ (where $x\leq n/2$ is included in the condition), which indicate that boxes of smaller sizes might possibly lead to more number of ways.

For example, if I have 3 items and 100 boxes all with size 2, then I have 0 ways to achieve my goal. However, if one of the boxes have size 3, then I would have one way to achieve my goal. If I have 3 items and 100 boxes all with size 1, then I have ${100\choose 3}$ ways to achieve my goal.

I suspect that the upper bound for the number of ways of allocating those $x$ items into $n$ boxes such that all boxes must be fully filled is ${n\choose x}$, which is the case where all boxes have size 1. I'm having trouble arguing this rigorously. Is there a way to show this rigorously or is the claim wrong?

答:
Just to flesh out the details of the proof alluded to in the comments:

Let $\mathcal A$ be the collection of $S$ for which $\sum_{i\in S}y_i=x$. Then $\mathcal A$ is an antichain in the poset of subsets of $\{1,\dots,n\}$. Let $a_k$ be the number of subsets of size $k$ in $\mathcal A$.
The LYM inequality implies
$$
\sum_{k=0}^n \frac{a_k}{\binom{n}k}\le 1.\tag{*}
$$
Therefore,
$$
|\mathcal A|
= \sum_{k=0}^n a_k
\stackrel{1}=\sum_{k=0}^{x}a_k
= \binom{n}{x}\sum_{k=0}^{x}\frac{a_k}{\binom{n}x}
\stackrel{2}\le \binom{n}{x}\sum_{k=0}^{x}\frac{a_k}{\binom{n}k}
\stackrel{*}\le \binom{n}x\cdot 1.
$$

We must have $a_k=0$ for $k>x$, since if there were more then $x$ full boxes, the total number of items would be more than $x$.

Here, we use the fact that $x<n/2$, so that $\binom{n}k\le \binom{n}x$ for all $k\le x$.


问:
Residual sum of squares and variance.
So, from the ANOVA table one can get an output of residual sum of squares. Should the RSS be divided by degrees of freedom in ANOVA + 1 or should it be divided by df in ANOVA - 1? Because the residual variance formula is: $\frac{\left((\hat{y}_i - y_i) - \overline{(\hat{y}_i - y_i)}\right)^2 }{n-2}$. Thanks in advance.

答:
The RSS in ANOVA should be divided by the number observations $n$ minus the number of groups $K$, i.e.,
$$
MSE = \frac{RSS}{n-K},
$$
the reason is that RSS is the weighted sum of the variances of each group.  For every group you estimate its mean $\mu_j$ with $\bar{x}_{n_j\cdot}$, hence you "lose" one degree of freedom every estimator, i.e.,  $RSS = \sum_{j=1}^K (n_j-1) S^2_j$, where $S_j^2 = \frac{1}{n_j -1}\sum_{i=1}^{n_j}(x_{ij} - \bar{x}_{n_j\cdot})^2$

Title: Residual sum of squares and variance.


问:
Question about Do Carmo's notation
Let M be a differentiable manifold. On page 26 of Riemannian Geometry by Do Carmo, he makes this statement:

Observe that if $\varphi:M\longrightarrow M$ is a diffeomorphism, $v\in T_pM$ and $f$ is a differentiable function in a neighborhood of $\varphi(p)$, we have
$$(d\varphi(v)f)\varphi(p)=v(f\circ\varphi)(p).$$
Indeed, let $\alpha:(-\varepsilon,\varepsilon)\longrightarrow M$ be a differentiable curve with $\alpha'(0)=v$, $\alpha(0)=p$. Then
$$(d\varphi(v)f)\varphi(p)=\left.\frac{d}{dt}(f\circ\varphi\circ\alpha)\right|_{t=0}=v(f\circ\varphi)(p).$$

I understand what he means by the right hand side. Namely, a vector $v$ in the tangent space at $p$ acting as a derivation on the differentiable function $f\circ\varphi$ and then evaluating the resultant function at $p$. However, I'm unsure if I'm understanding what the object on the left hand side is supposed to be. You take the differential of the map $\varphi$ (not at any point in particular?, I assume there's supposed to be a subscript $p$?) and then act this Jacobian on the vector $v$. This is fine, and it should give you a vector in the tangent space at $\varphi(p)$. You then treat this as a derivation and so it maps $f$ to some other differentiable function which you then evaluate at $\varphi(p)$. Is this correct? If so, I still can't see where the middle step comes from.

答:
I think do Carmo's equation does not make sense.

do Carmo introduces tangent vectors at $p \in M$ as tangent vectors at $t = 0$ of some curce $\alpha : (-\epsilon, \epsilon) \to M$ with $\alpha(0) = p$, i.e. as $\alpha'(0)$. Here $\alpha'(0)$ is defined as the function $\alpha'(0) : \mathcal D_p(M) \to \mathbb R, \alpha'(0)(f) = (f \circ \alpha)'(0)$. Note that $\mathcal D_p(M)$ is the set of functions $f : M \to \mathbb R$ which are differentiable at $p$ so that $f \circ \alpha : (-\epsilon, \epsilon) \to \mathbb R$ is a function which is differentiable at $0$ in the sense of elementary calculus and $(f \circ \alpha)'(0)$ is its usual derivative.

If $\varphi : M \to N$ is differentiable, do Carmo introduces the differential at $p$ as follows: Each $v  \in T_pM$ has the form $v = \alpha'(0)$ as above and he defines
$$d\varphi_p : T_pM \to T_{\varphi(p)}N, d\varphi_p(v) = (\varphi \circ \alpha)'(0) .$$

The claim on p.26 occurs in section 5. "Vector fields, brackets. Topology of manifolds". The equation
$$(d\varphi (v) f)\varphi(p) = v(f \circ \varphi)(p)$$
does not make sense as it stands. We have $v \in  T_pM$ and $f$ differentiable in a  neighborhood of $\varphi(p)$. Then $f \circ \varphi$ is differentiable in a  neighborhood of $p$ and $v(f \circ \varphi) \in \mathbb R$. It remains unclear what $v(f \circ \varphi)(p)$ should be. The same issue is on the LHS. However, if we omit the arguments $\varphi(p)$ and $p$, we get
$$d\varphi (v) f =  v(f \circ \varphi) .$$
It seems that the LHS stands for $d\varphi_p(v)(f)$. But then, if $v = \alpha'(0)$ for some curve $\alpha$ through $0$, we have by the above definitions $d\varphi_p(v) = (\varphi \circ \alpha)'(0)$ and
$$d\varphi_p(v)(f)  = (\varphi \circ \alpha)'(0)(f) = (f \circ \varphi \circ \alpha)'(0) = \frac{d}{dt}(f \circ \varphi \circ \alpha) \mid_{t=0}$$
and
$$(f \circ \varphi \circ \alpha)'(0) = \alpha'(0)(f \circ \varphi) = v(f \circ \varphi) .$$
Therefore
$$d\varphi_p(v)(f) = \frac{d}{dt}(f \circ \varphi \circ \alpha) \mid_{t=0} = v(f \circ \varphi)$$
which makes sense. The formula
$$d\varphi_p(v)(f) = v(f \circ \varphi)$$
is correct for any differentiable $\varphi : M \to N$; it is not limited to diffeomorphisms $\varphi : M \to M$.

Update:

As Moishe Kohan says in a comment, it is poor notation and makes sense if we give it the correct interpretation.

Indeed we have to consider a vector field $X$ on $M$ instead of a single $v \in T_pM$. Then the equation reads as
$$d\varphi(X)(f)(\varphi(p)) = X(f \circ \varphi)(p) .$$
Let $X_p = X(p) = v \in T_pM$. Then we have $X(f \circ \varphi)(p) = X_p(f \circ \varphi) = v(f \circ \varphi)$ for the RHS. On the LHS we have, as will be explained below, $d\varphi(X)(f)(\varphi(p)) = d\varphi_p(X_p)(f) = d\varphi_p(v)(f)$.

The context is this: do Carmo states on p. 25 that a vector field $X  :M \to TM$ can also be regarded as function
$$X : \mathcal D (M) \to \mathcal F (M)$$
where $\mathcal F (M)$ denotes the set of functions $M \to \mathbb R$ and $\mathcal D (M)$ is the subset of differentiable functions. For the sake of proper distinction let us temporarily  write $\bar X : \mathcal D (M) \to \mathcal F (M)$ for this alternative representation of $X$.

Then $X : M \to TM$ is differentiable iff $\bar X(\mathcal D (M)) \subset \mathcal D (M)$. Equation (5) can be written without using a coordinate parameterization as
$$\bar X(f)(p) = X(p)(f)  =X_p(f) .$$

A diffeomorphism $\varphi : M \to N$ induces isomorphisms $\varphi^* : \mathcal D (N)  \to \mathcal D (M)$ and $\varphi^* : \mathcal F (N)  \to \mathcal F (M)$ given by $\varphi^*(f) = f \circ \varphi$. Thus it assigns to each vector field $\bar X$ a new vector field $\bar X^\varphi$ on $N$ given by
$$\mathcal D (N) \stackrel{\varphi^*}{\to}  \mathcal D (M)  \stackrel{\bar X}{\to} \mathcal F (M) \stackrel{(\varphi^*)^{-1}}{\longrightarrow} \mathcal F (N) .$$
This means
$$\bar X^\varphi(f) = \bar X(f \circ \varphi) \circ \varphi^{-1} .$$
For $q = \varphi(p) \in N$ this reads as
$$\bar X^\varphi(f)(\varphi(p)) = \bar X(f \circ \varphi)(p) .$$
This is do Carmo's equation and he writes $\bar X^\varphi = d\varphi(X)$ because, as we have seen above, $\bar X^\varphi(f)(\varphi(p)) = \bar X(f \circ \varphi)(p) = \bar X_p(f \circ \varphi) = d\varphi_p(X_p)(f)$ with $d\varphi_p(X_p) \in T_{\varphi(p)}N$.

Title: Question about Do Carmo's notation


问:
Why is the generating function of this combinatorial problem $\prod \frac {1}{1-set(i)*set(j)}$?
Divide the 14 elements {A, B, C, C, C, C, D, D, D, D, E, E, E, E} into 7 groups (one group all have two elements), and I want to find out how many kinds of methods there are without repetition.

I already know a mathematical way to solve the problem:

$$
\begin{array}{c}
set=\{a,b,c,d,e\} \\
G=\prod \frac {1}{1-set(i)*set(j)} \\=\frac{1}{\left(1-a^2\right) \left(1-b^2\right)
   \left(1-c^2\right) \left(1-d^2\right) \left(1-e^2\right)
   (1-a b) (1-a c) (1-a d) (1-a e) (1-b c) (1-b d) (1-b e)
   (1-c d) (1-c e) (1-d e)}
\end{array}
$$

Then we expand the generating function g at the point $\{0,0,0,0,0\}$ into a multivariate Taylor series.

Where the index of $a$ is 1, the index of $b$ is 1, the index of $c$ is 4, the index of $d$ is 4, and the index of $e$ is 4, the coefficient of this term 96 is the desired result.

```<code>set = {a, b, c, d, e};
f = Times @@
  DeleteDuplicates[
   Flatten[Table[1/(1 - set[[i]]*set[[j]]), {i, 1, 5}, {j, 1, 5}]]]
SeriesCoefficient[f, {a, 0, 1}, {b, 0, 1}, {c, 0, 4}, {d, 0, 4}, {e,
  0, 4}]
</code>```

But I don't understand why its generating function is in this form. I hope you can give a detailed explanation and other related examples.

There may be some grammatical errors in the above problems. Please correct them.

答:
The non-rigorous intuition for what's going is:

Let $S=\{AA,AB,AC,...,DD,DE,EE\}$ be the set of 15 different 2-element groups, and call a multiset of elements of $S$ (such as a particular division of the given 14 elements into 2-element groups) an "arrangement". We can completely describe an arrangement by saying how many of each element of $S$ it has. E.g. $(AC,BC,CC,DD,DD,EE,EE)$ has 0 ABs, 1 AC, 0ACs, ..., 2 DDs, 0 DEs, 2 EEs.

Now imagine expanding the product
$$\left((aa)^0+(aa)^1+(aa)^2+(aa)^3+\dots\right)\left((ab)^0+(ab)^1+(ab)^2+(ab)^3+\dots\right)\cdots\left((ee)^0+(ee)^1+(ee)^2+(ee)^3+\dots\right)$$
by just "multiplying it out" into a big sum. Each term of the result will correspond to choosing exactly one term from each of the 15 factors. This is the same as specifying an arrangement by choosing how many of each kind of group it has, and the term we end up with consists of the elements that the corresponding arrangement uses. For example, the arrangement $(AC,AC,BC,DD)$ corresponds to the term $(aa)^0(ab)^0(ac)^2(ad)^0(ae)^0(bb)^0(bc)^1(bd)^0(be)^0\cdots(dd)^1(de)^0(ee)^0=a^2bc^3d^2$.

If we then "combine like terms" of our big sum, we're effectively counting arrangements according to which multiset of underlying elements they use. In particular, the coefficient on $abc^4d^4e^4$ is the number of arrangements that use exactly the 14 elements in question.

The $G$ in your question is exactly the product above, after applying the geometric series formula $x^0+x^1+x^2+\ldots=\frac1{1-x}$ to each factor.


问:
Joint Probability function in expected value of sum of 2 random variable and product of 2 random variable
I realized that same joint probability $f_{xy}(x,y)$ is used in calculation of expected value for sum $E[X+Y]$ and product of 2 random variables $E[XY]$.
How can we use same joint probability in totaly 2 different cases? What is the intuition here?

In case of calculation of sum, I was expecting to use $$f_{X+Y}(x,y)$$ and for calculation of product $$f_{XY}(x,y)$$

As you can see from the pictures, $f_{XY}(x,y)$ is used for both calculations.

Thank you

答:
By the law of the unconscious statistician,

$$E(g(X,Y))=\int_{-\infty}^\infty\int_{-\infty}^\infty g(x,y)f_{X,Y}(x,y)dxdy$$

So in the first case

$$\begin{split}E(X+Y)&=\int_{-\infty}^\infty\int_{-\infty}^\infty(x+y)f_{X,Y}(x,y)dxdy\\
&=\int_{-\infty}^\infty\int_{-\infty}^\infty xf_{X,Y}(x,y)dxdy+\int_{-\infty}^\infty\int_{-\infty}^\infty yf_{X,Y}(x,y)dxdy\end{split}$$

Note that $\int_{-\infty}^\infty f_{X,Y}(x,y)dy=f_X(x)$, the marginal pdf of X, so that

$$\begin{split}\int_{-\infty}^\infty x\left[\int_{-\infty}^\infty f_{X,Y}(x,y)dy\right]dx+\int_{-\infty}^\infty y\left[\int_{-\infty}^\infty f_{X,Y}(x,y)dx\right]dy
&=\int_{-\infty}^\infty xf_X(x)dx+\int_{-\infty}^\infty yf_Y(y)dy\\
&=E(X)+E(Y)\end{split}$$

For the second one,

$$E(XY)=\int_{-\infty}^\infty\int_{-\infty}^\infty xy f_{X,Y}(x,y)dxdy$$

Hypothetically, if X and Y are independent, then $f_{X,Y}(x,y)=f_X(x)f_Y(y)$ so that

$$\begin{split}\int_{-\infty}^\infty\int_{-\infty}^\infty xy f_{X,Y}(x,y)dxdy
&=\int_{-\infty}^\infty x\left[\int_{-\infty}^\infty y f_{Y}(y)dy\right]f_X(x)dx\\
&=\int_{-\infty}^\infty xE(Y)f_X(x)dx\\
&=E(Y)\int_{-\infty}^\infty xf_X(x)dx\\
&=E(Y)E(X)\end{split}$$


问:
Proof ODE approximator isn't universally stable

Given the scalar ODE $y'(t) = \lambda y(t)$ we're given an
approximation $$ y_{n+1} = y_n + P(\lambda h)y_n $$ where $$P(\lambda
 h) = \sum_{k=1}^m \alpha_k \cdot (\lambda h)^k.$$ Note that $h >
 0$ since its the step size and $P$ isn't constant. Now we want to show that for $\lambda <0$,
that $y_n \to 0$ does not hold for all $h$ thus showing the ODE
approximation isn't universally stable.

I've tried a few different approaches and I understand that if $\lambda < 0$ the $exp(-\lambda t)$ goes to 0 as $t$ goes to infinity. If we write
$$
y_n = \left(1 + P(\lambda h)\right)^n y_0
$$
then we see that
$$
| 1 + P(\lambda h) | \geq 1
$$
for $y_n$ to not tend to 0 although I'm not sure how to show this for $\lambda < 0$.

答: You can select $h$ so large that $|P(h\lambda)| \geq 2$ and then $|1+P(h\lambda)| \geq 1$ from the triangle inequality. This is the generalization of the familiar situation that the Euler method for $y'=-y$ diverges if $h>2$.


问:
How can one manipulate an equation within variance?
I encountered the following equation to find the variance of, given that $V(\hat\theta_1)=\sigma^2_1$,and $V(\hat\theta_2)=\sigma^2_2$.

I considered that for the variance of $\mathcal F=\alpha\hat\theta_1+(1-\alpha)\hat\theta_2$:
$$V(\mathcal F)=V(\alpha\hat\theta_1+(1-\alpha)\hat\theta_2)=V(\alpha\hat\theta_1+\hat\theta_2-\alpha\hat\theta_2)=\alpha^2V(\hat\theta_1)+V(\hat\theta_2)+\alpha^2V(\hat\theta_2)=\alpha^2\sigma^2_1+\sigma^2_2+\alpha^2\sigma^2_2$$

The professor considered:
$$V(\mathcal F)=\alpha^2\sigma^2_1+(1-\alpha)^2\sigma^2_2\color{red}{\stackrel{\mathtt{expansion}}{=}\alpha^2\sigma^2_1+(1-2\alpha+\alpha^2)\sigma^2_2=\alpha^2\sigma^2_1+\sigma^2_2-2\alpha\sigma^2_2+\alpha^2\sigma^2_2}$$

I do not see how these two statements are equivalent, presuming both ways are correct in the application of the variance properties.

答:
$$V(\mathcal F)=V(\alpha\hat\theta_1+(1-\alpha)\hat\theta_2)=V(\alpha\hat\theta_1+\hat\theta_2-\alpha\hat\theta_2)\color{red} \neq\alpha^2V(\hat\theta_1)+V(\hat\theta_2)+\alpha^2V(\hat\theta_2)$$

Could you see the reason? $\hat\theta_2$ and $\alpha\hat\theta_2$ are not uncorrelated.

Please see Basic properties of Variance. In particular, check the formula for $$\text{Var} \left(\sum_{i=1}^N a_i X_i \right)$$


问:
How to prove this $\tilde{\mathbb{P}}$ is a probability measure
I am learning the change of measure recently, and the textbook defines a probability measure $\tilde{\mathbb{P}} $ in this following way:

Let (Ω, $\mathbb{F} $, $\mathbb{P} $) be a probability space. Let $f $ be a nonnegative random variable with $\mathbb{E}^{\mathbb{P}}[f]=1 $. For any set $A \in \mathbb{F} $, define the quantity

$\qquad \qquad \tilde{\mathbb{P}}(A) = \mathbb{E}^{\mathbb{P}}[\mathbb{1}_Af]= \int_Af(ω)d\mathbb{P}(ω) $

Then it states this $\tilde{\mathbb{P}} $ is a probability measure without any explanations.

I want to prove this is a probability measure. I recalled the definition of a probability measure:

A probability measure is a map $\mathbb{P}:\mathbb{F} \to [0,1] $ such that

For countable ${A_n} $ such that $A_n \cap A_m = \phi $ for $n \ne m $, $\mathbb{P}(\bigcup_nA_n)=\sum_n\mathbb{P}(A_n)$.

$\mathbb{P}(\phi)=0 $ and $\mathbb{P}(\Omega)=1 $.

Here, the second definition is easy to prove. Here is the proof:

The indicator $\mathbb{1}_{\phi}=0$, $\mathbb{1}_{\Omega}=1$

$\qquad \tilde{\mathbb{P}}(\phi)= \mathbb{E}^{\mathbb{P}}[\mathbb{1}_\phi f]= 0, \quad \tilde{\mathbb{P}}(\Omega)= \mathbb{E}^{\mathbb{P}}[\mathbb{1}_\Omega f]= 1$.

However my question is, how to prove the first definition? Or there is other ways to prove this is a probability measure using definitions?

Many thanks.

答: (1) can be proved in a straightforward manner by just plugging things in. Assume you have a countable family of pairwise disjoint sets $A_i$. Just use the fact (or prove) that for such a family $1_{\cup_i A_i}(\omega)=\sum_i 1_{A_i}(\omega)$


问:
Injective function $f: A \rightarrow B$
I'm trying to prove that $|B| \geq |A|$. Given that $B$ is $[0,1]$ (the real interval) and $A$ is $\mathbb P$ (the set of all primes).

I know I need to define an injective function $f: A \rightarrow B$ and I need to use the injection rule to make my proof.

I need help after this though. What should $f$ be and how should I go from there, especially considering $A$ and $B$ have infinite cardinalities.

答:
You are correct: you must find an injective function (i.e., which satisfies that, if $f(x_{1})=f(x_{2})$, then $x_{1}=x_{2}$, the "injection rule" you mention, better known simply as injectivity of the function) of the form $f:\mathbb{P}\rightarrow[0,1]$.

As Brian M. Scott points out in a comment, the map $f:\mathbb{P}\rightarrow[0,1]$ such that $f(p)=1/p$ is a very natural choice:

the function is well-defined, since for any prime number $p\in\{2, 3, 5, 7, ...\}$, $1/p$ is a real number (strictly) between $0$ and $1$;

the function is injective: take prime numbers $p_{1}$ and $p_{2}$ and assume $f(p_{1})=f(p_{2})$; then
$$\frac{1}{p_{1}}=\frac{1}{p_{2}};$$
by multiplying both sides by $p_{1}p_{2}$ (what you can do since $p_{1}p_{2}$ is not zero, given both $p_{1}$ and $p_{2}$ are prime numbers), you get $p_{2}=p_{1}$, what means $f$ is injective,

and there you go, you presented an injective function, although both sets are infinite.

Other possible functions satisfying what you want are:

$f$ taking a prime number $p$ to $1/\alpha^{p}$, for a fixed real number $\alpha>1$;

if $p$ is the $n$-th prime number, $f$ taking $p$ to $1/n$;

if $p$ is the $n$-th prime number, $f$ taking $p$ to $1/\alpha^{n}$, for a fixed real number $\alpha>0$;

if $p$ is written, in decimal notation, as $a_{1}\cdots a_{n}$, for digits $a_{1}$ through $a_{n}$, $f$ taking $p$ to $0.a_{1}\cdots a_{n}$;

etc...

Title: Injective function $f: A \rightarrow B$


问:
Construction of a vector consisting of entries with a peculiar pattern
Note: A similar problem was asked in this MO post, but no importance was attached to order of nonzero numbers and zeros

Consider constructing a $n$-tuple of numbers $v=(a_1,a_2,\ldots,a_n)$ consisting of nonnegative integers  such that it consists of $m$ nonzero entries, $a_1=1$ and, if $a_j$'s are nonzero, then $a_j\equiv a_{n-j+2}+j-1 \pmod m $, ; with the additional constraint that all nonzero $a_i$'s are distinct modulo $m$. The places of the nonzero numbers  (the $j$s) and zeros are chosen in a fixed predetermined order. For example, suppose $n=20$, $m=11$ and the order of nonzero numbers is $\{1,*,0,*,*,*,0, *,0,0,0,0,0,*,0,*,*,*,0,*\}$, then we can find the numbers satisfying the congruence as $(1,11,0,8,7,9,0,2,0,0,0,0,0,6,0,4,3,5,0,10)$. If $n=7$, $m=5$, and the order of nonzero numbers and zeros is $\{1,*,*,0,0,*,*\}$, we can find it as $(1,4,2,0,0,5,3)$   Note that  the number $m$ itself appears once as a nonzero number satisfying the above congruence.

Is it always possible to construct such a vector for any given $n, m$ and order of nonzero numbers and zeros? I think this should be possible if $m\le n$ is odd.  It is easy to construct if the order of nonzero and zeros were consecutive  (within $\lfloor\frac{n}{2}\rfloor$) . But, in other cases of nonzero and zero patterns, it is not clear as to how to proceed with the construction. Any hints? Thanks beforehand.

答:
The problem can be posed as a maximum independent set problem.

Let $D$ be the set of values for differences $a_j-a_{n-j+2}$ allowed  by the given pattern.
Consider a graph $G$ with the vertex set:
$$V = \{ (a,b) \in \{2,3,\dots,m\}^2\ :\ ((a-b)\bmod m)\in D \}$$
and two vertices $(a_1,b_1)\ne (a_2,b_2)$ are connected with an edge whenever $\{a_1,b_1\}\cap\{a_2,b_2\}\ne\emptyset$ or $a_1-b_1\equiv a_2-b_2\pmod{m}$.

Let $S$ be the maximum independent set. It is clear that $|S|\leq |D|$. When $|S|=|D|$, $S$ contains exactly one pair $(a,b)$ for each difference from $D$, and these pair are disjoint as sets. Hence, they can be placed to the corresponding positions in the pattern to deliver a solution.

Here is a sample SageMath code that for a given $m$ and a difference set $D$, computes the corresponding disjoint pairs of elements.

For a given example with $m=11$ and $D = \{ 1, 3,4,5, 7\}$ it gives the following corresponding disjoint pairs:

```<code>{1: (3, 2), 7: (4, 8), 4: (9, 5), 3: (10, 7), 5: (11, 6)}
</code>```

In some cases the problem allows explicit solution. For example, when $D\cap \{ (-d)\bmod m\ :\ d\in D\}=\emptyset$, a solution can be derived from my construction for the previous question.

Body:
The problem can be posed as a maximum independent set problem.

Let $D$ be the set of values for differences $a_j-a_{n-j+2}$ allowed  by the given pattern.
Consider a graph $G$ with the vertex set:
$$V = \{ (a,b) \in \{2,3,\dots,m\}^2\ :\ ((a-b)\bmod m)\in D \}$$
and two vertices $(a_1,b_1)\ne (a_2,b_2)$ are connected with an edge whenever $\{a_1,b_1\}\cap\{a_2,b_2\}\ne\emptyset$ or $a_1-b_1\equiv a_2-b_2\pmod{m}$.

Let $S$ be the maximum independent set. It is clear that $|S|\leq |D|$. When $|S|=|D|$, $S$ contains exactly one pair $(a,b)$ for each difference from $D$, and these pair are disjoint as sets. Hence, they can be placed to the corresponding positions in the pattern to deliver a solution.

Here is a sample SageMath code that for a given $m$ and a difference set $D$, computes the corresponding disjoint pairs of elements.

For a given example with $m=11$ and $D = \{ 1, 3,4,5, 7\}$ it gives the following corresponding disjoint pairs:

{1: (3, 2), 7: (4, 8), 4: (9, 5), 3: (10, 7), 5: (11, 6)}

In some cases the problem allows explicit solution. For example, when $D\cap \{ (-d)\bmod m\ :\ d\in D\}=\emptyset$, a solution can be derived from my construction for the previous question.


问:
Derivative of a summation with variable in the upper limit
I wish to derive the following equation with respect to its upper limit, but i'm not sure how to proceed.

$${ d \over dv} \sum_{t=1}^{T-t_0(v)} f(t)$$

This link offered some insight, as well as this link, altough i did not quite understand the solution of this second one. I thought of a solution like this.

Create a function using the following integral

$$
G(v,T,t) =
\int_1^{T-t_0(v)} f(t) \, dt
$$

Then take the derivative of this function with respect to v:

$${ d \over dv} G(v,T,t)$$

This is correct?

答:
We can give a sound meaning to a sum with continuous (non-integral) bounds by the concept of Indefinite Summation.

This states that
$$
\eqalign{
  & f(x) = \Delta F(x) = F(x + 1) - F(x)\quad  \Rightarrow   \cr
  &  \Rightarrow \quad F(x) = \Delta ^{\left( { - 1} \right)} f(x) = \sum\nolimits_x {f(x)}
  = \sum\nolimits_{k = 0}^x {f(x)}  + c\quad  \Rightarrow   \cr
  &  \Rightarrow \quad \sum\limits_{k = a}^b {f(k)}  = \sum\nolimits_{k = a}^{b + 1} {f(k)}
  = F(b + 1) - F(a) \cr}
$$
which means that, if there is a  function $F(x)$ such that $F(x + 1) - F(x)=f(x)$, over $x \in \mathbb R$ or even $x \in \mathbb C$, then makes sense to put
that the sum of $f(x)$ between any real or complex bounds $a,b$ is $F(b)-F(a)$.

Which is a concept similar to the integral as anti-derivative.

Once you converted your sum into $F(b+1)-F(a)$, you can derivate wrt $b$ (or also wrt $a$).

Only to note that the traditional sum $\sum\limits_{k = a}^b$ converts in the new acception to $\sum\nolimits_{k = a}^{b + 1} $,
through the following steps
$$
\sum\limits_{k = a}^b {f(x)}  = \sum\limits_{a\, \le k\, \le \,b} {f(x)}
 = \sum\limits_{a\, \le k\, < \,b + 1} {f(x)}  \to \sum\nolimits_{k = 0}^{b + 1} {f(x)}
$$

Example:
$$
\eqalign{
  & \left( \matrix{  n + 1 \cr   3 \cr}  \right)
 = \left( \matrix{  n \cr   3 \cr}  \right) + \left( \matrix{  n \cr  2 \cr}  \right)
 \Rightarrow \Delta \left( \matrix{  n \cr  3 \cr}  \right)
 = \left( \matrix{  n + 1 \cr 3 \cr}  \right) - \left( \matrix{n \cr 3 \cr}  \right)
 = \left( \matrix{n \cr 2 \cr}  \right) = {{n\left( {n - 1} \right)} \over 2}  \cr
  & \sum\limits_{k = 4}^7 {k\left( {k - 1} \right)}
  = \sum\nolimits_{k = 4}^{7 + 1} {k\left( {k - 1} \right)}
  = 2\left( {\left( \matrix{ 7 + 1 \cr 3 \cr}  \right) - \left( \matrix{ 4 \cr 3 \cr}  \right)} \right) = 104  \cr
  & \sum\limits_{k = a}^b {k\left( {k - 1} \right)}
  = 2\left( {\left( \matrix{  b + 1 \cr 3 \cr}  \right) - \left( \matrix{ a \cr 3 \cr}  \right)} \right) \cr}
$$


问:
Integration of upper half of unit circle with fubini's theorem
So I have to calculate the integration of the upper half of the unit circle $K$ with Fubini´s theorem, where

$$K = \{(x,y)\in\mathbb{R}^{2} \mid (x^2 + y^2 \leq 1)\wedge(y\geq 0)\}$$

The given function is $f(x,y) = x^{2}y$.

I am thinking, is it possible that i use the parametrization coordinate, i.e $x = r\cos(t)$, $y = r\sin(t)$, and $r$ from $0$ to $1$, $t$ from $0$ to $\pi$ to calculate the integration function?

Would be glad if someone could help me out! Thanks!

答:
HINT

The integral you are interested in is given by
\begin{align*}
I = \int_{0}^{\pi}\int_{0}^{1}rf(r\cos(\theta),r\sin(\theta))\mathrm{d}r\mathrm{d}\theta
\end{align*}

Can you take it from here?


问:
On the notation of the likelihood function
Let $X$ be a random variable realized as the event $(X=x)$.
The corresponding likelihood function is given by

$$\mathcal{L}_x:\Theta\rightarrow[0,1]$$
$$\theta\mapsto P(X=x|\theta)$$

for a space $\Theta$ of parameter configurations $\theta$.

In the literature, $\mathcal{L}_x(\theta)$ is sometimes written as $\mathcal{L}(\theta|X=x)$. I assume this is done to emphasize that the event $(X=x)$ is 'given'.
However, this notation leads to confusion, since
it suggests that $\mathcal{L}_x$ is a probability density ('conditioning' on the event $X=x$), which appears to not be true in general (cf. second answer in this thread on math.overflow).

So my questions are:

Is $\mathcal{L}(\theta|X=x)$ just 'overloading' the notation $f(\cdot|\cdot)$, or is there some hidden meaning/analogy to conditional probability $P(\cdot|\cdot)$ which I am missing?

Are there other areas in mathematics where $f(\cdot|\cdot)$ is used? Could you provide an example?

Currently, I think $\mathcal{L}(\theta|X=x)$ is a bad notational choice because it caused confusion for me when trying to understand the likelihood function.
Especially since at any point $\theta$, one has $\mathcal{L}(\theta|X=x)=P(X=x|\theta)$

答:
In classical (frequentist) statistics $\theta$ is unknown constant, thus there is no sense in viewing $L(\theta|X)$ in any probabilistic manner. As you can see in the linked thread, $L(\theta|X)$ does not even need to be integrated (w.r.t. $\theta$) to one. Hence, the more common notation is $L(\theta; X=x)$ or its shorthand $L(\theta; X)$ or $L(\theta; x)$, which just designates the fact that we view it as a function of $\theta$ over the parametric space $\Theta$, and regard the $X$ as constant $X=x$.

In the case you are Bayesian, then you usually denote the posterior distribution of $\theta$ by $f(\theta|...)$ or $p(\theta|...)$, not to confuse it with the classical likelihood function. But to this notation to make sense you must assume a prior distribution for $\theta$, $f(\theta)$. That is, from the very beginning you regard $\theta$ as a random variable and not as a constant.


问:
Calculate the limit: $\lim_{n\to+\infty}\sum_{k=1}^n\frac{\sin{\frac{k\pi}n}}{n}$ Using definite integral between the interval $[0,1]$.
Calculate the limit: $\lim_{n\to+\infty}\sum_{k=1}^n\frac{\sin{\frac{k\pi}n}}{n}$ Using definite integral between the interval $[0,1]$.

It seems to me like a Riemann integral definition:

$\sum_{k=1}^n\frac{\sin{\frac{k\pi}{n}}}{n}=\frac1n(\sin{\frac{\pi}n}+...+\sin\pi)$

So $\Delta x=\frac 1n$ and $f(x)=\sin(\frac{\pi x}n)$ (Not sure about $f(x))$

How do i proceed from this point?

答:
I think it should be
$$
\int_{0}^{1}\sin \pi xdx=\frac{2}{\pi}
$$

EDIT OK, the point here is the direct implementation of Riemanns sums with $d x = \frac{b-a}{n}, \ b=1, \ a=0$ and $\frac{k}{n}=x$


问:
Prove by induction: $n(2n^2 − 3n + 1)$ is divisible by 6 for any n ∈ N
After a basis of n=1 and the induction step I have no idea where to go from here.</p>

Body:
Assume $f(n) = n(2n^2 - 3n +1) = 2n^3 - 3n^2 + n$ is divisible by $6$ for $n=k$

Let $f(k) = 6m$ for an integer $m$

Now $$f(k+1) - f(k) = 2(k+1)^3 - 3(k+1)^2 + (k+1) - (2k^3 - 3k^2 + k) = 6k^2$$
Or, $$f(k+1) = f(k) + 6k^2 = 6(m + k^2) = 6n$$ where $n ~ (= m+k^2)$ is an integer.

Hope you can do the rest!


答: Hint: Assume that $n(2n^2 − 3n + 1)$ is divisible by $6$ up to some $n$.  Then consider the expression with $n+1$: $$(n+1)(2(n+1)^2 - 3(n+1) + 1).$$  Try to manipulate the expression so that you end up with $$n(2n^2 − 3n + 1) + \text{expression clearly divisible by }6.$$

Title: Prove by induction: $n(2n^2 − 3n + 1)$ is divisible by 6 for any n ∈ N

Body: After a basis of n=1 and the induction step I have no idea where to go from here.</p>


答:
Without induction. Note that $$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} \in \mathbb{N},\tag{$\ast$}$$meaning that $6$ divides  $n(2n^2+3n+1)$. However, we also have that $$n(2n^2-3n+1) = n(2n^2+3n+1) - 6n^2,$$so $6$ must divide $n(2n^2-3n+1)$ as well.

Proof of $(\ast)$ without induction: start with $$\sum_{k=1}^n (k+1)^3 = \sum_{k=1}^n k^3 + 3\sum_{k=1}^n k^2 + \sum_{k=1}^n k + \sum_{k=1}^n1$$and do a change of variable $k+1 \mapsto k$ on the left hand side to obtain

$$\sum_{k=1}^n k^3 + (n+1)^3 - 1 = \sum_{k=1}^n k^3+3\sum_{k=1}^n k^2 + \frac{n(n+1)}{2} + n,$$cancel the sum of cubes and solve for the sum of squares in one shot. One can also prove that $\sum_{k=1}^n k = n(n+1)/2$ with the same strategy, starting from $\sum_{k=1}^n(k+1)^2$.


问:
Geometry question involving folding a piece of paper
A square piece of paper has side length 1 and vertices A, B, C, and D in that order. As shown in the figure, the paper is folded so that vertex C meets edge AD at point C', and edge BC intersects edge AB at point E. Suppose that C'D = 1/3. What is the perimeter of AEC'?

I already have the answer so I'm moreso looking for the solution. Any pictures would be greatly appreciated.

答:
Let's name the point between C and D, upon which the paper is folded, as point F.

$$ CF = C'F = x $$
$$ DF = CD - CF = 1-x$$

Using the Pythagoras's theorem in $\triangle C'DF$, we will see $x=\frac{5}{9}$.
Since $\triangle EAC'$ ~ $\triangle C'DF$, I can write:

NOTE: P(object) is the perimeter of the object

$$\frac{P(\triangle EAC')}{P(\triangle C'DF)} = \frac{AC'}{DF} $$
$$\frac{P(\triangle EAC')}{4/3} = \frac{2/3}{4/9} $$

$$ P(\triangle EAC') = 2 $$


问:
Left triangular matrix of the form $D(I + B)$?
I want to show that an invertible left triangular matrix $A$ can be represented as $A = D(I + B)$ where $B$ is nilpotent and $D$ is diagonal.
Could I perhaps multiply both sides by $(I + B)^{-1}$ as I know that can be represented as a series? Not sure if that would help.

答:
Let $A=D+L$ where the diagonal matrix $D$ agrees with $A$ on the diagonal and then $L$ is strictly lower triangular.

Since $A$ is invertible, then so is $D$. Therefore $A=D(I+D^{-1}L)$.

Since $D^{-1}L$ is  strictly lower triangular it is nilpotent, as required.


问:
If $B^{m\times m}$ is positive semidefinite, is $A^TBA$ is positive semidefinite?
If $B^{m\times m}$ is positive semidefinite, is it true that $A^TBA$ is positive semidefinite with $A^{m\times n}$? I think it is, because all of my counterexamples have failed, but I don't know how to prove it. I know that $x^TBx\ge 0$ if B is positive semidefinite. Is this the same? Do I need to prove it or is it well-known?

答:
First, note that $M := A^\top B A$ is an $n$-by-$n$ square matrix, and further it is symmetric:
$$
M^\top = (A^\top B A)^\top = A^\top B^\top A = A^\top B A = M
$$
since $B$ is symmetric. Now, indeed, a (real) symmetric $n$-by-$n$ matrix $M$ is psd if, and only if, $x^\top M x \geq 0$ for every $x\in\mathbb{R}^n$. In our case, we can check that, for every $x\in\mathbb{R}^n$,
$$
x^\top M x = x^\top A^\top B A x = (Ax)^\top B (Ax) \geq 0
$$
since $Ax\in\mathbb{R}^m$ and $B$ itself is psd, so $y^\top B y \geq 0$ for every $y\in \mathbb{R}^m$.


问:
How is it possible that we can solve some equations numerically, but can't solve them analytically?
For example, there is the Abel–Ruffini theorem that states there is no analytical solution to polynomial equations of degree five or higher, mentioning $x^5-x-1=0$ as the simplest example. Now if we take $$x^5-x-a=0$$ we can calculate $x$ numerically for every value of $a$ without any trouble. So clearly all the information necessary is contained in the equation. But despite this, we are not able to provide a formula like $x(a)=\dots$.

How is it possible that we have all the information necessary, but we still can not shape it into an explicit formula? What property is it that makes some equations solvable analytically, and others not?

答: What does it mean to solve a problem analytically? Finding "analytic" solutions to a cubic polynomial, I might have to write down numbers like $3^{1/3}$. What is $3^{1/3}$? Is that considered "analytic" to you? Radicals are just ways to write down a family of numbers not equal to the integers and not equal to a fraction of integers, just like how fractions are a way to write down numbers that are not equal to an integer. That theorem states that there are solutions to degree 5+ polynomials that cannot be stated in terms of integers, fractions, and radicals (read the first sentence of the wiki page). But like we did with radicals, we can invent a new way to write down this new class of numbers. I forgot what they are called but they are often written like $B(4)$, just like how $4^{1/4}$ is a thing. Writing down something "analytically" really depends on how you define analytic.


问:
Extension of operator $T$ from a subspace $Y$ of $X$ to $\mathbb{R^n}$ (incomplete answer)
I came across the same question in Carothers' A Short Course on Banach Space Theory as another poster (Extension of operator $T$ from a subspace $Y$ of $X$ to $\Bbb R^n$ without increasing the norm) and was stumped.

The question is as follows:

Suppose $Y$ is a subspace of a Banach space $X$ and let $T\in B(Y,\mathbb{R}^n)$ the space of bounded linear operators from $Y$ to $\mathbb{R}^n$. I am asked to show that there exists an extension $\tilde{T} \in B(X, \mathbb{R}^n)$ such that the operator norm does not increase, i.e. $\|\tilde{T}\| = \|T\|$.

But I do not find the posted solution to be satisfactory - the accepted answer proves that we can proceed if we use the supremum norm. This is all well and good, but can we solve the problem for any norm on $\mathbb{R^n}$? The phrasing of the question seems to indicate that the answer is "yes", but I am not so sure. As the commenter for the accepted solution points out, it's not clear that this will hold for any other norm on $\mathbb{R^n}$.

Further, looking at related posts, such as this one (Bounding the norm of an extension of a linear function from a subspace of a normed space to a finite dimensional normed space) there is a comment about extending bounded linear operators "norm preservingly" (only?) when taking the supremum norm.

In another related post (Extension of linear operator), the accepted response states without proof that if the finite-dimensional space of interest is over $\mathbb{C}$ then the result does not hold for any norms other than the supremum norm.

However, I have not been able to come up with a counter example myself. Any further insight is appreciated.

答:
In order not to emphasize any norm on ${\mathbb R}^n$,  let us replace it by a given finite dimensional normed space $Z$,  so   the
question becomes:

Question 1.  Let  $X$, $Y$ and $Z$ be normed spaces, with $Y\subseteq X$, and $\text{dim}(Z)<\infty $.  Given a bounded linear transformation $T:Y\to Z$, can one find an extension $\tilde T:X\to Z$, with
$\|\tilde T\| = \|T\|$?

In the special case in which $Z=Y$,  and $T$ is the identity operator,  any extension of $T$ to $X$ is effectively
a projection from $X$ to $Y$.  So,  should the above question have an affirmative answer, the follow would also be
answered affirmatively:

Question 2.  Let  $X$ be a normed space and let  $Y\subseteq X$ be a finite dimensional subspace.  Does there
exist a projection from $X$ to $Y$ with norm one?

Unfortunately the answer to (2) is negative, and hence the same goes for (1).

A  counter example is $X={\mathbb R}^3$, equipped
with the sup norm, namely
$$
  \|(x,y,z)\|_\infty  = \max\{|x|, |y|, |z| \},
  $$
and $Y$ being the two dimensional subspace given by
$$
  Y=\{(x, y, z) : x+y+z=0\}.
  $$
The best way to convince oneself that  there is no projection from $X$ to $Y$ with norm one, is to make a cardboard
model of this cube

cut it along the red line, place one of the two halves on top of the table with the red hexagon down,

and attempt to
shine a flashlight so that the shadow is restricted to within the  hexagon.  It is impossible!


问:
How to solve the irrational system of equations?
Solve the system of equations $$\begin{cases} \sqrt{x+2y+3}+\sqrt{9 x+10y+11}=10,&\\[10pt] \sqrt{12 x+13y+14}+\sqrt{28 x+29y+30}=20. \end{cases} $$</p>

答:
Hint: Try $t=x+y+1$ to get

\begin{cases} \sqrt{t+y+2}+\sqrt{9t +y+2}=10,&\\[10pt] \sqrt{12t+y+2}+\sqrt{28t+y+2}=20. \end{cases}

And solve by squaring.


问:
Linear Independence - Sequence
I am not sure how to prove this "we know that two polynomials are equal if and only if all their parameters are equal".

Please tell me if my answer is wrong too.

Show that $\{(1, l,l^2, l^3,... \}\in \mathbb{R}^\infty \colon l \in \mathbb{R}\}$ is linearly independent.

We need to prove that for all in the reals $a_1 \cdot 1 + a_2 \cdot l + a_3 \cdot l^2 + a_4 \cdot l^3 +\dotsc = 0$ iff $a_i = 0$ for all $i \in \mathbb{N}$. We can consider $a_1 \cdot 1 + a_2 \cdot l + a_3 \cdot l^2 + a_4 \cdot l^3 +\dotsc$ as a polynomial and we know that two polynomials are equal if and only if all their parameters are equal, therefore $a_i = 0$ for all $i\in \mathbb{N}$.

答:
You can start with by saying $l$ is not in the span of (1) since no linear combination of scalars can equal a degree one polynomial. So $(1,l)$ is a list of linearly independent vectors. Similarly, $l^2$ is not in the span of $(1,l)$ since no linear combination of polynomials of degree less than two can equal a polynomial of degree two. So $(1,l,l^2)$ is a list of linearly independent vectors. Next is $(1,l,l^2,l^3)$, and so on by induction

Also, you can modify the definition of linear independent to include infinite dimensional vector spaces. When we say that a vector space $V$ is generated by or spanned by an infinite set of vectors $\{v_1, v_2, . . . \}$, we mean that each vector $v$ in $V$ is a finite linear combination $\lambda_{i_1} v_{i_1} + · · · + \lambda_{i_n}v_{i_n}$ of the $v_i$ ’s. Likewise, an infinite set of vectors $\{v_1, v_2, . . . \}$ is said to be linearly independent if the only finite linear combination of the $v_i$’s that is zero is the trivial linear combination. So a set $\{v_1, v_2, v_3, . . . , \}$ is a basis for $V$ if and only if every element of $V$ can be be written in a unique way as a finite linear combination of elements from the set.

Title: Linear Independence - Sequence


问:
Intuitive understanding behind exterior algebra construction
I'm trying to get deeper intuition into the exterior algebra construction on a finite dimensional $\mathbb{R}$-vector space.

Our accustomed notion of volume given by measure is neither multi-linear nor anti-symmetric, so I don't buy construction of a 'volume function' as an a priori motivation for an exterior algebra.

It's great that $v_1 \wedge \cdots \wedge v_n = \alpha \ e_1 \wedge \cdots \wedge e_n$ computes the (signed) volume $\alpha$ of the parallelotope spanned by these vectors.

But this fact seems rather arbitrary and a priori unexpected.

It would be nice to have a narrative as to why constructing an exterior algebra on a vector space is just the natural thing to do. For instance, generalizing from metric spaces to topological spaces is very natural once we realize that metrics just generate open sets, and that continuity of functions can be characterized by their behavior on open sets alone.

Is there any reason why one would intuitively anticipate beforehand that constructing an alternating algebra on a vector space would give a device to compute volumes, detect linear dependence etc.?

Or should the recognition of these facts just be considered a random encounter in the process of experimentation with mathematical constructs?

答:
This is not rigorous, but hopefully addresses the intuitive understanding of the motivation for arriving at the exterior algebra.  I am putting non-standard terminology that I am inventing for this answer in quotes.

In an affine geometry, relative locations of points is naturally described by vectors.  For example, vectors giving the relative positions of a sequence of points naturally add together to give a resultant vector that gives the relative position of the last to the first point.  For my purpose here, it is sufficient to consider the case of a collinear sequence of points: when we add vectors from the same subspace, the original and final vectors all belong to the same one-dimensional subspace (are all scalar multiples of a single vector).  This holds independently of any length measure (e.g. a metric).

Similarly, in any plane of an $n$-dimensional affine space, we can have plane figures.  Any linear transformation of the space keeps them related, and in some sense there is a measure of relative "affine area" of parallel plane geometric figures that is preserved by affine transformations of the space.  A simple argument would show that any area must be a bilinear function of any two linearly independent vectors in the same plane that are transformed with the figure.  My crucial premise: the "affine area" of any two plane figures in parallel planes must be related by a scalar multiplication just as two vectors giving offsets on parallel lines are related, and the scalar value is unaffected by affine transformations of the space.  This holds for any higher-dimensional figures, such as the "affine volume" of parallel geometric figures in 3-dimensional affine subspaces: provided that the containing geometric $k$-flats containing the $k$-dimensional figures are parallel, the "affine $k$-measure" will be related by some scalar.

Multilinearity (bilinearity in the case of "affine area") together with this property of the "affine $k$-measure" of figures in parallel $k$-flats always being related by a scalar is sufficient to derive the exterior algebra as the unique algebra with the necessary properties.  We can see this as follows in the 2-dimensional case (but the argument generalizes to any number of dimensions).  Consider the "affine area" (i.e. "affine 2-measure") of a geometric figure in a plane (2-flat) in a possibly higher-dimensional affine space, along with two linearly independent vectors $u_1$ and $u_2$ and similarly of another pair $v_1$ and $v_2$ in the plane.  The "affine area" of the figure is a bilinear function of each pair of vectors: $a = k_1 f(u_1, u_2) = k_2 f(v_1, v_2)$ for some scalar constants $k_1$ and $k_2$.  However, being in the same plane, these two pairs of vectors are linearly related: $v_1 = c_{1,1}u_1 + c_{1,2}u_2$ and $v_2 = c_{2,1}u_1 + c_{2,2}u_2$.  This leads us to the relationship $k_1 f(u_1, u_2) = k_2 f(v_1, v_2) = k_2 f(c_{1,1}u_1 + c_{1,2}u_2, c_{2,1}u_1 + c_{2,2}u_2)$.  Using the bilinearity of the function $f$ and setting $u_2 = 0$, it follows that for all $u_1$, $f(u_1, u_1) = 0$.  Thus we have that any reasonable "affine area" function $f$ of two vectors that are directly related to this area is necessarily alternating, i.e. equal to zero when two of its arguments are equal.  I will not explicitly extend this to higher dimensions here.

Being an alternating multilinear function on its parameters is the defining characteristic of the exterior product: up to isomorphism, it is the unique "freest" or "most general" such function.  That is, the exterior algebra is unique in having the simple geometric properties that we demanded when working from geometric intuition.


问:
You roll a die 8 times. What is the probability of getting an odd number at least twice and an even number at least three times?
My initial idea was to do something such as $3^5 * 6^3$ for the possible cases. However, I believe this is not appropriate for this question, since simply multiplying is usually used in cases in which order matters (e.g. combination of a locker), and here order does not matter.
I'd imagine that maybe we should use something like Bernoulli trials or Combinations, but I'm not sure.

答:
Note: I'm not saying this the the answer to your question. I'm showing you the binomial distribution and a process for adding desired
terms. You need to decide which terms are
needed to answer your question.

The probability of an Odd result on any one roll is $p=1/2.$ The number $X$ of odd results in $n = 8$ rolls is
distributed $X \sim \mathsf{Binom}(n = 8, p = 1/2).$

Specifically, the probability that $X = k$ is
$$P(X = k) = {8\choose k}(1/2)^8,$$
for $k = 0,1,\dots,8.$

Suppose you want the probability $P(2 \le X \le 5).$ That's
$$P(2 \le X \le 5) = P(X \le 5) - P(X \le 1)\\
= \sum_{k=2}^5 {8\choose k}(1/2)^8 = 0.8203,$$
to four places. To get this probability, you have four terms to add.

In R statistical software this can be computed
as follows, where dbinom is a binomial PDF and pbinom is a binomial CDF.

```<code>k = 2:5; dbinom(k, 8, 1/2)
[1] 0.1093750 0.2187500 0.2734375 0.2187500
sum(dbinom(k, 8, 1/2))  # add 4 PDF terms
[1] 0.8203125

pbinom(1:5, 8, 1/2)
[1] 0.03515625 0.14453125 0.36328125 0.63671875 0.85546875
diff(pbinom(c(1,5), 8, 1/2))  # subtract 2 CDF values
[1] 0.8203125
</code>```

In the figure below, we are adding the heights of the four
maroon bars.

R code for figure above:

```<code>plot(0:8, dbinom(0:8, 8, 1/2), type="h", lwd=2,
     col="blue", ylab="PDF", xlab="k",
     main="PDF of BINOM(8, .5)")
 abline(h = 0, col="green2")
 lines(2:5, dbinom(2:5, 8, 1/2), type="h",
       lwd=2, col="maroon")
</code>```

Body:
Note: I'm not saying this the the answer to your question. I'm showing you the binomial distribution and a process for adding desired
terms. You need to decide which terms are
needed to answer your question.

The probability of an Odd result on any one roll is $p=1/2.$ The number $X$ of odd results in $n = 8$ rolls is
distributed $X \sim \mathsf{Binom}(n = 8, p = 1/2).$

Specifically, the probability that $X = k$ is
$$P(X = k) = {8\choose k}(1/2)^8,$$
for $k = 0,1,\dots,8.$

Suppose you want the probability $P(2 \le X \le 5).$ That's
$$P(2 \le X \le 5) = P(X \le 5) - P(X \le 1)\\
= \sum_{k=2}^5 {8\choose k}(1/2)^8 = 0.8203,$$
to four places. To get this probability, you have four terms to add.

In R statistical software this can be computed
as follows, where dbinom is a binomial PDF and pbinom is a binomial CDF.

k = 2:5; dbinom(k, 8, 1/2)
[1] 0.1093750 0.2187500 0.2734375 0.2187500
sum(dbinom(k, 8, 1/2))  # add 4 PDF terms
[1] 0.8203125

pbinom(1:5, 8, 1/2)
[1] 0.03515625 0.14453125 0.36328125 0.63671875 0.85546875
diff(pbinom(c(1,5), 8, 1/2))  # subtract 2 CDF values
[1] 0.8203125

In the figure below, we are adding the heights of the four
maroon bars.

R code for figure above:

plot(0:8, dbinom(0:8, 8, 1/2), type="h", lwd=2,
     col="blue", ylab="PDF", xlab="k",
     main="PDF of BINOM(8, .5)")
 abline(h = 0, col="green2")
 lines(2:5, dbinom(2:5, 8, 1/2), type="h",
       lwd=2, col="maroon")


答:
You must have at least $3$ even numbers, and as many as $6$

How many combinations for each number of odd rolls?

$\sum_\limits{n=3}^{6} \frac {{8\choose n}}{2^8}$

Title: You roll a die 8 times. What is the probability of getting an odd number at least twice and an even number at least three times?

Body:
My initial idea was to do something such as $3^5 * 6^3$ for the possible cases. However, I believe this is not appropriate for this question, since simply multiplying is usually used in cases in which order matters (e.g. combination of a locker), and here order does not matter.
I'd imagine that maybe we should use something like Bernoulli trials or Combinations, but I'm not sure.


问:
What do we call the negation of logical equivalence?
The statement that '$x$, $y$ and $z$ are equivalent' just means

all of $x$, $y$ and $z$ are false, or

all of $x$, $y$ and $z$ are true.

Now suppose its not the case that $x$, $y$ and $z$ are equivalent. That is, suppose:

at least one of $x$, $y$ and $z$ is false, and

at least one of $x$, $y$ and $z$ is true.

Under these circumstance, we say that $x$, $y$ and $z$ are what? Inequivalent? This seems wrong, because some of them will still be equivalent.

答:
There is no "special term": We can simply say "not all of $x, y, z$ are equivalent" or "$x, y, z$ are not all equivalent."

As you note, that's not the same thing as all of them being inequivalent, because two of them might still be equivalent, and in the case of logical equivalence, we would need one pair and not all pairs logically equivalent.

Negation of the statement of mutual logical equivalence involves DeMorgan's:

We can say "$x, y, z$" are all equivalent by stating $$(x \iff y) \land (y \iff z)\tag{A}$$ and by transitivity we have also conveyed in $(A)$ that $(x \iff z)$.

And so to negate an entire conjunction, we can arrive at the disjunction of negated dijunctions $$\lnot(x \iff y) \lor \lnot(y \iff z): \tag{$\lnot $ A}$$

$\qquad\qquad\qquad \qquad\qquad\qquad$

In disjunctive normal form $(\lnot A)$ equivalent to: $$(\lnot x \land y) \lor (\lnot y \land z) \lor (x \land \lnot z)\tag{DNF}$$

And in conjunctive normal form, as you describe in your bullets: we have that $(\lnot A)$ is equivalent to $$(x \lor y\lor z) \land (\lnot x \lor \lnot y \lor \lnot z)\tag{CNF}$$


问:
Existence of ''positive'' solution to a linear system implies existence of positive/negative basis element?
I'm encountering this statement and still haven't found out if it is true or not.

Assume $\left\{e_1,...,e_k\right\}$ is a basis of $\ker A$, where $A$ is a matrix of order $n$. Suppose there exists $x \in int(\mathbb{R}_{+}^n)$ such that $Ax=0$.

My question is whether we can conclude that there exists a basis element $e_i$ ($1\le i \le k$) such that $e_i \in int(\mathbb{R}_{+}^n) \cup int(\mathbb{R}_{-}^n)$ ?

Notation elaboration: $int(\mathbb{R}_{+}^n)/int(\mathbb{R}_{-}^n)$ denotes all points in $\mathbb{R}^n$ with all positive/negative components, respectively.

Any idea is highly appreciated.

答: The vectors $e_1=(-1,2)$ and $e_2=(2,-1)$ form a basis of $\mathbb R^2$. Now take $A=0$ so that $ker A=\mathbb R^2$.


问:
Degree of a map from the surface of a tetrahedron to a triangle with identified edges
This problem comes from Topology and Geometry of Bredon :

Consider the surface $S$ of a tetrahedron. Number its vertices in any order and number three vertices of a triangle $1,2,3$. Let $S'$ be the triangle with the sides collapsed to a point. Define a map $S \to S'$ by taking a triangle $T$ in $S$ and mapping its vertices to the points $1,2,3$ of the triangle in order of the numbering of the vertices of $T$. Extend that to an affine map of $T$ to the triangle and then collapse to get a map to $S'$. Find the degree of this map up to sign. Does the answer depend on the numbering?

The inverse image of the point at the center of the face $S'$ consists of 4 fours points, each of which is located on a different face of the tetrahedron. How can I compute the sign of the degree of the map? In order to do so, I think that one could say the degree is the sum of the local degree.

答:
Here is an intuitive visual argument. Picture the outward normal vector to each face of $S$ and picture the triangle, $\Delta$, flat on the $x, y$ plane in $\mathbb{R}^3$, with an upward-pointing normal vector. These vectors specify the orientations of these spaces (the faces of $S$ viewed as subspaces with induced orientation). Visualize taking each of these faces and placing them on $\Delta$ according to the numbering that you chose and see whether the normal vectors point in the same or opposite direction. If they point in the same direction, then the local degree around a point of the corresponding face is $+1$, otherwise, it is $-1$. If you draw a picture and label everything you'll see that two faces will have $+1$ and two $-1$ so the degree must be $0$.

Here is a sketch of a more rigorous approach using Corollary 7.5 in Chapter IV of Bredon.
Model $S$ as the boundary of the standard $3$-simplex:
$$ S = \partial\Delta_3 = \{(t_i)_{i=1}^4\in\mathbb{R}^4|\sum_{i=1}^4 t_i = 1, t_i = 0 \textrm{ for some }i\}.$$ Let $T_i\subset S$ be the subset of points for which $t_i = 0$. These are the faces of the tetrahedron. Let $v_i$ be the $i$-th vertex, which is also just the $i$-th standard basis vector in $\mathbb{R}^4$. Model the triangle $\Delta$ as just one of these faces, say $T_4$. We order the vertices of $S$ according to $i$ and similarly the vertices of $\Delta = T_4$. Now, the resulting map $f:S\to S'$ doesn't factor through a map into $\Delta$, but it does if we restrict to the interior of each face. Let $f_i:\textrm{int}(T_i)\to \Delta$ be the restriction. Then these maps are given by, for example, $f_1(0,t_2,t_3,t_4) = (t_2,t_3,t_4,0)$, and similarly defined for other $i$. Now, in order to compute the local degree, we need to take coordinates for the interior of each face. For int$(T_4) = \textrm{int}(\Delta)$, we can take for example
$$(0,1)\times (0,1)\xrightarrow{\phi_4}\textrm{int}(T_4).$$
$$(u,v)\mapsto (u,v,1-u-v,0) $$

Choose local coordinates for the rest of the $T_i$, keeping in mind that they should be rotations of the coordinates we chose for $T_4$. From there you can compute the Jacobian determinants of each of the maps $\phi_4^{-1}\circ f_i\circ\phi_i$ and find the same result as the visual argument above.


问:
permutation with restriction - no two zeroes can be together from an set consisting of 3 zeroes
I have this as data set - 0 1 0 1 0
now how to find out the number of permutations with restriction that no two zeroes can be together.

i.e. 00110, 00011, 11000 etc. -> all these are false cases

I was trying with 5!/2!.3! - (cases where two zeroes are always together).

But I am able to find out how to do this? Any help ?

答: For this case, and any time there is one less $1$ than $0$'s, there is only one-the one you show.  If you have more $1$'s, the number rises.  Let there be $a\ 1$'s and $b\ 0$'s.  Think of putting all the $0$'s in a line and scattering the $1$'s before, between, and after them.  You require at least one $1$ in each of the between spaces.  If you have a total of $c \  1$'s on the ends, you are looking for a composition of $a-c$ things into $b-1$ pieces, which is ${a-c-1 \choose b-2}$ times the ways of arranging the $c\ 1$'s on the ends, which is $1$ for $c=0$ and $2$ for $c \gt 0$.  The final answer is then $${a-1 \choose b-2}+\sum_{i=1}^{a-b+1}2{a-i-1 \choose b-2}$$</p>