问:
Singularities of $f(z)=z/\cos(z)$
Regarding complex functions (in complex variables), I was wondering why the function $g(z)= \cos(z)$ has a singularity at $z = \infty$ but $f(z)= \dfrac{z}{\cos(z)}$ does not.

I am a bit confused about the concept of singularities at $\infty$.

Thank you.

答:
So, to sum up after all these comments:

$g(z)$ indeed has an essential singularity at $z=\infty$ because the function $h(z)=g\big(\frac1z\big)$ has an essential singularity at $z=0$:
$$h(z) = \cos\left(\frac1z\right) = 1- \frac1{2!}\left(\frac1z\right)^2 + \frac1{4!}\left(\frac1z\right)^4+ \dots $$
has infinitely many terms in its Laurent expansion at $0$.

On the other hand, $f(z)$, all appearances to the contrary, does not have an essential singularity at $\infty$: $f(z)$ has poles at $z=z_n=n\pi + \pi/2$ and $z_n\to\infty$, and so $\infty$ is not an isolated singular point.

Title: Singularities of $f(z)=z/\cos(z)$


问:
What is the difference in the equations listed here
This is a question in economics where we create a model for GDP per capita. GDP per capita can be caculated in such formula where you take GDP/population. This formula (GDP/population) can be broken down in a way where you see what affects the GDP per capita formula "in the real world". I'll attach a picture down below so you understand it.

I have another formula where someone has changed this into something I can't understand really. So I was wondering if someone could explain this image below and why the changes were made. I'll explain more below the second picture.

Y = GDP

B = Population

H = Hours worked

N = Employed workers

B15-74 = Population which age bracket is eligible to work

I can't understand why you change the whole formula in this way as delta in the numerator and the regular expression in the denominator. The way this is explained in the question, I can write below with another attachment. This image to the left is suppose to describe the "relative change in hours worked per employee". So does this mean that the whole equation is the relative change in BNP per capita. But what does this mean "relative change" I want the mathematical explanation of why they would change the whole equation this way and remove the multiplication and instead add addition between the fractions. And also why they would change the fractions in such way, as delta in the numerator and the regular expression in the denominator.

答:
The equation for $\frac YB$ is the same as the equation for $\frac{GDP}{POP}$ except for renaming the terms.  You should be able to see that these equations are identities in that all the intermediate terms cancel.  Mathematically we could replace any of these terms with an unrelated variable, like replacing Employment with the number of flowers blooming in my yard.  As long as we do it in both the numerator and denominator it is fine.  They have chosen the intermediate terms to try to make the ratios meaningful.  They explain some about why they chose those variables in the text below the first equation.  $\frac {GDP}{Hours\ worked}$ is the average production per hour worked.  We can compare different countries to see if one is more efficient than another this way.  Incidentally, the $\frac YL$ they mention at the end would be $\frac YN$ in the terms of the second entry.

$\frac {\Delta (something)}{something}$ represents the fractional change in $something$.  If you have $1000$ cars and buy $50$ more we would say the fractional change in the number of cars you have is $\frac {\Delta \ cars}{cars}=\frac {1050-1000}{1000}=0.05=5\%$.  The last equation reflects a general result that if you have a quantity that is the product of a number of variables, the fractional change is (approximately) the sum of the fractional changes of the variables.  The proof goes through logarithms and calculus.  If you have $A=BC$ we can take the logarithm of each side, then take the derivative and get
$$A=BC\\ \log A=\log B + \log C\\ \frac {\Delta A}A=\frac {\Delta B}B+\frac {\Delta C}C$$  Your equation does the same with four terms.

ContentLicense: CC BY-SA 4.0


问:
Prove that Y is Hausdorff iff X is Hausdorff and A is a closed subset of X
In user642796's answer at Prove that $Y$ is Hausdorff iff $X$ is Hausdorff and $A$ is a closed subset of $X$, he writes "If $A \subseteq X$ is not closed, let $x \in \overline{A}\setminus A$. Show that the points $p(x,0),p(x,1)$ (which are distinct) cannot be separated in $Y$." But I do not see how to show that part - any assistance would be appreciated.

For context: the original problem statement. Let $X$ be a topological space and $A$ a subset of $X$. On $X\times\{0,1\}$ define the partition composed of the pairs $\{(a,0),(a,1)\}$ for $a\in A$, and of the singletons $\{(x,i)\}$ if $x\in X\setminus A$ and $i \in \{0,1\}$. Let $R$ be the equivalence relation defined by this partition, let $Y$ be the quotient space $[X \times \{0,1\}]/R$, and let $p:X \times \{0,1\} \to Y$ be the quotient map. Prove that $Y$ is Hausdorff if and only if $X$ is Hausdorff and $A$ a closed subset of $X$.

答:
Suppose that $U_0$ is an open nbhd of $p(\langle x,0\rangle)$ in $Y$. Then by definition $p^{-1}[U_0]$ is an open nbhd of $\langle x,0\rangle$ in $X\times\{0,1\}$, so there is an open nbhd $V_0$ of $x$ in $X$ such that $V_0\times\{0\}\subseteq p^{-1}[U_0]$. Similarly, if $U_1$ is an open nbhd of $p(\langle x,1\rangle)$ in $Y$, there must be an open nbhd $V_1$ of $x$ in $X$ such that $V_1\times\{1\}\subseteq p^{-1}[U_1]$.

Now $x\in\operatorname{cl}_XA$, and $V_0\cap V_1$ is an open nbhd of $x$ in $X$, so there is a point $a\in V_0\cap V_1\cap A$. Then $\langle a,i\rangle\in V_i\times\{i\}\subseteq p^{-1}[U_i]$ for $i\in\{0,1\}$, so $p(\langle a,0\rangle)\in U_0$, and $p(\langle a,1\rangle)\in U_1$. But $a\in A$, so $\langle a,0\rangle\mathrel{R}\langle a,1\rangle$, and therefore $p(\langle a,0\rangle)=p(\langle a,1\rangle)=\{\langle a,0\rangle,\langle a,1\rangle\}\in Y$. In other words,

$$\{\langle a,0\rangle,\langle a,1\rangle\}\in U_0\cap U_1\,,$$

and $U_0$ and $U_1$ are not disjoint. Since they were arbitrary open nbhds of $p(\langle x,0\rangle)$ and $p(\langle x,1\rangle)$, respectively, those two points cannot be separated by disjoint open sets in $Y$, and $Y$ is not Hausdorff.


问:
Solving a system of differential equations by substitution
I am being asked to solve a system of differential equations. We are provided with a hint to use substitution to simplify the process, but I don't follow how to use this process to move forward.

Define the vectors $\vec y$ = [$y_1$, $y_2$], $\vec y$' = $\left[ \frac{dy_1}{dt}, \frac{dy_2}{dt}\right]$ and the coefficient matrix A = $\begin{bmatrix} a & -b\\ b & a\end{bmatrix}$.

Then, $\vec y$' = Ay defines a system of linear homogeneous differential equations. Note that A has complex eigenvalues.

Hint: To solve the system of equations, write $z = y_1 + iy_2$ and calculate $z'$. For some $u$ defined in terms of $a$ and $b$ you should find that $z'=uz$, which is easily solved. Now, write the solution in terms of the original variables $y_1$ and $y_2$. You will need to use the identity $e^{itb}=cos tb+isin tb.

I've reached out to my instructor but it consistently takes over a week to get a response.

My attempt so far is below.

Finding $z'$:

$z' = \frac{dz}{dt} = \frac{d}{dt}\left(y_1 + iy_2\right) = \frac{dy_1}{dt} + i\frac{dy_2}{dt}$

Let us write all of this in a different form:

$\frac{dy_1}{dt} = ay_1 - by_2$

$\frac{dy_1}{dt} = by_1 + ay_2$

$\frac{dz}{dt} - (ay_1-by_2) + i(by_1 + ay_2)$

$z' = (a+ib)y_1 + (-b+ia)y_2$

Now we set $z' = uz$

$uz = u(y_1 + iy_2) = (a+ib)y_1 + (-b+ia)y_2$

$u = \frac{(a+ib)y_1+(-b+ia)y_2}{y_1+iy_2}$

My problem is I'm not really sure where I'm going at this point.
Any advice would be greatly appreciated!

答:
$$\frac{dz}{dt} = (ay_1-by_2) + i(by_1 + ay_2)$$
$$\frac{dz}{dt} = a(y_1+iy_2) + ib(y_1 +i y_2)$$
This can be rewritten as:
$$\frac{dz}{dt} = (a+ib)(y_1+iy_2) $$
$$\frac{dz}{dt} = (a+ib)z $$
$$\frac{dz}{dt} = uz $$
Where $u=a+ib$. Since $u$ is a function of $a,b$.


问:
Can I solve this blood question by re-arranging?
I believe the answer to the question in the attached image is the 4th bubble. Be careful not to accidentally miss the formula under the worded explanation, at the bottom of the image.

The unit litre/gram is equivalent to C = something, so rearrange the equation. My reasoning:

F = Q/C

Multiply both sides by C.

CF = Q

Divide by F to get C on its own.

C = Q/F

Could someone please check if my thought process was correct, and if I have done the algebra in the optimal way?

Blood equation

答:
Yes, your answer is correct.

We can check by using the units to see if we get back the correct equation.

Starting with

$$\dfrac {\text {liter}}{\text {minute}} = \dfrac {\text {grams/minute}}{\text {grams/liter}}$$

Multiply by $\dfrac {\text {grams/liter}}{\text {grams/minute}}$  to get $$\text {liter}\cdot \text {grams/liter} = \text {minute}\cdot \text {grams/minute}$$

Divide by $\text {liter}$ to get $$\dfrac {\text {grams}}{\text {liter}} = \dfrac {\text {grams}}{\text {minute}} \cdot \dfrac {\text {minute}}{\text {liter}}$$

Transform to a division problem by inverting the $\dfrac {\text {minute}}{\text {liter}}$ to get $$\dfrac {\text {grams}}{\text {liter}} = {\dfrac {\text {grams}}{\text {minute}} \div \dfrac {\text {liter}}{\text {minute}}}$$

Since $C$ is $\text {grams/liter}$, $Q$ is $\text {grams/minute}$ and $F$ is $\text {liter/minute}$, we have $$C = \dfrac QF$$ as desired.


问:
Why can a $2k$-regular graph be $k$-factored?
From Wikipedia:

If a connected graph is $2k$-regular it may be $k$-factored, by choosing
  each of the two factors to be an alternating subset of the edges of an
  Euler tour.

I don't understand why those alternating subsets form $k$-factors.

答:
In a connected $2k$-regular graph $G=(V,E)$, you can find a path $\pi=v_0 e_1 v_1 e_2\dots e_n v_0$, taking each edge $e_i$ of the graph exactly once (Eulerian path). Then if you look at the partitions $E_1=\{e_1,e_3,\dots\}$ and $E_2=\{e_2,e_4,\dots\}$, they form a $k$-factorization of $G$. Indeed, each vertex $v$ is adjacent to exactly $k$ edges from $E_1$, and $k$ edges from $E_2$, because every time an edge from $E_1$ enters $v$ in $\pi$, then the next one is from $E_2$ and exits $v$, and vice-versa.
However this only works if $e_n\in E_2$, otherwise the above statement is not true for $v_0$. This means that we have to additionally require that $|E|$ is even. $K_3$ is counter-example for the odd case.</p>

ContentLicense: CC BY-SA 3.0


问:
Persistant homology - Point data sets from images
I have been reading about topological data analysis techniques and specifically about Persistent Homology. The examples I have seen so far use point clouds as the data sets. But what if we have, say, a set of images? How do we take our images and go to point clouds?</p>

答:
One way to take an image to a point cloud is if you want to study the statistics of the small local patches in that image. For example, you can study $3 \times 3$ pixel patches by cutting a black-and-white image up into many $3 \times 3$ pixel patches, and then considering each patch as a vector in 9-dimensional space (as each patch is a list of 9 different pixel intensity values). The papers "On the Local Behavior of Spaces of Natural Images" (https://link.springer.com/article/10.1007/s11263-007-0056-x) and "On the Nonlinear Statistics of Range Image Patches" (https://www.math.colostate.edu/~adams/research/OnTheNonlinearStatisticsOfRangeImagePatches.pdf) take this approach, for example. To try out example computations along these lines, see the image patch datasets at https://github.com/ds4m/topological-data-analysis/wiki or at Section 6 of https://www.math.colostate.edu/~adams/research/javaplex_tutorial.pdf.

It is important to note that you can instead transform an image into persistence barcodes (not passing through


问:
$\frac{d}{dx}[\int_{2x^2}^6 \sin(y^3)dy]$
I'm trying to evaluate the following:
$$\frac{d}{dx}[\int_{2x^2}^6 \sin(y^3)dy]$$

I'm really unsure as to how I'm supposed to evaluate these, I was thinking in the direction of well, what ever the outcome is, the upper bound is not going to stick around since that would just be a constant which you then differentiate, but for the lowerbound I was thinking it would just be $-sin((2x^2)^3)$ but that doesn't seem to be correct.

Would appreciate some help, thanks!

答:
Let $f(x) = \sin(x^3)$ and let $F(x)$ be an antiderivative of $f$, then

\begin{align*}
\frac{d}{dx} \int_{2x^2}^6 \sin(y^3) \,dy
&= \frac{d}{dx} \int_{2x^2}^6 f(y) \,dy
\overset{(*)}{=} \frac{d}{dx} \left[\,F(6) - F(2x^2) \,\right]\\
&= 0 - \tfrac{d (2x^2)}{dx} F'(2x^2) = - 4xf(2x^2)\\
&= -4x\sin(8 x^6)
\end{align*}

where (*) is the fundamental theorem of calculus


问:
Binomial formula in $GF(2^m)$
there is a binomial formula:

$$(x+y)^n=\displaystyle\sum_{k=0}^n \binom{n}{k} x^{n-k} y^k$$

When operations are done in $GF(2^m)$ then all positive integers are reduced $\bmod2$, so binomial formula for $n=2^i$ in $GF(2^m)$ is:

$$(x+y)^{2^i}=x^{2^i} + y^{2^i} $$

So now the question. If there are reduced all binomial coefficients $\binom{n}{k}$, then why exponents like $2^i$ are not reduced?

答:
Regard the simplest non-trivial example to get a better understanding.

For example $\{0,1,x,x+1\}$ with $2=0$ and $x^2=x+1$.

Title: Binomial formula in $GF(2^m)$


问:
Is an odd function differentiable at zero if it is differentiable elsewhere (and has no vertical tangent)?
Let $f: \mathbb{R} \to \mathbb{R}$ be an odd function that is differentiable everywhere except possibly at $x = 0$. Suppose that $f$ does not have a vertical tangent at $x = 0$ (so, an example like $y = x^{1/3}$ doesn't count). Does it follow that $f$ is differentiable at zero?

If that is true, how about if we weaken the hypothesis of differentiability on $\mathbb{R} \backslash \{0\}$ to just continuity on $\mathbb{R} \backslash \{0\}$?

Edit: I meant to assume that the function is also continuous at $0$.

答:
I’ll assume you want to ask the function to be continuous at $0$; otherwise, the example by @player3236 gives a counterexample because the lack of continuity at $0$ messes you up even before you get started.

Let $f(x)$ be defined by
$$f(x) = \left\{\begin{array}{ll}
|x|\sin\left(\frac{1}{x}\right) &\text{if }x\neq 0,\\
0&\text{if }x=0.
\end{array}\right.$$
Then for any $x\neq 0$ we have
$$f(-x) = |-x|\sin\left(\frac{1}{-x}\right) = -|x|\sin\left(\frac{1}{x}\right) = -f(x),$$
so $f$ is odd. It is also differentiable everywhere other than $0$, and is continuous at $0$. However,
$$\lim_{x\to 0^+}\frac{f(x)-f(0)}{x} = \lim_{x\to 0^+}\sin\left(\frac{1}{x}\right)$$
does not exist, so it does not have a derivative at $x=0$.


答: Hint: consider $\operatorname{sgn}(x) := \begin{cases}1 &\text{for } x > 0\\0 &\text{for } x = 0\\-1 &\text{for } x < 0\end{cases}$.

Title: Is an odd function differentiable at zero if it is differentiable elsewhere (and has no vertical tangent)?

Body:
Let $f: \mathbb{R} \to \mathbb{R}$ be an odd function that is differentiable everywhere except possibly at $x = 0$. Suppose that $f$ does not have a vertical tangent at $x = 0$ (so, an example like $y = x^{1/3}$ doesn't count). Does it follow that $f$ is differentiable at zero?

If that is true, how about if we weaken the hypothesis of differentiability on $\mathbb{R} \backslash \{0\}$ to just continuity on $\mathbb{R} \backslash \{0\}$?

Edit: I meant to assume that the function is also continuous at $0$.


问:
subdifferential of a linear function which has a non-negative domain
I am trying to find the sub-derivative of the function $f(x) = \lambda x, x\ge0$. I am familiar of the sub-derivative of the absolute value but not sure how to find the sub-derivative of this function at $x=0$. Applying the same logic of the absolute value sub-derivative, I got the following expression:

$$\partial f(x)|_{x=0} = k, k\in(-\infty, \infty) - \{\lambda\}$$

My reasoning is that you can draw a line that touches $f(0)$ with 360 degrees rotations (except when the slopes equals to $\lambda$). Am I right? If yes, how to generalize it to a function with a domain $\in \mathbb{R}^n$?

答: The derivative at $0$ is not defined since right and left limits are not equal. Left limit does not exist, since the function is not defined for values less than zero. Right limit equals to $\lambda$.


问:
Prove for $Pr[A] - Pr[B] \leq Pr[A \setminus B]$
I have events $A$ and $B$

I want to prove that for $Pr[A] - Pr[B] \leq Pr[A \setminus B]$.

I need help getting started. Are there any axioms of probability I could use to rewrite this?

答:
To begin with, I would start with noticing that $A\cup B = (A\cap B^{c})\cup B$. Indeed, we have
\begin{align*}
(A\cap B^{c})\cup B = (A\cup B)\cap(B\cup B^{c}) = (A\cup B)\cap \Omega = A\cup B
\end{align*}
where $\Omega$ is the sample space.

Moreover, we do also have that $A\subseteq A\cup B$. Since $A\cap B^{c}$ and $B$ are disjoint, the following result holds due to the monotonicity of the probability measure as well as the finite additivity property:
\begin{align*}
\mathbb{P}(A) \leq \mathbb{P}(A\cup B) & = \mathbb{P}((A\cap B^{c})\cup B) = \mathbb{P}(A\cap B^{c}) + \mathbb{P}(B) \Rightarrow \mathbb{P}(A) - \mathbb{P}(B) \leq \mathbb{P}(A\cap B^{c})
\end{align*}
and we are done.

Hopefully this helps!


问:
Partial derivatives where implicit equations are involved
I have a function:
$$f\bigl(P,R\bigr)=h\bigl(P,R,N(P,R)\bigr)$$
where $P$ and $R$ are independent variables and $N$ is the solution of the equation:
$$P-g(R,N)=0$$
I am trying to find $\frac{\partial f}{\partial P}$ and $\frac{\partial f}{\partial R}$.

答:
To obtain $\frac{\partial f}{\partial P}$, we compute $$\frac{\partial f}{\partial P}(P,R) = \frac{\partial h}{\partial P}(P,R,N(P,R)) + \frac{\partial h}{\partial N}(P,R,N(P,R))\cdot \frac{\partial N}{\partial P}(P,R).\tag{1}$$ Now, to compute $\frac{\partial N}{\partial P}$ we use the chain rule and the equation $P-g(R,N)=0$.  Differentiating with respect to $P$: $$1  - \frac{\partial g}{\partial N}(R,N)\cdot \frac{\partial N}{\partial P}(P,R) = 0.$$  This gives us $$1 = \frac{\partial g}{\partial N}(R,N)\cdot \frac{\partial N}{\partial P}(P,R) \implies \frac{\partial N}{\partial P}(P,R) = \frac{1}{\frac{\partial g}{\partial N}(R,N)}.$$  In all, then, by $(1)$ we get $$\boxed{\frac{\partial f}{\partial P}(P,R) = \frac{\partial h}{\partial P}(P,R,N(P,R)) + \frac{\frac{\partial h}{\partial N}(P,R,N(P,R))}{\frac{\partial g}{\partial N}(R,N)}.}$$

For the second part, $\frac{\partial f}{\partial R}$, we compute $$\frac{\partial f}{\partial R}(P,R) = \frac{\partial h}{\partial R}(P,R,N(P,R)) + \frac{\partial h}{\partial N}(P,R,N(P,R))\cdot \frac{\partial N}{\partial R}(P,R).\tag{2}$$  Now, as before, we differentiate $P-g(R,N)=0$ with respect to $R$ to obtain $\frac{\partial N}{\partial R}$: $$-\frac{\partial g}{\partial R}(R,N) - \frac{\partial g}{\partial N}(R,N)\cdot \frac{\partial N}{\partial R}(P,R) = 0 \implies \frac{\partial N}{\partial R}(P,R) = -\frac{\frac{\partial g}{\partial R}(R,N)}{\frac{\partial g}{\partial N}(R,N)}.$$  In all, then, by $(2)$ we get $$\boxed{\frac{\partial f}{\partial R}(P,R) = \frac{\partial h}{\partial R}(P,R,N(P,R)) - \frac{\partial h}{\partial N}(P,R,N(P,R))\cdot \frac{\frac{\partial g}{\partial R}(R,N)}{\frac{\partial g}{\partial N}(R,N)}.}$$


问:
The dollar menu at your favorite tax-free fast food restaurant has 10 items. You have $15 to spend.
Full Question:

The dollar menu at your favorite tax-free fast food restaurant has 10 items. You have $15 to spend. How many different meals can you buy if you spend all your money and:
a. Purchase at least one of each item.
b. Possibly skip some items.
c. Don’t get more than 2 of any particular item.

I've determined that a is:

C(14,5)

and that b is:

C(24,9)

I'm having trouble figuring out what I'm doing wrong on part c. Here's what I'm trying.

C(24,9) - [ C(12,1) C(21,11) - C(12,2) C(18,11) + C(12,3) C(15,11) - C(12,4) C(12,11) ]

My thinking is that we have the total meals to spend as C(24,9), and then since that is over-counting we subtract the meals for which we order 3 or more of an item.

答:
After looking over this, and with help from @JMoravitz, it does appear that I was thinking about 12 items instead of 10 for some reason.

For future references, I was able to get the answer by doing this:

C(24,9) - [ C(10,1) C(21,9) - C(10,2) C(18,9) + C(10,3) C(15,9) - C(10,4) C(12,9) + C(10,5) C(9,9)]


问:
rank of matrix created from 2 vectors
Take two vectors $a,b\in \mathbb{R}^{12}$ and create a 12x12 matrix $C$ so that $c_{ij}=(a_i-b_j)^2$. What are the possible values of rank(C)?

Not sure how to proceed. One idea I had was to express C as a product of two other matrices and then I might put together an estimate of some sort using $rank(AB)\leq min \{ rank(A),rank(B) \}$. Though having tested it out on a couple of concrete 3x3 matrices I found they always had full rank. So maybe that's the answer?

Either way, appreciate any help.

答:
I believe the rank is no greater than $3$. Notice that by expanding out $c_{ij}=(a_i-b_j)^2 = a_i^2 + b_j^2 - 2a_ib_j$, we can write

$$ C = (a^2) \boldsymbol{1}^T + \boldsymbol{1} (b^2)^T - 2ab^T$$

Where $\boldsymbol{1}$ is the vector of all ones and $a^2$ is the vector whose entries are the squared entries of $a$. Hence there are at most three linearly independent columns: $a,a^2,$ and $\boldsymbol{1}$. When $a,a^2$, and $\boldsymbol{1}$ are all linearly independent (and analogously for the rows $b,b^2,\boldsymbol{1}$) the rank is 3. We can also obtain rank $2,1,$ or $0$. Choosing $a=a^2\neq \boldsymbol{1}$ gives rank $2$. Choosing $\boldsymbol{1} = a = a^2$ gives rank $1$. And $a=a^2=b=b^2=0$ trivially gives rank zero.

Title: rank of matrix created from 2 vectors


问:
Matching in Bipartite Graphs - Saturation
I have the following question in my homework and I'm not sure how to go about proving it. Hall's Theorem states that the graph contains a matching saturating X is and only if |N(S)| >= |S| for all S _( X. First of all in this question it says that the first part is only greater and the second part is strictly a subset. Am I correct in interpreting the following: Since S is a non-empty subset of X, and N(S) > S for all S, doesn't that mean that every vertex in X has at least 2 edges going to Y? Because I could make S any single vertex and then I would have N(S) > S so there must be more than one edge going from X to Y right? If this is true, how does this help me make my proof?

答:
Your interpretation is correct, but I'm not sure if it is useful.

Take your graph and pick some edge $e \in E$. Then remove $e$ along with the two vertices it connects. Look for a matching in this new graph. Do you see a connection between a matching in this new graph and a matching in the original graph? Can you conclude that $e$ belongs to some matching of the original graph?


问:
Given the sequence $\{x_n\}$, prove $\{x_{2n}\}$ and $\{x_{2n-1}\}$ converge to the same limit
Let $\{a_n\}_{n=1}^{\infty}$ be a decreasing sequence of positive numbers with limit 0. I define a new sequence $\{x_n\}_{n=1}^{\infty}$ as follows:
\begin{align*}
    x_1 &= a_1 \\
    \forall n \in \mathbb{Z}^{+}, x_{n+1} &= x_n + (-1)^{n}a_{n+1}
\end{align*}
I have to prove that $\{x_{2n}\}_{n=1}^{\infty}$ and $\{x_{2n-1}\}_{n=1}^{\infty}$ are both convergent to the same limit $L$.

I have tried writing out the first few terms of the sequence:
\begin{align*}
    x_1 &= a_1 \\
x_2 &= a_1 - a_2 \\
x_3 &= a_1 - a_2 + a_3 \\
x_4 &= a_1 -a_2 + a_3 - a_4
\end{align*}
So $x_{2n} = x_{2n-1} - a_{2n}$. But I am stuck where to proceed from here. How should I show that $\{x_{2n}\}_{n=1}^{\infty}$ and $\{x_{2n-1}\}_{n=1}^{\infty}$ are both convergent to the same limit using the provided fact that $\{a_n\}_{n=1}^{\infty}$ has limit 0 and decreasing?

答: First show that $x_{2n}$ has a limit. You can do so by observing that through telescoping: $x_{2n}-x_1 = \sum_{k=2}^{2n}(-1)^ka_k$. The right side converges by Dirichlet's test. Now use $x_{2n}-a_{2n}=x_{2n-1}$ consider the limit of the left side.


答: Look at the limit of $\{x_{2n+1} - x_{2n}\}$. If $\{x_{2n+1} - x_{2n}\} \to 0$ and $\{x_{2n}\}$,$\{x_{2n+1}\}$ are both convergent, then you can conclude they have the same limit. You can use Monotone Convergence to prove $\{x_{2n}\}$ and $\{x_{2n+1}\}$ are convergent.

Title: Given the sequence $\{x_n\}$, prove $\{x_{2n}\}$ and $\{x_{2n-1}\}$ converge to the same limit

Body:
Let $\{a_n\}_{n=1}^{\infty}$ be a decreasing sequence of positive numbers with limit 0. I define a new sequence $\{x_n\}_{n=1}^{\infty}$ as follows:
\begin{align*}
    x_1 &= a_1 \\
    \forall n \in \mathbb{Z}^{+}, x_{n+1} &= x_n + (-1)^{n}a_{n+1}
\end{align*}
I have to prove that $\{x_{2n}\}_{n=1}^{\infty}$ and $\{x_{2n-1}\}_{n=1}^{\infty}$ are both convergent to the same limit $L$.

I have tried writing out the first few terms of the sequence:
\begin{align*}
    x_1 &= a_1 \\
x_2 &= a_1 - a_2 \\
x_3 &= a_1 - a_2 + a_3 \\
x_4 &= a_1 -a_2 + a_3 - a_4
\end{align*}
So $x_{2n} = x_{2n-1} - a_{2n}$. But I am stuck where to proceed from here. How should I show that $\{x_{2n}\}_{n=1}^{\infty}$ and $\{x_{2n-1}\}_{n=1}^{\infty}$ are both convergent to the same limit using the provided fact that $\{a_n\}_{n=1}^{\infty}$ has limit 0 and decreasing?


答:
HINT: Use the fact that $x_{2n}=x_{2n-1}-a_{2n}$ for each $n\in\Bbb Z^+$ to show that

$$\lim_{n\to\infty}x_{2n}=\lim_{n\to\infty}x_{2n-1}-\lim_{n\to\infty}a_{2n}\,,$$

and use that to show that if either of the sequences $\langle x_{2n}:n\in\Bbb Z^+\rangle$ and $\langle x_{2n-1}:n\in\Bbb Z^+\rangle$ converges, then both converge to the same limit.

The alternating series test tells you that $\sum_{n\ge 1}(-1)^{n+1}a_n$ converges, and $x_n$ is the $n$-th partial sum of this series, so ... ?


答:
Use the following syntax:

$p(X) \equiv $ probability of event $X$ occurring

$(XY) \equiv $ the event that events $X$ and $Y$ both occur

$(\neg X) \equiv $ the event that event $X$ does not occur.

Clearly $(B)$ can be partitioned into two mutually exclusive events:
$[B(\neg A)]$ and $(BA).$

Thus, $p(B) = p[B(\neg A)] + p(BA).$

Similarly, $p(A) = p[A(\neg B)] + p(BA).$

Therefore:
$p(A) - p(B) = p(A) - \{p[B(\neg A)] + p(BA)\} \leq p(A) - p[BA] = p[A(\neg B)].$

Title: Prove for $Pr[A] - Pr[B] \leq Pr[A \setminus B]$

Body:
I have events $A$ and $B$

I want to prove that for $Pr[A] - Pr[B] \leq Pr[A \setminus B]$.

I need help getting started. Are there any axioms of probability I could use to rewrite this?


问:
Why isn't this an axiom of choice?
The standard axiom of choice can be expressed in the first order language of ZF by $$\forall X\big((X\neq\emptyset\wedge\emptyset\notin X)\implies\exists f:X\to\bigcup X\ \forall Y\in X(f(Y)\in Y)\big).$$ Naively, a simpler version would seem to be $$\forall X(X\neq\emptyset\implies\exists f:\{X\}\to X)$$ since we then have that $\forall A\in\{X\}(f(A)\in A)$, as specified explicitly in the choice axiom. This version is provable in ZF according to the discussion on this answer, however, and is thusly not an axiom of choice. How exactly do we lose choice moving from the first formulation to the second?

答:
This is not the axiom of choice, since you're only requiring that given a single set, there is a function that selects an element from that set. It's just how we interpret quantifiers: first we replace $X$ with an instance, then we replace $f$ with an instance; but there's no requirement that these replacements are coherent.

Therefore, this is quite literally existential instantiation. This follows from the rules of logic, so it is provable in any weak theory you'd like (as long as you can discuss choice function in that theory).

If you switch the order of the quantifiers, $\exists f\forall X$, then you'd get something more akin to global choice, presuming you're working in second-order set theory.

Title: Why isn't this an axiom of choice?


问:
Proving arctan x as the limit of a sequence
First semester real analysis - I'm currently preparing for an exam and going through exercises in a book. I came across this problem where I can't seem to get off the ground at all.

The task is to prove that given the recursive sequence with $x_{0}=x$ and $x_{n+1}=\frac{x_{n}}{1+\sqrt{1+x_{n}^2}}$,

$$
\lim_{n\to\infty}2^{n}x_{n}=\arctan x
$$

I've worked with recursive sequences a bit, but hadn't come across one with a variable starting point, and am a bit dumbstruck by having to look at the limit of the product of these two sequences where one is recursive and one is explicit.

Obviously, as $n$ goes to infinity, $2^{n}$ goes to infinity as well, while the only thing I managed to see wrt the recursive sequence so far is that

The denominator is always greater than one and the numerator just the last term of the sequence, which leads me to conclude that we retain the original sign of $x$.

The denominator is always larger than the numerator, meaning both that our value for each $x_n$ is less than one and that our sequence should be strictly monotonically approaching 0 from above or below depending on the sign of $x$.

Together, if I'm not mistaken, this means I can say that $0<|x_{n+1}|<|x_{n}|<1$ for all $n\geq1$ (which also means extreme values for $x$ are immediately reduced to lie within $]-1,1[$ if my conclusion is correct).

Of course, this leads me to assume that the sequence converges to $0$ - and given that the numerator is just $x_{n}$, when I substitute $0$ I of course get $0$ for the next term - so it is a fixed point.

Now I have an idea that the recursive sequences converges to $0$ while $2^{n}$ goes to infinity - but I haven't proved that, nor do I know how to bring these two together.

After writing down the first two terms of $x_{n}$, it quickly gets overwhelming with nested fractions, and not obvious how to potentially extract factors to look for an explicit formulation.

About the arctan, I only know that it is the inverse of tan, that it maps $\mathbb{R}$ into the interval $]-\frac{\pi}{2},\frac{\pi}{2}[$, reaching $\pm\frac{\pi}{4}$ at $\pm1$ and that it approaches the limits of the interval asymptotically as $x$ goes to $\pm\infty$ - but have no explicit formulation nor (as far as I can see) any other information that could help me.

What are specific strategies and tactics I can use when faced with such a problem setup - and what are some specific techniques to use in this particular case?

答:
Use substitution $x_n = \tan\theta_n$

$$\tan\theta_{n+1} = \frac{\tan\theta_n}{1+\sec\theta_n} =\frac{\sin\theta_n}{\cos\theta_n+1}$$

$$=\frac{2\sin\frac{\theta_n}{2}\cos\frac{\theta_n}{2}}{2\cos^2\frac{\theta_n}{2}} = \tan\frac{\theta_n}{2} \implies \theta_{n+1} = \frac{\theta_n}{2}$$

Of course the last implication requires that the sequence is decreasing, monotonic, etc, which I will leave to you to prove. This means we have that

$$\theta_n = \frac{\arctan x}{2^n} \implies \lim_{n\to\infty}2^n\tan\frac{\arctan x}{2^n} = \arctan x$$

by the standard $\frac{\tan x}{x}$ limit or by series expansion.


问:
What sequence is this $\prod_{i=1}^n \frac{2i}{2i + 1} = $?
I'm trying to find a closed form solution for $\prod_{i=1}^n \frac{2i}{2i + 1} $, but I'm confused what kind of sequence this is. Does it have a closed form?

答:
You can simplify it.
\begin{align*}
\prod_{i=1}^n \frac{2i}{2i + 1} &=\frac{2^n n!}{3\cdot5\cdot7\cdot\ldots\cdot(2n+1)}\\
&=\frac{2^{2n} (n!)^2}{(2n+1)!}
\end{align*}


问:
Given $2\sin A + \sin B = 2\sin C$ in a triangle find minimum of $\frac{5}{\sin A} + \frac{9}{\sin C}$
Given $2\sin A + \sin B = 2\sin C $ where $A,B,C$ are angles in a triangle find minimum of $\frac{5}{\sin A} + \frac{9}{\sin C}$

so $\sin C - \sin A = \frac{ \sin B}{2}$ so $2\cos \frac{C+A}{2} \sin \frac{C-A}{2} = \frac{ \sin B}{2} = 2\sin \frac{B}{2} \sin \frac{C-A}{2} $  so $\sin \frac{C-A}{2} = \frac{1}{4}$

I feel like I am awfully close here, but how do I finish it?

答:
In terms of the sides and circumradius:

$$2\sin A + \sin B = 2\sin C\iff \dfrac{a}{R} + \dfrac{b}{2R} = \dfrac{c}{R}\iff 2c=b+2a.$$

Then, you can turn your expression into a function of two variable, say $a,b$, using the same Law of Sines and take derivatives, using the formula:

$$R = \dfrac{abc}{4S},$$
where $S$ is the area.


问:
Def. of Homology Cell
I would like an explanation for the following definition.

"A metric space is an $\ homology \ cell $ if it is nonempty and homologically trivial (acyclic) in all dimensions."

What does "homologically trivial (acyclic)" mean?

What are intuitive examples of homology cells?

答:
From your question, I'm guessing you don't know what homology is, so answering this will be challenging. The first homology $H_1(X)$ is easy to define. It is the abelianization of the fundamental group. So in particular the fundamental group of a homology cell must be perfect (equal to its own commutator subgroup.) The definitions of $H_2$ and higher homology groups require a fair bit of machinery.

A simple example of a homology cell is any contractible space. (This is actually a homotopy cell, which is a stronger condition.) A less trivial example arises by taking the Poincare homology sphere (Google it!) and removing the interior of a small ball. This space has a nonzero fundamental group so it cannot be a homotopy ball, but all of its (reduced) homology groups vanish.

Title: Def. of Homology Cell


问:
Short question about $|\int_{a}^{b}f(x)dx|=\int_{a}^{b}|f(x)|dx$
I have 2 very similar questions and short questions here:

If $|\int_{a}^{b}f(x)dx|=\int_{a}^{b}|f(x)|dx$, must $f(x) \geq 0$ for $x \in (a,b)$? I have an intuition that this is false, but can not prove it. Maybe the use of a counterexample like $cos(x)$ would work?

Similarly, if $\int_{a}^{c_2}f(x)dx+\int_{c_1}^{b}f(x)dx=\int_{a}^{b}f(x)dx$ for $c_1, c_2 \in (a, b)$ with $c_1< c_2$, then must $f(x)=0$ for $x \in (a,b)$? On the surface, this looks alot like one of the inherent properties of integrals, where to get the integral from $[a,b]$ you can choose a midpoint $c$ between a and b and break it into 2 parts.

答:
$f$ can also be always negative. However, these are the only possibilities. Indeed, put $f_+=\max(f,0)$ and $f_-=\max(-f,0)$. Then $f=f_+-f_-$ and $\lvert f\rvert=f_++f_-$, so $\int f=\int f_+-\int f_-$ whereas $\int\lvert f\rvert=\int f_++\int f_-$ (all integrals on the interval $[a,b]$) and noting that these are sums/differences of nonnegative numbers, the claim follows that either $\int f_+=0$ or $\int f_-=0$.

Not necessarily, this is just saying that $\int_{c_1}^{c_2}f=0$, for instance $f=\cos$, $c_1=0$, and $c_2=2\pi$.


答:
No, let $f(x) = -1,$ then $$\left| \int_a^b -1dx\right|= \left| -\int_a^b 1dx\right|=\int_a^b|-1|dx$$

No, let $f(x)\mid_{[c_1,c_2]}=0$, and let $f(x)$ be your favorite, non-zero integrable function on the rest of $(a,b).$

Title: Short question about $|\int_{a}^{b}f(x)dx|=\int_{a}^{b}|f(x)|dx$

Body:
I have 2 very similar questions and short questions here:

If $|\int_{a}^{b}f(x)dx|=\int_{a}^{b}|f(x)|dx$, must $f(x) \geq 0$ for $x \in (a,b)$? I have an intuition that this is false, but can not prove it. Maybe the use of a counterexample like $cos(x)$ would work?

Similarly, if $\int_{a}^{c_2}f(x)dx+\int_{c_1}^{b}f(x)dx=\int_{a}^{b}f(x)dx$ for $c_1, c_2 \in (a, b)$ with $c_1< c_2$, then must $f(x)=0$ for $x \in (a,b)$? On the surface, this looks alot like one of the inherent properties of integrals, where to get the integral from $[a,b]$ you can choose a midpoint $c$ between a and b and break it into 2 parts.


问:
Calculate $\displaystyle\lim_{x\rightarrow 0}\frac{1}{x}\int_{0}^{x}e^{t^{2}}{\rm d}t$

Calculate $\displaystyle\lim_{x\rightarrow 0}\frac{1}{x}\int_{0}^{x}e^{t^{2}}{\rm d}t$

An answer given online that uses fundamental theorem of calculus is that lets $\displaystyle F\left ( x \right )= \int_{0}^{x}e^{t^{2}}{\rm d}t,$ then
$$\lim_{x\rightarrow 0}\frac{1}{x}\int_{0}^{x}e^{t^{2}}{\rm d}t= {F}'\left ( 0 \right )= e^{0}= 1$$
But this doesn't quite make sense to me. Where do the $1/x$ and the limit go ?

答:
We may think as $x> 0.$ According to mean-value theorem$,\quad\exists c\in\left ( 0, x \right )$ so that
$$\frac{1}{x}\int_{0}^{x}\exp t^{2}{\rm d}t= \exp c^{2}\rightarrow\exp 0= 1\,{\rm as}\,x\rightarrow 0^{+}$$


问:
$\lim _{n\to\infty}\frac{\prod_{z^n=-1}P(z)^n}{\prod_{z^n=1}P(z)^n},$
I am trying to calculate the limit of the following quantity $$\lim _{n\to\infty}\frac{\prod_{z^n=-1}P(z)^n}{\prod_{z^n=1}P(z)^n},$$ where $P$ is a polynomial of one complex variable. I guess the limit is 1, but I don't know how to prove.

Some one can help me? Thanks in advance!

Body:
If $z_1>1$, then given any $c>0$, we have $1+\frac{2}{z_1^n-1}<1+\frac{c}{n}$ for large $n$, so $\lim_{n\to \infty}(1+\frac{2}{z_1^n-1})^n\le\lim_{n\to \infty} (1+\frac{c}{n})^n= e^c$. Let $c\to 0$, we obtain $\lim_{n\to \infty}(1+\frac{2}{z_1^n-1})^n\le 1$, hence $\lim_{n\to \infty}(1+\frac{2}{z_1^n-1})^n= 1$.

This is simply
$$
\lim_{n\to \infty}(1+\frac{2}{z_1^n-1})^n=\lim_{n\to \infty}(1+\frac{2}{z_1^n-1})^{\frac{z_1^n-1}{2} \frac{2n}{z_1^n-1}}=e^0=1,
$$
which holds for $|z_1|>1$.

If $0<z_1<1$, $(1+\frac{2}{z_1^n-1})^n=(-1)^n\bigg[1+\frac{2}{\frac{1}{z_1^n}-1}\bigg]^n$, which diverges.

So the limit might not exist.


问:
If $f_{n}\left ( x \right )= \frac{x}{\left ( x+ 1 \right )^{n}}$ show that $\sum f_{n}\left ( x \right )$ is uniformly convergent
I am stuck trying to solve the problem below.

Suppose $f_{n}:\quad\left [ 1, 2 \right ]\rightarrow\mathbb{R}$ be $f_{n}\left ( x \right )= \dfrac{x}{\left ( x+ 1 \right )^{n}}$

Show that $\sum f_{n}\left ( x \right )$ is uniformly convergent on $B= \left [ 1, 2 \right ]$

I think we should use the Weierstrass M-test but I'm having trouble choosing which function to compare it to.

Any hints would be appreciated.

Body: For $x \in B$, $$\frac{x}{(x+1)^n} \leq \frac{2}{2^n} = \frac{1}{2^{n-1}}.$$


问:
Negative factors of a number
Can a factor of a number be negative?

Is $-5$ a factor of $25$ or $-25$?

A number is said to be prime if it has two factors : $1$ and the number itself. So if $-5$ can be a factor of $5$, how to define the prime numbers?

Thanks.

答:
You raise an interesting question, which has repercussions in various forms of more advanced algebra. When we are dealing with the positive integers there is no difficulty with definitions - a prime is a positive integer, but in wider contexts there may be no natural order to say whether a number is positive or negative. For example $i=\sqrt {-1}$ cannot be distinguished algebraically from $-i$.

The one point of interest is whether we should count $1$ as a prime - and the general convention is that it is counted not as a prime, but a unit. $-1$ is also a unit - we can add $(-1)^2$ to any factorisation without changing its value.

Numbers which are obtained from each other by multiplication by a unit are often known as associates, so $-5, 5$ are associates in the integers, and it can be proper to call them both (associated) primes.

There are certain things we can only do "up to multiplication by units" - and factorisation and the identification of prime numbers are two of these.

However, back to the beginning - with the integers, associated primes come in pairs, and it is natural and convenient to work with the positive member of each pair.

Title: Negative factors of a number


问:
Why is 23 this summation's upper limit?
Can someone explain why $12$ and $23$ are the lower and upper limits for the series equation in the answer to this question:

BlackJack Card Probability when Counting Cards

答:
Short answer: $47-2c$ should not get negative.

Longer answer: the index $c$ is the number of high cards. If you have $c=23$ high cards, then you have $28$ low cards, which is possible. But $24$ high cards means there are $29$ low cards, but there are only $52$ cards in total.

Title: Why is 23 this summation's upper limit?


问:
Let $X$ be a metric space and $A$ a subset of $X$. Show that $bdry(A)$ is closed in $X$
Working through my first topology book, trying to teach myself.

Abridged definitions given are:

Closed set: $F$ is closed if $F^c$ is open.

Boundary point: Let $X$ be a metric space and $A⊆X$.
$x_0$ is a boundary point if, for every $r > 0$, $B(x_0,r) ∩ A$ is non-empty and $B(x_0,r) ∩ A^c$ is non-empity. The boundary is the set of all boundary points.

Interior point: Let $X$ be a metric space and $A⊆X$. $x_0$ is an interior point if there exists an $r > 0$ such that $B(x_0,r) ⊆ A$. $A$ is open if every point of $A$ is an interior point.

My answer:

By definition, $bdry(A)$ is closed if $bdry(A)^c$ is open.

So, we want to show that $B(a,r) ∩ bdry(A) = ∅$, for every $a ∈ X - bdry(A)$.

We have two cases:

I) if $a ∈ A$, choose $r$ such that $B(a,r) ⊆ A$ ⟹ $B(a,r) ∩ A^c = ∅$ ⟹ $B(a,r) ∩ bdry(A) = ∅$.

II) if $a ∈ X - A$, choose $r$ such that $B(a,r) ⊆ A^c$ ⟹ $B(a,r) ∩ A= ∅$ ⟹ $B(a,r) ∩ bdry(A) = ∅$.

Thus, $a$ is an interior point of $bdry(A)^c$ ⟹ $bdry(A)^c$ is open ⟹ $bdry(A)$ is closed.

Is this correct? Anything I could improve?

答:
For part (I), it should start $a \in \textrm{Int}(A)$. This would give us an $r$ such that $B(a,r) \subseteq A$. If we just pick any $a \in A$, such an $r$ might not exist because $a$ is not necessarily an interior point of $A$. Same thing with (II), we want $a \in \textrm{Int}(X-A)$. Moreover, this is sufficient because $X - bd(A) = \textrm{Int}(A) \cup \textrm{Int}(X-A)$, so we have checked that every $a \in X - A$ is an interior point.

Note, $\textrm{Int}(A)$ denotes the interior of $A$, which is the set of all interior points of $A$.


问:
Find vertical asymptotes of $\frac{-2 + \sqrt{3x^2-23}}{x-3}$
So Im trying to see if the following expression has any vertical asymtotes (algebraically without looking at the graph): $$\frac{-2 + \sqrt{3x^2-23}}{x-3}$$

Now after a bit of playing around with factoring stuff out, I figured this would be the equivalent of

$$\frac{-3x-9}{-2-\sqrt{3x^2-23}}$$

But this doesn't really seem all that useful, I managed to cancel the $x-3$ out but I still have a denominator that looks like it might have a vertical asymptote (which it doesn't).

Any help would be appreciated, thank you.

答: Look inside the square root.  $3x^2 - 23$ is nonnegative if and only if $|x| \ge \sqrt{23/3}$, which by comparison with $\sqrt{27/3} = 3$, is just a bit smaller in magnitude.  Therefore, the domain of the function $$f(x) = \frac{-2 + \sqrt{3x^2 - 23}}{x - 3}$$ is the set $$\{x : |x| \ge \sqrt{23/3} \cap x \ne 3\}.$$  In this region, of course there is no issue with $x > 3$.  For $x \le -\sqrt{23/3}$, the function is also smooth, but abruptly stops on the boundary point; we cannot call that a vertical asymptote since there is no unbounded behavior there; its value is simply $2/(\sqrt{23/3} + 3)$.  Similarly, there is no asymptote at $x = \sqrt{23/3}$; the function is bounded there.  The only question is whether $$\lim_{x \to 3} f(x)$$ is well defined, and it turns out that the answer is yes:  from your algebra, which is correct, we have $$f(x) = \frac{3(x+3)}{2 + \sqrt{3x^2 - 23}}$$ whenever $x \ne 3$.  So the limiting behavior is $$\lim_{x \to 3} f(x) = \frac{3(3+3)}{2 + \sqrt{3(3^2) - 23}} = \frac{9}{2},$$ meaning that there is also no vertical asymptote at $x = 3$.  The function has no vertical asymptotes anywhere.


问:
Limit and Average Value of Integral
This is a limit-esque question:
\begin{align}
&\mbox{If there is a}\ \epsilon > 0\
\mbox{such that}
\left\vert\operatorname{f}\left(x\right) -\operatorname{g}\left(x\right)\right\vert \leq \epsilon\
\mbox{for every}\ x \in \left(a,b\right),
\\[2mm] &\
\mbox{then does}\
\frac{1}{b-a}\,
\left\vert\,\int_{a}^{b}
\operatorname{f}\left(x\right)\,{\rm d}x -\int_{a}^{b}\operatorname{g}\left(x\right)
\,{\rm d}x\,\right\vert \leq \epsilon
\end{align}
Note that $\operatorname{f}\left(x\right), \operatorname{g}\left(x\right)$ are functions integrable and differentiable on $\left[a,b\right]$.

I'm not sure if I'm interpreting the question correctly, but what I see is a question asking if there's a limit for $\operatorname{f}\left(x\right) -\operatorname{g}\left(x\right)$, then does the average value of the absolute value of the function also smaller than the limit $?$.

This is an especially difficult question for me since it connects limits and integrals closely together, which I've previously not seen before.

答: Hint: $$\left|\int_a^b f(x)\mathrm{d}x - \int_a^b g(x)\mathrm{d}x\right| = \left|\int_a^b f(x)-g(x)\mathrm{d}x\right| \leq \int_a^b \left|f(x)-g(x)\right| \mathrm{d}x$$ This question is unrelated to limits.

Title: Limit and Average Value of Integral


问:
Prove that the language $A= \left \{ x\in\left \{ 0, 1 \right \}^{*}\mid\#\left ( 0, x \right )= \#\left ( 1, x \right ) \right \}$ is not regular.
I initially tried using the pumping lemma, with little success, albeit I still think it's possible. I've been directed that using the Myhill-Nerode theorem would be optimal, but I'm struggling. I need to show that $A$ has infinitely many equivalence classes. I also need to know what those equivalence classes are. Anything to help me better understand this problem would be greatly appreciated.

答:
HINT: For the Myhill-Nerode approach, define

$$d:\{0,1\}^*\to\Bbb Z:w\mapsto|w|_0-|w|_1\,.$$

Thus, for instance, $d(01001)=3-2=1$, while $d(111)=0-3=-3$. Show that $x$ and $y$ are in the same Myhill-Nerode equivalence class if and only if $d(x)=d(y)$.


问:
Product of non-zero eigenvalues (pseudo-determinant) of a singular matrix
Consider a real symmetric matrix $P$ that is partitioned as
$$
P = \begin{bmatrix}
A & AB \\
B^\top A & C+B^\top A B
\end{bmatrix},
$$
where P is singular, while $A$ is positive definite, and $C$ is a non-zero matrix.

I am wondering if the following relation is true:
\begin{equation}
Det(P) = \det(A) Det(C),  \quad (1)
\end{equation}
where $Det$ denotes the pseudo determinant, i.e. $Det(P)$ is the product of all the non-zero eigenvalues of $P$.

The above relation is straightforward when only the determinant is concerned, in which case we can apply the following formula:
$$
\det(P) = \det(A) \det[(C+B^\top A B) - (B^\top A) A^{-1} (AB) ] = \det(A) \det(C).
$$
However, I do not know if $(1)$ holds.

答:
This doesn't hold. E.g. when $A=B=1$ and $C=0$, we have
$$
P=\pmatrix{1&1\\ 1&1}
$$
and hence
$$
Det(P)=2\ne 1=\det(A)Det(C).\tag{2}
$$

This counterexample can be easily extended to the case where $C$ is nonzero or even entrywise nonzero. First, by appending $1$s to $A,B$ and $C$, i.e. when $A=B=I_2$ and $C=\operatorname{diag}(0,1)$, the new $P$ is (by a simultaneous permutation of rows and columns) similar to $\pmatrix{1&1\\ 1&1}\oplus\pmatrix{1&1\\ 1&2}$. Therefore $Det(P)=2$ and the inequation $(2)$ still holds. Next, to make $C$ entrywise nonzero, we can simply replace $C$ by $QCQ^T$ for some appropriate real orthogonal matrix $Q$. In particular, the inequation $(2)$ is still true when $A=B=I_2$ and
$$
C=\left[\frac{1}{\sqrt{2}}\pmatrix{1&1\\ 1&-1}\right]\pmatrix{0&0\\ 0&1}\left[\frac{1}{\sqrt{2}}\pmatrix{1&1\\ 1&-1}\right]=\frac12\pmatrix{1&-1\\ -1&1}.
$$


问:
Need clarification on the complex form of Fourier series
I wish to ask you guys to fill in a few steps for the derivation of complex form of Fourier series.
This is taken from "Fourier series" of Tolstov (Dover publication).

$$f(x)\sim c_0+\sum_{n=1}^{m}(c_ne^{inx}+c_{-n}e^{-inx})=\sum_{\color{red}{n=-m}}^{m} c_ne^{inx}{\tag1}$$

The author said: "Therefore it is natural to write:

$$f(x)\sim\sum_{-\infty}^{+\infty}c_ne^{inx}\tag{2}$$

$c_n=\dfrac{a_0}{2}$, $c_n=\dfrac{a_n-ib_n}{2}$, $c_{-n}=\dfrac{a_n+ib_n}{2}$

I am totally confused with the summation index. Why does $n=-m$ lead to the exclusion of $c_{-n}e^{-inx}$. I have watched some youtube videos on the complex form of Fourier series but they don't write the standard Fourier series as a partial sum. I think the author skips some steps already.

Isn't $\displaystyle\sum_{n=-m}^{m}c_ne^{inx}$ means $f(x)\sim c_0+\displaystyle\sum_{-\infty}^{+\infty}(c_ne^{inx}+c_{m}e^{imx})$

Also, how can you transform $(1)$ into $(2)$

答:
\begin{eqnarray*}
 c_0+\sum_{n=1}^{m}(c_ne^{inx}+c_{-n}e^{-inx}) &=& c_0+(c_1 e^{ix}+c_{-1} e^{-ix}) +\cdots + (c_me^{imx}+c_{-m}e^{-imx})
\end{eqnarray*}
Now rearrange the order of the terms to
\begin{eqnarray*}
 c_0+\sum_{n=1}^{m}(c_ne^{inx}+c_{-n}e^{-inx}) &=& c_{-m}e^{-imx} +\cdots +c_{-1} e^{-ix} +  c_0+c_1 e^{ix} +\cdots + c_me^{imx} \\
 &=& \sum_{\color{red}{n=-m}}^{m} c_ne^{inx}.
\end{eqnarray*}
How do we get from $(1)$ to $(2)$ ? ... What does the symbol $ \sim$ mean ?


答: The bounds on $c$ are due to counting neutral cards. If $c\gt23$ then $47-2c\lt0$, and if $c\lt12$, then $47-2c\gt23$, but there are at most $24$ neutral cards.

Title: Why is 23 this summation's upper limit?

Body:
Can someone explain why $12$ and $23$ are the lower and upper limits for the series equation in the answer to this question:

BlackJack Card Probability when Counting Cards


问:
Game Theory: draw sets using a program?
I am asked to draw two different sets, S1 and S2 using a program to determine if the sets are equal. But for starters, I don't know how am I supposed to "draw" a set with a program and I have never heard a thing like that. Do you know any good resources to understand more about the topic?

The problem I am talking about:

Given a bimatrix $(A, B)$ we can define two sets:
$S_{1},$ the set of all outcomes that can be reached when the players coordinate their strategies and $S_{2}$, the set of outcomes that can be reached when the players cannot coordinate their strategies. (It is the set $S_{1}$ we use in Nash's bargain solution.) If we take the bimatrix

$$
\left(\begin{array}{ll}
(2,1) & (1,0) \\
(0,1) & (1,2)
\end{array}\right)
$$

Then $S_{1}=\{\alpha(2,1)+\beta(1,0)+\gamma(0,1)+\delta(1,2)\}$ where $0 \leq \alpha, \beta, \gamma, \delta \leq 1$, $\alpha+\beta+\gamma+\delta=1$.

The set $S_{2}$ is

$$
\left\{\left(\bar{x}\left(\begin{array}{ll}
2 & 1 \\
0 & 1
\end{array}\right) \bar{y}^{T}, \bar{x}\left(\begin{array}{ll}
1 & 0 \\
1 & 2
\end{array}\right) \bar{y}^{T}\right)\right\}
$$

where $\bar{x}, \bar{y}$ are probability vectors. Draw both sets (you can use any program or tool). Are the sets equal?

答:
Both sets are subsets of $\ \mathbb{R}^2\ $.  I would take the  instruction to "draw" these sets as asking you to construct diagrams to represent them, just as you do when you draw the graph of a function.  If I were to ask you to draw the set
$$
S_3=\left\{x_1(0,0)+x_2(0,1)+x_3(1,0)\,\bigg|\,0\le x_i, \sum_{i=1}^3x_i=1\right\}\ ,
$$
for instance, I would expect you to produce a diagram something like the one below, not necessarily done with
a graphing app like GeoGebra, which I used.  A similar hand-drawn diagram should be perfectly acceptable.

Title: Game Theory: draw sets using a program?


问:
What is the complexity of this formula?
Keeping in mind the classification of (arithmetical) formulas in the arithmetical hierarchy, and given decidable predicates $P(x)$ and $Q(a,b,c)$, what is the complexity of

$\forall x P(x) \wedge \exists a \forall b \exists c Q(a,b,c)$?

My thinking is that, since the variables are independent, it could just as easily be $\Pi^0_4$:

$(\forall x  \exists a \forall b \exists c [P(x) \wedge Q(a,b,c)])$

as it could be $\Sigma^0_4$:

$( \exists a \forall b \exists c \forall x[P(x) \wedge Q(a,b,c)])$,

but I can’t pick one.

答:
It is in fact guaranteed to be $\Sigma^0_3$. Remember that each arithmetical class is closed under (finite) conjunctions and disjunctions, and here the left conjunct is $\Pi^0_1$ hence a fortiori $\Sigma^0_3$ while the right conjunct is $\Sigma^0_3$.

To see this explicitly, note that the formula is equivalent to $$\exists a\forall b\forall x\exists c(P(x)\wedge Q(a,b,c)).$$

(And as usual you can avoid all syntactic manipulations entirely by using Shoenfield's limit lemma(s) to translate to a question about computational complexity; although that doesn't actually make things materially easier, it may be intuitively simpler - it was for me.)

Title: What is the complexity of this formula?


问:
Product of two first countable sets is first countable
Prove that the product of two first countable spaces $X$ and $Y$ is first countable.

Attempt:Let $X$ and $Y$ be two first countable spaces. Let $(x,y) \in X \times Y$. Then $x \in X$ and $y \in Y$. Then $x$ has a countable neighborhood basis, $\beta_x$.Also $y$ has a countable neighborhood basis $\beta_y$. Then the set $\beta_x \times \beta_y=\{B_x \times B_y|B_x \in \beta_x,B_y \in \beta_y\}$ is a countable neighborhood basis at $(x,y)$. The set is countable since the cardinality is $|\beta_x \times \beta_y|$ and the cartesian product of two countable sets is countable. Each basic open set $U \times V$ containing $(x,y)$ contains some $B_x \times B_y$, since if $x \in U$ and $y \in V$, there is some $B_x$ with $x \in B_x\subset U$ and some $B_y$ with $y \in B_y \subset V$.So $(x,y) \in B_x \times B_y \subset U \times V$.

Comment: I looked at the proof of this for countable products and noticed the proof to be much more demanding, which makes me unsure if this attempt suffices.However I noticed this is generally the case for proofs involving infinite/finite products. Is this ok, if not how could I change/improve it?Also I am unsure if it appropriate to use the notation $\beta_x \times \beta_y$ for the set $\{B_x \times B_y|B_x \in \beta_x,B_y \in \beta_y\}$.

答:
The issue is that an open set in $X \times Y$ doesn't necessarily have the form $U \times V$ for open $U \subset X$ and open $V \subset Y$. Consider an open ball in $\mathbb{R} \times \mathbb{R}$. This set is certainly open, but it cannot be expressed in the form $U \times V$.

The key here is realizing that $\{U \times V: U \subset X \textrm{ open}, V \subset Y \textrm{ open}\}$ is a basis for the topology on $X \times Y$.


问:
smallest filter
Below in the block quotes is what I am trying to prove. Below the block quotes is my attempt for the first part of part (b). I guess I don't see what makes part a and part b different. For part a could I just change the way I define the filter?

Let $I$ be a nonempty set and let $\mathscr{C}$ be a nonempty collection of subsets of $I$.

(a) Prove that if $\mathscr{C}$ is closed under finite intersections and $\emptyset \notin \mathscr{C}$, then there exists $\mathscr{F} \supseteq \mathscr{C}$ such that $\mathscr{F}$ is the smallest filter containing $\mathscr{C}$.

(b) Prove that if $\mathscr{C}$ has the finite intersection property, i.e., if any finite intersection of elements of $\mathscr{C}$ is nonempty, then there exists $\mathscr{F} \supseteq \mathscr{C}$ such that $\mathscr{F}$ is the smallest
filter containing $\mathscr{C}$.

Proof (b): Define a filter $\mathscr{F}$ to be, $$\mathscr{F}=\{F : A_{1} \cap \dots \cap A_{n} \subset F \text{ for some } n \in \mathbb{N} \text{ and } A_{i} \in \mathcal{C}\}$$

First we want to show that the empty set is not in $\mathscr{F}$.

Suppose the empty set is contained in $\mathscr{F}$ such that for some $j \in \mathbb{N}$, $A_{1} \cap \dots \cap A_{j} \subseteq \emptyset$ which implies $A_{1} \cap \dots \cap A_{j} = \emptyset$. Hence a contradiction.
So $\emptyset \notin \mathscr{F}$

Now we want to show that if $A$ is an element of the filter and $A$ is a subset of $B$, then $B$ is also an element of the filter.

Let $A \in \mathscr{F}$ with $A$ a subset of $B$. Since $A \in \mathscr{F}$, there exists an $m \in \mathbb{N}$ such that $A_{1} \cap \dots \cap A_{m} \subseteq A$, but $A \subseteq B$, so $A_{1} \cap \dots \cap A_{m} \subseteq B$.
Therefore B is an element of $\mathscr{F}$.

Now we will show that if A and B are elements of the filter, then $A \cap B$ is an element of the filter.
Let $A, B \in \mathscr{F}$. Then there exists an $n_{1} \in \mathbb{N}$ such that $A_{1} \cap \dots \cap A_{n_{1}} \subseteq A$. And there exists and $n_{2} \in \mathbb{N}$ such that $A_{1} \cap \dots \cap A_{n_{2}} \subseteq B$.
Let $k=max\{n_{1}, n_{2}\}$ so that $$A_{1} \cap \dots \cap A_{k} \subseteq A_{1} \cap \dots \cap A_{n_{1}} \subseteq A$$ and $$A_{1} \cap \dots \cap A_{k} \subseteq A_{1} \cap \dots \cap A_{n_{2}} \subseteq B$$
Hence $A_{1} \cap \dots \cap A_{k} \subseteq A \cap B$.

Therefore $A \cap B \in \mathscr{F}$.

Lastly we will show that $\mathcal{C}$ is contained in the filter, $\mathscr{F}$.
Let $A_{j}$ be any element of $\mathcal{C}$. Then we have that $A_{1} \cap A_{j}$ is an element of $\mathcal{C}$ and $A_{1} \cap A_{j} \cap A_{2} \subseteq A_{j}$. Thus $A_{j} \in \mathscr{F}$.
Hence, $A_{j}$ being an element of $\mathcal{C}$ implies $A_{j}$ is an element of $\mathscr{F}$ and thus $$\mathcal{C} \subseteq \mathscr{F}$$

Therefore $\mathscr{F}$ is a filter on $I$ with $\mathcal{C} \subseteq \mathscr{F}$.

答:
The difference between (a) and (b) is that in (a) we assume that $\bigcap\mathscr{B}$ is actually a member of $\mathscr{C}$ for each finite $\mathscr{B}\subseteq\mathscr{C}$, while in (b) we assume only that these intersections are non-empty. The hypothesis of (a) implies that of (b), so once you’ve proved (b), you’ve also proved (a).

Your argument showing that $\mathscr{F}$ is a filter is basically correct but does have a couple of flaws. You write:

Let $A, B \in \mathscr{F}$. Then there exists an $n_{1} \in \mathbb{N}$ such that $A_{1} \cap \dots \cap A_{n_{1}} \subseteq A$. And there exists and $n_{2} \in \mathbb{N}$ such that $A_{1} \cap \dots \cap A_{n_{2}} \subseteq B$.
Let $k=max\{n_{1}, n_{2}\}$ so that $$A_{1} \cap \dots \cap A_{k} \subseteq A_{1} \cap \dots \cap A_{n_{1}} \subseteq A$$ and $$A_{1} \cap \dots \cap A_{k} \subseteq A_{1} \cap \dots \cap A_{n_{2}} \subseteq B$$
Hence $A_{1} \cap \dots \cap A_{k} \subseteq A \cap B$.

First, you haven’t said what the sets $A_k$ are: it is important that they are elements of $\mathscr{C}$. That could be simply an oversight, but you also appear to be assuming that one of the collections $\{A_k:k=1,\ldots,n_1\}$ and $\{A_k:k=1,\ldots,n_2\}$ is a subset of the other. It is possible to show that they can be chosen in such a way that this is true, but it is not necessary to do so, and in any case you’ve not done so. I would just say that there are finite $\mathscr{C}_A,\mathscr{C}_B\subseteq\mathscr{C}$ such that $\bigcap\mathscr{C}_A\subseteq A$ and $\bigcap\mathscr{C}_B\subseteq B$. Then $\bigcap(\mathscr{C}_A\cup\mathscr{C}_B)\subseteq A\cap B$, and $\mathscr{C}_A\cup\mathscr{C}_B$ is a finite subset of $\mathscr{C}$, so $A\cap B\in\mathscr{F}$.

Finally, you worked much too hard in showing that each $\mathscr{C}\subseteq\mathscr{F}$. All that’s needed is that for each $A\in\mathscr{C}$, $\{A\}$ is a finite subset of $\mathscr{C}$, so $A=\bigcup\{A\}\in\mathscr{F}$.

To complete (b), just show that if $\mathscr{G}$ is a filter on $I$ such that $\mathscr{C}\subseteq\mathscr{G}$, then $\mathscr{F}\subseteq\mathscr{G}$.

Title: smallest filter


问:
Find the probability density function on a unit disk
Let $D=\{(x,y)\in\mathbb{R}^2:x^2+y^2\leq1\}$ be a unit disk and $(X,Y)$ be uniformly distributed on D. Find the probability density function of W for

$W=X+Y$

$W=Y/X$

I think I need to find a cdf and then differentiate it to find the pdf. However, I have never tried finding a cdf for a circle.

Should I split the circle into four parts and then find the pdf for each of them so that each part os either strictly increasing or stringly decreasing?

答:
The joint probability density function is $f_{\small X,Y}(x,y)= \dfrac{1}{\pi}\mathbf 1_{x^2+y^2\leqslant 1}$.

Your solution requires integrating wrt $x$ and $y$ then deriving wrt the composite variable so... use the fundamental theorem of calculus to avoid some unnecessary calculations.

Thus
$$\begin{align}f_{\small X+Y}(w) &= \frac{\mathrm d ~~}{\mathrm d w}\iint_{x^2+y^2\leq 1 , x+y\leqslant w}f_{\small X,Y}(x,y)\mathrm d x\mathrm d y\\&=\mathbf 1_{-\surd 2\leqslant w\lt \surd 2}\int_{x^2+(w-x)^2\leq 1}\left\lvert\dfrac{\partial (w-x)}{\partial w}\right\rvert f_{\small X,Y}(x,w-x)\,\mathrm d x\\&=\mathbf 1_{-\surd 2\leqslant w\lt\surd 2}\int_{(w-\sqrt{2-w^2})/2}^{(w+\sqrt{2-w^2})/2}\frac 1\pi\,\mathrm d x\\&~~\vdots\end{align}$$

Likewise
$$\begin{align}f_{\small Y/X}(w) &= \frac{\mathrm d ~~}{\mathrm d w}\iint_{x^2+y^2\leq 1 , y/x\leqslant w}f_{\small X,Y}(x,y)\mathrm d x\mathrm d y\\&=\mathbf 1_{-1\leqslant w\lt 1}\int_{x^2+(xw)^2\leq 1}\left\lvert\dfrac{\partial xw}{\partial w}\right\rvert f_{\small X,Y}(x,xw)\,\mathrm d x\\&=\mathbf 1_{-1\leqslant w\lt 1}\int_{x^2\leq 1/(1+w^2)} \lvert x\rvert\cdot\frac 1\pi\,\mathrm d x\\&~~\vdots\end{align}$$


问:
Understanding open set in product topology
Let $X,Y$ are topological spaces and $U\times V\subset X\times Y$ be a subset where $X\times Y$ with product topology. If $U\times V$ is open then $U$ and $V$ are open sets in $X, Y$ respectively?

I know that if $U\times V$ is open then $U\times V =\bigcup_{i\in I}U_i\times V_i$ where $U_i\subset X$: basic open set and $V_i\subset Y$: basic open set. But I'm not sure this implies the above statement. Could you help?

答: Yes, it’s true. Let $x\in U$; we’ll show that $x$ has an open nbhd contained in $U$. To do so, let $y\in V$; then $\langle x,y\rangle\in U\times V$, so there are basic open sets $B_x$ in $X$ and $B_y$ in $Y$ such that $\langle x,y\rangle\in B_x\times B_y\subseteq U\times V$. Thus, $B_x$ is an open nbhd of $x$ contained in $U$. If we do this for each $x\in U$, we see that $U=\bigcup_{x\in U}B_x$, so $U$ is open in $X$. The argument for $V$ is similar.


问:
How we can pick an initial value for fixed point iteration to converge?
For some fixed point iterations, we get the third order convergence given by
the following error relation:

$|\alpha − x_{n+1}| = \frac14(|\alpha − x_n|)^3$

For what initial values of $x_0$ will these iterations converge? Show this range of possible values of $x_0$ in terms of interval around $\alpha$.

My initial thought is that $|\alpha − x_0|$ should be less than $1$ to make the iteration converge. Therefore, $-1+\alpha<x_0<1+\alpha$

Is my thought is correct solution to this problem?

答:
$|\alpha - x_{n+1}|=\frac14(|\alpha-x_{n}|)^3$

$\frac12|\alpha - x_{n+1}|=\frac18(|\alpha-x_{n}|)^3$

$\frac12|\alpha - x_{n+1}|=(\frac12|\alpha-x_{n}|)^3$
In order to have the iteration converge, $\frac12|\alpha-x_{n}|<1$

Therefore,

$\frac12|\alpha-x_{0}|<1$

$|\alpha-x_{0}|<2$

$\alpha+2>x_{0}>\alpha-2$


问:
Help $\lim_{k \to \infty} \frac{1-e^{-kt}}{k}=? $
what is the lime of $ \frac{1-e^{-kt}}{k}$ as $k \to \infty$?

Is that just equal $\frac{1}{\infty}=0$?

Does any one can help, I am not sure if We can apply L'Hopital's rule. S

答:
HINT:

If $t=0,e^{-kt}=1$

If $t>0, \lim_{k\to\infty}e^{-kt}=0$

If $t<0, t=-r^2$(say), $\lim_{k\to\infty}\frac{1-e^{-kt}}k=\lim_{k\to\infty}\frac{1-e^{kr^2}}k=\frac\infty\infty $

So, applying L'Hospitals' rule, $\lim_{k\to\infty}\frac{1-e^{kr^2}}k=-r^2\cdot\lim_{k\to\infty}\frac{e^{r^2t}}1=-r^2\cdot\infty=-\infty$


问:
$\int_{a}^{b}f'(x)g(x)h(x)dx$ equality
Let $f(x), g(x), h(x)$ be functions integrable on $[a, b]$ and differentiable on $(a, b)$. Does $\int_{a}^{b}f'(x)g(x)h(x)dx=f(x)g(x)h(x)|_{a}^{b}-\int_{a}^{b}f(x)g'(x)h(x)dx-\int_{a}^{b}f(x)g(x)h'(x)dx$? This question with the numerous derivatives of functions seems to apply the Leibniz Formula for derivatives of an integral, though I know it is not since we're not finding the derivative here. I also notice a patter of each equation only having one function that is a derivative, for example, $f'(x), g'(x), h'(x)$. The real problem is I don't have a method of checking whether this equation holds true or not, since the functions are unknown.

Title: $\int_{a}^{b}f'(x)g(x)h(x)dx$ equality

Body:
Hint:

$$ \int_{a}^{b}f(x)g'(x)h(x)dx+\int_{a}^{b}f(x)g(x)h'(x)dx  =
\int_{a}^{b}f(x)(g(x)h(x))'dx
$$


问:
How do you inductively prove a conditional?
I'm very lost here. I thought I understood inductive proofs, but then I was hit with this: 'Prove by induction that if n≥4, then n!≥2^n.' How should I approach solving this?

答:
First start with the base case, i.e. show it is true for $n=4$. For the inductive case, assume $n! \geq 2^n$ and show $(n+1)! \geq 2^{n+1}$. Hint: $(n+1)! = (n+1)n!$

Don't think of this problem as a conditional, think of it as you're proving some inequality holds for all $n \geq 4$.


问:
Finding a generating function for Quicksort Algorithm
I've been trying to find a generating function for the Quicksort Algorithm from the recurrence $$(n+1)b_{n+1}-(n+2)b_n=2n$$

I take the generating function as $B(x)=\sum_{n\geq 0}b_nx^n$, and, as usual, try to find this sum on the recurrence once I multiply by $x^i$ and add all of them. But I've never worked with exercises like this one involving the index and I don't find a way to get $B(x)$ from expressions like $\sum_{n\geq 1}(n+1)b_{n+1}x^n$.

答:
Let
\begin{eqnarray*}
B(x) = \sum_{n=1} ^{\infty} b_n x^n.
\end{eqnarray*}
Differentiating gives
\begin{eqnarray*}
B'(x) = \sum_{n=1} ^{\infty} n b_n x^{n-1}.
\end{eqnarray*}
Now multiply your equation by $x^{n}$ and the sum $n$ from $1$ to $ \infty$
\begin{eqnarray*}
 \sum_{n=1} ^{\infty} (n+1) b_{n+1} x^{n} -  \sum_{n=1} ^{\infty} (n+2) b_{n} x^{n} =  2 \sum_{n=1} ^{\infty} n x^{n} \\
 B'(x)-b_1  - (x B'(x)+2B(x))= \frac{2x}{(1-x)^2}\\
\end{eqnarray*}
Now you just need to solve this differential equation.


问:
Given $2\sin A + \sin B = 2\sin C$ in a triangle find minimum of $\frac{5}{\sin A} + \frac{9}{\sin C}$
Given $2\sin A + \sin B = 2\sin C $ where $A,B,C$ are angles in a triangle find minimum of $\frac{5}{\sin A} + \frac{9}{\sin C}$

so $\sin C - \sin A = \frac{ \sin B}{2}$ so $2\cos \frac{C+A}{2} \sin \frac{C-A}{2} = \frac{ \sin B}{2} = 2\sin \frac{B}{2} \sin \frac{C-A}{2} $  so $\sin \frac{C-A}{2} = \frac{1}{4}$

I feel like I am awfully close here, but how do I finish it?

答:
\begin{align*}
\sin B &= 2 (\sin C - \sin A) \\
&=4 \cos \left(\frac{A+C}{2}\right) \sin \left(\frac{C-A}{2}\right)
\end{align*}
Moreover
\begin{align*}
\sin B &= \sin(\pi-(A+C)) \\
&=2 \sin \left(\frac{A+C}{2}\right) \cos \left(\frac{A+C}{2}\right)
\end{align*}
Therefore
\begin{align*}
\sin \left(\frac{A+C}{2}\right) &= 2 \sin \left(\frac{C-A}{2}\right) \\
\sin \left(\frac{A}{2}\right)\cos \left(\frac{C}{2}\right)+\sin \left(\frac{C}{2}\right)\cos \left(\frac{A}{2}\right)&=2\sin \left(\frac{C}{2}\right)\cos \left(\frac{A}{2}\right)-2\sin \left(\frac{A}{2}\right)\cos \left(\frac{C}{2}\right) \\
3\tan \left(\frac{A}{2}\right)&=\tan \left(\frac{C}{2}\right)
\end{align*}
Let $c=\tan \left(\frac{C}{2}\right)$ and $a=\tan \left(\frac{A}{2}\right)$. We have both $a$ and $c$ in the interval $(0,\infty)$. Using $\sin(2x)=\frac{2\tan x}{1+\tan^2 x}$, we have
\begin{align*}
\frac{5}{\sin A} + \frac{9}{\sin C}&= 5\left(\frac{1+a^2}{2a}\right)+9\left(\frac{1+c^2}{2c}\right)\\
&= \frac{5}{2a}+\frac{5a}{2}+\frac{9}{2c}+\frac{9c}{2}\\
&=\frac{5}{2a}+\frac{5a}{2}+\frac{3}{2a}+\frac{27a}{2}\\
&=16a+\frac{4}{a}
\end{align*}
By the AM-GM inequality, $a=\frac{1}{2}$ minimizes this function in $(0,\infty)$ and the minimum value is $16$.


问:
How does (x1,x2,x3) go to the origin?
Consider the simple matrix equation $X'=AX$ where A is a matrix with $0's$ everywhere and $\lambda_1,\lambda_2, \lambda_3$ on the main diagonal. All the lambda's are negative and $\lambda_3 <\lambda_2< \lambda_1$ .
The solution clearly tends to $(0,0,0)$ as $\lambda_3$ dominates but I want to answer to describe how the solution through an arbitrary
initial value tends to the origin

答: The general solution is $X=(c_1e^{\lambda_1t},c_2e^{\lambda_2t},c_3e^{\lambda_3t})$, where the $c_i$ reflect the initial condition. Assuming $c_1\ne0$, we can write this as $X=c_1e^{\lambda_1t}(1,(c_2/c_1)e^{(\lambda_2-\lambda_1)t},(c_3/c_1)e^{(\lambda_3-\lambda_1)t})$. The vector part approaches $(1,0,0)$ as $t\to\infty$, confirming your belief.

Title: How does (x1,x2,x3) go to the origin?


问:
Prove for $Pr[A] - Pr[B] \leq Pr[A \setminus B]$
I have events $A$ and $B$

I want to prove that for $Pr[A] - Pr[B] \leq Pr[A \setminus B]$.

I need help getting started. Are there any axioms of probability I could use to rewrite this?

答: Start with expressing $A$ and $B$ as unions of disjoint sets. $$A=(A\smallsetminus B)\cup(A\cap B)\\ B =  (B\smallsetminus A)\cup(A\cap B)$$Now the additive property of probability can be applied.


答:
It is known that

$$
\zeta(1-s)=2^{1-s}\pi^{-s}\cos\left(\pi s\over2\right)\Gamma(s)\zeta(s)
$$

Taking logarithmic derivative on both side, we have

$$
-{\zeta'\over\zeta}(1-s)=-\log2\pi-\frac\pi2\tan\left(\pi s\over2\right)+{\Gamma'\over\Gamma}(s)+{\zeta'\over\zeta}(s)
$$

When $s=2n\ge0$, we have $\tan(n\pi)=0$, so

$$
-{\zeta'\over\zeta}(1-2n)=-\log2\pi+{\Gamma'\over\Gamma}(2n)+{\zeta'\over\zeta}(2n)
$$

For the Gamma term, plug in Weierstrass product that

$$
\Gamma(s)={e^{-s\gamma}\over s}\prod_{k=1}^\infty\left(1+\frac sk\right)^{-1}e^{s/k}
$$

we have

$$
{\Gamma'\over\Gamma}(s)=-\gamma-\frac1s+\sum_{k=1}^\infty\left(\frac1k-{1\over s+k}\right)
$$

so that

$$
\begin{aligned}
{\Gamma'\over\Gamma}(2n)
&=-\gamma-{1\over2n}+\sum_{k=1}^\infty\left(\frac1k-{1\over2n+k}\right) \\
&=-\gamma-{1\over2n}+\sum_{k=1}^{2n}\frac1k=-\gamma+H_{2n-1}
\end{aligned}
$$

Plugging this into the original equation, we get

$$
-{\zeta'\over\zeta}(1-2n)=-\log2\pi-\gamma+H_{2n-1}+{\zeta'\over\zeta}(2n)
$$

Rearranging this equation gives

$$
{\zeta'\over\zeta}(2n)=-H_{2n-1}-{\zeta'\over\zeta}(1-2n)+\gamma+\log2\pi
$$

Similar procedure can be done for $n<0$ by using

$$
\zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)\Gamma(1-s)\zeta(1-s)
$$

Title: Questions on evaluation of $\frac{\zeta'(s)}{\zeta(s)}$ at integer values of s

Body:
Question (1): Evaluation at non-negative even integers

Is it true that

$$\frac{\zeta '(2 n)}{\zeta (2 n)}=f(n)\tag{1}$$

where

$$f(n)=-H_{2 n-1}-\frac{\zeta'(1-2 n)}{\zeta(1-2 n)}+\gamma+\log(2 \pi)\tag{2}$$

and $n\in \mathbb{Z}\land n\geq 0$?

Question (2): Evaluation at negative odd integers

Is it also true that

$$\frac{\zeta '(-2 n-1)}{\zeta (-2 n-1)}=g(n)\tag{3}$$

where

$$g(n)=-H_{2 n+1}-\frac{\zeta '(2 n+2)}{\zeta (2 n+2)}+\gamma +\log (2 \pi )\tag{4}$$

and $n\in \mathbb{Z}\land n\geq 0$?

The last column in the following table illustrates the error in formula (1) above associated with the evaluation of $\frac{\zeta'(s)}{\zeta(s)}$ at non-negative even integers. At $n=0$ the function $f(n)$ must be evaluated as $\underset{\epsilon\to 0}{\text{lim}}f(\epsilon)$.

$$\begin{array}{ccccc}
 n & 2 n & \frac{\zeta'(2 n)}{\zeta(2 n)} & f(n) & f(n)-\frac{\zeta'(2 n)}{\zeta(2 n)} \\
 0 & 0 & \log (2 \pi ) & \log (2 \pi ) & 0. \\
 1 & 2 & \frac{6 \zeta '(2)}{\pi ^2} & 12 \left(\frac{1}{12}-\log (A)\right)-1+\gamma+\log (2 \pi ) & 0. \\
 2 & 4 & \frac{90 \zeta '(4)}{\pi ^4} & -120 \zeta '(-3)-\frac{11}{6}+\gamma +\log (2 \pi ) & \text{5.551115123125783$\grave{ }$*${}^{\wedge}$-16} \\
 3 & 6 & \frac{945 \zeta '(6)}{\pi ^6} & 252 \zeta '(-5)-\frac{137}{60}+\gamma +\log (2 \pi ) & \text{2.220446049250313$\grave{ }$*${}^{\wedge}$-16} \\
 4 & 8 & \frac{9450 \zeta '(8)}{\pi ^8} & -240 \zeta '(-7)-\frac{363}{140}+\gamma +\log (2 \pi ) & \text{3.0531133177191805$\grave{ }$*${}^{\wedge}$-16} \\
 5 & 10 & \frac{93555 \zeta '(10)}{\pi ^{10}} & 132 \zeta '(-9)-\frac{7129}{2520}+\gamma +\log (2 \pi ) & \text{3.3306690738754696$\grave{ }$*${}^{\wedge}$-16} \\
 6 & 12 & \frac{638512875 \zeta '(12)}{691 \pi ^{12}} & -\frac{32760}{691} \zeta '(-11)-\frac{83711}{27720}+\gamma +\log (2 \pi ) & \text{6.661338147750939$\grave{ }$*${}^{\wedge}$-16} \\
 7 & 14 & \frac{18243225 \zeta '(14)}{2 \pi ^{14}} & 12 \zeta '(-13)-\frac{1145993}{360360}+\gamma +\log (2 \pi ) & \text{1.2212453270876722$\grave{ }$*${}^{\wedge}$-15} \\
 8 & 16 & \frac{325641566250 \zeta '(16)}{3617 \pi ^{16}} & -\frac{8160 \zeta '(-15)}{3617}-\frac{1195757}{360360}+\gamma +\log (2 \pi ) & \text{1.9984014443252818$\grave{ }$*${}^{\wedge}$-15} \\
 9 & 18 & \frac{38979295480125 \zeta '(18)}{43867 \pi ^{18}} & \frac{14364 \zeta '(-17)}{43867}-\frac{42142223}{12252240}+\gamma +\log (2 \pi ) & \text{3.1086244689504383$\grave{ }$*${}^{\wedge}$-15} \\
 10 & 20 & \frac{1531329465290625 \zeta '(20)}{174611 \pi ^{20}} & -\frac{6600 \zeta '(-19)}{174611}-\frac{275295799}{77597520}+\gamma +\log (2 \pi ) & \text{2.6645352591003757$\grave{ }$*${}^{\wedge}$-15} \\
\end{array}$$

The last column in the following table illustrates the error in formula (3) above associated with the evaluation of $\frac{\zeta'(s)}{\zeta(s)}$ at negative odd integers.

$$\begin{array}{ccccc}
 n & -2n-1 & \frac{\zeta'(-2n-1)}{\zeta(-2n-1)} & g(n) & g(n)-\frac{\zeta'(-2n-1)}{\zeta(-2n-1)} \\
 0 & -1 & -12 \left(\frac{1}{12}-\log (A)\right) & -\frac{6 \zeta '(2)}{\pi ^2}-1+\gamma+\log (2 \pi ) & 0. \\
 1 & -3 & 120 \zeta '(-3) & -\frac{90 \zeta '(4)}{\pi ^4}-\frac{11}{6}+\gamma +\log (2 \pi ) & \text{5.551115123125783$\grave{ }$*${}^{\wedge}$-16} \\
 2 & -5 & -252 \zeta '(-5) & -\frac{945 \zeta '(6)}{\pi ^6}-\frac{137}{60}+\gamma +\log (2 \pi ) & \text{2.220446049250313$\grave{ }$*${}^{\wedge}$-16} \\
 3 & -7 & 240 \zeta '(-7) & -\frac{9450 \zeta '(8)}{\pi ^8}-\frac{363}{140}+\gamma +\log (2 \pi ) & \text{3.0531133177191805$\grave{ }$*${}^{\wedge}$-16} \\
 4 & -9 & -132 \zeta '(-9) & -\frac{93555 \zeta '(10)}{\pi ^{10}}-\frac{7129}{2520}+\gamma +\log (2 \pi ) & \text{3.3306690738754696$\grave{ }$*${}^{\wedge}$-16} \\
 5 & -11 & \frac{32760}{691} \zeta '(-11) & -\frac{638512875 \zeta '(12)}{691 \pi ^{12}}-\frac{83711}{27720}+\gamma +\log (2 \pi ) & \text{6.661338147750939$\grave{ }$*${}^{\wedge}$-16} \\
 6 & -13 & -12 \zeta '(-13) & -\frac{18243225 \zeta '(14)}{2 \pi ^{14}}-\frac{1145993}{360360}+\gamma +\log (2 \pi ) & \text{1.2212453270876722$\grave{ }$*${}^{\wedge}$-15} \\
 7 & -15 & \frac{8160 \zeta '(-15)}{3617} & -\frac{325641566250 \zeta '(16)}{3617 \pi ^{16}}-\frac{1195757}{360360}+\gamma +\log (2 \pi ) & \text{1.9984014443252818$\grave{ }$*${}^{\wedge}$-15} \\
 8 & -17 & -\frac{14364 \zeta '(-17)}{43867} & -\frac{38979295480125 \zeta '(18)}{43867 \pi ^{18}}-\frac{42142223}{12252240}+\gamma +\log (2 \pi ) & \text{3.1086244689504383$\grave{ }$*${}^{\wedge}$-15} \\
 9 & -19 & \frac{6600 \zeta '(-19)}{174611} & -\frac{1531329465290625 \zeta '(20)}{174611 \pi ^{20}}-\frac{275295799}{77597520}+\gamma +\log (2 \pi ) & \text{2.6645352591003757$\grave{ }$*${}^{\wedge}$-15} \\
 10 & -21 & -\frac{276 \zeta '(-21)}{77683} & -\frac{13447856940643125 \zeta '(22)}{155366 \pi ^{22}}-\frac{18858053}{5173168}+\gamma +\log (2 \pi ) & \text{8.881784197001252$\grave{ }$*${}^{\wedge}$-15} \\
\end{array}$$


问:
Show that $L_A$ is a reflection and find the line of reflection
Question:

Let $\phi \in \mathbb{R}$, and consider the matrix:

\begin{equation}
A =
\begin{pmatrix}
\cos \phi & \sin \phi\\
\sin \phi & - \cos \phi
\end{pmatrix}
\end{equation}

Prove that:

$L_A$ is a reflection

Find the axis in $\mathbb{R}^2$ about which $L_A$ reflects

My attempt:

So I did solve this question, but only after having found what the line of reflection in $\mathbb{R}^2$ is online.

The line is given by:

\begin{equation*}
    L = \{t \begin{pmatrix}
\cos \phi + 1\\
\sin \phi\\
\end{pmatrix} : t \in \mathbb{R}\}
\end{equation*}

So I worked backwards and said that given the orthonormal basis :

\begin{equation*}
    \beta_N = \left\{\frac{1}{c_1}
    \begin{pmatrix}
\sin \phi\\
- \cos \phi - 1\\
\end{pmatrix}
,\frac{1}{c_2}
\begin{pmatrix}
\cos \phi + 1\\
\sin \phi\\
\end{pmatrix}
\right\}
\end{equation*}

(for some constants $c_1, c_2$ that make these vectors unit length), we can show that

$[L_A]_{\beta_N} = \begin{pmatrix}
-1 & 0\\
0 & 1\\
\end{pmatrix}$

which by definition means that $L_A$ is a reflection. And then in $\mathbb{R}^2$, the line of reflection is the span of the second vector (i.e. $L$)

However, I was only able to solve this because I knew what the line of reflection was before hand. Without this, how would one solve this question?

Initially, I thought we could use that if $\mathbb{F} = \mathbb{R}$, then we just need to show that det$(A) = -1$. However, nowhere does it say that the field is necessarily real.

So how would we find a suitable orthonormal basis from scratch, or otherwise solve these questions?

Thank you!

答: Pick a point, any point, in the plane (except the origin), call it $P$. Work out where $A$ takes $P$, call the answer $Q$. Then the line of reflection (assuming $A$ really is a reflection) is just the perpendicular bisector of the line segment joining $P$ and $Q$.


问:
Logic problem - what kind of logic is it?
I would be most gratefull, if someone could verify my solution to this problem.

```<code> 1) Of Aaron, Brian and Colin, only one man is smart.

 Aaron says truthfully:
    1. If I am not smart, I will not pass Physics.
    2. If I am smart, I will pass Chemistry.

 Brian says truthfully:
    3. If I am not smart, I will not pass Chemistry.
    4. If I am smart, I will pass Physics.

 Colin says truthfully:
    5. If I am not smart, I will not pass Physics.
    6. If I am smart, I will pass Physics.

 While
   I. The smart man is the only man to pass one particular subject.
   II. The smart man is also the only man to fail the other particular subject.

 Which one of the three men is smart? Why?
</code>```

I would say that it could have been any one of them, as the implications in every statement are not strong enough to disprove the statements I,II. But I'm not sure if my solution is enough, as I'm not sure what kind of logic it is.

答:
Let use:

$X_S$ to mean $X$ is smart. $X_{\bar S}$ to mean $X$ is not smart.

$X_P$ to mean $X$ passes physics. $X_{\bar P}$ to mean $X$ does not pass physics.

$X_C$ to mean $X$ passes Cemistry. $X_{\bar C}$ to mean $X$ does not pass Chemistry.

1) Of Aaron, Brian and Colin, only one man is smart.
translates to :
$A_S\lor B_S \lor C_S \equiv T$

$(A_S \land B_S) \lor (A_S \land C_S) \lor (B_S \land C_S) \equiv F$

What Aaron,Brian and Colins said truthfully can be stated as:

$\left( A_S \land A_C \land A_{\bar P}\right ) \lor \left( A_{\bar S} \land A_{\bar P} \right ) \tag{I}$

$\left( B_S \land B_{\bar C} \land B_P \right ) \lor \left( B_{\bar S} \land B_{\bar C} \right ) \tag{II}$

$\left( C_S \land C_{\bar C} \land C_P\right ) \lor \left( C_{\bar S} \land C_{\bar P} \right ) \tag{III}$

While
   1. The smart man is the only man to pass one particular subject. can be stated as $X_S \land X_{P} \land Y_{P} \equiv X_S \land X_{C} \land Y_{C} \equiv F $

   2. The smart man is also the only man to fail the other particular subject.
$X_S \land X_{\bar P} \land Y_{\bar P} \equiv X_S \land X_{\bar C} \land Y_{\bar C} \equiv F $

$I \land II \land III \equiv B_S \land B_P \land B_{\bar C} \land A_{\bar S} \land A_{\bar P} \land C_{\bar S} \land C_{\bar P} \tag{IV}$

Which one of the three men is smart? Why?
Brian is the smart one because of $IV$.

I think this is simple case of combinatorial logic.


问:
How prove this $\frac{\sqrt{2}}{2}-\frac{1}{15}<F\left(\frac{\pi}{2}\right)<\frac{\sqrt{2}}{2}$

Let $$f(x)=\sum_{n=1}^{\infty}\dfrac{\cos{nx}}{\sqrt{n^3+n}}$$
  and $F(x)=\int_{0}^{x}f(t)\,\mathrm dt,F(0)=0$.

Show that: $$\dfrac{\sqrt{2}}{2}-\dfrac{1}{15}<F\left(\dfrac{\pi}{2}\right)<\dfrac{\sqrt{2}}{2}$$

Find the value $F\left(\dfrac{\pi}{2}\right)$.

答:
Since the series $f$ converges normally on $\mathbb{R}$, we can integrate term-by-term. This gives
$$
F\left(\frac{\pi}{2}\right)=\sum_{n\geq 1}\frac{1}{\sqrt{n^3+n}}\int_0^{\pi/2}\cos (nx)dx=\sum_{k\geq 0}\frac{(-1)^k}{\sqrt{2}(2k+1)^\frac{3}{2}(2k^2+2k+1)^\frac{1}{2}}.
$$
The series converges absolutely to $S$. But this is an alternating series which satisfies Leibniz criterion (the relevant function is indeed decreasing on $(0,+\infty)$). So the sequence of partial sums $S_n$ alternates about $S$ and satisfies
$$
S_{2n+1}< S< S_{2n} \qquad \forall n\geq 0.
$$
In particular, we get
$$
\frac{\sqrt{2}}{2}-\frac{1}{15}<\frac{\sqrt{2}}{2}-\frac{1}{3\sqrt{5}\sqrt{6}}=S_1<S<S_0=\frac{\sqrt{2}}{2}.
$$
This answers 1.

I don't know if there is a closed form for $S$ and I am not alone.


问:
Problem on filters
$\mathcal{F}$ is filter on $\mathcal{I}$, but not ultrafilter.

Prove that $\exists$$\mathcal{X, Y}\notin\mathcal{F}$ | $\forall\mathcal{Z}\in\mathcal{F}$ $\mathcal{X}\cap\mathcal{Z}\neq$ $\mathcal{Y}\cap\mathcal{Z}$

Thank you for your time.

答:
Do you know that a filter $\mathcal F$ on a set $I$ is an ultrafilter if and only if for any $X\subseteq I$ exactly one of the sets $X$ and $I\setminus X$ belongs to $\mathcal F$?

Using this fact we get that if $\mathcal F$ is not an ultrafilter, then there exists a set $X$ such that $X\notin\mathcal F$ and $I\setminus X\notin\mathcal F$. What could be a good candidate for $Y$ in such situation?

Title: Problem on filters


问:
Method of Undetermined Coefficients Using Complex Arithmetic
I just got done reading through the section on the technique of undetermined coefficients using complex arithmetic in the fourth edition of Nagle and Saff's book, and I'm having trouble applying what I read to problems. Here's the section I read:

I'm currently trying to use the procedure in the reading to determine the particular solutions to the following ODEs:

(a) $$y''+y=e^{-x}[\cos(2x)-3\sin(2x)]$$

(b) $$y''-2y'+10y=xe^x\sin(3x)$$

I'm stuck though. For the first problem I have the auxiliary equation $r^2+1=0\Rightarrow r=\pm i$, but I don't understand where to go from here. It would be nice if someone could work out an example, not necessarily these ones, but perhaps one such as

$$y''-3y'+15y=te^t\cos(5t)$$

I would really appreciate that.

答:
For (a), a particular solution is of the form $e^{-x}(A\cos(2x)+B\sin(2x))$, with $A,B$ to be determined.

For (b), ordinarily it would be $(Ax+B)e^x(C\cos(3x)+D\sin(3x))$, but $e^x\cos(3x)$ and $e^x\sin(3x)$ are solutions of the homogeneous equation, so you have to use $(Ax^2+Bx)e^x(C\cos(3x)+D\sin(3x))$. Well, try it, see if it works, it has been a while since I last taught this material.


问:
Average Value of a function $f_{avg}$
Let $f_{avg}$ be the average value of a continuous function f(x) on the interval [a,b]. Assume that there's only one c $\in$ [a , b] for which $f(c)=f_{avg}$. The question is, if this is the case, then must the function be either increasing or decreasing on [a, b]? This questions seems to jump from the assumption that $f(c)=f_{avg}$ for 1 value of c to the conclusion above, and I'm having trouble making sense of the question - obviously this is about the average value of a function, which is denoted by $\frac{1}{b-a}\int_{a}^{b}f(x)dx$, but how does it relate to whether the function is increasing or decreasing? One way I know of finding out is by taking the derivative of f(x), but we don't know what f is in the first place. So...plz help :D

答:
Not sure if you are asking for the proof and how can you come up with it.

In case you ask for intuitive understanding, if there is only one $c$ for which $f$ is the average value of the function, it means that even the slightest and the smallest change in the derivative of $f$ locally at some point that is not $c$ won't change the average value and therefore $c$ will still be the single argument for such $f$

$$\exists_{c\in(a, b)} \forall_{x\in[a, b]} \; \lim_{x\rightarrow k}F(x)=F(k),f(c)=\frac{F(b)-F(a)}{b-a} \implies f'(x)\geq 0 \;(f'(x) \leq 0)$$

The thing is that you might assume that there is such $k \in [a,b]$ for which the implication is wrong
 What left is to check if the assumption doesn't refutes what we are starting with

$$\exists_{c\neq d\in(a, b)} \forall_{x\in[a, b]} \; 0\in[f'(d),f'(c)] \subset\mathbb{R} \quad\lim_{\Delta x \rightarrow 0}\Delta F=0, \;f(c)=\frac{\left.F(x)\right|_a^b}{\left.x\right|_a^b} \implies f'(x)\geq 0 \;(f'(x) \leq 0)$$

$$\text{while}$$

$$\lim_{\Delta x\rightarrow 0}\Delta F=\lim_{\Delta x\rightarrow 0}f(x)\Delta x \quad \text{but $f$ is continuous in the domain [a,b], unaffected by $d$ for which the derivative is defined.}$$

$$\text{While we know that $F$ is integrable by definition of the average value:} \\f'(d)\leq0\leq f'(c)=\frac{\left.f(x)\right|_a^b}{\left.x\right|_a^b}$$

Which is unaffected by, $d\neq c$ and all is kept as it is
 Now if you assume that $d$ as an argument of $f$ doesn't makes it an average value, all that needed isn't refuted, and therefore the implication is wrong, and that's what we wanted to show, it can all be showed in short and very formally well written proof.

Title: Average Value of a function $f_{avg}$


答:
Another approach after @Donald Splutterwit's answer.

If you have $$(n+1)b_{n+1}-(n+2)b_n=2n \qquad \text{with} \qquad b_0=a$$which is not the simplest reccurence equation, you have
$$b_n=\left(a+2 \psi (n+1)+2 \gamma\right)+\left(a+2 \psi (n+1)+2 \gamma -4\right)n$$ from which
$$\sum_{n=0}^\infty b_n \,x^n=\frac{a-2 x-2 \log (1-x)}{(1-x)^2}$$

Title: Finding a generating function for Quicksort Algorithm

Body:
I've been trying to find a generating function for the Quicksort Algorithm from the recurrence $$(n+1)b_{n+1}-(n+2)b_n=2n$$

I take the generating function as $B(x)=\sum_{n\geq 0}b_nx^n$, and, as usual, try to find this sum on the recurrence once I multiply by $x^i$ and add all of them. But I've never worked with exercises like this one involving the index and I don't find a way to get $B(x)$ from expressions like $\sum_{n\geq 1}(n+1)b_{n+1}x^n$.


问:
any physical interpretation for these two matrices?
I know that the physical interpretation of $\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin\theta & \cos \theta \end{pmatrix}$ is the rotation matrix.

But what are the physical interpretations of the matrices $\begin{pmatrix} \sin \theta & \cos \theta \\ -\cos\theta & \sin \theta \end{pmatrix}$ and $\begin{pmatrix} \sec \theta & \tan \theta \\ \tan\theta & \sec \theta \end{pmatrix}$ ?

I have tried with a couple of values but was unable to arrive at a conclusion.

答:
$$\begin{pmatrix} \sin \theta & \cos \theta \\ -\cos\theta & \sin \theta \end{pmatrix} = \begin{pmatrix} \cos \left(\theta-\frac{\pi}{2}\right) & -\sin \left(\theta-\frac{\pi}{2}\right) \\ \sin \left(\theta-\frac{\pi}{2}\right) & \cos \left(\theta-\frac{\pi}{2}\right) \end{pmatrix}$$

So it is also a rotation matrix.

A physical interpretation for the other is not so clear to me. But the diagonalization may help.

$$\begin{pmatrix} \sec \theta & \tan \theta \\ \tan\theta & \sec \theta \end{pmatrix}=\begin{pmatrix} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}  \\ \frac{1}{\sqrt{2}}  & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} \sec \theta - \tan \theta & 0  \\ 0  & \sec \theta + \tan \theta \end{pmatrix}\begin{pmatrix} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}  \\ \frac{1}{\sqrt{2}}  & \frac{1}{\sqrt{2}} \end{pmatrix}
$$


答:
You aren't proving: For every $n\in \mathbb N$; we have $n \ge 4 \implies n!\ge 2^n$.

You are proving:  For every $n \ge 4; n \in \mathbb N$; we have $n!\ge 2^n$.

Okay.... admittedly those two statements are logically equivalent but you seem to be thrown you are expected to prove a condition within the induction step.  And you don't need to do any such thing.  At least no more than any other induction prove.

(After all, in a "normal" induction proof, you are not proving: for all $n$ then $n\in \mathbb N\implies P(n)$; you are just proving: for all $n\in \mathbb N$ we have $P(n)$.)

Here we are simply offsetting our base case and ignoring $n =1,2,3$ (the statement just isn't true for those) and are simply offsetting you base case.

Your base case is $n=4$ and $4! = 24 \ge 2^4 = 16$.

Now in your in your induction step you probably will not refer $n \ge 4$ but even if you did, that's fine.  You know that is true as you are only considering natural numbers at least $4$.

Anyway indcution step: If $n=k$ and we assume $k! \ge 2^k$.  Then $(k+1)! = (k+1) k! > (k+1)2^k \ge2\cdot 2^k = 2^{k+1}$.

And that's it.

Title: How do you inductively prove a conditional?

Body: I'm very lost here. I thought I understood inductive proofs, but then I was hit with this: 'Prove by induction that if n≥4, then n!≥2^n.' How should I approach solving this?


问:
If $X_i$ follows $ U( \theta, \theta+1)$ and n is even. How do I find the probability distribution of median?

I have used the following transformation to find the joint pdf: $u= X_\frac{n}{2}$ and $ v = \frac{X_\frac{n}{2}+X_{\frac{n}{2}+1}}{2}$

The joint pdf I have found is like below:
$f_{(U,V)}(u,v)= \frac{2n!}{(\frac{n}{2}-1)!*(\frac{n}{2}-1)!}(u-\theta)^{\frac{n}{2}-1}(\theta+1-2v+u)^{\frac{n}{2}-1}$ and the support for u , v is $\theta<u<\theta+1$ and $u<v<\frac{u+\theta+1}{2}$

I am trying to integrate the joint pdf with respect to u to obtain the marginal pdf of v=Median. But things are getting complicated. How should I approach this problem. Please help me!

答:
Comment: A simulation illustrates the approximate
shape of the density function of the median--in the
special case $n = 10, \theta = 5.$

```<code>set.seed(303)
h = replicate(10^5, median(runif(10, 5,6)))
hist(h, prob=T, ylim=c(0,3), col="skyblue2",
     main="n=10: Dist'n of Medians of Uniform Data")
 curve(dnorm(x, mean(h), sd(h)), add=T, lwd=2, col="orange")
</code>```

There is a 'Central Limit Theorem for Medians'. It requires
that the population density be positive at the population median ($f(5.5)>0$ in my example).

In my example, $n = 10$ is yields sample medians that
are 'nearly' normal, but the fit it not perfect.
If you run the simulation with $n = 100,$ you'll get
a better fit to normal.

Ref: The Irwin-Hall
distribution may be relevant.

Body:
Comment: A simulation illustrates the approximate
shape of the density function of the median--in the
special case $n = 10, \theta = 5.$

set.seed(303)
h = replicate(10^5, median(runif(10, 5,6)))
hist(h, prob=T, ylim=c(0,3), col="skyblue2",
     main="n=10: Dist'n of Medians of Uniform Data")
 curve(dnorm(x, mean(h), sd(h)), add=T, lwd=2, col="orange")

There is a 'Central Limit Theorem for Medians'. It requires
that the population density be positive at the population median ($f(5.5)>0$ in my example).

In my example, $n = 10$ is yields sample medians that
are 'nearly' normal, but the fit it not perfect.
If you run the simulation with $n = 100,$ you'll get
a better fit to normal.

Ref: The Irwin-Hall
distribution may be relevant.


问:
How should I deal with zero values in a list of values to be summed/averaged?
I work in a warehouse where we take components and put them together to make a finished good product.

We have these values for each component:

Quantity Required

Quantity Used

Variance (Quantity Used- Quantity Required)

Overage ((Variance/Quantity Required)) * 100%

So basically these values tell me how much we needed, how much we used, and how much we wasted.

At the bottom of the report that shows all these values, I take the sum of Quantity Required, Quantity Used, and Variance. I also take the average of Quantity Required and Quantity Used and subtract these averages to find the average Variance.

The problem: Sometimes the Quantity Required value for a component is zero. How should I adjust my summations/averages accordingly?

What I've tried: I've simply made the summations/averages ignore the rows where the Quantity Required is zero, but I think this is wrong. If my boss asks me "What is the average overage for Item X for this year?" my method will ignore the rows with zeros in the Quantity Required column, thus ignoring an amount in Quantity Used that is technically still an amount to consider.

Sample Data of 1 item being used across multiple jobs:

```<code>                   Quantity Required    Quantity Used           Variance       Overage

Job A              1000                 1100                    100            10%
Job B              900                  800                    -100           -11.1%
Job C              0                    300                     300            (N/A)

    Summations:    1900                 1900                    0              0%
    Averages:      950                  950                     0              0%
</code>```

答:
As mentioned in other answers to this question, it depends on what the 0 value represents.

If a sample is missing, then there is no way to impute its values, so the zero sample can be excluded from mean and average calculations.

However, if you have a situation where you started with 10 items and used 8 of them, then a remainder of 0 indicates an error of some sort. In such a case, the mean and average is meaningless. You first need to account for the reason(s) for the discrepancy. Then you will likely have a different data set on which to compute the mean and average.


问:
Did the computer misgrade my answer for this conditional probability question?
So I recently finished an examination (so this isn't cheating!) and this was one of the questions asked. This exam was autograded by a computer so I want to see if this is a mistake by me or the computer.

You have three cards. One is black on both sides. One is red on both sides. One is red on one side and black on the other.

You pick a card randomly. Given that you have a card whose upper side is red. What's the probability that the other side is black?

I said the answer is $1/2$ since there are only two cards that are red and since you know one side is Red already, then the probability that the other side is black is $1/2$ since there's only one card of the two reds that has black on the other side.

But the computer said that this is wrong. Who is right and why?

Thank You

答:
You are wrong.  Rather than cards you should focus on sides.

Each side is equally likely to have been seen.

You have seen a red side.  There are three of these but only one has black on the flipside.


问:
What sequence is this $\prod_{i=1}^n \frac{2i}{2i + 1} = $?
I'm trying to find a closed form solution for $\prod_{i=1}^n \frac{2i}{2i + 1} $, but I'm confused what kind of sequence this is. Does it have a closed form?

答:
After @Lemoine's answer, you can also write
$$P_n=\prod_{i=1}^n \frac{2i}{2i + 1}=\frac{\sqrt{\pi }\,\, \Gamma (n+1)}{2\,\, \Gamma \left(n+\frac{3}{2}\right)}$$ Take logarithms, use Stirling approximation twice and continue with Taylor series to obtain
$$\log(P_n)=-\frac{1}{2} \log \left(\frac{4 n}{\pi }\right)-\frac{3}{8 n}+\frac{1}{8 n^2}+O\left(\frac{1}{n^3}\right) $$
$$P_n=e^{\log(P_n)}=\frac{\sqrt{\pi }}{2 \sqrt{n}}\,\,\exp\Big[-\frac{3}{8 n}+\frac{1}{8 n^2}+O\left(\frac{1}{n^3}\right) \Big]$$

Try it for $n=10$; the exact value is
$$P_{10}=\frac{262144}{969969}\sim 0.270260$$ while the above truncated expansion gives $0.270272$.