问:
Prove that $\sum_{n=1}^{\infty}\left( 1-\frac{1}{n^2+1}\right)^{n^3}$ is convergent or divergent.
I have the series $$\sum_{n=1}^{\infty}\left( 1-\frac{1}{n^2+1}\right)^{n^3}.$$ How can I prove the convergence or divergence of it?

I tried to use the comparison test and claim that:

$$\sum_{n=1}^{\infty}\left( 1-\frac{1}{n^2+1}\right)^{n^3}\leq \sum_{n=1}^{\infty}\left( 1+\frac{1}{n^2}\right)^{n^2}$$ but the right series divergent because the limit of the sequence is e and therefore the series is divergent.

答:
Using $\log(1+x)\le x$ for $x\ge -1$, we have for $n\ge 1$

$$\begin{align}\left(1-\frac1{n^2+1}\right)^{n^3}&=e^{n^3 \log\left(1-\frac1{n^2+1}\right)}\\\\
&\le e^{-n^3/(n^2+1)}\\\\
&\le e^{-n/2}
\end{align}$$

And you can conclude now.


答:
Hint: Use the inequality
$$
1-t\le e^{-t}.
$$
Now you can compare your series to the series
$$
\sum \exp\left(-\frac{n^3}{n^2+1}\right)
$$
which in turn can be compared to the series
$$
\sum e^{-cn}
$$
for an appropriate constant $c>0$.

ContentLicense: CC BY-SA 4.0

Title: Prove that $\sum_{n=1}^{\infty}\left( 1-\frac{1}{n^2+1}\right)^{n^3}$ is convergent or divergent.

Body:
I have the series $$\sum_{n=1}^{\infty}\left( 1-\frac{1}{n^2+1}\right)^{n^3}.$$ How can I prove the convergence or divergence of it?

I tried to use the comparison test and claim that:

$$\sum_{n=1}^{\infty}\left( 1-\frac{1}{n^2+1}\right)^{n^3}\leq \sum_{n=1}^{\infty}\left( 1+\frac{1}{n^2}\right)^{n^2}$$ but the right series divergent because the limit of the sequence is e and therefore the series is divergent.


问:
Probability of A and not B
I'm studying Introduction to probability and currently, I'm stuck with the following problem. Given:

$P(A)=0.7$,
$P(B)=0.5$,
$P(A\cap B)=0.45$

What is the probability of A and not B?

I've checked this similar question but I don't understand the answers. Also I've asked my instructor and she told me that $1-P(A\cap B)(P(B^c))$ is the answer (As the answers suggested, this result is not correct). Why is that? She does not provided me a completely explanation.

Update 1: The original problem is the following

In a multiplex cinema, there are two different rooms, $A$ and $B$, working simultaneously. Let $SA$ be the event that, during a certain showing, room $A$ becomes full before the film begins, and let $SB$ be the event that, during the same showing, room $B$ becomes full before the beginning of the movie. We know that $P(SA)=0.7$; $P(SB)=0.5$ and $P(SA∩SB)=0.45$

Calculate the probability that room $A$ will become full and room $B$ will not.

Did I state the problem correctly?

Update 2: Add a Venn diagram. Following the advice of Ethan Bolker. Here is the Venn diagram that I made.

答:
I would imagine A to be a line segment of length 0.7 and B to be a line segment of length 0.5 that overlap by a distance of 0.45.

For example A could be [0, 0.7] and B [0.25, 0.75].  Then A union "not B" is [0, 0.25] so has probability 0.25.

Title: Probability of A and not B


问:
What are 'contexts' actually called?
Consider the following argument by contradiction.

\begin{array}{|l}
\mbox{We wish to deduce A.} \\
{\begin{array}{|l}
\mbox{Suppose not A.} \\
\hline \\
\mbox{Then B. Thus C. Therefore, contradiction. }
\end{array} } \\
\mbox{Thus, A.}
\end{array}

So to actually perform the argument by contradiction, we had to open a new ‘context’ wherein additional assumptions hold. What are these ‘contexts’ actually called? And which deductive systems actually utilize them?

答:
In general, natural deduction systems are most commonly associated with what you call "contexts". People call them different things, but I often hear "assumptions" or "suppositions" used, and the proofs that follow are "subproofs". When you move outside the assumption, you are discharging the assumption. An example of a natural deduction system is the Fitch system, which has resemblance to what you present. Tree proofs are also generally natural deduction systems.

Two classes of systems which don't utilize this idea are Hilbert axiomatic systems and sequent calculi. They have different ways of handling the idea of making assumptions. Hilbert systems, for instance, just force your assumptions to end up in the antecedents of conditionals one way or another, which makes the proofs much more tedious.

ContentLicense: CC BY-SA 3.0

Title: What are 'contexts' actually called?


问:
Normally distributed and uncorrelated does not imply independent wiki page
In https://en.wikipedia.org/wiki/Normally_distributed_and_uncorrelated_does_not_imply_independent, it states

To say that the pair $(X,Y)$ of random variables has a bivariate normal distribution means that every linear combination $aX+bY$ of $X$ and $Y$ for constant (i.e. not random) coefficients $a$ and $b$ has a univariate normal distribution. In that case, if $X$ and $Y$ are uncorrelated then they are independent.[1] However, it is possible for two random variables $X$ and $Y$ to be so distributed jointly that each one alone is marginally normally distributed, and they are uncorrelated, but they are not independent; examples are given below.

What does the "to be SO distributed jointly" part mean?

答:
This is a grammatical error on Wikipedia. It should read (in my opinion):

However, it is possible for two random variables $X$ and $Y$ to be jointly distributed and such  that each one alone is marginally normally distributed, and they are uncorrelated, but they are not independent; examples are given below.

The statement is trying to say that if $X$ and $Y$ are defined on the same probability space, so that they have a joint distribution, $\mathrm{corr}(X,Y) = 0$ does not imply $X,Y$ are independent.

This is important because of a subtle technicality with normally distributed random variables. If $(X,Y)$ is normally distributed (that is "$X$ and $Y$ are jointly normally distributed") then  $\mathrm{corr}(X,Y) = 0$ implies $X,Y$ are independent. However, if $X$ and $Y$ are both normally distributed, it is actually possible that $(X,Y)$ is not normally distributed, in which case $\mathrm{corr}(X,Y) = 0$ does not imply that $X,Y$ are independent.


问:
A solution space for a consistent homogeneous system which has 5 equations and 4 variables
A solution space for a consistent homogeneous system (which has 5 equations and 4 variables) is not a subspace from R5:

A) True

B) False

Answer given is true but I think it should be false.
I took an analogous condition, instead of R5 i took R3 and then took three equation:

1)x=y

2)2x=2y

3)3x=3y

All these three equation are homogeneous consistent equations.

Since x=y forms plane that pass through origin and so it's a subspace of R3, and therefore I think that if there are 4 variable and 5 equation then that should form subspace of R5. Please correct me, if I am wrong.
Thank You.

答:
The solution space is a subspace of the domain (the space that you are taking the variables from). What is the domain here? It is $\mathbb{R}^4$.

In general, a system of $m$ equations in $n$ variables can be represented by $Ax = b$, where $A$ is an $m \times n$ matrix, which we can think of as a linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^m$, and $b$ is a vector in $\mathbb{R}^m$. So the solution space is a subspace of $\mathbb{R}^n$. (In this case, $m = 5, n = 4$.)

Now, your intuition is not really wrong, in the sense that you could think of a subspace of $\mathbb{R}^4$ as embedded in $\mathbb{R}^5$, so that it is also a “subspace” of $\mathbb{R}^5$. But usually when we say that a set is a subspace of $\mathbb{R}^n$, we mean firstly that it is a subset of $\mathbb{R}^n$, and I think this is what the question was trying to get at.


问:
Sketch of 1-norm and ∞-norm
Why do we say a sketch of unit ball of Norms ||.||1 and ||.||∞?? Actually as seen in sketch it is a square.

答: The term "unit-ball" is widely used to indicate the set of elements (in any normed space) whose norm is at most $1$.

Title: Sketch of 1-norm and ∞-norm


问:
Finding a vector in Euclidian space that minimizes a loss function subject to some constraints
I'm trying to solve the following minimization problem, and I'm sure there must be a standard methodology that I could use, but so far I couldn't find any good references. Please let me know if you have anything in mind that could help or any references that you think would be useful for tackling this problem.

Suppose you are given $K$ points, $p_i \in R^n$, for $i \in \{1,\ldots,K\}$. Assume also that we are given $K$ constants $\delta_i$, for $i \in \{1,\ldots,K\}$. We want to find the vector $x$ that minimizes:

$\min_{x \in R^n} \sum_{i=1,\ldots,K} || x - p_i ||^2$

subject the following $K$ constraints:

$\frac{ || x - p_i ||^2 } { \sum_{j=1,\ldots,K} ||x - p_j||^2} = \delta_i$

for all $i \in {1,\ldots,K}$.

Any help is extremely welcome!

Bruno

edit: also, we know that $\sum_{i=1,\ldots,K} \delta_i = 1$.

答: You can try using Lagrange multiplier method - see wikipedia.</p>


答:
Hint
$$
\left( 1-\frac{1}{(n^2+1)}\right)^{n^3}= \frac{1}{\left( 1+\frac{1}{n^2}\right)^{n^3}}
$$

Now, for $n \geq 2$ you have
$$
\left( 1+\frac{1}{n^2}\right)^{n^3} \geq \binom{n^3}{2}(\frac{1}{n^2})^2=\frac{n^3-1}{2n}
$$

Therefore
$$
\sum_{n=1}^\infty \left( 1-\frac{1}{(n^2+1)}\right)^{n^3} \leq \frac{1}{2}+\sum_{n=2}^\infty \frac{2n}{n^3-1}
$$

It is trivial to show that the sum on the RHS is convergent.


答: It's called ball because for the Euclidean norm $||\cdot||_2$ it is a disk in $\mathbb{R}^2$ and a sphere in $\mathbb{R}^3$. The name is used by extension to other norms.

Title: Sketch of 1-norm and ∞-norm

Body: Why do we say a sketch of unit ball of Norms ||.||1 and ||.||∞?? Actually as seen in sketch it is a square.


问:
Scaling a compact gives a compact in a locally compact vector space
While wondering about why locally compact vector space are important, I was told that in a locally compact vector space if one scales a compact set by some number, than we still have a compact set, while this might fail in a vector space that is not locally compact. I was trying to see this with the space $\ell^2$ (space of real sequences with finite sum of squares) which is according to this post not locally compact. But then I realized that I don't even know what compact space look like in this vector space.

答: Since scaling is a continuous map of any Topological Vector Space to itself, the image of any compact subset of any TVS under scaling will also be compact.</p>


答:
Render

$(1-\frac{1}{n^2+1})^{n^3}=(\frac{n^2}{n^2+1})^{n^3}=(1+\frac{1}{n^2})^{-n^3}$ (**)

Then as $n\to\infty, (1+\frac{1}{n^2})^{-n^\color{blue}{2}}\to1/e$, so the terms of the series $\in O(2^{-n})$.  Therefore convergence by comparison with a geometric series.

(**)  You may have missed the minus sign in the exponent when you set up your comparison test.


问:
Convergence of $A^n v$ for a matrix $A$ and a vector $v$
In this question, Martin Sleziak shows that for a complex square matrix $A$, we have $$\lim A^n = 0 $$ if and only if the spectral radius of $A$ is less than $1$. I was wondering what happens if we are interested only in the convergence with respect to a given vector.

To be more specific, suppose now that all eigenvalues of $A$ have modulus $\ge 1$ and let v be some given vector such that $$\lim A^n v = 0. $$ Does it imply that $ v = 0 $?

I suspect the answer is yes, but I don't know how to deal with the case where two distinct eigenvalues of $A$ have the same modulus.

Edit: The comments encouraged me to add more details. Let $J$ be the Jordan form of $A$ and $P$ such that $A=PJP^{-1}$. Then
$$A^n =
P\begin{pmatrix}
J_1^n & \;     & \; \\
\;  & \ddots & \; \\
\;  & \;     & J_m^n
\end{pmatrix}P^{-1} $$
where $J_i$ is a Jordan block of size $k_i\times k_i$, so that

$$ J_i^n = \begin{pmatrix}
\lambda_i^n & \binom{n}{1}\lambda_i^{n-1} & \binom{n}{2}\lambda_i^{n-2} & \cdots & \cdots & \binom{n}{k_i-1}\lambda_i^{n-k_i+1} \\
 & \lambda_i^n & \binom{n}{1}\lambda_i^{n-1} & \cdots & \cdots & \binom{n}{k_i-2}\lambda_i^{n-k_i+2} \\
 &  & \ddots & \ddots & \vdots & \vdots\\
 &  & & \ddots & \ddots & \vdots\\
 &  & &  & \lambda_i^n & \binom{n}{1}\lambda_i^{n-1}\\
 &  &  &  &  & \lambda_i^n
\end{pmatrix}. $$
Thus every entry of $A^n v$ has the form
$$\tag{1}\label{sum}\sum_{i=1}^m p_i(n)\lambda_i^n$$
where $p_i$ is a complex polynomial such that $\deg p_i< k_i$ for $i=1,\dots,m$.

So essentially what I'm asking is this:

Suppose $|\lambda_i|\ge1$ for $i=1,\dots,m$ and that \eqref{sum} converges to $0$ (when $n\to\infty$). Does it imply that the sequence \eqref{sum} is identically zero?

答: By Schur triangulation, we may assume that $A$ is lower triangular. Hence all diagonal elements of $A$ (which are the eigenvalues of $A$) have moduli greater than $1$. Since $A$ is lower triangular, the first entry of $A^nv$ is $a_{11}^nv_1$. By assumption, $|a_{11}|\ge1$ and $\lim_{n\to\infty}a_{11}^nv_1=0$. Hence $v_1$ is necessarily zero. But then the second entry of $A^nv$ is $a_{22}^nv_2$. Using a similar argument to the above, we can show that $v_2=0,v_3=0,\ldots$ and so on. Hence $v=0$.


问:
Mapping reflections in 3D
I am working on a program in python that takes a point $\begin{pmatrix} x_i \\y_i\\z_i \end{pmatrix} $, generates a displacement vector $\begin{pmatrix} x_d \\y_d \\z_d\end{pmatrix} $ and adds it to the initial pont to determine the next point$\begin{pmatrix} x_f \\y_f \\z_f\end{pmatrix} = \begin{pmatrix} x_i \\y_i\\z_i \end{pmatrix}+\begin{pmatrix} x_d \\y_d\\z_d \end{pmatrix} $

My trajectory is confined to a rectangular box $\begin{bmatrix} x_0, x_1 \end{bmatrix} \times \begin{bmatrix} y_0, y_1 \end{bmatrix} \times \begin{bmatrix} z_0, z_1 \end{bmatrix} $
So when the trajectory hits a wall it bounces off. My question is how can I construct a function that takes the initial point and the displacement vector (and the boundaries of the box) and returns the point where my trajectory will end up after bouncing off said wall?

答:
There is a great way to do this, that works whenever the confinement volume is convex; and boxes and cuboids are definitely convex.

Define each of your walls as a plane (or half-space) using its unit normal vector $\hat{n}$ and its signed distance $d$ from origin.  Have $\hat{n}$ point in the direction of the bounce.

For an axis-aligned box $(x_1, y_1, z_1) - (x_2, y_2, z_2)$, where $x_2 \gt x_1$, $y_2 \gt y_1$, $z_2 \gt z_1$, those planes are
$$\begin{array}{ll}
\hat{n}_1 = (1, 0, 0), & d_1 = x_1 \\
\hat{n}_2 = (-1, 0, 0), & d_2 = -x_2 \\
\hat{n}_3 = (0, 1, 0), & d_3 = y_1 \\
\hat{n}_4 = (0, -1, 0), & d_4 = -y_2 \\
\hat{n}_5 = (0, 0, 1), & d_5 = z_1 \\
\hat{n}_6 = (0, 0, -1), & d_6 = -z_2 \\
\end{array}$$

Point $\vec{p}$ signed distance $s_i$ from wall $i$ is
$$s_i = \vec{p} \cdot \hat{n}_i - d_i$$
This is positive whenever the point is within the confined volume, and negative otherwise.  Note that if your particle is a sphere of radius $r$, just use
$$s_i = \vec{p} \cdot \hat{n}_i - d - r$$

To calculate the step for a point currently at $(x_c, y_c, z_c)$ with velocity $(x_\Delta, y_\Delta, z_\Delta)$, we start by calculating the next location for this point at $(x_n, y_n, z_n) = (x_c + t_\Delta x_\Delta, y_c + t_\Delta y_\Delta, z_c + t_\Delta z_\Delta)$.  Note that if you use fixed time steps, then you can use $t_\Delta = 1$.  However, if the particle velocities are not capped to the shortest distance between opposing walls, the particle may escape.  One way to avoid this is to use (for each time step, for all particles in that time step)
$$t_\Delta = \min\left(\tau_\max, ~ \frac{\Delta_\max}{\max\left(\lvert x_{\Delta, i} \rvert, \lvert y_{\Delta, i} \rvert, \lvert z_{\Delta, i} \rvert\right)}\right) \text{over all particles } i$$
which caps the maximum time step at $\tau_\max$, but will yield a smaller time step when necessary so that no particle is displaced by more than $\Delta_\max$ along an axis (so step distance is at most $\sqrt{3} \, \Delta_\max \approx 1.733 \, \Delta_\max$).  If you choose $L$ based on the size of your confinement box, say $L = 0.5 \min(x_2 - x_1, y_2 - y_1, z_2 - z_1)$, then no matter what the velocities of your particles, they won't travel so fast they can escape (by colliding with the same wall twice in a time step - as this implementation can only bounce once from each wall per time step).

Then, we enter a loop that fixes $(x_n, y_n, z_n)$ for any bounces.  This is because a single time step may involve more than one bounce, say near a corner.  The loop maintains a list of planes we have already bounced from once (or say a flag per plane), and those planes are ignored from the following iterations.

Within the loop, you calculate $s_i$ for all planes we haven't yet bounced from in this timestep for this particle, and find the smallest signed distance, i.e $k$ so that $s_k = \min(s_i)$.  If $s_k \gt 0$, there are no bounces, and we can exit the loop for the current particle, setting its location to $(x_n, y_n, z_n)$.  Otherwise, we bounce the particle from wall $k$, and mark that wall for "bounced" so that we won't consider it in any following iterations (for this particle in this time step).

Calculate $c_k$, the signed distance of the particles current position $(x_c, y_c, z_c)$ to plane $k$.  Note that $c_k \gt 0$, or it has already escaped; and $s_k \lt 0$ because we already checked.  Because the trajectory during the time step is linear, we know that the wall divides the trajectory as $c_k:-s_k$, i.e. it encounters the wall at fraction $f$ of the distance (or time step), with the fraction $b$ reflected from the wall,
$$f = \frac{c_k}{c_k - s_k}, \quad b = \frac{-s_k}{c_k - s_k}, \quad f + b = 1$$
Calculate the position $(x_w, y_w, z_w)$ exactly at the wall contact point, and the vector corresponding to the remaining displacement that we need to reflect $(x_d, y_d, z_d)$:
$$\left\lbrace\begin{aligned}
x_w &= b x_c + f x_n \\
y_w &= b y_c + f y_n \\
z_w &= b z_c + f z_n \\
\end{aligned}\right., \quad \left\lbrace\begin{aligned}
x_d &= x_n - x_w \\
y_d &= y_n - y_w \\
z_d &= z_n - z_w \\
\end{aligned}\right.$$
Note that this displacement vector is the one we need to reflect with respect to plane $k$, to find where the particle (tentatively) should end up at the end of current time step.  This is simple: to reflect vector $\vec{v}$ with respect to a plane having unit normal $\hat{n}$ (unit meaning $\lVert\hat{n}\lVert = 1$),

$$\vec{v}_r = \vec{v} - 2\left(\hat{n}\cdot\vec{v}\right)\hat{n}$$
Note that we need to reflect both the remaining displacement as well as the velocity vector for this particle.  If we use $(x_k, y_k, z_k)$ for the normal $\hat{n}_k$ of this plane, we calculate
$$\left\lbrace\begin{aligned}
g_1 &= 2 ( x_k x_d + y_k y_d + z_k z_d ) \\
g_2 &= 2 ( x_k x_\Delta + y_k y_\Delta + z_k z_\Delta ) \\
\end{aligned}\right.$$
and then $$\left\lbrace\begin{aligned}
x_n &= x_w + x_d - g_1 x_k \\
y_n &= y_w + y_d - g_1 y_k \\
z_n &= z_w + z_d - g_1 z_k \\
\end{aligned}\right., \quad \left\lbrace\begin{aligned}
x_c &= x_w \\
y_c &= y_w \\
z_c &= z_w \\
\end{aligned}\right., \quad \left\lbrace\begin{aligned}
x_\Delta &= x_\Delta - g_2 x_k \\
y_\Delta &= y_\Delta - g_2 y_k \\
z_\Delta &= z_\Delta - g_2 z_k \\
\end{aligned}\right.$$

At this point, we have reflected the particle from plane $k$, but it might have reflected from the other planes as well.  That's why we need to iterate the loop for this particle during this time step, until there are either no more walls to check (in which case we could just reset to checking all walls again!), or the smallest signed distance we find is positive, $s_k \gt 0$.

Also note that during this iteration, we essentially split the time step into smaller parts, at each collision with a wall.

I wonder if I should restate the "rules" in basic Vector Algebra notation, with links to some details (like Wikipedia pages)? This answer is already rather long.  Let me know in a comment, if you think it is necessary or useful.

Title: Mapping reflections in 3D


问:
Does $\int_0^1 \frac{1}{f(x)} \, dx$ converge?
If $f(x)$ is a continuous function, then if $\int_1^\infty f(x)\,dx$ converges, does $\int_0^1\frac 1 {f(x)} \, dx$ converge? I answer is yes, because $\frac{1}{x^2}$ converges, but $1\frac{1}{\frac{1}{x^2}}$ also converges. in the range $[0,1].$ And in general its true.

答:
No, consider the functon $f(x)$ that's $x$ in $[0,1]$, $e^{1-x}$ everywhere else.

Another simple one: $f(x) = \exp(-x - 1/x)$.


问:
binary-decimal primes
Converting an integer $N$ into binary and then reading the result as a base-10 number $N_2$, the prime $N=3$ gives $N_2=11$ (which is also prime) and $N=5$ gives $N_2=101$. Are $3$ and $5$ the only prime numbers whose binary expansions, read as decimals, also give primes?  If so, how is it possible to prove it?</p>

答:
A naive analysis suggests that there should be infinitely many of these: there are $2^{n-2}$ numbers of $n$ (binary) digits to test (note that every $n$-digit number must start with $1$, and it has to end with $1$ or else it'll be even).  An 'average' number of size approximately $10^n$ is prime with probability $\approx\dfrac1{\ln(10^n)}=\dfrac1{n\ln10}$ by the PNT, and likewise one of size roughly $2^n$ is prime with probability $\approx\dfrac1{n\ln2}$.  Assuming that these two events are independent, then the probability that a length-$n$ bitstring is prime in both representations is $\dfrac1{\ln2\ln10}\cdot\dfrac1{n^2}$ and so we should expect about $\dfrac{2^n}{4\cdot\ln2\cdot\ln10\cdot n^2}$ such 'double primes' of $n$ digits.  Since $2^n\gg n^2$, this should remain positive for every $n$; in fact, it increases exponentially, so summing over $n$ and letting $n\to\infty$ we should expect an infinite number of these primes.

On the other hand, it's almost a certainty that no one has proved there are infinitely many, and it would be a shock if anyone were able to: the current machinery of number theory has proven ill-equipped to deal with 'exponential' questions about primes (for instance, the questions of whether there are infinitely many Mersenne primes or Fibonacci primes), and likewise even seemingly simple questions about representations of numbers in multiple bases (for instance, 'does the base-3 expansion of $2^n$ contain a digit 2 for all $n\gt8$?') are not only unsolved but even without any reasonable known 'plan of attack'.  Since your question lies at the intersection of these two bumpy roads, it's fair to say that it would take a revolution in our knowledge of number theory to definitively answer it.

Title: binary-decimal primes


问:
How to evenly scale up measurements in a recipe to meet certain weight by gram?
I have a recipe that has 9 ingredients, with varying weights that produces a total of 191grams.

For example:

```<code>ingredient   weight(grams)
a              15
b              10
c              5
d              5
e              145
f              8
g              1
h              1
i              1
</code>```

I need to scale up the recipe to make total of 120 oz or 3401.94 grams.

How do I calculate a mulitplier for all the ingredients that evenly scales up? Do I divide 191/3401.94?

答: It's the other way around - you're scaling the recipe up, so your multiplication factor should be greater than one. You need to scale by 3401/191, multiplying all your ingredient weights by 17.8.


问:
Geometric Distribution - Probability
How we can proof that if $X$~$Geo(p)$

Than: $Pr(X>a) = (1-p)^a$

How can we proof it formally? There is an example for that to understand it intuitively?

答:
The geometric distribution gives the probability that the first occurrence of success requires $a$ independent trials, each having a success probability of $p$. Then the probability that the $a$th trial is the first success is $$P(X=a)=(1-p)^{a-1}p, $$
by independence.
Thus, \begin{align}
P\left(X>a\right)&=\sum_{k=a+1}^\infty P(X=k)\\ &=\sum_{k=a+1}^\infty (1-p)^{k-1}p \\ &=\sum_{k=1}^\infty (1-p)^{k-1}p \ - \sum_{k=1}^a(1-p)^{k-1}p  \\ &= \frac{p}{1-(1-p)}-\frac{p\left(1-(1-p)^a\right)}{1-(1-p)} =(1-p)^a.
\end{align}

Title: Geometric Distribution - Probability


答:
To proof it formally you should know the formula for the partial sum of a geometric series. The pmf of X is

$$P(X= a)= p(1-p)^{a-1}= pq^{a-1} \quad \forall \ \ a=1,2, \dots$$

And the cdf is

$$P(X \leq a) = p\sum_{i=1}^a q^{i-1}\quad \forall \ \ a=1,2, \dots$$

For a discrete random variable $X$ we have $P(X>a)=1-P(X\leq a)$.

Thus $P(X>a)=1-p\sum\limits_{i=1}^a q^{i-1}=1-\frac{p}q\sum\limits_{i=1}^a q^{i}$

The sum is a partial sum of a geometric series: $\sum\limits_{i=1}^a q^{i}=q\cdot \frac{q^a-1}{q-1}$. Use this and the fact that $p=1-q$ to obtain $P(X>a)=(1-p)^a$

Title: Geometric Distribution - Probability

Body:
How we can proof that if $X$~$Geo(p)$

Than: $Pr(X>a) = (1-p)^a$

How can we proof it formally? There is an example for that to understand it intuitively?


答:
The other answers here are correct.

Another way to understand the use of the word "ball" is to think from the point of view of someone living in the space. For them the distance from the origin to the edge of the unit ball is $1$ in any direction, so "ball" is a good name for the set. For those two norms the unit ball looks like a square only when you view it from outside, with Euclidean geometry instead of its intrinsic geometry.

Title: Sketch of 1-norm and ∞-norm


问:
Double integrate the following function
$$ \int_{0}^{a} \int_{0}^{\sqrt{a^2-y^2}} xy(x^2+y^2)^\frac{3}{2}dxdy
  $$
any way to evaluate the integrand tried this with multiple substitution unable to evaluate

答:
Without using polar coordinates,

$\displaystyle \int_{0}^{a} \int_{0}^{\sqrt{a^2-y^2}} xy(x^2+y^2)^{3/2} \ dx \ dy$

For first integral, substitute $t = x^2+y^2 \implies \frac{1}{2}dt = x dx$

Integrate $t^{3/2}$ wrt $dt$ and substitute $t$ back with $x^2+y^2$ and you get,

$\displaystyle \frac{1}{5}\int_{0}^{a} y \ \bigg[(x^2+y^2)^{5/2}\bigg]_{x=0}^{x=\sqrt{a^2-y^2}} \ dy = \frac{1}{5}\int_{0}^{a} y (a^5 - y^5) \ dy = \frac{a^7}{14}$

Title: Double integrate the following function


问:
Find a subset of $(\mathbb{Q}, <)$ isomorphic to $Exp(\omega, \omega)$ as well ordered set
I have tried to find an isomorphism $\phi$ in this way:
Let $f \in \operatorname{Exp}(\omega, \omega)$ and let the $k$ pairs $(n_i,f(n_i))$ such that for every $0<i<k+1$, $f(n_i)\ne 0$.
I'd like to close $\phi(f)$  between $n_k$ and $n_k +1$.
For example, if exist an only pairs with $f(n)\not=0$, $\phi(f)=n+1-
 \frac{1}{f(n)}$. I think it's possible to build $ \phi $ for countable recursion but I can not figure out how.</p>

答:
The following is the best I could find.

The idea is to construct the images $S_n:=\text{Im}\left(\phi|_{[0,\omega^n]}\right)$ recursively, so that:

$\phi(0)=0$ and $\phi(1)=1$;

for any $n \ge 0$, if $\phi(\omega^n)=a_n$ (so that $S_n \setminus \{ a_n \}=\text{Im}(\phi) \cap [0,a_n)$ is isomorphic to $\omega^n$), the set $S_{n+1}$ is obtained by concatenating $\omega$ "rescaled copies" of $S_n \setminus \{ a_n \}$, each of which is contained in an interval of the form $\left[2a_n(1-2^{-k}),2a_n(1-2^{-k-1})\right]$, and finally $\phi(\omega^{n+1})=2a_n=a_{n+1}$.

This actually satisfies $\phi(f)\in [2^{n_k},2^{n_k+1}]$ (with the same notation you used), but it should have some other nice properties, like the following (I didn't prove all of them, but I strongly believe all of them to be true):

$\phi(\omega^n)=2^n$;

$\phi(\omega \cdot \alpha)=2 \phi(\alpha)$;

$\phi$ is limit-preserving;

if $\alpha=\omega^{n_1}+\omega^{n_2}+...+\omega^{n_r}$ with $n_1 \ge n_2 \ge ... \ge n_r$, then $\phi(\alpha)=\sum\limits_{i=1}^r 2^{n_i+1-i}$.


问:
Optimization : Solve $4$ unknowns using $4$ equations
Suppose we have 4 points say A($x_1,y_1,z_1$), B($x_2,y_2,z_2$),C($x_3,y_3,z_3$), D($x_4,y_4,z_4$). Where $x_1,x_2,x_3,x_4,y_1,y_2,y_3,y_4$ are known points and rest $z_1,z_2,z_3,z_4$ are unknown points .

$4$ points will make a rectangle we want to find the corresponding Z - Coordinate using optimization .

We can make 4 equations as

$AD^2 = AB^2$

$$(z_4- z_1)^2 - (z_3 - z_2)^2 = (x_3 - x_2)^2 + (y_3 - y_2)^2 - (x_4 - x_1)^2 - (y_4 - y_1)^2 $$

$AB^2 = CD^2$

$$(z_2- z_1)^2 - (z_3 - z_4)^2 = (x_3 - x_4)^2 + (y_3 - y_4)^2 - (x_2 - x_1)^2 - (y_2 - y_1)^2 $$

(AB perpendicular to AD)
$$(z_2- z_1) (z_4 - z_1) = -[(x_2 - x_1)(x_4 - x_1) + (y_2 - y_1)(y_4 - y_1)] $$

(BC perpendicular to CD)
$$(z_2- z_3) (z_4 - z_3) = -[(x_2 - x_3)(x_4 - x_3) + (y_2 - y_3)(y_4 - y_3)] $$

Objective equation can be to minimize the area of rectangle i.e.
$$Minimize (AB.AD) $$
$$i.e (Length* breadth)$$

How to solve this optimization so that we can we find other $4$ z-coordinate (i.e. $z_1, z_2, z_3, z_4$) corresponding to other 4 point .

答:
Given four points $p_1,p_2,p_3,p_4$ if they pertain to a rectangle then they are contained into the same plane $\pi$. Also the coordinates $x,y$ in each point, represent the $XY$ plane projection which should be a parallelogram. Assuming this projection is not a line (degenerate case) from this projection we can select a sequence to form a rectangle. From here we assume that the sequence $p_1\to p_2\to p_3\to p_4\to p_1$ gives us a rectangle. Curiously, only three of $z_1,z_2,z_3,z_4$ are independent so we assume $z_4 = z_1+z_3-2z_2$. If the objective is to determine the minimum perimeter rectangle then the optimization problem

$$
\min_{z_i}|p_1-p_2|+|p_2-p_3|+|p_3-p_4|+|p_4-p_1| \ \ \ \text{s. t.} \ \ \cases{(p_1-p_2)\cdot(p_2-p_3)=0\\
(p_2-p_3)\cdot(p_4-p_3)=0\\
(p_4-p_3)\cdot(p_4-p_1)=0
}
$$

comes with the solution.

Follows a MATHEMATICA script to accomplish with the calculations

```<code>p1 = {0, 0, z1};
p2 = {1, 2, z2};
p3 = {5, 3, z3};
p4 = {4, 1, z1+z3-2z2};
obj = Norm[p2 - p1] + Norm[p3 - p2] + Norm[p4 - p3] + Norm[p1 - p4]
sol = NMinimize[{obj, (p1 - p2).(p2 - p3) == 0, (p2 - p3).(p4 - p3) == 0, (p3 - p4).(p4 - p1) == 0}, {z1, z2, z3}]
quad = {p1, p2, p3, p4} /. sol[[2]]
Graphics3D[Polygon[quad]]
</code>```

NOTE

To obtain the minimum area use instead

```<code>obj = Norm[p2 - p1] Norm[p3 - p2]
</code>```

In python a possible solution script could be

```<code>import numpy as np
import math
from scipy.optimize import minimize

x0 = [-1,1,-3]

def dif(x,y):
    z = np.subtract(x,y)
    return z

def obj(z):
    p1 = [0,0,z[0]]
    p2 = [1,2,z[1]]
    p3 = [5,3,z[2]]
    obj = math.sqrt(np.dot(dif(p1,p2),dif(p1,p2)))*math.sqrt(np.dot(dif(p2,p3), dif(p2,p3)))
    return obj

def restr1(z):
    p1 = [0,0,z[0]]
    p2 = [1,2,z[1]]
    p3 = [5,3,z[2]]
    return np.dot(dif(p1,p2), dif(p2,p3))

def restr2(z):
    p2 = [1,2,z[1]]
    p3 = [5,3,z[2]]
    p4 = [4,1,z[0]+z[2]-2*z[1]]
    return np.dot(dif(p2,p3),dif(p4,p3))

def restr3(z):
    p1 = [0,0,z[0]]
    p3 = [5,3,z[2]]
    p4 = [4,1,z[0]+z[2]-2*z[1]]
    return  np.dot(dif(p4,p3), dif(p4,p1))

cons = ({'type': 'eq', 'fun': restr1},
        {'type': 'eq', 'fun': restr2},
        {'type': 'eq', 'fun': restr3})

res = minimize(obj,x0,constraints=cons)
print (res.fun)
print(res.x)

# Those prints show the residual form zero, to each
# restriction

print(restr1(list(res.x)))
print(restr2(list(res.x)))
print(restr3(list(res.x)))

def plane(p):
    p1 = [0,0,res.x[0]]
    p2 = [1,2,res.x[1]]
    p3 = [5,3,res.x[2]]
    n = list(np.cross(dif(p2,p1),dif(p3,p2)))
    residue = np.dot(n,dif(p,p1))
    return residue

# The following prints show the residual from zero
# at each vertex regarding the plane containing the rectangle

print(plane([0,0,res.x[0]]))
print(plane([1,2,res.x[1]]))
print(plane([5,3,res.x[2]]))
print(plane([4,1,res.x[0]+res.x[2]-2*res.x[1]]))
</code>```

Body:
Given four points $p_1,p_2,p_3,p_4$ if they pertain to a rectangle then they are contained into the same plane $\pi$. Also the coordinates $x,y$ in each point, represent the $XY$ plane projection which should be a parallelogram. Assuming this projection is not a line (degenerate case) from this projection we can select a sequence to form a rectangle. From here we assume that the sequence $p_1\to p_2\to p_3\to p_4\to p_1$ gives us a rectangle. Curiously, only three of $z_1,z_2,z_3,z_4$ are independent so we assume $z_4 = z_1+z_3-2z_2$. If the objective is to determine the minimum perimeter rectangle then the optimization problem

$$
\min_{z_i}|p_1-p_2|+|p_2-p_3|+|p_3-p_4|+|p_4-p_1| \ \ \ \text{s. t.} \ \ \cases{(p_1-p_2)\cdot(p_2-p_3)=0\\
(p_2-p_3)\cdot(p_4-p_3)=0\\
(p_4-p_3)\cdot(p_4-p_1)=0
}
$$

comes with the solution.

Follows a MATHEMATICA script to accomplish with the calculations

p1 = {0, 0, z1};
p2 = {1, 2, z2};
p3 = {5, 3, z3};
p4 = {4, 1, z1+z3-2z2};
obj = Norm[p2 - p1] + Norm[p3 - p2] + Norm[p4 - p3] + Norm[p1 - p4]
sol = NMinimize[{obj, (p1 - p2).(p2 - p3) == 0, (p2 - p3).(p4 - p3) == 0, (p3 - p4).(p4 - p1) == 0}, {z1, z2, z3}]
quad = {p1, p2, p3, p4} /. sol[[2]]
Graphics3D[Polygon[quad]]

NOTE

To obtain the minimum area use instead

obj = Norm[p2 - p1] Norm[p3 - p2]

In python a possible solution script could be

import numpy as np
import math
from scipy.optimize import minimize

x0 = [-1,1,-3]

def dif(x,y):
    z = np.subtract(x,y)
    return z

def obj(z):
    p1 = [0,0,z[0]]
    p2 = [1,2,z[1]]
    p3 = [5,3,z[2]]
    obj = math.sqrt(np.dot(dif(p1,p2),dif(p1,p2)))*math.sqrt(np.dot(dif(p2,p3), dif(p2,p3)))
    return obj

def restr1(z):
    p1 = [0,0,z[0]]
    p2 = [1,2,z[1]]
    p3 = [5,3,z[2]]
    return np.dot(dif(p1,p2), dif(p2,p3))

def restr2(z):
    p2 = [1,2,z[1]]
    p3 = [5,3,z[2]]
    p4 = [4,1,z[0]+z[2]-2*z[1]]
    return np.dot(dif(p2,p3),dif(p4,p3))

def restr3(z):
    p1 = [0,0,z[0]]
    p3 = [5,3,z[2]]
    p4 = [4,1,z[0]+z[2]-2*z[1]]
    return  np.dot(dif(p4,p3), dif(p4,p1))

cons = ({'type': 'eq', 'fun': restr1},
        {'type': 'eq', 'fun': restr2},
        {'type': 'eq', 'fun': restr3})

res = minimize(obj,x0,constraints=cons)
print (res.fun)
print(res.x)

# Those prints show the residual form zero, to each
# restriction

print(restr1(list(res.x)))
print(restr2(list(res.x)))
print(restr3(list(res.x)))

def plane(p):
    p1 = [0,0,res.x[0]]
    p2 = [1,2,res.x[1]]
    p3 = [5,3,res.x[2]]
    n = list(np.cross(dif(p2,p1),dif(p3,p2)))
    residue = np.dot(n,dif(p,p1))
    return residue

# The following prints show the residual from zero
# at each vertex regarding the plane containing the rectangle

print(plane([0,0,res.x[0]]))
print(plane([1,2,res.x[1]]))
print(plane([5,3,res.x[2]]))
print(plane([4,1,res.x[0]+res.x[2]-2*res.x[1]]))


答:
What does $\{X>a\}$ mean? It means that the first $a$ trials were failures. Also you know that the trials were independent. Therefore

$$
Pr(X>a) = (1-p)^a.
$$

If you want to derive it from the probability mass function: $Pr(X=k)=p(1-p)^{k-1}$ for $k=1,2,\ldots$, then

$$
Pr(X>a) =p \sum_{k=a+1}^{\infty}(1-p)^{k-1} = (1-p)^a.
$$

Title: Geometric Distribution - Probability


答:
The function $f(x)\equiv 0$ is continuous, $\int_0^\infty f(x)\,dx$ converges, but $\frac{1}{f(x)}$ is not even defined.

Edit:

After the edits done to the question:

$\displaystyle \int_1^\infty f(x)\,dx$ and $\displaystyle\int_0^1\frac{1}{f(x)}\,dx$ evaluates $f$ at different points with the only exception of $x=1$. The values of $f$ at $(1,\infty)$ in no way can influence the behavior of $f$ in the interval $[0,1)$, the only condition is that the function must glue well at $1$. The function
$$f(x) = \begin{cases}
x & x\in [0,1]\\
1/x^2 & x \in [1,\infty)
\end{cases}$$

is continuous, $\displaystyle \int_1^\infty f(x)\,dx$ converges but $\displaystyle\int_0^1\frac{1}{f(x)}\,dx$ diverges.

This function is not strictly greater than $0$ because $f(0) = 0$.

Another edit:

If you want the additional requirement that $f(x)> 0$ for all $x$ then $\frac{1}{f(x)} > 0$ and is continuous in $[0,1]$. Then $\int_0^1\frac{1}{f(x)}\,dx$ converges just because $\frac{1}{f(x)}$ is a continuous function in $[0,1]$, it has nothing to do with the convergence of $\int_1^\infty f(x)\,dx$.

Title: Does $\int_0^1 \frac{1}{f(x)} \, dx$ converge?

Body: If $f(x)$ is a continuous function, then if $\int_1^\infty f(x)\,dx$ converges, does $\int_0^1\frac 1 {f(x)} \, dx$ converge? I answer is yes, because $\frac{1}{x^2}$ converges, but $1\frac{1}{\frac{1}{x^2}}$ also converges. in the range $[0,1].$ And in general its true.


问:
What is the branch of mathematics that combines discrete mathematics and linear algebra known as?
Many different branches of mathematics are combined together.
For example, algebra and topology combine to give Algebraic Topology; algebra and geometry give rise to Algebraic Geometry, etc.

What is the name of the branch which combines Discrete Mathematics and Linear Algebra, if such a branch exists? If not, which subfields of discrete mathematics make significant use of linear algebra and do they have specific names?

答:
I believe what you are looking for is Discrete Geometry. I say so because;

At the beginning of Discrete Geometry, you define notions like "affine combination" and "convex hull" which are inherited from Linear Algebra. For this you can check any Discrete Geometry textbook, and if you need recommendations I would definitely suggest Lectures on Discrete Geometry by Jiří Matoušek. It is a dense book but easily readable if you are familiar with things like compactness.

As you have it in the name, Discrete Geometry deals with combinatorial properties of objects like convex sets, vertices, hyperplanes and edges.

I hope this helps!


问:
Let $f:(0,\infty)\to \mathbf{R}$ and $L,L' \in \mathbf{R}$. If $\lim_{x\to\infty}f(x)=L$ and $\lim_{x\to\infty}f'(x)=L'$, show that $L'=0$

Let $f:(0,\infty)\to \mathbf{R}$ be differentiable on $(0,\infty)$ and $L,L' \in \mathbf{R}$. Suppose that $\lim_{x\to\infty}f(x)=L$ and $\lim_{x\to\infty}f'(x)=L'$. Prove that $L'=0$. I know that there are a few posts on this statement, but I didn't see my approach. So, I would like to know if my proof holds, please. In the following proof I will distinguish 2 cases: (1) $L'>0$ and (2) $L'<0$

(1) Suppose that $L'>0$. As  $\lim_{x\to\infty}f'(x)=L'$ exists, then there $\exists M>0 \ \forall x\ge M:0<\frac{B}{2} \le f'(x)$. Thus, $f$ is strictly increasing on $[\frac{B}{2},\infty)$.

Let $[x,x+a]\subset [\frac{B}{2},\infty)$ with $a>0$. As $f$ is differentiable on $(0,\infty)$, it is differentiable on $[x,x+a]$ as well and so continuous. By MVT we have then that there $\exists c_x \in ]x,x+a[$ such that:

$f'(c_x)=\frac{f(x+a)-f(x)}{a}$ and by inserting a limit on RHS and LHS we have $\lim_{x\to\infty}f'(c_x)=\lim_{x\to\infty}\frac{f(x+a)-f(x)}{a}=0$. As $f$ is increasing and $\lim_{x\to\infty}f'(x)$ exists, we can conlude that $\lim_{x\to\infty}f'(x)=0$.

(2) The proof is similar to the previous one, but this time we have that $\exists M>0 \ \forall x\ge M$: $f'(x)<-2B<0$ and we have that $f$ is decreasing on $[M, \infty)$.

答: I think you do not even need to divide two cases. Since you already know that a limit of $f'(x)$ with $x\rightarrow +\infty$ exists you can write that $L'=\lim\limits_{x\rightarrow +\infty}f'(x) =\lim\limits_{x\rightarrow +\infty} f'(c_x)=\lim\limits_{x\rightarrow +\infty} \frac{f(x+a)-f(x)}{a}=\frac{1}{a}(L-L)=0$. The second equality is true because $c_x\in (x,x+a)$ converges to $+\infty$ if $x$ converges to $+\infty$ and because of the fact that any subsequence of converging sequence converges to the same limit. The last step you can explain in different ways, I named just one of way.


问:
Does a formula exist that gives a function of only solutions that are multiples of any given integer m up to n multiples?
I feel like this is a stupid question but does a formula exist that gives a continuous function of x such that its only solutions are at integer multiples of m up to n? The domain of the function is any real value x and the range is any real value y for the sake of argument, but could be a wave function, polynomial, etc or any continuous function. This is the working graph I have at m = 3 except that solutions should lie under each bell curve at y = 0 instead of at y = 1, I for some reason I lost the formula that does the former. I have an interesting zeta form of the equations that yield actual solutions at t = m but I'm still playing around with it.

I started playing around with a set of formulas one day after finding a strange relationship with Young modulus, Ramanujan congruencies, Collatz function, and logistic chaos equations I was working with at the time--as far as I could tell what I was looking at was a kernel for the zeta function for the given s expression I had, but I'm no mathematician so don't crucify me, ha. I would share the formulas but I don't know math jax or whatever that is, I only know latex. But other than what my question suggests I don't think the answer will be surprising, it's just that I've seen some weird stuff with this math which is simple so why would it do such as prime factorization and prime maps I didn't learn in school, so I've toyed with it for over a year on occasions, but mostly can't make sense of it.

答:
Assuming that $n$ is itself a multiple of $m$, the function
$$
\frac{\Gamma(1+(n-x)/m)}{\Gamma(1-x-m)}
$$
(where $\Gamma$ is the Euler Gamma function)
has zeros precisely at $x=m,2m,\dots,n$ and is continuous (indeed, infinitely differentiable) everywhere.

Note that this is simply a polynomial
$$
\frac{\Gamma(1+(n-x)/m)}{\Gamma(1-x-m)} = \frac{(x-m)(x-2m)\cdots(x-n)}{m^{n/m}},
$$
but perhaps writing it in this fancy way makes it clearer that it's a well-behaved function.


答: Your proof is technically correct (modulo the typo $[B/2, \infty)$ should be $[M, \infty)$ everywhere). I think there is a simpler approach. Note that $f'(x) \geq B/2$ doesn't just imply that $f$ is increasing, it actually implies (just integrate both sides) that $\lim_{x \to \infty} f(x) = \infty$, a contradiction. If this is for homework and you have not learned about integration yet, then I would stick with your original approach though.

Title: Let $f:(0,\infty)\to \mathbf{R}$ and $L,L' \in \mathbf{R}$. If $\lim_{x\to\infty}f(x)=L$ and $\lim_{x\to\infty}f'(x)=L'$, show that $L'=0$

Body:
Let $f:(0,\infty)\to \mathbf{R}$ be differentiable on $(0,\infty)$ and $L,L' \in \mathbf{R}$. Suppose that $\lim_{x\to\infty}f(x)=L$ and $\lim_{x\to\infty}f'(x)=L'$. Prove that $L'=0$. I know that there are a few posts on this statement, but I didn't see my approach. So, I would like to know if my proof holds, please. In the following proof I will distinguish 2 cases: (1) $L'>0$ and (2) $L'<0$

(1) Suppose that $L'>0$. As  $\lim_{x\to\infty}f'(x)=L'$ exists, then there $\exists M>0 \ \forall x\ge M:0<\frac{B}{2} \le f'(x)$. Thus, $f$ is strictly increasing on $[\frac{B}{2},\infty)$.

Let $[x,x+a]\subset [\frac{B}{2},\infty)$ with $a>0$. As $f$ is differentiable on $(0,\infty)$, it is differentiable on $[x,x+a]$ as well and so continuous. By MVT we have then that there $\exists c_x \in ]x,x+a[$ such that:

$f'(c_x)=\frac{f(x+a)-f(x)}{a}$ and by inserting a limit on RHS and LHS we have $\lim_{x\to\infty}f'(c_x)=\lim_{x\to\infty}\frac{f(x+a)-f(x)}{a}=0$. As $f$ is increasing and $\lim_{x\to\infty}f'(x)$ exists, we can conlude that $\lim_{x\to\infty}f'(x)=0$.

(2) The proof is similar to the previous one, but this time we have that $\exists M>0 \ \forall x\ge M$: $f'(x)<-2B<0$ and we have that $f$ is decreasing on $[M, \infty)$.


问:
Dual vector space, proof of inclusion in a span
Question:
V is a vector space (of finite dimension) over F. We assume $\alpha, \beta \in V^*$  (The dual space) and they satisfy: $\forall v: (\alpha (v)=0 \Rightarrow \beta (v)=0$)
prove that $\beta \in sp(\alpha)$.

What I thought:
What I understand is that I need to prove that $\beta$ is actually $\alpha$ multiplied by a scalar. I also think that $\operatorname{Ker}\alpha \subseteq \operatorname{Ker}\beta$ (from what we were given).
Would love some hints at first.

答:
Your hypothesis is indeed $\ker(\alpha) \subseteq \ker(\beta)$.

There are two possibilities. Either $\beta = 0$, and then...

Otherwise $\beta \ne 0$. Then the kernels of $\alpha$ and $\beta$ are two subspaces of the same dimension, one contained in the other, and then...

Finally, extend a basis of $\ker(\alpha)$ to a basis of $V$, and see how $\alpha$ and $\beta$ act on this basis.


答: Your idea is good, but you can see that in (1) since $\exists M>0$ such that $\forall x\geq M$ $: f'(x)>B/2>0$ if you integrate the last inequality you have that $f(x)-f(M)>B/2(x-M)$ $ \forall x\geq M$ and taking $x\to\infty$ you obtain that $L=\infty$ and contradiction. For (2) is the same reasoning.


问:
Maximum modulus theorem for holomorphic function on an unbounded region
Suppose $\Gamma$ is the boundary of an unbounded region $\Omega$, $f \in H(\Omega)$, $f$ is continuous on $\Omega \cup \Gamma$, and there are constants $B < \infty$ and $M < \infty$ such that $ |f| \leq M$ on $\Gamma$ and $ |f| \leq B $ in $\Omega$. Prove that we then actually have $ |f| \leq M $ in $\Omega$.

Proof:

According to the suggestion,
fix $z_0 \in \Omega$ and let n be a large integer. Let $V$ be a large disc with center at $0$, and apply the maximum modulus theorem to the function $f(z)^n/z$ in the component of $V \cap \Omega$ which contains $z_0$.

Then
$$
\left|\frac{f(z_0)^n}{z_0}\right| \leq \max\left(\frac{B^n}{r}, M^n\right)
$$
r is the radius of $V$. If $r$ is large enough this means $|\frac{f(z_0)^n}{z_0}| \leq M^n$, for $n \to \infty$ it means $|f(z_0)|<M$, as $z_0$ is arbitrary  $ \in \Omega$, this means $|f(z)| \leq M$.

Is my proof correct?

答:
There is a problem with your proof: The function $f(z)^n/z$ is not holomorphic in $\Omega$ if $0 \in \Omega$, and it is not bounded on the boundary of $V \cup \Omega$ if $0 \in \Gamma$.

If $0 \in \Bbb C \setminus \overline \Omega$ then $\delta = \operatorname{dist}(0, \Gamma) > 0$, and your approach works with the estimate
$$
\left|\frac{f(z_0)^n}{z_0}\right| \leq \max\left(\frac{B^n}{r}, \frac{M^n}{\delta}\right) \, .
$$
If there is any other point $z_1$ in $\Bbb C \setminus \overline \Omega$ then the same idea works by considering $f(z)^n/(z-z_1)$ instead.

But all this fails if $\overline \Omega = \Bbb C$, i.e. if there is no open disk in the complement of the domain. In that case one has to replace $1/z$ by a different function, see below.

Also note that the statement is wrong if the boundary of $\Omega$ is empty, i.e. if $\Omega = \Bbb C$. Then the condition “$|f|\le M$ on $\Gamma$” holds for arbitrary values of $M$ (as a “vacuous truth”), but that does not imply that $|f| \le M$ in $\Omega$.

A correct formulation of the statement would be:

Let $\Omega \subset \Bbb C$ be a domain with non-empty boundary $\Gamma$, and $f: \Omega \cup \Gamma \to \Bbb C $ a continuous function which is holomorphic in $\Omega$. Suppose there are real constants $M$ and $B$ such that

$|f(z)| \le M$ for all $z \in \Gamma$, and

$|f(z)| \le B$ for all $z \in \Omega$.

Then $|f(z)| \le M$ for all $z \in \Omega$.

(The following proof is taken from Bak J., Newman D.J. (2010) Maximum-Modulus Theorems for Unbounded Domains. In: Complex Analysis. Undergraduate Texts in Mathematics. Springer, New York, NY. https://doi.org/10.1007/978-1-4419-7288-0_15 .)

We may assume that $f$ is not constant, because the statement is trivially true for constant functions. (Why?)

Fix $z_0 \in \Bbb \Omega$ and a positive integer $n$. We choose any $z_1 \in \Omega$ with $f(z_1) \ne f(z_0)$ and define the function $g$ on $\Omega \cup \Gamma$ as
$$
 g(z) = \begin{cases}
\frac{f(z)-f(z_1)}{z-z_1} & \text{ if } z \ne z_1 \, , \\
f'(z_1) & \text{ if } z = z_1 \, .
\end{cases}
$$

$g$ replaces the function $1/z$ from your proof. It has the following properties:

$g$ is holomorphic in $\Omega$ and continuous on $\Omega \cup \Gamma$.

$\lim_{z \to \infty} g(z) = 0$ (because $f$ is bounded).

$g$ is bounded on $\Omega \cup \Gamma$.

We set $K = \sup \{z \in \Omega \cup \Gamma : |g(z)| \}$ and choose $r> 0$ so large such that

$|z_0| < r$, and

$|g(z)| < K M^n/B^n$ for all $z \in \Omega$ with $|z| = r$.

Now we can apply the maximum modulus theorem to $h(z) = f(z)^n g(z)$ on $\Omega \cap B_r(0)$. On boundary points $z$ with $|z| = r$ we have
$$
 |f(z)^n g(z)| \le B^n |g(z)| \le M^n K
$$
and on boundary points $z \in \Gamma$
$$
 |f(z)^n g(z)| \le  M^n K
$$
holds as well. It follows that
$$
 |f(z_0)| \le M \left( \frac{K}{|g(z_0)|}\right)^{1/n} \, ,
$$
and for $n \to \infty$ we conclude that $|f(z_0)| \le M$.

Question to the reader: Where does the proof use that the boundary of the domain is not empty?


问:
How to solve linnear approximation - a calculus problem
Let $f(x) = 2x^3+(0)x^2+ (-6)x.$ Find $x=c$ so that the linear approximation $g(x)$ to $f(x)$ in $x = c$ for $g(0) = -108$

I really can't solve this problem and it has been bugging me for hours. Can someone please explain how to solve this?

答:
If $f(x)=2x^3-6x$, its linear approximation around $x=c$ is given by
$$
g(x)= f(c) + f'(c)(x-c) = 2c^3-6c + (6c^2-6) (x-c).
$$

Se $g(0)=-108$, you have that
$$
2c^3-6c-6c^3+6c = -108 \Leftrightarrow c^3 = 27 \Leftrightarrow c = 3.
$$


问:
Why does $v \in \ker(A- \lambda I)^m$ imply that $Av$ is contained in $\ker(A- \lambda I)^m$?
Given an $n$ dimensional vector space $V$ and an $n \times n$ matrix $A$,

Then why does $v \in \ker(A- \lambda I)^m$ imply that $Av$ is contained in  $\ker(A- \lambda I)^m$?

I'm not seeing it.

答: It is easier here to go for a more general fact: the kernel of any polynomial $P$ of $A$ is stable under $A$. Or even more general and closer to what you need in the proof: the kernel of any linear operator commuting with $A$ is stable under $A$. The same is also true if you replace "kernel" by "image" (the proof is slightly different, but equally straightforward).


问:
Maximum and minimum values of unions and intersections
If P(A) = 0.3 and P(B) = 0.9, find
(a) the maximum possible value of P(A ∪ B);
(b) the minimum possible value of P(A ∪ B);
(c) the maximum possible value of P(A ∩ B);
(d) the minimum possible value of P(A ∩ B).

Is it correct to say that max value of P(A ∪ B) = 1? I know P(A ∪ B) = P(A) + P(B), but 0.9 + 0.3 = 1.2 and this isn't possible.

Minimum value of P(A ∪ B) = P(A) + P(B) - P(A ∩ B), 0.9 + 0.3 - 0.27 = 0.93

Maximum value of P(A ∩ B) =  P(A) x P(B) = 0.9 x 0.3 = 0.27

Minimum value of P(A ∩ B) = 0 if A and B are disjoint and can't occur together

I'd really appreciate any feedback on whether I have this correct. Thank you

答:
Recall the following:

$\Pr(A\cup B)\geq \Pr(A)$ by monotonicity

$\Pr(A\cap B) \leq \Pr(A)$ also by monotonicity

$\Pr(A\cup B) = \Pr(A)+\Pr(B) -\Pr(A\cap B)$ by inclusion-exclusion

$0\leq \Pr(A)\leq 1$

For your specific problem, you are told $\Pr(A)=0.3$ and $\Pr(B)=0.9$.

$0.9=\Pr(B)\leq \Pr(A\cup B)$ so $\Pr(A\cup B)$ is bounded below by $0.9$.  This is a tight lower bound since it is possible that $A\subseteq B$ in which case $\Pr(A\cup B)$ is exactly equal to $0.9$.  For instance, where we have the uniform distribution over $\{1,2,3,\dots,10\}$ and $A=\{1,2,3\}$ and $B=\{1,2,3,\dots,9\}$.

$\Pr(A\cup B)\leq 1$ since it is a probability.  This is a tight upper bound since it is possible that $\Pr(A\cup B)=1$.  For instance, where we have the uniform distribution over $\{1,2,3,\dots,10\}$ and $A=\{1,2,3\}$ and $B=\{2,3,4,\dots,10\}$.

$\Pr(A\cap B)\leq \Pr(A)=0.3$ by monotonicity.  This is a tight upper bound since it is possible that $\Pr(A\cap B)=0.3$.  For instance, where we have the uniform distribution over $\{1,2,\dots,10\}$ and $A=\{1,2,3\}$ and $B=\{1,2,3,\dots,9\}$.

$\Pr(A\cap B)\geq 0.2$, this one is a bit trickier to see.  Yes, it is bounded below by zero as you note, but that bound is not tight.  Instead, here we have $\Pr(A\cap B)=\Pr(A)+\Pr(B)-\Pr(A\cup B)$ by rearranging the inclusion-exclusion identity and note that $\Pr(A\cap B)$ is minimized when $\Pr(A\cup B)$ is maximized.  We have that $\Pr(A\cup B)$ is maximized when it is equal to $1$ so it follows that $\Pr(A\cap B)$ is minimized at $0.3+0.9-1=0.2$.  This bound is tight, for example where we have the uniform distribution over $\{1,2,\dots,10\}$ and $A=\{1,2,3\}$ and $B=\{2,3,\dots,10\}$.

As alluded in the comments above, $\Pr(A\cup B)$ is not necessarily equal to $\Pr(A)+\Pr(B)$.  That is only guaranteed true when $A$ and $B$ are disjoint which is not generally the case.  Instead, the general identity is $\Pr(A\cup B)=\Pr(A)+\Pr(B)-\Pr(A\cap B)$.

Similarly, $\Pr(A\cap B)$ is not necessarily equal to $\Pr(A)\times\Pr(B)$.  That is true if and only if $A$ and $B$ are independent events which is not generally the case.  Instead, the general identity is $\Pr(A\cap B) =\Pr(A)\times \Pr(B\mid A)$


问:
Solving the homogeneous recurrence relation $y[n]=\frac{1}{2}y[n-1]$
We have that $y[n]-\frac{1}{2}y[n-1]=0\implies y[n]=\frac{1}{2}y[n-1]$. Observe that :
\begin{align*}
y[n+1]-y[n]&=\frac{1}{2}y[n]-\frac{1}{2}y[n-1]\\
&=\frac{1}{2}(y[n]-y[n-1])
\end{align*}
\begin{align*}
y[n+2]-y[n+1]&=\frac{1}{2}y[n+1]-\frac{1}{2}y[n]\\
&=\frac{1}{2}\left(y[n+1]-y[n]\right)\\
&=\frac{1}{4}(y[n]-y[n-1])
\end{align*}
\begin{align*}
y[n+3]-y[n+2]&=\frac{1}{2}y[n+2]-\frac{1}{2}y[n+1]\\
&=\frac{1}{2}\left(y[n+2]-y[n+1]\right)\\
&=\frac{1}{8}(y[n]-y[n-1])
\end{align*}
$$
\vdots
$$
$$
y[n+k]-y[n+k-1]=\frac{1}{2^{k}}(y[n]-y[n-1])
$$
I am currently stuck here and I should end up with $y[n]=A\left(\frac{1}{2}\right)^n$ but I don't know how to proceed.

答:
Just successively insert the recursion relation i.e:
$$y[n]=1/2 y[n-1] = 1/2^2 y[n-2]=\ldots =1/2^n y[0]$$

You can reassure that this solves the recurrence relation by induction.


问:
Bound determined by Coherence of matrix using SVD for Matrix Completion
So I'm referring to the paper Exact Matrix Completion via Convex Optimization: https://arxiv.org/pdf/0805.4471.pdf

If you look on page 6 of this paper,

The coherence of a matrix $\mu(U) = \frac{n}{r} \max_{1 \le i \le n} \|P_Ue_i\|_2^2$ where $i$ is the standard basis vector with a $1$ in the $i$-th entry and $U$ is a subspace of $\mathbb{R}^n$ of dimension $r$.

Now we are considering some matrix $M \in \mathbb{R}^{n_1 \times n_2}$ which is expressed as $\sum_ {1 \le k \le r} \sigma_k u_kv_k^*$ (SVD). Now there is a property A0 which states that the coherence matrices obey $\max(\mu(U), \mu(V)) \le \mu_0$.

Now if you look a bit below they show that $\left|\sum_k u_{ik}v_{jk} \right| \le \sqrt{\sum_k|u_{ik}^2|}\sqrt{\sum_k|v_{jk}^2|}$ by cauchy schwarz ( this part is obvious to me)...but then they say that this whole thing is $\le \frac{\mu_0 r}{\sqrt{n_1n_2}}$. Now how they jumped to this makes no sense to me I do notice that $\mu(U) \le \mu_0 \implies \max_i \|P_Ue_i\|_2^2 \le \frac{\mu_0r}{n_1}$. Hence if you multiply the lengths of the maximum projections onto $U$ and $V$ you get the quantity $\frac{\mu_0 r}{\sqrt{n_1n_2}}$. Notice for example $u_{ik}$ is the $i$th entry of the $k$th left singular vector and similarly for $v_{jk}$. If someone could help me prove this result it would be much appreciated.

答:
Note that $\max_{i} \| P_U e_i \|_2^2 \leq \frac{\mu_0 r}{n_1}$ implies that $\max_{i} \| u_{i} \|_2^2 \leq \frac{\mu_0 r}{n_1}$, where $u_i$ is the $i^{\text{th}}$ row of the matrix of left singular vectors $U$. This is because $P_U e_i = UU^{\mathsf{T}} e_i$ and thus $\| UU^{\mathsf{T}} e_i \|_2 = \| U^{\mathsf{T}} e_i \|_2 = \|u_i\|_2$ for each $i$. The same can be shown to hold for $P_V$: $\max_{i} \|v_i\|_2^2 \leq \frac{\mu_0 r}{n_2}$.

Now using these identities, observe that

$$
\sqrt{\sum_{k} u_{ik}^2} \leq \max_{i} \sqrt{\sum_{k} u_{ik}^2} = \max_{i} \| u_i\|_2 \leq \sqrt{\frac{\mu_0 r}{n_1}}, \\
\sqrt{\sum_{k} v_{jk}^2} \leq \max_{j} \sqrt{\sum_{k} v_{jk}^2} = \max_{j} \| v_j\|_2 \leq \sqrt{\frac{\mu_0 r}{n_2}}. \\
$$

Multiplying the two gives the claimed result.


问:
Distribution of payoff in a card game
Consider a pack of $10$ cards marked $1,2,3,3,4,4,5,6,7,8$ ($3$ and $4$ are repeated).
There are two players that chose $3$ cards each (a total of $6$ cards without
replacement are drawn). The player with the lowest total wins and the loser
pays the winner an amount equal to the smallest card the winner is holding.
If the totals are equal there is no winner. So if player $A$ has drawn $2,3$ and
5 and player $B 1,4$ and $7, A$ is the winner and B pays her an amount of
2. Let $Y$ be the amount that $A$ wins or loses by after each turn. Find the
distribution of $Y$ .

Continuing from the previous question, the rules are modified a bit. Player
$A$ has the right to choose one of the three cards in advance and the other $5$
cards are drawn from the remaining pack. Choosing $1$ is tempting because
it maximises the chances of winning but then every time she wins she will
only win by 1. She is thinking of choosing card 2 as the chances of winning
are still good and the winning amount will be 2 more often than 1. Also the
possible losing payments will be a bit lower. Can she do even better?
The game is neither fair nor symmetric any more.

I am pretty new to probability and stochastic processes. I came across this problem and am not sure on how to go about it. Any thoughts and hints will be appreciated. Thanks.

Edit: I am trying to write a Matlab code for the question but I am stuck on how to go ahead with the code. From what I can understand, there will be around $4200$ cases to check which cannot be done mathematically.

答:
Because the numbers on these cards are rather arbitrary, I believe that it will be very hard to solve this problem by hand. Writing a piece of code to do this job is straightforward if you're used to it. I wrote mine in SageMath, since it has easy tools for combinatorics. This gave the following results:

For the first part of the question: the distribution of $Y$ is of course symmetrical. The chances of winning are as follows: $P(Y=0)=0.0628$, $P(Y=1)= 0.2259$, $P(Y=2)= 0.1329$, $P(Y=3)= 0.0985$, $P(Y=4)= 0.0111$.

For the second part, the results are more interesting. Since after A has picked his card, no more 'strategy' is left for either player, it suffices to calculate the expected outcome of the game conditional on A's pick (call it $X$). With very similar code, I found the following values:
\begin{align}
E(Y|X=1)&=-0.01389\\
E(Y|X=2)&=0.2069\\
E(Y|X=3)&=0.0888\\
E(Y|X=4)&=-0.2931\\
E(Y|X=5)&=-0.6333\\
E(Y|X=6)&=-0.9208\\
E(Y|X=7)&=-1.328\\
E(Y|X=8)&=-1.549\\
\end{align}

This means that your intuition was right: picking 2 is the best choice! If you pick 1, your chance of winning is high, but if you lose, it will mostly be by -2.  When you pick 2, you expect to win often by 2 and by 1, giving a higher sum on average.

At request: here is the code I used for the first part. The second part is very similar.

```<code>cards = [1,2,3,3,4,4,5,6,7,8]; n = len(cards)
distr = {x:0 for x in [-4..4]} #The distribution of the payoffs

for Ci1 in Combinations(n,3):
    C1 = [cards[x] for x in Ci1]
    leftOver = [cards[x] for x in range(n) if x not in Ci1]
    for Ci2 in Combinations(n-3,3):
        C2 = [leftOver[x] for x in Ci2]

        if sum(C1)>sum(C2):
            distr[-min(C2)] += 1
        elif sum(C1)<sum(C2):
            distr[min(C1)] += 1
        else:
            distr[0] += 1
</code>```

Title: Distribution of payoff in a card game

Body:
Because the numbers on these cards are rather arbitrary, I believe that it will be very hard to solve this problem by hand. Writing a piece of code to do this job is straightforward if you're used to it. I wrote mine in SageMath, since it has easy tools for combinatorics. This gave the following results:

For the first part of the question: the distribution of $Y$ is of course symmetrical. The chances of winning are as follows: $P(Y=0)=0.0628$, $P(Y=1)= 0.2259$, $P(Y=2)= 0.1329$, $P(Y=3)= 0.0985$, $P(Y=4)= 0.0111$.

For the second part, the results are more interesting. Since after A has picked his card, no more 'strategy' is left for either player, it suffices to calculate the expected outcome of the game conditional on A's pick (call it $X$). With very similar code, I found the following values:
\begin{align}
E(Y|X=1)&=-0.01389\\
E(Y|X=2)&=0.2069\\
E(Y|X=3)&=0.0888\\
E(Y|X=4)&=-0.2931\\
E(Y|X=5)&=-0.6333\\
E(Y|X=6)&=-0.9208\\
E(Y|X=7)&=-1.328\\
E(Y|X=8)&=-1.549\\
\end{align}

This means that your intuition was right: picking 2 is the best choice! If you pick 1, your chance of winning is high, but if you lose, it will mostly be by -2.  When you pick 2, you expect to win often by 2 and by 1, giving a higher sum on average.

At request: here is the code I used for the first part. The second part is very similar.

cards = [1,2,3,3,4,4,5,6,7,8]; n = len(cards)
distr = {x:0 for x in [-4..4]} #The distribution of the payoffs

for Ci1 in Combinations(n,3):
    C1 = [cards[x] for x in Ci1]
    leftOver = [cards[x] for x in range(n) if x not in Ci1]
    for Ci2 in Combinations(n-3,3):
        C2 = [leftOver[x] for x in Ci2]

        if sum(C1)>sum(C2):
            distr[-min(C2)] += 1
        elif sum(C1)<sum(C2):
            distr[min(C1)] += 1
        else:
            distr[0] += 1


问:
Is Right hand, left hand, and middle riemann sum formula the same?
Are the right hand, left hand, and middle Riemann sum formulas for $\int_{a}^{b}f(x)\,dx$ the same?
$$\lim_{n\to\infty}\sum_{i=1}^{n}f(x_{i})\Delta x,$$ where $\Delta x =\frac{b-a}{n}$ and $x_i=a+i\Delta x$.  Is this correct?

答:
No, the formulas are not the same, but they are similar.  Consider an interval $[a,b]$ and a partition of the interval $a = x_{1} < x_{2} < \dotsb < x_{n} = b$, where $\Delta x = \frac{b-a}{n}$ is the spacing between points.  Then the formulas for the Riemann sums will change, depending on which point (right/left/middle) in the subinterval is chosen:
\begin{align}
\text{Right:}\quad \int_{a}^{b}f(x)\,dx &= \lim_{n\to\infty}\sum_{i=2}^{n}f(x_{i})\Delta x\\
\text{Left:}\quad \int_{a}^{b}f(x)\,dx &= \lim_{n\to\infty}\sum_{i=1}^{n-1}f(x_{i})\Delta x\\
\text{Middle:}\quad \int_{a}^{b}f(x)\,dx &= \lim_{n\to\infty}\sum_{i=1}^{n-1}f\left(\frac{x_{i+1}-x_{i}}{2}\right)\Delta x\\
&=\lim_{n\to\infty}\sum_{i=2}^{n}f\left(\frac{x_{i}-x_{i-1}}{2}\right)\Delta x
\end{align}

It is easiest to see why this is the case if you sketch out a graph and label some points.  You should be able to see why the indices on the sums change depending on whether you pick the right or left point on each sub-interval.


问:
Is the following Statement always True?
Let $\sum\limits_{n=1}^{\infty}a_n$ be a series such that $a_n\to 0$. Suppose that for any $ n\geq 1$ we have that $a_n = \frac{1}{2^{k_n}}$, where $k_n$ is some positive integer. Then the series is always convergent. Is the statement True or Flase?

答: False. Take $k_n = \log_2  n > 0, \forall n \ge 2\implies a_n = 2^{-k_n} = 2^{-\log_2 n} = \dfrac{1}{n} \to 0$. But $\displaystyle \sum_{n=1}^\infty a_n  = \displaystyle \sum_{n=1}^\infty \dfrac{1}{n} = \infty$  i.e divergent !

Title: Is the following Statement always True?


问:
Is there a way to tell if a number is divisible by 4 if the figure is of 2 digits
I'm trying to help my daughter learn maths. She is struggling with factors, which is to work out what numbers go into a larger number (division).

I've already learned that by summing numbers, if they make 3, it can divide by 3. I also know the rules for 2, 5, 6, 9 and 10.

I'm trying to see if there is a rule for 4. I'm thinking not.

https://www.quora.com/Why-does-the-divisibility-rule-for-the-number-4-work shows the following

The divisibility rule for 4 is in any large number, if the digits in tens and units places is divisible by then the whole number is divisible by 4.

This doesn't make sense. 56 divides by 4. However, the 2 numbers add to 11, and so can't be divided by 4.

It may very well get a "no" answer, but is there any pattern/method I can use for determining if a  number can be divided by 4 if it is less than 100 (and greater than 4)

答:
There is a rule of divisibility for the number $4$. Here it is:

To figure out if a number is divisible by four, you first need to look at the last two digits, and if they're divisible by four together, you can assume that the whole number is divisible by $4$.

Why does this work? Well, $100$ is divisible by four, any number value in the place values greater than the hundreds place is a multiple of $100$. For example, in the number $2,375$, the $2$ in the thousands place stands for $2,000$, and $100\times 20=2,000$, so $2,000$ is a multiple of $100$. If we then add the two digits below the hundreds place, we can say that if all the digits above the ones place, and the tens place are divisible by four, if the two digits left are also divisible by four, it doesn't change anything!

Hopefully this helped you with your question.


问:
Find the total number of positive integral solutions for $x,y,z$ such that $xyz=24$
I recognize that this is related to the stars and bars algorithm, but I am just not able to apply it here

The general method is factor the number, in this case is $2^3\times 3$, but what next?

It’s explained using some weird and hopelessly arbitrary box analogy, which all goes over my head.

How is this question solved?

答:
Imagine you have $3$ boxes. You may place a $2,2,2,3$ among the boxes, and when you multiply all the boxes together, you get $24$. First, spread the $2$s. There are $3$ of these twos, and $3$ boxes, so that means you need $2$ dividers.

It should look like this o|o|o, where | is divider, and o represents a $2$. You can place $2$ dividers among the $5$ spots.

Thus, the possible ways to spread the $2$s are equal to $ 5 \choose 2$ or $10$ ways. Now, find the number of ways to spread the $3$. The $3$ can go in any of the $3$ boxes, so $3$ ways. Now you have $3*10$ or $30$ possible ways.

NOTE: A box without any numbers will just have a value of $1$, adding numbers into the box multiplies the numbers in the box together.

You can also find all the combinations easily by listing out each triplet from least to greatest, then rearranging to find the rest.

The combinations are:

$(1,1,24): 3$ ways

$(1,2,12)$ : $6$ ways

$(1,3,8)$ : $6$ ways

$(1,4,6)$ : $6$ ways

$(2,2,6)$ : $3$ ways

$(2,3,4)$ : $6$ ways

This also gives us $30$ ways total


问:
Brutal Estimate for multinomial expression
For $a,b,c\in \mathbb{R}_{+}$ and $1<p<2$ does the following estimate hold:

$$(a+b+c)^{p-1} \leq C (a^{p-1} + b^{p-1} + c^{p-1})$$

for a large enough constant?

I tried to prove this in the following way:

$(a+b)^{p-1}\leq 2^{p-1}(a^{p-1}+ b^{p-1})$

and so

$$(a+b+c)^{p-1} \leq 2^{p-1} ( (a+b)^{p-1} + c^{p-1}) \leq 2^{2(p-1)}(a^{p-1}+b^{p-1} + c^{p-1}).$$

Is this reasoning correct?

答:
You can get an even better estimate: Consider $g: [0, \infty)^2 \rightarrow \mathbb{R}$ where
$$
g(x, y) := (x+y)^{p-1}-x^{p-1}-y^{p-1}.
$$
Note, since $p-2<0$:
$$
\partial_x g(x, y) = (p-1)((x+y)^{p-2} - x^{p-2}) \leq (p-1)(x^{p-2} - x^{p-2}) = 0
$$
You can show the same way that $\partial_y g(x, y) \leq 0$. So $g$ is decreasing in $x$ and $y$ and $g(0, 0) = 0$. Therefore $g$ is nonpositive everywhere. It follows that
$$
(x+y)^{p-1} \leq x^{p-1} + y^{p-1}.
$$
You can iterate this argument to receive
$$
(a+b+c)^{p-1} \leq a^{p-1} + b^{p-1} + c^{p-1}.
$$


问:
Integral eigenvectors and eigenvalues
I need to find the eigenvalues e eigenvectors of this integral.

a)
$$\int_{0}^{2\pi}(\cos^2(x+y)+1/2)\phi (y)dy$$

b)- Solved thanks
$$\int_{0}^{1}(x^2y^2-2/45)\phi (y)dy$$

答:
First, let me try to rephrase part of the question in a way that is likely to be more palatable for most folks on math.SE:

Edited question

I would like to find the eigenvalues and eigenvectors of the following integral:

$$\int_{0}^{1}(x^2y^2-2/45)\phi (y)dy$$

I tried to find a matrix so I could compute the determinant, but I'm not sure what matrix I should use.

Discussion on the edit

Note that we are at least indicating some attempt to indicate our thoughts.  Even if this is minimal, it demonstrates to folks who are donating there time to adress your question where you are.  Also, never use URGENT in all caps in your title; I promise that it won't go over well.

Finally, it's generally considered polite to upvote an answer that gives reasonable help and to accept the best answer whenever appropriate.  I notice that you have 14 questions with only one accept.

A partial answer

At any rate, after that, someone is likely to politely explain that this is not a problem involving matrices.  Rather, you are looking for a function $\phi(x)$ and a non-zero, real number $\lambda$ such that

$$\int_{0}^{1}(x^2y^2-2/45)\phi (y)dy = \lambda \phi(x).$$

My recommendation is that you guess and refine.  If you let $\phi(x)=ax^2+bx+c$ and compute the integral, you should find that

$$\int_{0}^{1}(x^2y^2-2/45)(ay^2+by+c)dy = -\frac{2 a}{135}-\frac{2 c}{45}-\frac{2 b \
x}{45}+\left(\frac{a}{5}+\frac{c}{3}\right) x^2+\frac{b x^3}{3}.$$

From there, the question is simply, can you find $a$, $b$, and $c$ to make this work?  It's reasonably clear that $b=0$, right?  I bet you can make some progress from here.

One more hint: My solutions for $\lambda$ have a 45 in the denominator. I also solved for $a$ in terms of $c$, which is to be expected in an eigenvalue problem.

Good luck!

Title: Integral eigenvectors and eigenvalues


问:
What is the average distance between two randomly chosen points on a Sierpiński triangle?
Since a Sierpiński triangle is made up of three smaller versions of itself, an algorithm made to pick random points on the triangle might go like this:

Select one of the three Sierpiński triangles inside the larger Sierpiński triangle

Repeat step 2 with the smaller Sierpiński triangle

It will continue getting more specific and will eventually narrow down to a point. If you continue picking two points like this, what will be the average distance? (Euclidian Metric)

答:
I don't know how to do this analytically, but I did a simulation, assuming that the original triangle is equilateral of side $1$.  With one million trials, I got a $99\%$ confidence interval of $$(0.42238756083753565, 0.4234628881174311)$$

I started with $$(1,0,0), (0,1,0), (0,0,1)$$ as the vertices of the triangle in barycentric coordinates.  Then $30$ times, I chose one of the coordinates uniformly at random, and computed the vertices of the smaller triangle as the average of the chosen vertex and the old vertices.  For example, if the first vertex chosen were $(1,0,0)$ the new triangle would have coordinates $$(1,0,0),\left(\frac12,\frac12,0\right),\left(\frac12,0,\frac12\right)$$
After $30$ samples, I took the average of the vertices as the chosen point.

For each trial, I chose two points, and computed the Euclidean distance between them using the formula from Wikipedia.  Since I assumed that the side lengths of the triangle are all $1$, this is simply$$d=\sqrt{\frac{x^2+y^2+z^2}2}$$ where $x,y,z$ are the coordinates of the difference between the two chosen points, treated as vectors.

It seemed to me at first that one should be able to do this analytically.  If the coordinates of the $n$th triangle are $(X_n, Y_n, Z_n)$ then $X_{n+1}$ is equally likely to be any of $X_n,\ \frac{X_n+Y_n}2,\ \frac{X_n+Z_n}2$.  I haven't made any progress, though.  This may just be because I'm not thinking clearly, but as I'm typing this it occurs to me that, since the Sierpiński gasket is a fractal, perhaps the distributions functions of X,Y,Z are singular, so that we can't hope for analytic expressions.  I'm going to compute the distribution of $X_n$ for small $n$ and see if that gives me any ideas.

For completeness, here's the simulation script:

```<code>from sys import argv
from random import choices
from math import sqrt

def main(trials, samples):
    total = 0
    total2 = 0
    for _  in range(trials):
        x = pickPoint(samples)
        y = pickPoint(samples)
        # Triangle side length are all 1
        v =  [(z[0]-z[1])**2 for z in zip(x,y)]
        d = sqrt(sum(v)/2)  # Euclidean distance
        total += d
        total2 += d*d
    mean = total/trials
    variance = total2/trials - mean**2
    sigma = sqrt(variance/trials)
    return mean, sigma

def pickPoint(samples):
    V = [(1,0,0), (0,1,0), (0,0,1)]
    s = choices(range(3), k=samples)
    for k in s:
        v = V[k]
        V = [tuple((z[0]+z[1])/2 for z in zip(v,V[k])) for k in range(3)]
    return [(V[0][k]+V[1][k]+V[2][k])/3 for k in range(3)]

if __name__ == '__main__':
    trials = int(argv[1])
    samples = int(argv[2])
    mu, sigma = main(trials, samples)
    print(f'Mean = {mu}')
    print(f'sigma = {sigma}')
    print(f'{trials} trials')
    print(f'Sample depth = {samples}')
    delta = 2.576*sigma
    print(f'99% confidence interval: ({mu-delta}, {mu+delta})')
</code>```

When I ran this with parameters 1000000 30 it produced the output

```<code>Mean = 0.4229252244774834
sigma = 0.00020872035712257242
1000000 trials
Sample depth = 30
99% confidence interval: (0.42238756083753565, 0.4234628881174311)
</code>```

Body:
I don't know how to do this analytically, but I did a simulation, assuming that the original triangle is equilateral of side $1$.  With one million trials, I got a $99\%$ confidence interval of $$(0.42238756083753565, 0.4234628881174311)$$

I started with $$(1,0,0), (0,1,0), (0,0,1)$$ as the vertices of the triangle in barycentric coordinates.  Then $30$ times, I chose one of the coordinates uniformly at random, and computed the vertices of the smaller triangle as the average of the chosen vertex and the old vertices.  For example, if the first vertex chosen were $(1,0,0)$ the new triangle would have coordinates $$(1,0,0),\left(\frac12,\frac12,0\right),\left(\frac12,0,\frac12\right)$$
After $30$ samples, I took the average of the vertices as the chosen point.

For each trial, I chose two points, and computed the Euclidean distance between them using the formula from Wikipedia.  Since I assumed that the side lengths of the triangle are all $1$, this is simply$$d=\sqrt{\frac{x^2+y^2+z^2}2}$$ where $x,y,z$ are the coordinates of the difference between the two chosen points, treated as vectors.

It seemed to me at first that one should be able to do this analytically.  If the coordinates of the $n$th triangle are $(X_n, Y_n, Z_n)$ then $X_{n+1}$ is equally likely to be any of $X_n,\ \frac{X_n+Y_n}2,\ \frac{X_n+Z_n}2$.  I haven't made any progress, though.  This may just be because I'm not thinking clearly, but as I'm typing this it occurs to me that, since the Sierpiński gasket is a fractal, perhaps the distributions functions of X,Y,Z are singular, so that we can't hope for analytic expressions.  I'm going to compute the distribution of $X_n$ for small $n$ and see if that gives me any ideas.

For completeness, here's the simulation script:

from sys import argv
from random import choices
from math import sqrt

def main(trials, samples):
    total = 0
    total2 = 0
    for _  in range(trials):
        x = pickPoint(samples)
        y = pickPoint(samples)
        # Triangle side length are all 1
        v =  [(z[0]-z[1])**2 for z in zip(x,y)]
        d = sqrt(sum(v)/2)  # Euclidean distance
        total += d
        total2 += d*d
    mean = total/trials
    variance = total2/trials - mean**2
    sigma = sqrt(variance/trials)
    return mean, sigma

def pickPoint(samples):
    V = [(1,0,0), (0,1,0), (0,0,1)]
    s = choices(range(3), k=samples)
    for k in s:
        v = V[k]
        V = [tuple((z[0]+z[1])/2 for z in zip(v,V[k])) for k in range(3)]
    return [(V[0][k]+V[1][k]+V[2][k])/3 for k in range(3)]

if __name__ == '__main__':
    trials = int(argv[1])
    samples = int(argv[2])
    mu, sigma = main(trials, samples)
    print(f'Mean = {mu}')
    print(f'sigma = {sigma}')
    print(f'{trials} trials')
    print(f'Sample depth = {samples}')
    delta = 2.576*sigma
    print(f'99% confidence interval: ({mu-delta}, {mu+delta})')

When I ran this with parameters 1000000 30 it produced the output

Mean = 0.4229252244774834
sigma = 0.00020872035712257242
1000000 trials
Sample depth = 30
99% confidence interval: (0.42238756083753565, 0.4234628881174311)


Title: Is the following Statement always True?

Body: Let $\sum\limits_{n=1}^{\infty}a_n$ be a series such that $a_n\to 0$. Suppose that for any $ n\geq 1$ we have that $a_n = \frac{1}{2^{k_n}}$, where $k_n$ is some positive integer. Then the series is always convergent. Is the statement True or Flase?

Body:
Hint: Try
$$ \frac{1}{2^1} + \frac{1}{2^1} + \frac{1}{2^2} + \frac{1}{2^2} +\frac{1}{2^2} +\frac{1}{2^2} + \frac{1}{2^3} +\frac{1}{2^3} +\frac{1}{2^3} +\frac{1}{2^3} +\frac{1}{2^3} +\frac{1}{2^3} +\frac{1}{2^3} +\frac{1}{2^3} + \ldots$$


问:
Uniform continuity general case
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function defined on (all of) the real axis. Assume that $$f(x) \rightarrow a, \quad \text{for } x \rightarrow -\infty$$ $$f(x) \rightarrow b, \quad \text{for } x \rightarrow \infty$$ for some real numbers $a,b$. Show/explain that for every $\epsilon >0$ there exists a $M >0$ such that $$\left | f(x)-f(y) \right |<\varepsilon$$ given $x,y \geq M$. Furthermore, show that $g$ is uniformly continuous.

Attempt (or rather thoughts)

I am fairly sure that I have gotten the solution down for the first part of the question as it can be solved directly form the definition of a limit at infinity.

Now, I have absolutely no idea on how to show that the function is uniformly continuous as I am not confident at all in the approach of these sort of problems. I would have to use my the fact given above that I have shown but, again, no idea how to approach such a problem as I seem to be missing a general approach and lack of properties. I really need some explanation as to what is going on.

答:
$$
\forall\varepsilon>0 \, \exists \delta>0\, \forall x,y \, \Big( |x-y|<\delta \longrightarrow |f(x)-f(y)|<\varepsilon\Big).
$$
The point is that $\delta$ does not depend on $x$ and $y.$

Given $\varepsilon>0,$ there exists $M_1$ such that for $x,y>M_1$ you have $|f(x)-f(y)|<\varepsilon,$ and for the same reason there exists $M_2$ such that for $x,y<M_2$ you have $|f(x)-f(y)|<\varepsilon.$

A theorem asserts that a continuous function on a compact set is uniformly continuous. And $[M_2,M_1]$ is a compact set.

Title: Uniform continuity general case


问:
Critique my geometrical understanding of the complex derivative
I wrote the following bit on the geometry of the complex derivative based on my understanding from reading several books and from some side comments my lecturer has made. I wanted to put it all in one place. Could you critique it according to correctness, completeness, pedantism (if that's the correct word to use here), conciseness and how didactic it is?

As with real functions, a geometrical understanding of the derivative is possible with complex differentiation. Given $X \subset \mathbb{C}$ and $f : X \rightarrow \mathbb{C}$ differentiable at $a \in X$, we have
$$f'(a) = \lim_{z \to a} \frac{f(z) - f(a)}{z - a}.$$
We now introduce an error term $\epsilon \in \mathbb{C}$ such that
$$\frac{f(z) - f(a)}{z - a} = f'(a) + \epsilon.$$
Clearly, $\epsilon$ is a function of $z$. Hence, we will write $\epsilon = \epsilon(z)$. Moreover, we must have
$$\lim_{z \to a} \epsilon(z) = 0.$$
Isolating $f(z)$ on the left-hand side with
$$f(z) = f(a) + f'(a)(z-a) + \epsilon(z)(z - a),$$
and then factoring,
$$f(z) = f(a) + (f'(a) + \epsilon(z))z - (f'(a) + \epsilon(z))a.$$
Thus, we have obtained an alternate expression for the values of the function $f$ in terms of its derivative at a single point and an error function which we know becomes arbitrarily small near that point. This expression becomes interesting and carries geometrical significance when $f'(a) + \epsilon(z) \approx f'(a)$. This is easily accomplished by letting $z$ approach $a$. Suppose $z$ is close enough to $a$ that we can make the above approximation. Then,
$$f(z) \approx f(a) + f'(a)z - f'(a)a.$$
Note that this expression is linear in $z$. Thus, we refer to it as the first order approximation, or linearization, of $f$ at $a$. The fact that the approximation is linear provides us with an intuitive geometrical understanding of the derivative of a function when it exists. To see this, consider the linear expression $w = az + b$, where $w, a, b, z \in \mathbb{C}$. The multiplication of $a$ and $z$ can be viewed geometrically as rotating and dilating the vector $(\operatorname{Re}(z), \operatorname{Im}(z)) \in \mathbb{R}^2$. The term $b$ simply translates a vector in the plane. Thus, the linear function $w$ will map every point on the unit circle to a new circle of radius $|a|$ centered at $b$. Going back to our linearization of $f$, we see that, if $z$ is sufficiently near $a$, the existence of the derivative of $f$ at $a$ signifies, geometrically, that $f$ maps circles onto circles. This can be understood to mean that $f$, in a sense, preserves shapes near $a$. The error term $\epsilon(z)$ has the effect of distorting these shapes, such that a circle may not map to another circle in the plane. Let $f : X \rightarrow \mathbb{C}$ be a function such that
$$f(z) = z^2 + 2z.$$
Since $f$ is a polynomial, it is differentiable on it domain, with derivative
$$f'(z) = 2z + 2.$$
At $z = 0$, $f'(z) = 2$. Then, the linearization of $f$ at $0$ is
$$f(z) \approx f(0) + f'(0)z - f'(0) \cdot 0= 2z.$$
Thus, if the domain is a small circle centered at $0$, the image should be a concentric circle of twice the radius. In the first image of the figure below, the inner circle is the domain $X = \{z \ ; \ |z| = 0.01\}$ and the outer circle is the image $f(X)$. Visually, the linearization seems to accurately predict the image. In the second image below, with the domain $X = \{z \ ; \ |z| = 0.99\}$, the distortion is, visually speaking, significant, and our linear approximation seems to break down. We can evaluate the error term directly to see whether our assessment above is valid or not.
$$\epsilon(z) = \frac{f(z) - f(0)}{z - 0} - f'(0) = z \quad (z \neq 0 \text{ in either case})$$
Indeed, in our first case, $|f'(0)| = 2 \gg |\epsilon(z)| = |z| = 0.01$. In our second case however, $|\epsilon(z)| = |z| = 0.99$ becomes comparable to $|f'(0)|$.

答:
There's one very important thing I think you should have mentioned. While real and complex derivatives are both limits, the former exists when two one-sided limits exist and are equal, but the latter requires existent and equal directional limits from (uncountably) infinitely many different directions. Differentiability is therefore a very stringent constraint on complex functions (with consequences that take a lot of work to flesh out, but let's limit our scope here.) This is why, for example, $z^\ast$ isn't differentiable:$$\lim_{r\to0^+}\frac{(z+re^{i\theta})^\ast-z^\ast}{re^{i\theta}}=e^{-2i\theta}$$is direction-dependent. But it's not too stringent; you need only verify two equations to prove a derivative exists, as it does for e.g. all polynomials in $z$. It's common to write functions in terms of $z,\,z^\ast$: the rule of thumb is if $z^\ast$ shows up, there's no derivative.

Here's another perspective on this, albeit one you probably needn't include in your notes. The univariate chain rule $df=\frac{\partial f}{\partial x}dx$ for real-to-real functions has a more complicated variant in number systems of some dimension $n>1$ over $\Bbb R$. If $f$ is a function from one such number to another - in your case, $n=2$ - the $i$th part of $f(z)$ satisfies$$df_i=\frac{\partial f_i}{\partial z_j}dz_j,$$where $z_j$ is the $j$th part of $z$ and I've summed over $j$. The display-line equation above looks like $n$ multivariate chain rules for functions from $\Bbb R^n$ to $\Bbb R$, doesn't it?

But that's not the whole story. For $df=g(z)dz$ to be true in this number system, with $g=f^\prime$, we need $g$ to be a number, not a matrix. There are only $n$ degrees of freedom in that number, but they have to do the work of $n^2$ DOFs. If we write numbers as $a=a_ie_i$ and multiplication obeys $a_ie_ib_je_j=c_{ijk}a_ib_je_k$ (and feel free to compute the eight $c_{ijk}$ for multipication on $\Bbb C$), a differentiable $f$ satisfies$$df_k=c_{ijk}g_idz_j\implies\frac{\partial f_k}{\partial z_j}=c_{ijk}g_i.$$Needless to say, not all matrices of derivatives will be of this form. (If you did the exercise I suggested a moment ago, you can verify the equations in my first hyperlink emerge from applying this logic to complex analysis.) More generally, the $n^2$ values $\frac{\partial f_k}{\partial z_j}$ have to be linear combinations of $n$ values $g_i$, which impose $n(n-1)$ constraints which, if satisfies, allow $n$ DOFs in the value of $g(z)$ to do the work of $n+n(n-1)=n^2$ DOFs.


Title: I can't figure out my mistake re-arranging this series

Body:
Define:$$\varphi(s)=e^{-1}+e^{-2^s}+\cdot\cdot\cdot$$

and then

$$ e^{-2^s}\varphi(s)=e^{-(2^s+1)}+e^{-(2^{s+1})}+e^{-(2^s+3^s)}+e^{-(2^s+4^s)}+e^{-(2^s+5^s)}+\cdot\cdot\cdot $$

$$ e^{-2^s}\varphi(s)=e^{-(2^s+1)}+e^{-(2^{s+1})}+e^{-2^s}\big( \varphi(s)-e^{-1}-e^{-2^s}-e^{-3^s} \big)$$

$$ e^{-2^s}\varphi(s)=e^{-2^s}\varphi(s)-e^{-3^s} $$

$$ 0=-e^{-3^s} $$

I'm not sure what I've done wrong.

Body: When you wrote$$ e^{-2^s}\varphi(s)=e^{-(2^s+1)}+e^{-(2^{s+1})}+e^{-2^s}\big( \varphi(s)-e^{-1}-e^{-2^s}-e^{-3^s} \big)$$you should have just written$$ e^{-2^s}\varphi(s)=e^{-(2^s+1)}+e^{-(2^{s+1})}+e^{-2^s}\big( \varphi(s)-e^{-1}-e^{-2^s} \big).$$


问:
Holomorphic vs differentiable (in the real sense).
Why a holomorphic function is infinitely differentiable just because of satisfying the Cauchy Riemann equations, but on the other side, a two variable real function that is twice differentiable is not infinitely differentiable?

I'm asking this for two reasons:

1) $\mathbb{R}^2$ is somehow an analog to the complex plane (they are isomorphic).

2) Holomorphic means that is differentiable in any direction of the complex plane, so why if a function is differentiable in any direction of the real plane then it is not infinitely differentiable?

答: This is one of the beautiful features of complex analysis. The answer is that holomorphic functions can be represented by an integral formula (the Cauchy integral formula), whereas most continuously differentiable functions (even smooth ones) on $\mathbb R^2$ cannot. </p>


问:
Is the exterior square of a projective faithful representation faithful?
Let $G$ be a matrix group. (I am interested in $G=O_n$ in particular). Then we can define the projective version $PG$ of $G$ to be the group consisting of the tensor products
$$PG=\{A\otimes A\mid A\in G\}$$
It is easy to show that this corresponds to the maybe more standard definition of the projective version as $PG=G/(\pm I)$. Indeed, take some $A\in G$ and suppose for example that $A^1_1\neq 0$ (using the "physics notation" $A^i_j$ for the entries of $A$). Then we have $[A\otimes A]^{i1}_{j1}=A^i_jA^1_1$ and in particular $[A\otimes A]^{11}_{11}=(A^1_1)^2$, so we can recover all the entries of $A$ from $A\otimes A$ up to a global sign.

Now, my question is, what if we take the antisymmetrization of the tensor product. That is, consider the set of matrices $\widetilde{PG}:=\{A\wedge A\mid A\in G\}$ with $[A\wedge A]^{ik}_{jl}=A^i_jA^k_l-A^i_lA^k_j$. Surely this is a group (a quotient of $PG$). What kind of group is this? Does it coincide with $PG$? In other words, is the map $A\otimes A\mapsto A\wedge A$ injective?

答:
Let $V$ be a real vector space. Since $PGL(V)$ is a simple group, the kernel of $GL(V) \to GL(\wedge^2 V)$ is contained in the scalar matrices. The scalar $a$ acts by $a^2$ on both $V^{\otimes 2}$ and $\wedge^2 V$, so $a$ acts trivially on $\wedge^2 V$ if and only if it does on $V \otimes V$.

In general, each complex irreducible representation of $GL(V)$ is contained in $V_{\mathbb C}^{\otimes n} \otimes \mathrm{det}^{-m}$ for some $n, m \geq 0$, where $\mathrm{det}$ is the determinant character and $V_{\mathbb C}$ is the complexification of $V$. The kernel of $GL(V)$ on such a representation is exactly the $(n-(\dim V)m)$th roots of unity, by the same argument as above.


问:
I'm teaching a college geometry course. What should I cover?
I've been asked to teach "Foundations of Geometry" at the University of South Carolina. Apparently, professors in the past have all done very different things, and I have a lot of choice in the matter. What course would you like to see?

Some constraints:

I want to say at least a little something about (1) the axiomatic approach (e.g., Euclid's axioms), (2) the modern approach (cool theorems such as Ceva's theorem, the nine point circle, etc., etc.), (3) the constructive approach, with straightedge and compass.

My students differ in background and motivation. Most of them are prospective high school math teachers. Typically they have seen some proofs, but not a lot; for example, they may have proved that the sum of two odd numbers is even. Most of the students won't yet have taken analysis, or algebra, or any other course obliging them to work really hard.

It has been a long time since I have seriously dealt with the subject, so I will definitely want a good book (or books) or other materials to follow.

I won't say anything about projective or other non-Euclidean geometry, because there's another course for that.

I asked a similar question on MathOverflow, but I was a little bit spooked by the answers. I'm sure that the books of Hartshorne and Hadamard are excellent, but I suspect these might be better for stronger students. (And the Hartshorne book discusses "geometry over fields", etc., and most of my students won't have taken abstract algebra.) Some respondents rolled their own solutions -- but are there really not good source materials out there?

My colleague taught a course out of Isaacs' book -- but he really went the extra mile (more like the extra ten miles). He said that the exercises were a bit too difficult, and that he was constantly having to give the students hints, and he slaved over writing up complete solutions. I believe this would not work as well for me as it would for him: he has a very approachable demeanor that I haven't (yet) been able to duplicate, and I'm afraid the students would be unlikely to come to my office hours no matter how much I encouraged them.

I've also looked at other books -- Coxeter and Greitzer is very cool, but perhaps more naturally suited to hotshot high school kids. Posamentier looks promising, a review copy is on its way. Clark's book seems to be somewhat off the beaten path (where I'm not quite sure I want to follow), but it looks very interesting and I ordered a review copy of that also. Other suggestions, big or small, would be very welcome. Thank you so much! --Frank

答:
Something that was never emphasized to me (and I didn't figure it out until later) was why we spent so much time on proofs, etc.

I only realized later it was to show how mathematical reasoning can go from axioms to proven knowledge, with a subject matter that is intuitively grasped.  I've emphasized this now as a homeschooling dad, and wish more teachers would do this.


问:
Can I take the gradient of a function like this?
Let us say I have a vector function

$$r(u,v) = \langle u,v,1000+u^2+v^2 \rangle.$$

I also have,

$$p(x,y,z) = 5e^{(x + y + z^2)}.$$

Let, $Q(u,v) = (P \circ r)(u,v).$ Then,

$$Q(u,v) = 5e^{(u+v+(1000+u^2+v^2)^2)} $$

Would
$$\frac{\partial Q}{\partial u} = \frac{\partial}{\partial u} (5e^{(u+v+(1000+u^2+v^2)^2)}) ? $$

答:
It is correct. You could also use the chain rule. Denoting $r(u,v) = \bigl(r_1(u,v), r_2(u,v),r_3(u,v)\bigr)$:
$$ \frac{\partial Q}{\partial u} = \frac{\partial r_1}{\partial u} \frac{\partial P}{\partial z}\bigl(r(u,v)\bigr) + \frac{\partial r_2}{\partial u}\frac{\partial P}{\partial y}\bigl(r(u,v)\bigr)+ \frac{\partial r_2}{\partial u}\frac{\partial P}{\partial z}\bigl(r(u,v)\bigr) = P\bigl(r(u,v)\bigr)\bigl(1+4(1000+u^2+v^2)\bigr).
$$


问:
Suggest a method to solve $y + y' + y'' = ry - ry^2$
I'm stuck with the following differential equation:

$$y+\frac{d^2y}{dx^2} + \frac{dy}{dx} = ry - ry^2$$

where $r$ is a real positive constant and $x,y \in \mathbb{R}$.

Since it is for a small research purpose, I wish to get a proper analytic solution that is not a series expansion.

I tried multiplying with $\frac{dy}{dx}$ and simplifying. This gave:

$$\frac{d}{dx} {\left(\frac{1}{2}{y'}^2-\frac{r-1}{2}y^2+ \frac{r}{3}y^3\right)}=-{y'}^2$$

But I'm unable to continue. Just to mention I also tried solving the first order equation:

$$y+ \frac{dy}{dx} = ry - ry^2$$

and tried similar solutions. That didn't work too.

答:
Maple does not find a closed-form solution.  This makes it quite likely that the general solution can't be expressed in closed form.  There are of course two constant solutions:
$y = 0$ and $y = 1 - 1/r$.


问:
Each Weyl group orbit in the character lattice of $V$ contains exactly one dominant weight
Let $V = \mathbb{C}^3 \otimes \mathbb{C}^3$ be a representation of $G = SL_3(\mathbb{C})$.
The weights of this representation is the set of $\varepsilon_i + \varepsilon_j$ for $i, j = 1, 2, 3$, where $\varepsilon_i$ takes $\text{diag}[h_1, h_2, h_3] \in \mathfrak{h}$ to $h_i$.

The Weyl group is $W = \{ 1, s_1, s_2, s_1s_2, s_2s_1, s_1s_2s_1 \}$.
For a simple root $\alpha_i$, the coroot $h_i$ is simply the matrix $E_{ii} - E_{i+1, i+1}$.
Then the pairing $$ \langle \varepsilon_j, h_i \rangle \alpha_i = \begin{cases} -\alpha_i & \text{ if } i = j, \\ \alpha_i & \text{ if } i = j -1, \\ 0 & \text{ else }. \end{cases}$$
By the defining equation of root reflections, $$ s_i (\beta) = \beta - \langle \beta, h_i \rangle \alpha_i$$ for $\beta \in \mathfrak{h}^*$, we have $$ s_i(\varepsilon_j) = \begin{cases} \varepsilon_j-\alpha_i & \text{ if } i = j, \\ \varepsilon_j + \alpha_i & \text{ if } i = j -1, \\ \varepsilon_j & \text{ else }. \end{cases}$$
Using this last part, the $W$-orbit of the weight $\varepsilon_1 + \varepsilon_2$ is the set $\{ \varepsilon_1 + \varepsilon_2, \varepsilon_1 + \varepsilon_3, \varepsilon_2 + \varepsilon_3 \}$.

Dominant weights are non-negative integral linear combinations of the fundamental weights, $m\varpi_1 + n \varpi_2$.
Expanding this out in terms of the $\varepsilon_i$, I get $$ m \varpi_1 + n \varpi_2 = \frac{1}{3} \left( 2(m+n) \varepsilon_1 + (2n-m) \varepsilon_2 - (m+n) \varepsilon_3 \right).$$
Equating coefficients shows that none of the set of weights $\{ \varepsilon_1 + \varepsilon_2, \varepsilon_1 + \varepsilon_3, \varepsilon_2 + \varepsilon_3 \}$ are dominant weights, contradicting what I am supposed to show.
What have I done wrong?

答:
First, your coefficients for $m\varpi_1+n\varpi_2$ look wrong. As far as I know,

$$\varpi_1 = \frac13 (2\varepsilon_1 -\varepsilon_2 -\varepsilon_3), \varpi_2 = \frac13 (\varepsilon_1 +\varepsilon_2 -2\varepsilon_3)$$

and hence

$$m\varpi_1+n\varpi_2 = \frac13 \left((2m+n)\varepsilon_1 +(-m+n)\varepsilon_2 +(-m-2n)\varepsilon_3\right).$$

Now at first that looks like running into the same problem as with your coeffcients. But now remember that because of the determinant-$1$-condition in $SL$ (or the trace-$0$-condition in $\mathfrak{sl}$) we actually must have $\varepsilon_1 +\varepsilon_2 +\varepsilon_3 =0$ or in other words, we can assume $\varepsilon_3=-\varepsilon_1-\varepsilon_2$. This turns the above into

$$\varpi_1 = \varepsilon_1 , \qquad \varpi_2 = \varepsilon_1 +\varepsilon_2$$

and I'm sure you can take it from there.


问:
Modular Arithmetic Congruences
Can someone please help me on this problem in modular Arithmetic Congruences I ran into this problem on aops alcumus

If $3x+7\equiv 2\pmod{16}$, then $2x+11$ is congruent $\pmod{16}$ to what integer between $0$ and $15$, inclusive?

答: Hint: $6 \cdot 3 \equiv 2 \bmod 16$

Title: Modular Arithmetic Congruences

Body: Hint: $6 \cdot 3 \equiv 2 \bmod 16$


答:
Hint:

Solve the initial congruence $\;3x\equiv -5\mod 16$. As $3$ is coprime to $16$, it is invertible $\bmod16$, and the inverse is deduced from a Bézout's relation $\;3u+16v=1$, which shows the inverse of $3\bmod 16$ is the congruence class of $u$.

Title: Modular Arithmetic Congruences

Body:
Can someone please help me on this problem in modular Arithmetic Congruences I ran into this problem on aops alcumus

If $3x+7\equiv 2\pmod{16}$, then $2x+11$ is congruent $\pmod{16}$ to what integer between $0$ and $15$, inclusive?


问:
A discrete subset $Y$ of a topological space $X$ which is Haussdorff is closed?
Question is in the title. Let $X$ be a topological space which is Hausdorff, and let $Y \subset X$ be a subset, so that the induced topology on $Y$ is discrete. Is it then true that $Y$ is closed in $X$?

I have the impression that this is only true if $X$ is locally compact, but I feel like I could be missing something.

答: Hint: consider a convergent sequence in $\mathbb R$.


问:
Why the sum of residuals equals 0 when we do a sample regression by OLS?
That's my question, I have looking round online and people post a formula by they don't explain the formula. Could anyone please give me a hand with that ? cheers</p>

答:
I want to provide a more general answer from the statistical sense of the word "residual". I figured this out in my quest to understand degrees of freedom and Bessels's correction in statistics.

A residual in statistics means the difference between a variable's value and the sample mean (not the true, usually unknowable average).

So if $x_i, i \in \{1, ..., N\}$ represents a sample value:

$$
\sum_i r_i = \sum_i x_i - \mu \\
= \sum_i (x_i - \frac{1}{N}\sum_i x_i) \\
= \sum_i x_i - N \frac{1}{N}\sum_i x_i \\
= 0
$$

Which is the basis behind the argument behind Bessel's Correction, which is the practice of dividing the sum of squared residuals by $N-1$ rather than $N$.

$$ \sigma^2 = \frac{1}{N - 1}\sum(x_i - \mu)^2$$

The idea is (according to Wikipedia) that the residuals are not independent because they sum to zero, therefore you subtract one. I do not actually understand this statement (how do we know that the span of the residuals is $N-1$ and not $N-2$ for example?).
However, this nice proof explains the intuition behind the correction from a functional/applied point of view.


问:
Definite integration by substitution problem with the substitution $x=e^u−e^{−u}$

I'm stuck at the following integration problem:

a) Given that $x = e^u - e^{-u}$, find an expression for $e^u$ in terms of $x$.

b) Use the substitution $x = e^u - e^{-u}$ to find the exact value of $\int_0^1 \frac {1} {\sqrt {4+x^2}}\, dx$.

My textbook says that the solutions are:

a) $e^u=\frac {x+\sqrt{x^2+4}}{2}$

b) $\ln\frac {1+\sqrt5}{2}$

I have tried some algebraic manipulation in task a) but this did not bring me far. The textbook solutions looks pretty much like the result of the quadratic formula but I do not know if this is a coincidence or not.

I did not understand b) either since I lack the understanding of task a). I might be able to solve it because you could use the substitution $u =\frac x 2 $ because this nicely leads to $\int_0^1 \frac 1 {\sqrt{u^2+1}}\,du $ and then I one could trig-substitution but I cannot solve it with the given substitution.

Could somebody tell me I can solve or even approach the problem?

答:
Multiplying by $e^u$, you get
$$
x = e^u-e^{-u}\Leftrightarrow x e^u = (e^u)^2-1\Leftrightarrow (e^u)^2-x e^u-1=0
$$

solving the quadratic equation, you conclude that

$$
e^u = \frac{x \pm \sqrt{x^2+4}}{2}
$$

but the minus sign does not yield any solution and you finally arrive at
$$
e^u = \frac{x + \sqrt{x^2+4}}{2}.
$$

Regarding the integral,

\begin{align*}
\int_0^1 \frac{1}{\sqrt{x^2+4}}dx = \int_0^{\log \frac{1+\sqrt{5}}{2}} (e^u+e^{-u})\frac{1}{e^u+e^{-u}} du = \log \frac{1+\sqrt{5}}{2}
\end{align*}

Note:
$$
e^u = \frac{x+\sqrt{x^2+4}}{2} \Rightarrow \sqrt{x^2+4} = 2e^u -x = 2e^u-(e^u-e^{-u}) = e^u+e^{-u}
$$