问:
Calculate the integral $ \iint_R\sqrt{y/x}\,e^{\sqrt{xy}}\,dA$
The $R$ is the region of integration, described in the following image.

To solve this integral we make a change of coordinates with $u=\sqrt{xy}$ and $v=\sqrt{\frac{y}{x}}$. Furthermore, in the region of integration we have $x>0$ and $y>0$. Then $u^2=xy$ y $v^2=\frac{y}{x}$, so we have:
\begin{align*}
    v^2 &=\frac{y}{x}\\
    x^2v^2 &= xy\\
    x^2v^2 &= u^2.
\end{align*}
\begin{align*}
    x^2 &= \frac{u^2}{v^2}\\
    x &= \frac{u}{v}.
\end{align*}
Observation: $x=-\frac{u}{v}$ is not taken, because $x>0$ in the $R$ region.

With the above we can solve for $y$.
\begin{align*}
    v^2& = \frac{y}{x}\\
    v^2& = \frac{y}{\frac{u}{v}}\\
    uv & = y.
\end{align*}
On the other hand, in the $R$ region we have the following equations $xy = 1$, $y = x$ and $y = 2$. Now with all of the above, let's pass these equations to the coordinates $u$ and $v$.
\begin{align*}
    xy = 1 &\Rightarrow u  = \sqrt{1} \Rightarrow u = 1\\
    y = x  &\Rightarrow uv = \frac{u}{v} \Rightarrow v^2 = 1 \Rightarrow v = 1\\
    y = 2  &\Rightarrow 2 = uv.
\end{align*}
Observation: By definition of $u$ and $v$, we have to be positive about the values taken by the $R$ region, therefore $v = -1$ is not taken.

Using the above equations, we have that the region $R$ is transformed into the region $S$, shown below.

Now using $x = \frac{u}{v}$ and $y = uv$ we calculate the Jacobian to use the change of variable in the original integral.
\begin{align*}
    \frac{\partial x}{\partial u}&=\frac{1}{v} &\frac{\partial y}{\partial u} =v\\
    \frac{\partial x}{\partial v}&=-\frac{u}{v^2} &\frac{\partial y}{\partial v} =u
\end{align*}
Thus, the Jacobian is:
\begin{align*}
    \begin{vmatrix}
    \frac{1}{v}    &v\\
    -\frac{u}{v^2} &u
    \end{vmatrix}
    =\frac{u}{v}+\frac{u}{v}=\frac{2u}{v}.
\end{align*}
So using all the information developed we have:
\begin{align*}
    \iint\limits_R\sqrt{\frac{y}{x}}e^{\sqrt{xy}}dA &= \iint\limits_Sve^u\left(\frac{2u}{v}\right)dA\\
    &= 2\int_{1}^{2}\int_{1}^{2/u}ue^u\,dv\,du\\
    &= 2\int_{1}^{2}ue^uv\biggr\rvert_{1}^{2/u}\,du\\
    &= 2\int_{1}^{2}ue^u\left(\frac{2}{u}-1\right)du\\
    &= 2\int_{1}^{2}e^u(2-u)du=2(e^u(3-u))\biggr\rvert_{1}^{2}=2e(e-2).
\end{align*}

I think this is the correct solution, I await your comments. If anyone has a different solution or correction of my work I will be grateful.

答:
Your solution is correct. Still we can do without changing variables:
\begin{align*}
[\ldots]&=\int_1^2\int_{1/y}^y\sqrt\frac{y}{x}e^{\sqrt{xy}}\,dx\,dy
\\&=\int_1^2 2e^{\sqrt{xy}}\Bigg|_{x=1/y}^{x=y}\,dy
\\&=2\int_1^2(e^y-e)\,dy
\\&=2(e^y-ey)\Bigg|_{y=1}^{y=2}=2e(e-2).
\end{align*}


问:
General formula for the probability of a runner achieving their average time
The more I think about this, the less sense it makes, but I'll ask anyway since it could just be a simple "no.".

I run the 100-meters 50 times.

I get the mean of my times, let's say it's 13.00 seconds.

I also get some more information, like the minimum (12.50) and maximum (14.00) times.

I want to make a function using this data that gives me a pretty similar list of times when I run it 50 times.

Obviously, there's no way to say exactly what the distribution is, but since this sort of thing comes up a lot, I thought there might be some way to get a better estimate than just joining up the points.

Maybe some way of joining up the points, such that the more points you get, the more accurate your function gets?

答: Well, as you correctly stated, there is no standard way to determine which distribution these times would follow. However, in your example where the minimum is $12.5$ seconds and the maximum is $14$ seconds and the mean time is $13$ seconds, you could think of the following model: $\mathbb{P}(T \in [12.5, 13]) = \frac{2}{3}$ and $\mathbb{P}(T \in [13, 14]) = \frac{1}{3}$ and it is uniform in each of these two intervals. Other than this, without any additional relevant data, any model that meets your initial conditions (in the above example, minimum, maximum and mean) would do the trick, even though some models could be more sensible than others.

ContentLicense: CC BY-SA 4.0


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

答:
You won't have at least 2 odds if there are exactly 7 or 8 evens. You won't have at least 3 evens if there are exactly 6, 7, or 8 odds.

The probability of 8 odds = $\frac{1}{2^8}$ and same for 8 evens.

The probability for 7 odds (or evens) = $\frac{1}{2^8}$ with 1 degree of freedom for the even roll to appear anywhere between the odds, or $8*\frac{1}{2^8}$.

The probability for 6 odds (with 2 evens) = $\frac{1}{2^8}$ with 2 degrees of freedom for the 2 evens to appear anywhere. The way I think about this intuitively is to imagine the 6 odds on the ground, picking up the 2 evens for the time being, and there being a bucket between and at the beginning and end of the 6 odds for me to place the 2 evens. So, there would be 7 buckets I can place my evens in, but I need to be careful not to double count my choices because the order of the evens does not matter. In the first scenario the 2 evens will go in the same bucket, and there are 7 ways that can happen. In the second scenario, they will go to different buckets and there are $\frac{7*6}{2}$ ways for that to happen (need to divide by two because choosing the 1st bucket then the 3rd is the same as choosing the 3rd bucket then the 1st). So, we get 28 from $7+21$ and this probability being $28*\frac{1}{2^8}$.

Side note: if you want to say for the bucket choices it is $7*7=49$, to remove the double counting you're essentially removing duplicates from the set {(1,1),  (1,2), ..., (2,1), ..., (7,7)} which isn't as intuitive but would be $(49-7)/2+7$ (remove the same buckets, divide by 2, then add back in the same buckets).

So add them all up and take away from 1 to get:

$1 - 2*\frac{1}{2^8} - 2*8*\frac{1}{2^8} - 28*\frac{1}{2^8}$.

Also, 28 is also ${8}\choose{2}$ and most people will probably just automatically go with this format for this kind of problem but you can essentially derive it yourself if you just think about it.


问:
Trouble calculating a integral
I've difficulties calculating the following integral

$$\int_0^\infty y^k\frac{1}{\sqrt{2\pi}y}e^{\frac{-(\ln y)^2}{2}}[\sin (2\pi \ln y)]\,dy$$

This integral is a part of a density distribution of a random variable and I need to prove that this integral is equal to zero but I don't know how to do it. I've try to use the fact that $\sin$ is a even function but I don't know how to proceed any further than that. I appreciate any help. Thank you in advance.

答:
Using $y^k = e^{k\ln y} = \cosh(k\ln y) + \sinh(k\ln y)$ we have that the integral simplifies to

$$\int_0^{\infty}\sinh(k\ln y)\frac{1}{\sqrt{2\pi}y}e^{-\frac{(\ln y)^2}{2}}\sin(2\pi \ln y)\:dy$$

which is nonzero for $k\neq 0$

Title: Trouble calculating a integral


问:
Volumes by Slices
The base of a solid is the region in the xy-plane bounded by the curves $ y = \sqrt x, x=4,$ and $y = 0$. Cross-sections of the solid perpendicular to the x-axis are semicircles. Find its volume.

I solved the following integral, adding a coefficient of $\frac1 2$ because the cross section is a semicircle, thus a half rotation. I get a final answer of $4\pi$ as follows:

$$ V = \frac \pi2 \int_0^4 \sqrt x ^2 \,dx = \frac \pi2  \int_0^4 x dx = \frac \pi4 x^2 \bigg|_0^4 = 4\pi $$

The computer says this answer is incorrect. Am I misinterpreting the axis of rotation?

答: Since the solid has semicircular cross sections perpendicular to the $x$-axis, the area of one of these cross sections is $$\frac{1}{2}\pi r^{2} = \frac{1}{2}\pi \left(\frac{\sqrt{x}}{2}\right)^{2} = \frac{\pi x}{8}.$$  The total volume then is given by $$V = \int_{0}^{4}\frac{\pi x}{8}\,dx = \frac{\pi}{16}x^{2}\bigg|_{0}^{4} = \pi.$$  Your error was in that you said that the radius of the semicircles was $\sqrt{x}$, where in reality this is the diameter.

Title: Volumes by Slices


问:
Proof of $\sum_{k=0}^n{ \binom{2k}{k} 2^{2n-2k} } = (2n+1) \binom{2n}{n} = \frac{n+1}{2} \binom{2(n+1)}{n+1}$
I found some combinatorial identities in my old notebooks, but I cannot recall how I derived them. Can anyone help provide the most elementary/elegant proofs for the following identity? Specifically, the connection between the summation and either of the two closed-form expressions.

$$
\sum_{k=0}^n{ \binom{2k}{k} 2^{2n-2k} } = (2n+1) \binom{2n}{n} = \frac{n+1}{2} \binom{2(n+1)}{n+1}
$$

Some specific cases written out for some simple visualization: ($n=0$ through $n=3$)
$$
\begin{align}
\binom{0}{0} 2^{0} &= 1\binom{0}{0} = \frac{1}{2}\binom{2}{1} \\
\binom{0}{0} 2^{2} + \binom{2}{1} 2^{0} &= 3\binom{2}{1}  = \frac{2}{2}\binom{4}{2} \\
\binom{0}{0} 2^{4} + \binom{2}{1} 2^{2} + \binom{4}{2} 2^{0} &= 5\binom{4}{2} = \frac{3}{2}\binom{6}{3}\\
\binom{0}{0} 2^{6} + \binom{2}{1} 2^{4} + \binom{4}{2} 2^{2} + \binom{6}{3} 2^{0} &= 7\binom{6}{3} = \frac{4}{2}\binom{8}{4}
\end{align}
$$

EDIT: I should specify that I have already found an inductive proof, but it is not satisfying to me. I am specifically looking for a bijective/double-counting proof ideally.

答:
Proving the equality between the two closed form is easier:

$$\begin{align*}
(2n+1)\binom{2n}n &=(2n+1)\frac{(2n)!}{n!n!}\\
&= \frac{(2n+1)!}{n!n!}\\
&= \frac{(n+1)^2}{(2n+2)}\cdot\frac{(2n+2)!}{(n+1)!(n+1)!}\\
&= \frac{n+1}{2}\binom{2(n+1)}{n+1}
\end{align*}$$

By induction, consider the $n+1$ case of the summation form and the second closed-form,

$$\begin{align*}
\sum_{k=0}^{n+1} \binom{2k}{k}2^{2(n+1)-2k} &= 2^2\sum_{k=0}^{n}\binom{2k}{k}2^{2n-2k} + \binom{2(n+1)}{n+1}2^0\\
&= 2^2\cdot \underbrace{\frac{n+1}{2} \binom{2(n+1)}{n+1)}}_{\text{by induction hypothesis}} + \binom{2(n+1)}{n+1}2^0\\
&= [2(n+1)+1]\binom{2(n+1)}{n+1}
\end{align*}$$


问:
Need help in discerning whether or not this is a joke question or not?
Just a few moments ago, I have asked a question regarding a proof that my math teacher asked me to solve. But it turned out to be a joke; he was asking me to solve the ABC conjecture, which apparently is a known open question in the math community. Now I have another extra credit proof(my teacher gave 5 of them; hopefully they aren't all just jokes), but I'm having trouble figuring out if this is just another joke question or not. It is as follows:

Consider a sequence $(a_n)$ that has an initial value of $a_0=k$, where $k$ is a positive integer. Every term after that is then defined recursively as follows:$$a_n=\begin{cases}\frac{a_{n-1}}2&\text{if }a_{n-1}\equiv0\pmod2\\3a_{n-1}+1&\text{if }a_{n-1}\equiv1\pmod2\end{cases}$$Prove or disprove the statement that there are infinitely many values for $k$ such that $(a_n)$ will reach $1$ at some point.

My initial thought was to try out small values for $k$. I first tested $k=3$, which produces the following sequence:$$3,10,5,16,8,4,2,1$$
I noticed that if the sequence reaches any power of $2$, then it will eventually reach $1$. But after this, I don't know what I can do. I suppose I can try to find an explicit formula for this recursive sequence but I feel like that would be a hard task.

This problem is pretty easy to understand, yet the proof still eludes me. I didn't spend too much time on this, thinking it might be another joke, but I'm not sure if that's the case here. Any thoughts?

答: People aren't reading the question carefully.  This is NOT the Collatz conjecture, but a different problem about the Collatz sequence.   The question given is easy to solve:   Consider the infinite set $\{ 2^n | n = 1, 2, 3, \ldots\}.$


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

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

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

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

So my questions are:

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

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

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

答:
Consider data $x = 8$ sampled from a binomial population with success parameter $\theta$ (unknown and to be estimated) and a known number of Bernoulli trials $n = 12.$

PDF. If it happens that $\theta = 0.5,$ then the PDF gives
you the probability of getting $X = 8:$
$$f(x |n=12, \theta=0.5) = {n \choose x}\theta^x(1-\theta)^{n-x}\\ = {12 \choose x}.5^x(1-.5)^{12-x} = {12\choose x}(.5)^{12},$$ for $x = 0, 1, \dots, 12.$

```<code>plot(0:12, dbinom(0:12, 12, 1/2), type="h",
     lwd=2, col="blue", ylab="PDF", xlab="x",
     main="PDF of BINOM(12, 1/2)")
 abline(h = 0, col="green2")
</code>```

Likelihood. Now, if you have observed $x = 8$ and wish to find
the corresponding estimate $\hat \theta$ of $\theta,$
then the PDF, considered now as a likelihood function,
might be written as

$$\mathcal{L}(\theta|x,n) = \mathcal{L}(\theta|8,12)\propto \theta^x(1=\theta)^{n-x} = \theta^8(1-\theta)^4,$$
for $0 < \theta < 1,$
where the symbol $\propto$ (read 'proportional to') is
used as a reminder that the (now irrelevant) constant
${n\choose x}$ has been omitted.
Maximixing $\mathcal{L}(\theta|8,12)$ in $\theta$ we obtain the maximum likelihood estimate (MLE) $\hat\theta = x/n = 9/12 = 2/3.$

```<code>th = seq(0, 1, by=.01)
like = dbinom(8, 12, th)
plot(th, like, type= "l", ylab="Likelihood",
     xlab = "theta", lwd=2,
     main="Likelihood Function")
 abline(h=0, col="green2")
 abline(v=8/12, col="maroon")

mle = th[like==max(like)];  mle
[1] 0.67
</code>```

Comparison. It is simple to say that the PDF and the likelihood
function are the 'same thing'. But not exactly true.

The PDF is a function of $x,$ for given
parameters $n$ and $\theta.$ As in the first plot above.

The likelihood function is a function of
unknown $\theta$ for known values $n = 12, x = 8.$
As in the second plot.

It not not surprising that
this difference in viewpoint gets shown explicitly in the notation.

Body:
Consider data $x = 8$ sampled from a binomial population with success parameter $\theta$ (unknown and to be estimated) and a known number of Bernoulli trials $n = 12.$

PDF. If it happens that $\theta = 0.5,$ then the PDF gives
you the probability of getting $X = 8:$
$$f(x |n=12, \theta=0.5) = {n \choose x}\theta^x(1-\theta)^{n-x}\\ = {12 \choose x}.5^x(1-.5)^{12-x} = {12\choose x}(.5)^{12},$$ for $x = 0, 1, \dots, 12.$

plot(0:12, dbinom(0:12, 12, 1/2), type="h",
     lwd=2, col="blue", ylab="PDF", xlab="x",
     main="PDF of BINOM(12, 1/2)")
 abline(h = 0, col="green2")

Likelihood. Now, if you have observed $x = 8$ and wish to find
the corresponding estimate $\hat \theta$ of $\theta,$
then the PDF, considered now as a likelihood function,
might be written as

$$\mathcal{L}(\theta|x,n) = \mathcal{L}(\theta|8,12)\propto \theta^x(1=\theta)^{n-x} = \theta^8(1-\theta)^4,$$
for $0 < \theta < 1,$
where the symbol $\propto$ (read 'proportional to') is
used as a reminder that the (now irrelevant) constant
${n\choose x}$ has been omitted.
Maximixing $\mathcal{L}(\theta|8,12)$ in $\theta$ we obtain the maximum likelihood estimate (MLE) $\hat\theta = x/n = 9/12 = 2/3.$

th = seq(0, 1, by=.01)
like = dbinom(8, 12, th)
plot(th, like, type= "l", ylab="Likelihood",
     xlab = "theta", lwd=2,
     main="Likelihood Function")
 abline(h=0, col="green2")
 abline(v=8/12, col="maroon")

mle = th[like==max(like)];  mle
[1] 0.67

Comparison. It is simple to say that the PDF and the likelihood
function are the 'same thing'. But not exactly true.

The PDF is a function of $x,$ for given
parameters $n$ and $\theta.$ As in the first plot above.

The likelihood function is a function of
unknown $\theta$ for known values $n = 12, x = 8.$
As in the second plot.

It not not surprising that
this difference in viewpoint gets shown explicitly in the notation.


问:
Minimization of the SSE
I'm reading the demonstration in the Book of Wackerly about the minimization of the $\mathrm{SSE}$ for the least square method but there's is a step I don't understand why an $n$ appears where I highlighted after the differentiation. I'm completely lost. Thanks for your help

\begin{align*}
\frac{\partial\mathrm{SSE}}{\partial\hat\beta_0} & =\frac{\partial\left\{\sum_{i=1}^n[y_i-(\hat\beta_0+\hat\beta_1x_i)]^2\right\}}{\partial\hat\beta_0}=-\sum_{i=1}^n2[y_i-(\hat\beta_0+\hat\beta_1x_i)]\\
& =-2\left(\sum_{i=1}^ny_i-\color{red}{n\hat\beta_0}-\hat\beta_1\sum_{i=1}^nx_i\right)=0
\end{align*}
and
\begin{align*}
\frac{\partial\mathrm{SSE}}{\partial\hat\beta_1} & =\frac{\partial\left\{\sum_{i=1}^n[y_i-(\hat\beta_0+\hat\beta_1x_i)]^2\right\}}{\partial\hat\beta_1}=-\sum_{i=1}^n2[y_i-(\hat\beta_0+\hat\beta_1x_i)]x_i\\
& =-2\left(\sum_{i=1}^nx_iy_i-\hat\beta_0\sum_{i=1}^nx_i-\hat\beta_1\sum_{i=1}^nx_i^2\right)=0.
\end{align*}

Title: Minimization of the SSE

Body:
$$-\sum_{i=1}^n 2[y_i - (\hat{\beta_0} + \hat{\beta_1} x_i)]$$
$$= -2[(y_1 - (\hat{\beta_0} + \hat{\beta_1} x_1)) + (y_2 - (\hat{\beta_0} + \hat{\beta_1} x_2)) + \ldots + (y_n - (\hat{\beta_0} + \hat{\beta_1} x_n))]$$

Can you see the reason now?


问:
Density of smooth functions in $L^p$ space on Cantor Set
Let $\mu$ be the Cantor measure on the Cantor set $C$ in $[0,1]$. Is the space of functions obtained by restricting smooth compactly supported functions to $C$ dense in $L^p(\mu)$ for $1 \leq p < \infty$?

答:
Hint: There are two points to consider.

(1) $C([0,1])$ is dense in $L^{p}(\mu)$ for all $1 \leq p < \infty$.  This is true whenever $\mu$ is a finite Borel measure on $[0,1]$.

(2) In the case of the Cantor measure, $\mu(\{0\}) = \mu(\{1\}) = 0$.

(Also, in a sense, restricting the function to $C$ is only going to confuse your intuition slightly since $f \restriction_{C}$ and $f$ look the same to the Cantor measure.)


问:
$P(X+Y>t\mid aX+bY>u)\geq P(X+Y>t)$?
Let $X,Y$ be independent continuous random variables with support on some neighbourhood of infinity. Let $a, b\geq 0$, and $t,u\in\mathbb{R}$ be constants. Then it holds that $$P(X+Y>t\mid aX+bY>u)\geq P(X+Y>t). $$

Do you know how to prove this? it sounds very intuitive and easy, but all that I tried failed.

答:
For simplicity we assume $a>0$ rather than only $a\geq0$.

By Bayes' formula, we may equivalently show that
\begin{align*}
\mathbb{P}(aX + bY > u \, | \, X + Y > t) \geq \mathbb{P}(aX + bY > u).
\end{align*}
Using the tower property in both the numerator and the denominator,
\begin{align*}
\mathbb{P}(aX + bY > u \, | \, X + Y > t)
&=
\dfrac{\mathbb{P}(aX + bY > u, X + Y > t)}{\mathbb{P}(X+Y>t)} \\
&=
\dfrac{\mathbb{E}[\mathbb{P}(aX + bY > u, X + Y > t \, | \, Y)]}{\mathbb{E}[\mathbb{P}(X + Y > t \, | \, Y)]}.
\end{align*}
Similarly,
\begin{align*}
\mathbb{P}(aX + bY > u) = \mathbb{E}[\mathbb{P}(aX + bY > u \, | \, Y)].
\end{align*}
Since $X$ and $Y$ are independent and $a>0$,
\begin{align*}
\mathbb{P}(aX + bY > u, X + Y > t \, | \, Y)
&=
S_X(Y_1 \vee Y_2), \\
\mathbb{P}(X + Y > t \, | \, Y)
&= S_X(Y_2), \\
\mathbb{P}(aX + bY > u \, | \, Y)
&= S_X(Y_1),
\end{align*}
where $S_X(x) := 1 - \mathbb{P}(X \leq x)$ and $x_1 \vee x_2 = \max\{x_1,x_2\}$, while
\begin{align*}
Y_1 = \dfrac{u - bY}{a}, \hspace{5mm} Y_2 = t-Y.
\end{align*}
Note that $Y_1$ and $Y_2$ are functions of $Y$, and thus $S_X(Y_1 \vee Y_2)$, $S_X(Y_2)$, and $S_X(Y_1)$ are also random variables (more precisely, transformations of $Y$).

Collecting terms, we want to show that
\begin{align*}
\dfrac{\mathbb{E}[S_X(Y_1 \vee Y_2)]}{\mathbb{E}[S_X(Y_1)]\mathbb{E}[S_X(Y_2)]} \geq 1.
\end{align*}
Now obviously, $S_X(Y_1 \vee Y_2) \geq S_X(Y_1)S_X(Y_2)$. So it suffices to show that
\begin{align*}
\text{Cov}[S_X(Y_1),S_X(Y_2)] = \mathbb{E}[S_X(Y_1)S_X(Y_2)] - \mathbb{E}[S_X(Y_1)]\mathbb{E}[S_X(Y_2)] \geq 0.
\end{align*}
Note that $S_X(Y_1)$ is a non-decreasing function of $Y$ since $Y_1$ is a non-increasing as a function of $Y$ (recall $b\geq0$) and $S_X$ is a survival function and thus non-increasing. Similarly, $S_X(Y_2)$ is a non-decreasing function of $Y$. Consequently, see e.g. this question, we must have $\text{Cov}[S_X(Y_1),S_X(Y_2)] \geq 0$ as desired.

Title: $P(X+Y>t\mid aX+bY>u)\geq P(X+Y>t)$?


问:
Can somebody do this trigonometry compound angle question?
If $\cos(A) + \sin(B) = x$ and $\sin(B) + \cos(A) = y$, prove that

$$\sin(A + B) = \frac{x^2 + y^2 - 2}{2}$$

Body:
Assuming

$$x = \cos A + \sin B, y = \sin A + \cos B$$

We have

$$x^2 = \cos^2 A + \sin^2 B + 2 \cos A \sin B$$

$$y^2 = \sin^2 A + \cos^2 B + 2 \sin A \cos B$$

Adding, $$x^2 + y^2 = 2 + 2 \sin (A+B)$$

Or, $$\sin(A+B) = \frac{1}{2}(x^2 + y^2 - 2)$$


问:
Proof Concerning Indexed family of Sets and Intervals
I have been stuck on the following problem for some time.

I know to prove two things are equal, you use the proof where you show they are both subsets. I think I was able to show that the set on the left, (which I called A) is a subset of the interval through simple logic, but to show the converse I am having some issues.

I attached my working thus far for anyone to look at it, but please note that after the line " $ \Leftrightarrow \alpha \geq -1 = \alpha + 1 \geq 0 and  \alpha \leq -2 ....$" the rest is tata (lol- but an attempt to work towards what I know would be the end of the proof, showing that a belongs also to A)

I have tried using the hint, but then I end up with something along the lines of '$n > \alpha + 1$ so  $n + 1 > \alpha$', which does not bring me closer to either -x, or x+1 of the interval on the left.

Any help or suggestions would be greatly appreciated. Thanks in advance.

答:
You want to show that

$$\bigcap_{x>1}(-x,x+1]=[-1,2]\,.$$

Suppose first that $x>1$; then $x+1>2$, as you said. It’s not true, however, that $-x>-1$: rather, $-x<-1$, because the inequality is reversed when you multiply both sides by a negative number. It follows that $(-x,x+1]\supseteq[-1,2]$, and since this is the case for every $x>1$, we’ve shown that

$$\bigcap_{x>1}(-x,x+1]\supseteq[-1,2]\,.$$

Now we have to show that

$$\bigcap_{x>1}(-x,x+1]\subseteq[-1,2]\,;$$

this is the part for which the hint is intended. We want to show that if

$$y\in\bigcap_{x>1}(-x,x+1]\,,$$

then $y\in[-1,2]$, but the hypothesis that $y$ is in each of the intervals $(-x,x+1]$ for $x>1$ is hard to work with directly. It’s easier to show the contrapositive: if $y\notin[-1,2]$, then

$$y\notin\bigcap_{x>1}(-x,x+1]\,.$$

Suppose, then, that $y\notin[-1,2]$; then either $y<-1$, or $y>2$. I’ll do the first case and leave the second for you.

Suppose that $y<-1$, and let $u=-y$. Then $y=-u$, and $u>1$, so

$$y\notin(y,-y+1]=(-u,u+1]\supseteq\bigcap_{x>1}(-x,x+1]\,,$$

as desired, and it follows that

$$\bigcap_{x>1}(-x,x+1]=[-1,2]\,.$$


问:
Reflection matrix about a plane
How do you construct a matrix in $\mathbb{R}^3$ that reflects about the plane $y=z$? And is there a way to construct a reflection matrix about any plane in general?

答:
Any plane can be specified by a unit vector perpendicular to the plane. If the vector is $v \in\mathbb{R}^3$, then the matrix that reflects about the plane is

$$
R_v = I - 2 vv^T.
$$

It is easy to check that $R_v$ flips the sign of any vector which is a multiple of $v$ and acts as identity on any vector perpendicular to $v$. See Householder transformation for more details.

In particular, the plane $y=z$ is perpendicular to $(0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})^T \in \mathbb{R}^3$, so the matrix is

$$
R_{y=z}=\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix} - 2\cdot\begin{pmatrix}
0 & 0 & 0 \\
0 & \frac{1}{2} & -\frac{1}{2} \\
0 & -\frac{1}{2} & \frac{1}{2}
\end{pmatrix} = \begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{pmatrix}
$$

which leaves the $x$ coordinate unchanged and swaps the $y$ and $z$ coordinates as expected.

Title: Reflection matrix about a plane


问:
Having trouble understanding difference between polyhedron and polytope
Hi i´m reading a pdf about linear programming and i´m having trouble understanding the difference between a polyhedron and polytope between those two definitions

A polyhedron P ⊆ $R^{n}$
is the set of all points x ∈ $R^{n}$
that satisfy a finite set of linear inequalities.
Mathematically,
P = {x ∈ $R^{n}$: Ax ≤ b}
for some matrix A ∈ R
m×n and a vector b ∈ $R^{m}$.

And the definition of polytope is the next one

A polyhedron is called a polytope if it is bounded, i.e., can be enclosed in a ball of
finite radius.

So my problem isn´t the definition of polyhedron already bounded , and also are all polyhedron in these context convex?

And for example

{x ∈ $R^{n}$: Ax ≤ b} if i guide myself from the given definitions i can only assure this is a polyhedron but not a polytipe since it isn't technically bounded but is convex and {x|$x\in\mathbb{R}$} isn't any of those two.

Any insight would be helpful

答:
A polyhedron is not bounded in the sense that we might not be able to find a ball of finite radius to find it.

For example consider, $\{x \in \mathbb{R}^n : x \ge 0\}$, the first octant polyhedron, it is unbounded, it is a polyhedron but it is not a polytope. If you claim that it is bounded by a ball of radius $r$, consider the point $(r+1, 0, \ldots, 0)$, it is insider the polyhedron, but it is not contained inside the ball of radius $r$.

$\{x | x \in \mathbb{R}\}$ is a polyhedron since we can take $A=b=0$.


问:
Question about proofs using the converse
I'm reading the MCS textbook from MIT

where they state the following:

The valid formula of part (a) leads to sound proof method: to prove
that an implication is true, just prove that its converse is false.

For example, from elementary calculus we know that the assertion "If a
function is continuous, then it is differentiable" is false. This
allows us to reach at the correct conclusion that its converse, "If a
function is differentiable, then it is continuous" is true, as indeed
it is.

But wait a minute! The implication "If a function is differentiable,
then it is not continuous" is completely false. So we could conclude
that its converse "If a function is not continuous, then it is
differentiable", should be true, but in fact the converse is also
completely false.

So something has gone wrong here. Explain what.

Note: the formula of part (a) is "$(P \text{ implies } Q) OR (Q \text{ implies } P)$ is true"

Now I am having trouble solving this question. On the Wikipedia page for Converse it states that "the truth of the converse is generally independent from that of the original statement", so I initially thought the answer was, "the converse is not necessarily related to the original statement" and left it that. But then that is at odds with the formula of part (a) as well as the claim the textbook makes that you can prove something using the converse.

Any help would be very appreciated!

答:
This is a pretty confusing question to ask early in one’s logical education! What’s going on is that, if we want to think about the statement “$p$ implies $q$” and we can prove that the converse “$q$ implies $p$” is false, then in fact we know that $q$ is true and $p$ is false, which means $p$ does imply $q$. That is, symbolically, $$\neg(q\to p)\to (p\to q).$$
But this last implication is not an equivalence! That is, knowing $p\to q$ is true doesn’t mean $q\to p$ is false, or equivalently, knowing $q\to p$ is true doesn’t mean $p\to q$ is false.

Thus a statement is to some extent independent of its converse, but not perfectly so, as it’s impossible for both a statement and its converse to be false. This comes somehow from the asymmetry of the truth table for implication, which has three T’s out of four.


问:
Problem solving complex equation with high power
I have to solve :$$(i+z)^{16}-(7+7i)(i+z)^8+25i=0$$
I consider this equation like a quadratic equation and find its delta and solutions. But i can't solve two equations :$$(i+z)^8=4+3i$$
$$(i+z)^8=3+4i$$
I really need some helps to solve completely this equation. Thanks.

答:
If you don't want to use polar coordinates, the eighth roots can be found by repeated square root.

We get the square root
$$  (x+iy)^2 = a+ib $$
when
$$  x = \frac{1}{2} \left( \sqrt{ \sqrt{a^2+b^2} + b}  + \sqrt{\sqrt{a^2+b^2} - b} \right)  $$

$$  y = \frac{1}{2} \left( \sqrt{ \sqrt{a^2+b^2} + b}  - \sqrt{\sqrt{a^2+b^2} - b} \right)  $$

if the real parts $a,b$ are positive, so are the $x,y$ above are also positive. Switching $a,b$ to $b,a$ gives a different result.


答: Is this a valid counterexample?

Title: $P(X+Y>t\mid aX+bY>u)\geq P(X+Y>t)$?

Body:
Let $X,Y$ be independent continuous random variables with support on some neighbourhood of infinity. Let $a, b\geq 0$, and $t,u\in\mathbb{R}$ be constants. Then it holds that $$P(X+Y>t\mid aX+bY>u)\geq P(X+Y>t). $$

Do you know how to prove this? it sounds very intuitive and easy, but all that I tried failed.

Body: Is this a valid counterexample?


问:
Set of Power of a complex set
Let's say $K = \{(2,1),(2,3),(2,5),(4,1),(4,3),(4,5)\}$
what is the power of set of $K$, I know since it got 6 elements the cardinality should be 64. However, I can't seem to find the proper power of set, always finding around 38 elements.

Edit: I got the result of $K$ by a cartesian product $A\times B$.

Title: Set of Power of a complex set

Body:
As you observed the powerset $\mathcal{P}(K)$ contains $64$ elements. Here is the complete list:

$$
\mathcal{P}(K) = \{
\emptyset,
\{(2, 1)\},
\{(2, 3)\},
\{(2, 5)\},
\{(4, 1)\},
\{(4, 3)\},
\{(4, 5)\}, \\
\{(2, 3), (2, 1)\},
\{(2, 5), (2, 1)\},
\{(4, 1), (2, 1)\},
\{(2, 1), (4, 3)\},
\{(4, 5), (2, 1)\},
\{(2, 3), (2, 5)\},
\{(2, 3), (4, 1)\},
\{(2, 3), (4, 3)\},
\{(2, 3), (4, 5)\},
\{(2, 5), (4, 1)\},
\{(2, 5), (4, 3)\},
\{(4, 5), (2, 5)\},
\{(4, 1), (4, 3)\},
\{(4, 5), (4, 1)\},
\{(4, 5), (4, 3)\},
\{(2, 3), (2, 5), (2, 1)\},
\{(2, 3), (4, 1), (2, 1)\},
\{(2, 3), (2, 1), (4, 3)\},
\{(2, 3), (4, 5), (2, 1)\},
\{(2, 5), (4, 1), (2, 1)\},
\{(2, 5), (2, 1), (4, 3)\},
\{(4, 5), (2, 5), (2, 1)\},
\{(4, 1), (2, 1), (4, 3)\},
\{(4, 5), (4, 1), (2, 1)\},
\{(4, 5), (2, 1), (4, 3)\},
\{(2, 3), (2, 5), (4, 1)\},
\{(2, 3), (2, 5), (4, 3)\},
\{(2, 3), (4, 5), (2, 5)\},
\{(2, 3), (4, 1), (4, 3)\},
\{(2, 3), (4, 5), (4, 1)\},
\{(2, 3), (4, 5), (4, 3)\},
\{(2, 5), (4, 1), (4, 3)\},
\{(4, 5), (2, 5), (4, 1)\},
\{(4, 5), (2, 5), (4, 3)\},
\{(4, 5), (4, 1), (4, 3)\},
\{(2, 3), (2, 5), (4, 1), (2, 1)\},
\{(2, 3), (2, 5), (2, 1), (4, 3)\},
\{(2, 3), (4, 5), (2, 5), (2, 1)\},
\{(2, 3), (4, 1), (2, 1), (4, 3)\},
\{(2, 3), (4, 5), (4, 1), (2, 1)\},
\{(2, 3), (4, 5), (2, 1), (4, 3)\},
\{(2, 5), (4, 1), (2, 1), (4, 3)\},
\{(4, 5), (2, 5), (4, 1), (2, 1)\},
\{(4, 5), (2, 5), (2, 1), (4, 3)\},
\{(4, 5), (4, 1), (2, 1), (4, 3)\},
\{(2, 3), (2, 5), (4, 1), (4, 3)\},
\{(2, 3), (4, 5), (2, 5), (4, 1)\},
\{(2, 3), (4, 5), (2, 5), (4, 3)\},
\{(2, 3), (4, 5), (4, 1), (4, 3)\},
\{(4, 5), (2, 5), (4, 1), (4, 3)\},
\{(2, 1), (4, 3), (2, 3), (2, 5), (4, 1)\},
\{(2, 1), (2, 3), (4, 5), (2, 5), (4, 1)\},
\{(2, 1), (4, 3), (2, 3), (4, 5), (2, 5)\},
\{(2, 1), (4, 3), (2, 3), (4, 5), (4, 1)\},
\{(2, 1), (4, 3), (4, 5), (2, 5), (4, 1)\},
\{(4, 3), (2, 3), (4, 5), (2, 5), (4, 1)\},
K
\}.
$$

Perhaps, it is helpful in finding out what elements were missing from your list of $38$.


问:
Unique geodesic between two points
Suppose $(X,g)$ is a simply connected complete Riemannian manifold, and $X$ has negative sectional curvature everywhere. Is it true in such a case, for any two points in this space, namely $A$ and $B$, there exists a unique geodesic between them?

I know it's true for  $dim(X) = 2$ since we can use Gauss Bonnet. How should I approach for higher dimension? If it's not true, are there some counterexamples?

答: This is a direct consequence of the Cartan-Hadamard theorem. Pretty much every Riemannian Geometry textbook will have it. Note that you only need the sectional curvature to be $\le 0$ for the exponential map $\exp_p$ to be a diffeomorphism (for each $p\in M$).

Title: Unique geodesic between two points


问:
How would I prove that if 11n-5 is odd, then n is an even number using only a direct proof?
I've been able to prove this statement through contraposition and contradiction but I'm struggling to prove it through a direct proof. It seems I always get it in the form where 11n=2(k+3).</p>

答: Since the right-hand side has $2$ as a factor, and $2$ and $11$ are coprime, the left-hand side also has $2$ as a factor. This relies on the fundamental theorem of arithmetic, where every positive integer can be represented as a unique product of prime numbers.


答:
If you translate $[A \implies B]~~$ to $~~[(\neg A) \vee B]~~$ then the assertion that (in general) $~~(P \implies Q)~~$ must hold when $~~\neg(Q \implies P)~~$ ::: is a true assertion, demonstrated as follows:

The only way that the statement $(Q \implies P)$ can be false is if

$$(\neg P) \wedge Q.\tag1$$

The only way that the statement $(P \implies Q)$ can be false is if

$$(\neg Q) \wedge P.\tag2$$

If statement (1) above is true, then statement (2) above can not be true.

Title: Question about proofs using the converse

Body:
I'm reading the MCS textbook from MIT

where they state the following:

The valid formula of part (a) leads to sound proof method: to prove
that an implication is true, just prove that its converse is false.

For example, from elementary calculus we know that the assertion "If a
function is continuous, then it is differentiable" is false. This
allows us to reach at the correct conclusion that its converse, "If a
function is differentiable, then it is continuous" is true, as indeed
it is.

But wait a minute! The implication "If a function is differentiable,
then it is not continuous" is completely false. So we could conclude
that its converse "If a function is not continuous, then it is
differentiable", should be true, but in fact the converse is also
completely false.

So something has gone wrong here. Explain what.

Note: the formula of part (a) is "$(P \text{ implies } Q) OR (Q \text{ implies } P)$ is true"

Now I am having trouble solving this question. On the Wikipedia page for Converse it states that "the truth of the converse is generally independent from that of the original statement", so I initially thought the answer was, "the converse is not necessarily related to the original statement" and left it that. But then that is at odds with the formula of part (a) as well as the claim the textbook makes that you can prove something using the converse.

Any help would be very appreciated!


问:
Condition of Homogenous Differential Equation
We were be taught that a homogenous differential equation are those (for 1st order, 1st degree) that
$\frac{f(x,y)}{g(x,y)}$ should have same degree. Then we substitue y=vx and then reduce it to solve it by variable sepreable method

But in this question as i read in some text that he solved with homogenous method but the equation is not homogenous (or may be i am missing something). The equation is

$\frac{dy}{dx}=\frac{y}{x}+xsin(\frac{y}{x})$.

Can anyone help me on this?

答:
We need to solve a non linear equation given by $$y'=\frac{y}{x}+x\sin\left( \frac{y}{x}\right)$$
Let, $y=xv$, so we have that $$y'=xv'+v,$$then $$xv'+v=x\sin(v)+v \implies v'=\sin(v)$$Integrating both sides respect to $x$, we have that $$\int \csc(v)v'dx=\int dx \implies v=2\cot^{-1}(\exp(-x-c))$$
Therefore, $$y(x)=2x\cot^{-1}(c\exp(-x))$$


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

It seems to me like a Riemann integral definition:

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

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

How do i proceed from this point?

答:
HINT:

Using this, $$\sum_{1\le r\le n}\sin \frac{\pi r}n=\frac{\sin \left( n\cdot \frac{\pi}{2n}\right)}{\sin \frac \pi{2n}}\cdot \sin \left(\frac{2\frac\pi n+(n-1)\frac{\pi}n}2\right)=\frac{\sin\left(\frac{(n+1)\pi}{2n}\right)}{\sin \frac \pi{2n}}$$

Now, $\sin\left(\frac{(n+1)\pi}{2n}\right)=\sin\left(\frac\pi2+\frac\pi{2n}\right)=\cos \frac\pi{2n}\implies \lim_{n\to\infty}\sin\left(\frac{(n+1)\pi}{2n}\right)=1$

$$\lim_{n\to\infty}\frac 1{n\sin \frac \pi{2n}}=\lim_{h\to0}\frac {2h}{\pi\sin h }(\text{ putting } \frac \pi{2n}=h)$$

$$\implies \lim_{n\to\infty}\frac 1{n\sin \frac \pi{2n}}=\frac2\pi\cdot\lim_{h\to0}\frac h{\sin h }=\frac2\pi$$

ContentLicense: CC BY-SA 3.0


答:
Lets assume $11n-5$ is odd then there exists integer $k$ such that

$11n-5=2k+1 \implies 11n=2k+1+5 = 2k+6 = 2(k+3)$.

Now as $k$ is an integer we get $k+3$ is an integer. We can set $m = k+3$ then we get $11n = 2m$ and $2m$ is an even number which implies $11n$ is an even number. As we have found $11n$ to be an even number this implies that $n$ must be even since the product of two odd numbers will be odd.

Title: How would I prove that if 11n-5 is odd, then n is an even number using only a direct proof?

Body: I've been able to prove this statement through contraposition and contradiction but I'm struggling to prove it through a direct proof. It seems I always get it in the form where 11n=2(k+3).</p>


问:
Inequality for two complex numbers
If $a,b$ are any two complex numbers, then
$$2(|a|^2+|b|^2)= |a+b|^2+|a-b|^2$$ and therefore
$$2(|a|^2+|b|^2)\geq |a+b|^2.$$ My curiosity; Is there any genarization of the above inequality like

$$c(|a|^p+|b|^p)\geq |a+b|^p$$ where $c=f(p)>0$ and $p>1?$

答: Just the real number version would suffice, since $\lvert a\rvert+\lvert b\rvert\geq\lvert a+b\rvert$ gives $(\lvert a\rvert+\lvert b\rvert)^p\geq\lvert a+b\rvert^p$.  So any $c=f(p)\geq\inf_{x\in[0,1]} ((1-x)^p+x^p)^{-1}$ would do.

Title: Inequality for two complex numbers


问:
Supremum of a Sequence of S-measurable Functions
I'm working through Axler's "Measure, Integration, and Real Analysis" and have come across the following theorem:

Suppose $(X,S)$ is a measurable space and $f_1,f_2,\cdots$ is a sequence of $S$-measurable functions from $X$ to $[-\infty, \infty]$. Define $h:X\rightarrow[-\infty, \infty]$ by $h(x)=\sup{\{f_k(x):k\in \mathbf{Z}^+\}}$. Then $h$ is an $S$-measurable function.

He gives a proof using the following equation, which the states the reader should verify ($a \in \mathbf{R}$):
$$h^{-1}((a,\infty])=\bigcup_{k=1}^\infty f_k^{-1}((a,\infty])$$

I'm attempting to verify this. My attempted proof:

Let $x\in \bigcup_{k=1}^\infty f_k^{-1}((a,\infty])$. Then we know that $f_k(x)>a$ for some $k\in\mathbf{Z}^+$. By definition of the supremum, we know that $h(x)\geq f_k(x) \forall k \in \mathbf{Z}^+$. Since $h$ is always greater than or equal to all $f_k$, and there is at least one $f_k>a$, we have that $h(x)>a$, meaning that $x\in h^{-1}((a,\infty])$.
This shows that
$$ \bigcup_{k=1}^\infty f_k^{-1}((a,\infty]) \subseteq h^{-1}((a,\infty]) $$
For the opposite direction, let $x\in h^{-1}((a,\infty])$. Thus, $h(x)>a$. Now, by definition of the supremum, we know that $h(x)\geq f_k(x) \forall k \in \mathbf{Z}^+$. $\cdots$

and that's where I lose the thought process. I'm trying to show that $h^{-1}((a,\infty]) \subset \bigcup_{k=1}^\infty f_k^{-1}((a,\infty])$ thereby proving that the two sets are equal, but I'm missing the last step. If I can figure out how to show that $\exists k\in\mathbf{Z}^+$ such that $f_k(x)>a$, then I would have the proof completed.

答: It is sufficient to show that there is some $k \ge 1$ such that $f_k(x) > a$. To that end, assume to the contrary that $f_k(x) \le a$ for all $k \ge 1$. But then this implies $\sup_k f_k(x) \le a$ contradicting the fact that $\sup_k f_k(x) = h(x) > a$.


问:
Standardizing a binomial distribution
Let $Y$ be a binomial random variable with parameters $n = 72$ and $p = 1/3$.

a) Find the exact probability $P(22 ≤ Y ≤ 28)$.

b) Is the condition $n ≥ 9 \left( \frac{\max(p, 1−p)}{\min(p, 1−p)} \right)$ required to approximate the distribution of Y by a normal distribution (textbook p.380) satisfied? If yes, approximate
$P(22 ≤ Y ≤ 28)$ with a normal probability.

I found a. I got $0.6$ but when I apply the central limit theorem and standardize it, I get $8.485$ and $-4.2$.

This already looks really wrong but I put it in R anyways by doing pnorm(8.485) - pnorm(-4.2) and I get $.999$.

The weird thing is if i code

pnorm(28,24,4)-pnorm(22,24,4)  I get $.53$.
Then I remembered the .5 rule so I did pnorm(28.5,24,4)-pnorm(21.5,24,4)  which is $.65$ and doing pnorm(28,24,4)-pnorm(21,24,4) gave me $.61$ which is REALLY close now to the answer. So I'm pretty sure the normal distribution DOES work in this case but I do not know why everytime I standardize it I get a big mess.

答:
$$np=24$$

$$npq=16$$

CLT in the version of De Moivre- Laplace is

$$\frac{\Sigma_i X_i-np}{\sqrt{npq}}\sim \Phi$$

Thus

$$\mathbb{P}[22\leq Y\leq 28]=\mathbb{P}[21.5<Z<28.5]= $$

$$=\Phi\left[\frac{28.5-24}{4}\right]- \Phi\left[\frac{21.5-24}{4}\right] = 0.87-0.27=0.6 $$

Title: Standardizing a binomial distribution


问:
What makes the Var(x)+Var(y)=var(x+y) property important?
What makes the Var(x)+Var(y)=var(x+y) property important?

It was taught in my statistics class

答:
In the 18th century Abraham de Moivre considered this problem (his account of which you can find in his book The Doctrine of Chances: If you toss a fair coin $1800$ times, what is the probability that the number of heads in in a specified interval, for example at least $880$ but no more than $906$?  In the course of solving that problem, he first introduced the "bell-shaped curve"
$$
\varphi(x) = \text{constant}\cdot e^{-x^2/2}
$$
where the "constant" is chosen so as to make this a probability density function.  (De Moivre found the constant numerically, and James Stirling later showed that it's $1/\sqrt{2\pi\,{}}$.  This is the "normal distribution" with expectation $0$ and variance $1$.  It follows that the normal density with expectation $\mu$ and variance $\sigma^2$ has density
$$
x\mapsto \text{same constant}\cdot\frac1\sigma\varphi\left(\frac{x-\mu}{\sigma}\right).
$$
De Moivre found that the distribution of the number of heads can be made as close as desired to the normal distribution with the right values of $\mu$ and $\sigma$ by making the number of tosses large enough, and $1800$ was plenty.

But how do you know the right values of $\mu$ and $\sigma$?

In the case of $\sigma$, you just use the fact that the variance of the sum of independent random variable is the sum of the variances, i.e. the property of variances that you learned in your statistics class.  Other simple measures of statistical dispersion lack that property, and so cannot be applied to this coin-toss problem.

It's used in statistics all the time.  How do you know that if you take a sample of size $n$ from a population that's not necessarily normally distributed, you can use the two numbers
$$
\bar x \pm 1.96\frac{S}{\sqrt{n}}
$$
as endpoints of a $95\%$ confidence interval for the population mean, where $S$ is the sample standard deviation?  In particular, where does "$S/\sqrt{n}$ come from?  The answer is that it comes from that same property of variances that you were taught.


问:
Trouble calculating a integral
I've difficulties calculating the following integral

$$\int_0^\infty y^k\frac{1}{\sqrt{2\pi}y}e^{\frac{-(\ln y)^2}{2}}[\sin (2\pi \ln y)]\,dy$$

This integral is a part of a density distribution of a random variable and I need to prove that this integral is equal to zero but I don't know how to do it. I've try to use the fact that $\sin$ is a even function but I don't know how to proceed any further than that. I appreciate any help. Thank you in advance.

答:
You are considering $$I=\int_0^\infty y^k\frac{1}{\sqrt{2\pi}y}e^{\frac{-(\ln y)^2}{2}}[\sin (2\pi \ln y)]\,dy$$ Let $y=e^t$ to make
$$I=\frac{1}{\sqrt{2\pi}}\int_{-\infty} ^\infty e^{k t-\frac{t^2}{2}} \sin (2 \pi  t)\,dt=\frac{1}{\sqrt{2\pi}}\Im \int_{-\infty} ^\infty \exp\big[ (k+2 i \pi ) t-\frac{t^2}{2}\big]\,dt$$ Complete the square and you should arrive to the gaussian integral.

Edit

If you did what I suggested, you are suppose to have obtained

$$\int\exp\big[ (k+2 i \pi ) t-\frac{t^2}{2}\big]\,dt=-i \sqrt{\frac{\pi }{2}} e^{\frac{1}{2} (k+2 i \pi )^2} \text{erfi}\left(\frac{-i
   k+i t+2 \pi }{\sqrt{2}}\right)$$
$$\int_{-\infty} ^\infty \big[ (k+2 i \pi ) t-\frac{t^2}{2}\big]\,dt=\sqrt{2 \pi } e^{\frac{1}{2} (k+2 i \pi )^2}$$
$$\Im\Big[e^{\frac{1}{2} (k+2 i \pi )^2}\Big]=\sqrt{2 \pi } e^{\frac{k^2}{2}-2 \pi ^2} \sin (2 \pi  k)$$ So, finally, the integral is just
$$e^{\frac{k^2}{2}-2 \pi ^2} \sin (2 \pi  k)$$ which is zero when $k$ is zero or a multiple of half an integer.

Title: Trouble calculating a integral


问:
What is meant by, "the inclusion map X to Y 'extends' to an isometry of X-bar onto Y?
What is meant by, "the inclusion map X to Y extends to an isometry of X-bar onto Y?

I can answer this question mechanically by showing that one metric is equal to another, but I do not understand the language.  Is an inclusion map simply an injective mapping; yet it is said to extend to an "onto" mapping?  "Onto" is a word  usually used with bijective mapping.  In the present case, X-bar is the space of equivalence classes of X.  In what sense is "extend" used?  Can someone kindly elaborate?

答:
When $X$ is a subset (or sub"thing") of $Y$, the function $i: X \to Y$ given by $i(x)=x$ is typically called an inclusion.  So inclusions are injective, but not the other way around necessarily.

In your case, you have a construction called completion that takes $X$ and constructs a "desirable" $\overline{X}$ with good properties, and $X$ naturally lies inside $\overline{X}$. So, there is a natural map $e : X \to \overline{X}$ which comes about from the construction of $\overline{X}$.  (I've called it $e$ for ``embedding.")  To say that the inclusion $i: X \to Y$ extends to an isometry of $\overline{X}$ onto $Y$ means that there is an isometry $f: \overline{X} \to Y$ which is onto and has the property that $f \circ e = i$.


问:
Proof of $\sum_{k=0}^n{ \binom{2k}{k} 2^{2n-2k} } = (2n+1) \binom{2n}{n} = \frac{n+1}{2} \binom{2(n+1)}{n+1}$
I found some combinatorial identities in my old notebooks, but I cannot recall how I derived them. Can anyone help provide the most elementary/elegant proofs for the following identity? Specifically, the connection between the summation and either of the two closed-form expressions.

$$
\sum_{k=0}^n{ \binom{2k}{k} 2^{2n-2k} } = (2n+1) \binom{2n}{n} = \frac{n+1}{2} \binom{2(n+1)}{n+1}
$$

Some specific cases written out for some simple visualization: ($n=0$ through $n=3$)
$$
\begin{align}
\binom{0}{0} 2^{0} &= 1\binom{0}{0} = \frac{1}{2}\binom{2}{1} \\
\binom{0}{0} 2^{2} + \binom{2}{1} 2^{0} &= 3\binom{2}{1}  = \frac{2}{2}\binom{4}{2} \\
\binom{0}{0} 2^{4} + \binom{2}{1} 2^{2} + \binom{4}{2} 2^{0} &= 5\binom{4}{2} = \frac{3}{2}\binom{6}{3}\\
\binom{0}{0} 2^{6} + \binom{2}{1} 2^{4} + \binom{4}{2} 2^{2} + \binom{6}{3} 2^{0} &= 7\binom{6}{3} = \frac{4}{2}\binom{8}{4}
\end{align}
$$

EDIT: I should specify that I have already found an inductive proof, but it is not satisfying to me. I am specifically looking for a bijective/double-counting proof ideally.

答:
A Couple of "Common" Series

We have the Geometric Series
$$
\sum_{k=0}^\infty 4^kx^k=\frac1{1-4x}\tag1
$$
and the Generating Function for the Central Binomial Coefficients
$$
\sum_{k=0}^\infty\binom{2k}{k}x^k=\frac1{\sqrt{1-4x}}\tag2
$$
Equation $(2)$ is the Generalized Binomial Theorem applied to $\binom{-1/2}{k}=\left(-\frac14\right)^k\binom{2k}{k}$, which is shown in this answer.

Using the Cauchy Product

We get the product of the preceding functions by taking half the derivative of $(2)$ and adjusting the index:
$$
\sum_{k=0}^\infty\frac{k+1}2\binom{2k+2}{k+1}x^k=\frac1{\sqrt{1-4x}^3}\tag3
$$
The Cauchy Product Formula then says that
$$
\begin{align}
\sum_{k=0}^n\binom{2k}{k}4^{n-k}
&=\frac{n+1}2\binom{2n+2}{n+1}\tag4\\[3pt]
&=(2n+1)\binom{2n}{n}\tag5
\end{align}
$$


问:
Difficulty with proving tautologies
I have to prove a couple tautologies for a discrete math assignment but I'm stuck.

I've gotten this far:

$$\begin{align}& (x \to y) \wedge (x \to \neg y) \to \neg x
\\ = ~& (\neg x \vee y) \wedge ( x \to \neg y) \to \neg x &&
\tag{translate $\to$}
\\
=~&(\neg x \vee y) \wedge ( \neg x \vee \neg y) \to \neg x \tag{translate $\to$}\end{align}$$

..and I'm not sure where to go next. I think I have to use the distributive property to rearrange the problem, but I'm not entirely sure how.

Help?

答:
I've gotten this far: (ed: I've added the red brackets to show the implicit order of operations.)

$$\begin{align}& \color{crimson}((x \to y) \wedge (x \to \neg y)\color{crimson}) ~\to~ \neg x\\ = ~& \color{crimson}((\neg x \vee y) \wedge ( x \to \neg y)\color{crimson}) ~\to~ \neg x \tag{translate $\to$} \\ =~&\color{crimson}((\neg x \vee y) \wedge ( \neg x \vee \neg y)\color{crimson}) ~\to~ \neg x \tag{translate $\to$}\end{align}$$

..and I'm not sure where to go next. I think I have to use the distributive property to rearrange the problem, but I'm not entirely sure how.

Just distribute out the common disjunct in the antecedent of the conjunction. $$\begin{align}=~&\color{crimson}(\lnot x\lor(y\land\lnot y)\color{crimson})~\to~\lnot x\tag{distribution}\end{align}$$

The rest should be clear.

Title: Difficulty with proving tautologies


问:
What are some examples of function that approach a max but never exceed it?
I'm working on a project where I need to record and then score a number of problem areas for different organizations. Imagine an inspection or something where the number of "minor problems" is recorded, then a score up to a max I can set is assigned based on that number of problems.

This is not my image, but it's similar to what I need. I would like to be able to define the max Y value (even as a decimal), the x value at which the Y value reaches max, and if possible, the "flatness" of the curve so that I can refine how much the first few "problems" affect the score.

(I see that this example appears to be asymptotic to 100, I would like this one to actually reach that max at a value I can define)

Any help would be awesome! Thanks so much.

答:
What you are looking for is the Poisson distribution, which models how many times an event (in this case, minor problems) occurs in a given timeframe.

The cumulative distribution function for this is given by:

$$e^{-r} \sum_{i = 0}^{k} \frac{r^i}{i!}$$

where $r$ is the mean / average number of errors in the given timeframe, and this will give you the probability that $k$ errors or fewer will occur. You should be able to implement this using code.

Note that probabilities always range from $0$ to $1$, so multiply this by the maximum score.


答:
$11n - 5 \equiv 1\ (\textrm{mod}\ 2)$

$11n \equiv 0\ (\textrm{mod}\ 2)$

Since $11 \equiv 1\ (\textrm{mod}\ 2)$,

$11n \equiv 1 * n \equiv n \equiv 0\ (\textrm{mod}\ 2)$.


答:
Ideas: take the partition $\,P_n\,$ of the unit interval

$$P_n:=\left\{0=x_0\,,\,x_1=\frac1n\,,\ldots,\,x_k=\frac kn\,,\ldots,\,x_n=\frac nn=1\right\}$$

and choose points

$$c_i:=\frac in\;,\;\;1\le i\le n$$

Since we already know  that $\,\sin \pi x\,$ is continuous everywhere and Riemann integrable in any closed, finite interval, we can apply the Riemann integral definition for the particular partitions and particular points $\,c_i\,$ as above in each subinterval, and get:

$$\lim_{n\to\infty}\frac 1n\sum_{k=1}^n \sin\frac{\pi k}n=\int\limits_0^1 \sin\pi x\,dx=\ldots\ldots$$

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

It seems to me like a Riemann integral definition:

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

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

How do i proceed from this point?


问:
Proving $a\cos x+b\sin x=\sqrt{a^2+b^2}\cos(x-\alpha)$

Show that $a\cos x+b\sin x=\sqrt{a^2+b^2}\cos(x-\alpha)$ and find the correct phase angle $\alpha$.

This is my proof.

Let $x$ and $\alpha$ be the the angles in a right triangle with sides $a$, $b$ and $c$, as shown in the figure. Then, $c=\sqrt{a^2+b^2}$. The left-hand side is $a\cos x+b\sin x=\frac{ab}{c}+\frac{ab}{c}=2\frac{ab}{c}$. The right-hand side is $\sqrt{a^2+b^2}\cos(x-\alpha)=c\left(\cos{x}\cos{\alpha}+\sin{x}\sin{\alpha}\right)=c\left(\frac{ab}{c^2}+\frac{ab}{c^2}\right)=2\frac{ab}{c}$.

Is my proof valid? Is there a more general way to prove it?

For the second part of the question, I think it should be $\alpha=\arccos\frac{a}{\sqrt{a^2+b^2}}=\arcsin\frac{b}{\sqrt{a^2+b^2}}$. Is this correct?

答:
We can prove this by using properties of complex numbers:
$$ \begin{align}a\cos(x) + b\sin(x) &= a \cos(x) + b \cos\left(x-\frac{\pi}{2}\right)\\
  &=a \operatorname{Re}\left\{e^{ix}\right\} + b \operatorname{Re}\left\{e^{i(x-\pi/2)}\right\}\\
  &= \operatorname{Re}\left\{ae^{ix} + be^{i(x-\pi/2)}\right\}\\
  &= \operatorname{Re}\left\{e^{ix}(a-bi) \right\} \\
  &= \operatorname{Re}\left\{e^{ix}\sqrt{a^2+b^2}e^{i\cdot\operatorname{atan2}(-b,a)} \right\}\\
  &= \operatorname{Re}\left\{\sqrt{a^2+b^2}e^{i(x+\operatorname{atan2}(-b,a))}\right\}\\
  &= \sqrt{a^2+b^2} \cos(x+\operatorname{atan2}(-b,a))
\end{align}
$$

where $\operatorname{atan2}(y_0,x_0)$ is the angle of the complex number $x_0+iy_0$.

This formula works for all real numbers $a$ and $b$ except when $a=b=0$.

Generalization

Following the same procedure, we can find a general formula for the sum of two sinusoids with same frequency but (possibly) different phases:

$$ \begin{align}a\cos(x+\alpha) + b \cos(x+\beta) &= \operatorname{Re}\left\{ae^{i(x+\alpha)} + be^{i(x+\beta)}\right\}\\
&= \operatorname{Re}\left\{e^{ix}\left(ae^{i\alpha}+be^{i\beta}\right) \right\}\\
&= \operatorname{Re}\left\{e^{ix}ce^{i\gamma} \right\}\\
&= \operatorname{Re}\left\{ce^{i(x+\gamma)} \right\}\\
&= c \cos(x+\gamma)
\end{align}
$$

Where $ce^{i\gamma} = ae^{i\alpha}+be^{i\beta}$.

We have
$$\begin{align}ce^{i\gamma}&=a (\cos(\alpha)+i\sin(\alpha)) +b (\cos(\beta)+i\sin(\beta))\\
&= (a\cos(\alpha)+b\cos(\beta)) + (a\sin(\alpha)+b\sin(\beta))i.\\
\end{align}
$$

So
$$c=\sqrt{(a\cos(\alpha)+b\cos(\beta))^2+(a\sin(\alpha)+b\sin(\beta))^2}$$
and
$$\gamma=\operatorname{atan2}\big(a\sin(\alpha)+b\sin(\beta),a\cos(\alpha)+b\cos(\beta)\big).$$

Sidenotes:

This result is equivalent to
phasor addition.

We can get    similar formula involving $\sin$ by taking the imaginary
part.


答:
You are almost done!</p>
```<code>~x or (y and ~y) [undistribute]
~x or F [simultaneous conjunctivism]
~x [falsification disappearato]
</code>```

Title: Difficulty with proving tautologies

Body:
I have to prove a couple tautologies for a discrete math assignment but I'm stuck.

I've gotten this far:

$$\begin{align}& (x \to y) \wedge (x \to \neg y) \to \neg x
\\ = ~& (\neg x \vee y) \wedge ( x \to \neg y) \to \neg x &&
\tag{translate $\to$}
\\
=~&(\neg x \vee y) \wedge ( \neg x \vee \neg y) \to \neg x \tag{translate $\to$}\end{align}$$

..and I'm not sure where to go next. I think I have to use the distributive property to rearrange the problem, but I'm not entirely sure how.

Help?

Body:
You are almost done!

~x or (y and ~y) [undistribute]
~x or F [simultaneous conjunctivism]
~x [falsification disappearato]


问:
Intuition of the meaning of homology groups
I am studying homology groups and I am looking to try and develop, if possible, a little more intuition about what they actually mean. I've only been studying homology for a short while, so if possible I would prefer it if this could be kept relatively simple, but I imagine it is entirely possible there is no real answer to my query anyway.

As I said above, I want to gain a little deeper understanding of what the n-th homology group actually means: I can happily calculate away using Mayer-Vietoris but it doesn't really give me a great deal of intuition about what the n-th homology group actually means. For example, with homotopy groups, the fundamental group is in some sense a description of how loops behave on the object in question, and it is obvious to me why that is what it is for say, the torus or the circle. However, I have no idea what, if anything, I am actually saying about a triangulable object when I talk about it having 0-th homology group this or 1st homology group that.

The best I have been able to find online or in my limited book selection is the brief description "intuitively, the zeroth homology group counts how many disjoint pieces make up the shape and gives that many copies of $\Bbb Z$, while the other homology groups count different types of holes". What 'different types of holes' are there, roughly speaking? I appreciate that it may often be completely non-obvious what the low-order homology groups are for some complicated construction, but perhaps in simpler examples it might be more explicable. Are there (simple) cases where I could say, just from looking something like e.g. the torus, what its zero-th or first or second etc. homology group was based on the nature of the object? I guess in the zero-th case it is, as my source (http://teamikaria.com/hddb/wiki/Homology_groups) above says, related to the number of disjoint pieces. Can we delve deeper than this for the other homology groups?

Any book/website suggestions would be welcomed (preferably websites as I am nowhere near a library!) - I have Hatcher but not a great deal else, and I haven't gleaned as much as I wish to from that alone. Of course I know that there is a great deal we don't know about homology groups even today, so I don't expect some magical all-encompassing answer, but any thoughts you could provide would be appreciated. I hope this question is appropriate for SE Mathematics, apologies if not! -M

答:
"Cycles modulo boundaries" can get you surprisingly far. Recall that for a simplicial complex $X$, the $n^{th}$ homology $H_n(X)$ is $Z_n/B_n$, where $Z_n = \ker(d_n : C_n \to C_{n-1})$ is the group of cycles and $B_n = \text{im}(d_{n+1} : C_{n+1} \to C_n)$ is the group of boundaries. Some low-dimensional cases:

When $n = 0$, a cycle is a linear combination of $0$-simplices in $X$, and a boundary is a linear combination of $0$-simplices lying in the same connected component of $X$ such that the sum of their coefficients is zero. So $Z_0/B_0$ is precisely the free abelian group on the connected components of $X$.

When $n = 1$, a cycle is exactly what it sounds like: a linear combination of cycles in $X$ (closed paths made out of $1$-simplices). A boundary is also exactly what it sounds like: a linear combination of cycles in $X$ that bound $2$-simplices. So $Z_1/B_1$ describes the failure of $1$-cycles in $X$ to bound $2$-simplices (which is the precise sense in which it measures "$1$-dimensional holes").

For $n \ge 2$ I have trouble concisely describing what a cycle is beyond "a linear combination of $n$-simplices with zero boundary." I believe this is roughly like a linear combination of collections of $n$-simplices which together form an $n$-sphere in $X$, at least for sufficiently nice triangulations. A boundary is a linear combination of such things which bound $n+1$-simplices. So again $Z_n/B_n$ measures the failure of $n$-cycles to bound $n+1$-simplices, which is the precise sense in which it measures "$n$-dimensional holes."

$H_0$ and $H_1$ are probably easier to identify than the others in general, since connected components are intuitive and $H_1$ is just the abelianization of the fundamental group. If $X$ is a connected $n$-manifold, then $H_n \cong \mathbb{Z}$ if $X$ is orientable and $0$ otherwise, the idea being that an $n$-cycle has to involve all of the $n$-simplices in $X$ appropriately oriented so that their boundaries cancel, and such a linear combination is unique up to scalar multiplication and equivalent to providing an orientation for $X$. And if $X$ is a compact orientable $n$-manifold, then Poincaré duality indirectly relates $H_1$ to $H_{n-1}$.


问:
Connection between the $S_{mn}$-theorem and Gödel universal functions
I was wondering what is the connection between a Gödel universal function (defined below) and the S-m-n theorem (stated below)? I'm almost sure there must be some connection, but I can't figure out the details. Probably there's a connection between the $s$ in the definition of a Gödel universal function and the program $s_n^m$ from the statement of the S-m-n theorem?

Some textbooks use exclusively Gödel universal functions without ever mentioning the S-m-n theorem; other textbooks don't talk about Gödel universal functions but use the S-m-n theorem in those places where the textbooks of the first type would have used Gödel universal functions (e.g. in $m$-reducing one set to another). That's the reason I think there should be a connection.

A partial function $U: N\times N\to N$ is said to be a universal function for the class of computable functions of one argument if $U$ is computable and for any partial computable function of one argument $f$, there is $n\in N$ such that $U(n,x)=f(x)$ for all $x\in N$.

A Gödel universal function is a universal function $U$ with the following property: if $V:N\times N\to N$ is any partial computable function, then there exists a total computable $s:N\to N$ such that for all $x,n\in N$ one has $V(n,x)=U(s(n),x)$.

S-m-n theorem. There exists a natural number $s^m_n$ with the following property: for all natural $x,y_1,\dots,y_m,z_1,\
\dots, z_n$, $$[\![ [\![s^m_n ]\!](x,y_1,\dots,y_m) ]\!](z_1,\dots,z_n)=[\![x]\!](y_1,\dots,y_m,z_1,\dots,z_n).$$
(The notation $[\![ a ]\!](b)$, also written as $\varphi_a(b)$, stands for the result of the application of the program with number $a$ (in some computable (initailly fixed) enumeration of programs) to input $b$.)

Also see this question for another statement of the S-m-n theorem.

答:
Feel free to correct/criticize/improve this answer.

Alright, I think I got it, at least most of it. I think the S-m-n theorem (for $m=n=1$) can be used to prove the existence of Gödel universal functions.

From what I understand, $U(x,y)$, $\varphi_x(y)$, $[\![ x ]\!](y)$ are various notations for the same thing, so I'll use them interchangeably.

Let $U$ be a universal function (with respect to some fixed computable enumeration of programs). Let $V: N\times N \to N$ be a partial computable function. Then there exists a program that computes it, say it has number $e$ in the fixed numbering mentioned above. So $$V(x,y)=[\![e]\!](x,y)$$ for all $x,y$. Let's construct the function $s$ from the definition of a Gödel universal function. In the case $m=n=1$, the S-m-n theorem says $$[\![ [\![s^1_1 ]\!](e,x) ]\!](y)=[\![e]\!](x,y).$$
Define $s:N\to N$ by $x\mapsto [\![s^1_1 ]\!](e,x) $. Then we have $$[\![s(x)]\!](y)=[\![e]\!](x,y) $$ or in other notation $$U(s(x),y)=V(x,y).$$

The only thing that's unclear to me is why this $s:x\mapsto [\![s^1_1 ]\!](e,x) $ is total.


问:
Deducing $\sin x + \cos x \ge 1$ from $(\sin x + \cos x)^2 = 1 + 2 \sin x \cos x$
So in trig, say I have an acute angle $X$. And one can intuitively conclude that $\sin x + \cos x \ge 1$, but how does the fact that
$$(\sin x + \cos x)^2 = 1 + 2 \sin x \cos x$$
tell me that it is true that $\sin x + \cos x \ge 1$? I don't quite see the connection.

Thanks

答:
Let's use $\theta$ instead of $X$.

We are assuming that $0 < \theta < 90^\circ$. If we don't assume this, many of the things I say are not true.

Let $y = \sin \theta$ and $x = \cos \theta$

Now draw right triangle $ABC$ with

$m\angle B = \theta$

$x = BC$

$y = AC$

$1 = AB$

There is a theorem that says that the sum of the lengths of any two sides of a triangle exceeds the length of the third side. So you will always have $x+y> 1$. That is to say,
$\sin \theta + \cos \theta > 1$ for all $\theta$ in the first quadrant.

So you don't care if $(\sin \theta + \cos \theta)^2 = 1 + 2 \sin \theta \cos \theta$ because, if $\theta$ is an acute angle, then $\sin \theta + \cos \theta > 1$


问:
Tangent and normal space of the Stiefel manifold
Let $V_{n,p}$ be the real Stiefel manifold with parameters $n$ and $p$. Let $X \in V_{n,p}$ and $Y(t)$ curve on $V_{n,p}$ with $Y(0) = X$. Then, $\dot{Y}(0)$ is a tangent vector to $V_{n,p}$ at $X$.  With the standard knowledge on the Stiefel manifold. We know the following equation holds
\begin{equation}
\dot{Y}(0)^{T} X + X^{T}  \dot{Y}(0) = 0
\end{equation}

The projections, on the tangent space and the normal space, of a random vector $Z \in \mathbb{R}^{n\times p}$ at $X$  are respectively given by
\begin{equation}
\begin{aligned}
&T\doteq \dot{Y}(0) = X\operatorname{skew}(X^{T} Z) + (I - XX^{T}) Z\\
&N= X\operatorname{sym}(X^{T}Z)
\end{aligned}
\end{equation}
where
\begin{equation}
\begin{aligned}
&\operatorname{sym}(A) \doteq (A + A^{T})/2 \\
&\operatorname{skew}(A) \doteq (A - A^{T})/2
\end{aligned}
\end{equation}

The above equations are well-understood.
However, the above equations need to be modified to extend the above results to the complex manifold.

My numerical code shows there is something wrong to extend the above results to the complex case by simply replacing transposes of real matrices to conjugate transposes of complex matrices.

The third assertion (line 15) can not be passed, and it means the tangent vector and norm vector are not orthogonal.

Can anyone provide helps to extend the results to a complex manifold?  Thanks.

A reference to the real Stiefel manifold can be found in the paper

Edelman, A., Arias, T.A., Smith, S.T.: The Geometry of Algorithms with Orthogonality Constraints. SIAM Journal on Matrix Analysis and Applications 20(2), 303--353 (1998), Section 2.2

PDF: https://arxiv.org/pdf/physics/9806030.pdf

Big thanks for any help.

PS: Why I am interested in the complex Stiefel manifold? Because I think I can generalize the notion of Stiefel manifold over a novel commutative algebra called t-algebra. I need a complex manifold to generalize the Stiefel manifold over
T-Algebra.

答:
Some told me the Steifel manifold can never be complex. Even embedded in a complex space $\mathbb{C}^{nk}$, the Steifel manifold $V_{n, p}(\mathbb{C}) = \{X \in \mathbb{C}^{n\times p} | X^{*}X = I_p \}$ is still real rather than complex.

I finally accept the above result as accurate. It is interesting to know that the Stiefel manifold can never be complex.

Please correct me if I am wrong.


问:
Range and inverse of $f(x)=\sqrt{4x^2 + 1}$ on $[0,\infty)$

Consider the function $f(x) = \sqrt{4x^2 + 1}$ for $x \in [0,\infty)$. Denote the range of $f$ by $Y$. There is a function $g$ with domain $Y$ so that $g(f(x)) = x$
for every $x \in [0, \infty)$ and  $f(g(y)) = y$ for every $y \in Y$. Find an expression for $g(y)$

Also, how can I go about finding the range of $f$?

Any pointers would be much appreciated :).

答:
Such a $g$ is know as the inverse of $f$, usually denoted $f^{-1}$. To find it, set $x=f(y)$ and rearrange to find $y$ in terms of $x$.

The domain of $f$ ($^*f^{-1}$) is equal to the range of $f^{-1}$ ($^*f$)


问:
Finding all integer solutions to the Diophantine Equation: $ax+by=c$
I am not sure how to prove this theorem.

It states: Let $a,b,c$ be integers with $a$ and $b$ both no equal to $0$. If $x=x_0$ and $y=y_0$ is an integer solution to the equation $ax+by=c$, that is, $ax_0+by_0=c$, then for every integer $k$, the numbers $$x=x_0+\frac{kb}{(a,b)}\,\,\,\,\&\,\,\, y=y_0-\frac{ka}{(a,b)}$$ are integers that also satisfy the linear Diophantine equation $ax+by=c$. Moreover, every solution to the linear Diophantine equation $ax+by$ is of this form.

I understand how all solutions for a particular equation can be found using this equation, but I do not know how I can go about proving this theorem.

Example: Find all solutions to the equation $24x+9y=33$. The obvious solution to the equation is $x_0=1$ and $y_0=1$. Then, $x=1+\frac{9k}{3}$ and $y=1-\frac{24k}{3}$, so we have, $x=1+3k$ and $y=1-8k$ hence for all integer values of $k$, we have corresponding $x$ and $y$ values which satisfy the equation.

答:
Suppose that we have a solution of $ax+by=c$ given by $(x_0, y_0)$.
Let us assume that all the solutions are of the form $(x_k, y_k)=(x_0+s_k, y_0+t_k)$. Our goal is to find $s_k$ and $t_k$.

Since $(x_k, y_k)$ satisfy the original equation, we have:

$ax_k+by_k=c$

$=> ax_0 + as_k + by_0+bt_k=c$

$ => as_k +bt_k=0$  $(\because ax_0+by_0=c)$

Now, we have:

$ s_k = \frac{-bt_k}{a}$

Consider the $\frac{-bt_k}{a}$. For integer solutions, $a$ must divide $-bt_k$. Now let the irreducible of $\frac{b}{a}$ be given by $\frac{p}{q}$. Then we know that $p=\frac{b}{gcd(a,b)}$ and $q=\frac{a}{gcd(a,b)}$. Therefore, we are now required to solve for $\frac{-pt_k}{q}$.

As we know that $gcd(p,q)=1$ (since the fraction is irreducible), $q$ must divide $t_k$.

And so, $t_k= k q = \frac{ka}{gcd(a,b)}$

Substituting, $t_k$ we have:

$s_k = \frac{-pt_k}{q} =  \frac{-p(kq)}{q} = -pk = \frac{-kb}{gcd(a,b)}$

So, the final solution is given by:

$(x_k, y_k) = (x_k, y_k)=(x_0+s_k, y_0+t_k) = (x_0+ \frac{-kb}{gcd(a,b)}, y_0 +\frac{ka}{gcd(a,b)})$


问:
Inverse of the Third Axiom in the Definition of a Sigma Algebra
$\Sigma$ is a $\sigma$-algebra if it satisfies the following three properties:

$X$ is in $\Sigma$, and $X$ is considered to be the universal set in
the following context.

$\Sigma$ is closed under complementation: If $A$ is in $\Sigma$, then
so is its complement, $A^{c}=X-A$.

$\Sigma$ is closed under countable unions: If $A_{1}$, $A_{2}$,
$A_{3}$, ... are in $\Sigma$, then so is $A=A_{1}\cup A_{2}\cup
   A_{3}\cup\ldots$.

If the above three are our definitions.

Can we show that if  $A=A_{1}\cup A_{2}\cup A_{3}\cup\ldots$ and $A$ is in $\Sigma$ then each of the $A_{i}$ are in $\Sigma$?

Do we need any additional conditions to show this or would the three basic axioms suffice?

Notes:

I found some related discussions, but am unable to see if my doubts can be cleared using the pointers from these threads. Listing them below for completeness:

Can every member of a $\sigma$-algebra be represented by a countable union of disjoint members?

Show that any element of a sigma algebra is the union of disjoint sets

I was trying to understand the concept in the question asked since the corresponding property seemed to be required to show that If $E$ has $\sigma$-finite measure, then $E$ is inner regular (Rudin RCA, Definition 2.16, some related threads below):

If $E$ has $\sigma$-finite measure, then $E$ is inner regular

Rudin Real And Complex Definition 2.16

This point is clarified: If $E$ is in a sigma algebra $\Sigma$ and $E$ has $\sigma$-finite measure then it is a countable union of sets, $E_{i}$ with finite measure ($\mu\left(E_{i}\right)<\infty$). Which means the sets that make up the union of $E$ are already in $\Sigma$ since the measure $\mu$ is defined only on the members of the sigma algebra $\Sigma$. Hence the $E_{i}$ belong to the sigma algebra, $\Sigma$

Though my original doubt is clarified, I am still curios about the conditions under which the reverse of the third axiom hold. It is quite possible that I am missing something very basic. Happy to delete this question, if that is the case once my doubt is clarified.

答:
In the proof you're confused by, the $\sigma$-finite condition is used to construct a specific union with this property. This does not mean that an arbitrary union needs this property.

If $(X, \Sigma)$ is a measurable space, a measure $\mu$ is called $\sigma$-finite if there is some way to write $X = \bigcup_{n=1}^\infty X_n$ where $X_n \in \Sigma$ and $\mu(X_n) < \infty$ for all $n$.

It follows from this definition that for every $A \in \Sigma$, we can write $A = \bigcup_{n=1}^\infty A_n$ where $A_n \in \Sigma$ and $\mu(A_n) < \infty$ for all $n$. Just take $A_n = A \cap X_n$. (And this is what we mean by the set $A$ being $\sigma$-finite.)

Note that the definition specifically promises us some measurable sets. We don't need the much stronger and usually false property that every way of writing $A$ as a union $\bigcup_{n=1}^\infty A_n$ will result in $A_n \in \Sigma$ for all $n$.


问:
Laurent series expansion of $\frac{1}{z(e^z-1)}$
How would one go about determining the Laurent series of such a series as

$$\frac{1}{z(e^z-1)}$$

This is tricky since I know its a pole of second order at z = 0 but just unsure what to do past that point. I tried expanding the exponential but it just gives me an infinite series that I do not know how to deal with.

答:
Apply the Taylor series for $e^z$

$$\frac{1}{z(e^z-1)} = \frac{1}{z^2\left(1+\frac{z}{2}+\frac{z^2}{6}+\cdots\right)} = \frac{1}{z^2}\left(1-\frac{z}{2}-\frac{z^2}{6}-\cdots+\left(\frac{z}{2}+\cdots\right)^2-\cdots\right)$$

$$= \frac{1}{z^2}\left(1-\frac{z}{2}-\frac{z^2}{6}+\frac{z^2}{4}+O(z^3)\right) = \frac{1}{z^2}-\frac{1}{2z}-\frac{1}{12}+O(z)$$

by the geometric series formula for $\frac{1}{1+z}$


问:
the definition of a residue field in a scheme as a functor
We define a functorial scheme as in "Two functorial definitions of schemes". We would like to define a residue field of a scheme. The first idea I came up with was to use a notion of a structure sheaf. As is well known, the one of the way to define a residue field is to use the stalk of a structure sheaf. A structure sheaf can be defined as this MSE answer in a functorial scheme. However, to get the stalk require a set theoretical point, and so we don't know set theoretical points of a functor. This is why we can't get the stalk of a structure sheaf. Can we define the definition of a residue field in a scheme as a functor?</p>

答: If $X$ is a scheme and $p\in X(k)$ is a $k$-point for some field $k$, you can define the residue field at $p$ as the smallest subfield $k_0\subseteq k$ such that $p$ lifts to a point of $X(k_0)$ (with respect to the inclusion map $k_0\to k$).


问:
what is the group $\langle a,b,c,d,e\mid cde=1\rangle$?
As title, I'm doing a problem in algeraic topology, and met this  group, I need to figure out what it is. I met similar group such as $\langle a,b,c\mid c=1\rangle $ and $\langle a,b,c,d\mid cd=1\rangle$, I believe these two are both free groups, but not sure what happens when adding another generator.

答: As already pointed out by ahulpke, all you can say is that your group is a free group in 4 generators say $a,b,c,d$ and the fifth element namely $e$ can be written as a word in the form : $e=d^{-1}c^{-1}$ . In this case drawing a Cayley graph is easy:


答:
By Taylor
$$z(e^z-1)=z^2+\frac{z^3}{2}+\frac{z^4}{6}+\frac{z^5}{24}+\frac{z^6}{120}+\frac{z^7}{720}+\frac{z^8}{5040}+O\left(z^9\right)$$ Using the long division
$$\frac 1{z(e^z-1)}=\frac{1}{z^2}-\frac{1}{2
   z}+\frac{1}{12}-\frac{z^2}{720}+\frac{z^4}{30240}+O\left(z^5\right)$$

If you want the infinite series, just use the fact that
$$\frac 1{e^z-1}=\sum_{m=0}^\infty \frac{B_m }{m!}z^{m-1}$$ and divide by $z$.

Title: Laurent series expansion of $\frac{1}{z(e^z-1)}$

Body:
How would one go about determining the Laurent series of such a series as

$$\frac{1}{z(e^z-1)}$$

This is tricky since I know its a pole of second order at z = 0 but just unsure what to do past that point. I tried expanding the exponential but it just gives me an infinite series that I do not know how to deal with.


问:
The center of the circumcircle lies on a side of a triangle
Consider a triangle $ABC$. Let the angle bisector of angle $A$ be $AP,P\in BC$. $BP=16,CP=20$ and the center of the circumcircle of $\triangle ABP$ lies on the segment $AC$. Find $AB$.

 $$AB=\dfrac{144\sqrt5}{5}$$

By Triangle-Angle-Bisector Theorem $$\dfrac{BP}{PC}=\dfrac{AB}{AC}=\dfrac{16}{20}=\dfrac{4}{5}\\ \Rightarrow AB=4x, AC=5x.$$ The cosine rule on $ABC$ gives $$BC^2=AB^2+AC^2-2\cdot AB\cdot AC\cdot\cos\alpha \\ \iff 1296=41x^2-40x^2\cos\alpha,$$ where $\measuredangle A=\alpha.$ Is any of this helpful for the solution? Any help would be appreciated. Thank you in advance!

答:
Alternative solution using neither trigonometry nor power of a point:

You already figured out that $OP \parallel AB$, so draw two perpendicular lines $OQ$ and $BR$.

Let the radii $OA = OB = OP = 10x$.

Since $\triangle ABC \sim \triangle OPC$, we have $AB = \frac{20+16}{20} \cdot OP = 18x$, so $AQ = BQ = 9x$.

Therefore, $OQ = \sqrt{OA^2 - AQ^2} = \sqrt{19} x$, so $BR = \sqrt{19} x$.

$RP = OP - OR = 10x - 9x = x$.

$BP = \sqrt{BR^2 + RP^2} = 2 \sqrt{5} x$.

Given that $BP = 16$, we conclude that $x = \frac{16}{2\sqrt{5}} = \frac85 \sqrt{5}$.

Finally, $AB = 18 \cdot \frac85 \sqrt{5} = \frac{144}{5} \sqrt{5}$.


问:
Sum of rational numbers in (0,1)
Just out of curiousity , will the sum of all numbers belongs to rational numbers intercept open interval (0,1) equal to something?</p>

答:
Among the rational numbers between $0$ and $1$ are infinitely many number greater than $1/2.$

Therefore the sum is at least as big as what you get from adding up $1/2$ infinitely many times. So it is $+\infty.$

Title: Sum of rational numbers in (0,1)


问:
Prove existence of inverse of a bounded linear operator
Exercise 5.17 (Hunter and Nachtergaele) Suppose that $K: X \rightarrow X$ is a bounded linear operator on a Banach space $X$ with $\|K\|<1 .$ Prove that $I-K$ is invertible and
$$
(I-K)^{-1}=I+K+K^{2}+K^{3}+\ldots
$$
where the series on the right hand side converges uniformly in $\mathcal{B}(X)$.

My approach: I think we should use open mapping theorem in some form because it talks about inverses? But then we need to show the inverse exists instead of saying something about boundedness of the inverse, which is what open mapping theorem does, so I'm a bit confused.

答: Just verify that the series $\sum_n K^{n}x$ converges for each $x$ using completeness of $X$ and the fact that $\|K^{n}x\| \leq \|K\|^{n} \|x\|$. Then define $Tx =\sum_n K^{n}x$. Uniform Boundeness Principle gurantees that $T$ is a bounded operator: Apply the theorem to the sequence of operators $(T_n)$ where $T_n (x)= \sum\limits_{i=1}^{n} K^{i}x$. Now it is an easy matter to verify that $(I-K)T=T(I-K)=I$ so $T$ is indeed the inverse of $I-K$.


问:
Finding the number of integer solutions, why is this wrong?
The question is to find the number of solutions such that $(x, y)$ are integers: $(x-8)(x-10)=2^y$. Here's what I did: $u(u-2)=2^y$. From the quadratic formula, $u=1+\sqrt{1+2^y}$. This is where I think this question was meant to be solved by modular arithmetic here, but I don't know it so please try not to use it in your answers. I now constructed two right angle triangles. Triangle 1 has sides $1$, $2^{y/2}$,a hypotenuse of $\sqrt{1+2^y}$, and an angle $t$ such that $\sin(t)=\dfrac {1}{\sqrt{1+2^y}}$. Triangle 2 has sides of  $1$, $w$, a hypotenuse of $u$, and an angle $t$ such that $\cos t=\dfrac{1}{u}$. Now we want $\cot(t)+\dfrac1u=1$, or $\cot(t)+\cos(t)=1$. The problem now just reduces to finding the number solutions to that equation. However, this gives me an infinite number of solutions. Why is this wrong? Thanks!

答:
Using your idea of substitution but without applying quadratic equations stuff:

$$u,y\in\Bbb Z\;,\;\;u(u-2)=2^y\iff u, u-2\;\;\text{are powers of}\;\;2, $$

and the only possibility I see is

$$u=4\,,\,u-2=2\ldots\ldots$$


问:
On tangent spaces of Steifel Manifolds
I was trying to read Edelman et al.'s 1998 paper "The Geometry of Algorithms with Orthogonality Constraints" and since I don't have any differential geometry or much linear algebra background I am stuck at few places.

This is regarding the section 2.2.1 i.e tangent and normal spaces of the Steifel manifold. Here is the excerpt:

Let $Z$ be any n-by-p  matrix. Let $sym(A)$ denote $(A + A^T)/2$ and $skew(A)$ denote $(A - A^T)/2$ , then at $Y$ ,  $\pi_N(Z) = Ysym(Y^TZ)$ defines a projection of Z onto the normal space. Similarly at $Y$ , $\pi_T(Z) = Yskew(Y^TZ) + (I - YY^T)Z$

I tried deriving $\pi_N(Z)$ as $\pi_N(Z) = tr(N^TZ) ; N = {YS}$ where $S$ is any p-by-p symmetric matrix but couldn't arrive at the result. But using that result I could easily derive $\pi_T(Z)$ because $Z = \pi_T(Z) + \pi_N(Z)$

Next, they go on to say that tangent directions $\Delta$ at $Y$ then have the general form : $\Delta = YA + Y_{\perp}B$ where $A$ is p-by-p skew symmetric , $B$ is any (n-p)-by-p matrix and $Y_{\perp}$ is any n-by-(n-p) matrix such that $YY^T + Y_{\perp}{Y_{\perp}}^T = 1$

I couldn't derive the general form of the tangents either. Can someone please help me out on this?

Thanks

答: It is apparent that $\pi_{N}(Z) \neq tr(N^{T} Z)$ since $\pi_{N}(Z)$ should be an element in the ambient space $R^{n\times p}$, $tr(N^{T} Z)$ is only a real scalar, serving as the inner product between $N$ and $Z$.

Title: On tangent spaces of Steifel Manifolds


问:
How do I work out the number of combinations of a set of items?
If I have a set of 2 characters (1, 2) its quite simple to work out the number of combinations of those 2 characters given the following rules:

Each combination must be 2 characters

The order of the combination does matter i.e. 12 is not the same as 21

A character can be used more than once e.g. 11

The answer to the above would be:

11,12,21,22 = 4 combinations

Note: the formula below cant be used because it does not take into account rule #2 or rule #3

$\frac{n!}{(n-r)!(r)!}$

Now, what if the set of available characters increased to 4 (1, 2, 3, 4) and the same rules applied

The answer would be:

11,12,13,14,21,22,23,24,31,32,33,34,41,42,43,44 = 16 combinations

I am trying to determine a formula to work out the number of combinations given:

The number of available characters (4 in my example above)

The length of each combination (2 in my example above)

I need to use the formula to work determine how many combinations of 6 characters there are from a set of 36 available characters.

Any assistance would be most appreciated. I am feeling dumb.

答:
```<code>Slots 1-6:
_  _  _  _  _  _
36 36 36 36 36 36

{a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z,0,1,2,3,4,5,6,7,8,9}
some examples:
a9ehur
fh8209
fehiof
38924t
39u99r
</code>```

36 choices for the first character, 36 choices for the second, ..., 36 choices for the 6th.

```<code>36^6=2176782336
</code>```

2 billion.

Body:
Slots 1-6:
_  _  _  _  _  _
36 36 36 36 36 36

{a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z,0,1,2,3,4,5,6,7,8,9}
some examples:
a9ehur
fh8209
fehiof
38924t
39u99r

36 choices for the first character, 36 choices for the second, ..., 36 choices for the 6th.

36^6=2176782336

2 billion.


问:
Isoceles Triangle finding angle problem

In the diagram below $AB=AC=BD=DE$ and $\angle BAC=30^o$, what is the value of $\angle CED$ in degree?

答:
Let $AB=BD=DE=AC=x$ and $\angle CED=\alpha$. In rght angled triangle AFC we have:

$$CF=x \sin(30)=\frac x 2$$

$$\Rightarrow BC=\frac{CF}{ \sin(75)}=\frac x{ 2\sin(75)}$$

$\Rightarrow CD=BD-BC=x(1-\frac1{2\sin(75)})$

In triangle ECD due to sine rule we can write:

$$\frac x{\sin(105)}=\frac{x(1-\frac1{2\sin(75)})}{\sin(\alpha)}$$

$$\sin(\alpha)=\Big(1-\frac1{2\sin(75)}\Big)\sin(105)$$

$\sin(105)=\sin(180-75)=\sin(75)$

$\Rightarrow \sin(\alpha)=\frac 12[2\sin(75)-1]\approx 0.466$

Which gives $ \alpha\approx 27.8^o$

Title: Isoceles Triangle finding angle problem


问:
Queueing System - Probability
Consider a supermarket with three cash settling boxes and only one queue created from buyers. Four people, A, B, C and D enter the supermarket at almost the same time and in that order. A, B and C go directly into service while D waits for the first avaliable cash box. Suppose that the service time for each cash box is exponentially distributed with mean $5$.

How can I answer the following question:
What is the probability D is the last one to leave?

Am I wrong if I try
P(D last one) = P(D last | cbox 1 first)*P(cbox 1 first) + P(D last | cbox 2 first)*P(cbox 2 first) + P(D last | cbox 3 first)*P(cbox 3 first)

So,

P(D last one) = P(T1 > max{T2,T3})*P(T1 < min{T2,T3}) + P(T2 > max{T1,T3})*P(T2 < min{T1,T3}) + P(T3 > max{T1,T2})*P(T3 < min{T1,T2})

And for independence:

P(D last one) = P(T1 > T2)*P(T1 > T3)*P(T1 < T2)*P(T1 < T3) + P(T2 > T1)*P(T2 > T3)*P(T2 < T1)*P(T2 < T3) + P(T3 > T1)*P(T3 > T2)*P(T3 < T1)*P(T3 < T2)

$$(\frac{1/5}{1/5+1/5})^4+(\frac{1/5}{1/5+1/5})^4+(\frac{1/5}{1/5+1/5})^4 = = 3(\frac{1}{2})^4 = 3\frac{1}{16} = \frac{3}{16}$$

答:
Setting $X_i$ the rv's of the service time of A,B,C and $Y$ the one for D you have to calculate

$$\mathbb{P}[\min(X_i)+Y>\max(X_i)] = \mathbb{P}[Y>\max(X_i)-\min(X_i)]$$

Title: Queueing System - Probability


问:
Minimization of the SSE
I'm reading the demonstration in the Book of Wackerly about the minimization of the $\mathrm{SSE}$ for the least square method but there's is a step I don't understand why an $n$ appears where I highlighted after the differentiation. I'm completely lost. Thanks for your help

\begin{align*}
\frac{\partial\mathrm{SSE}}{\partial\hat\beta_0} & =\frac{\partial\left\{\sum_{i=1}^n[y_i-(\hat\beta_0+\hat\beta_1x_i)]^2\right\}}{\partial\hat\beta_0}=-\sum_{i=1}^n2[y_i-(\hat\beta_0+\hat\beta_1x_i)]\\
& =-2\left(\sum_{i=1}^ny_i-\color{red}{n\hat\beta_0}-\hat\beta_1\sum_{i=1}^nx_i\right)=0
\end{align*}
and
\begin{align*}
\frac{\partial\mathrm{SSE}}{\partial\hat\beta_1} & =\frac{\partial\left\{\sum_{i=1}^n[y_i-(\hat\beta_0+\hat\beta_1x_i)]^2\right\}}{\partial\hat\beta_1}=-\sum_{i=1}^n2[y_i-(\hat\beta_0+\hat\beta_1x_i)]x_i\\
& =-2\left(\sum_{i=1}^nx_iy_i-\hat\beta_0\sum_{i=1}^nx_i-\hat\beta_1\sum_{i=1}^nx_i^2\right)=0.
\end{align*}

答:
$$\begin{split}-2\sum (y-(\beta_0+\beta_1 x))&=-2\left(\sum y-\beta_0\sum -\beta_1\sum x\right)\text{ distribute}\\
&=-2\left(\sum y-\left(\underbrace{\beta_0+\beta_0+...+\beta_0}_{n}\right)-\beta_1\sum x\right)\\
&=-2\left(\sum y-n\beta_0-\beta_1\sum x\right)\end{split}$$

Title: Minimization of the SSE


问:
Solve the following Linear Congruences
Solve the following Linear Congruences:
 $$x\equiv 4\pmod {12}$$
 $$x\equiv 7\pmod {21}$$
 $$x\equiv 10\pmod {15}$$</p>

答:
Hint: $ $ if $\,(n,m)=1\,$ then $\, x\equiv a \pmod{nm}\ \iff \bigg\lbrace \begin{eqnarray} x&\equiv\,& a\pmod{n} \\ x &\equiv\,& a  \pmod{m}\end{eqnarray}$

so, $ $ if $\,(n_i,m)=1\,$ then $\, x\equiv a_i \pmod{n_i\, m}\ \iff \bigg\lbrace \begin{eqnarray} x&\equiv\,& a_i\pmod{n_i} \\ x\equiv a_1\equiv\cdots&\equiv\,& a_k  \pmod{m}\end{eqnarray}$

In your case the common factor of the moduli is $\, m = 3,\,$ and $\, n_i\mid a_i.$

Title: Solve the following Linear Congruences


问:
Constructing a circle which is tangent to a line and circle and passes through a point.
I am trying to solve a geometry problem. as first step I am trying to construct it on Geogebra:

To complete the construction I need to draw a circle which passes through the point $O$ and is tangent to the line $AB$ and also tangent inside the circle which passes through $AB$ (the bigger circle).

So the center of this circle is a point which has the same distance to the line and to the point $O$ and to the circle but I don't know how I can find this point.

答:
Consider the point of tangent of the circle to be constructed and link it to $O$. It is a radius of the big circle so it is perpendicular to the tangent line. Thus this line is also the diameter of the circle to be constructed. Knowing this, we know that the radius of this circle is half that of the large circle.

Now construct a circle with half the radius and with center $O$, along with a parallel line that is that same distance away (in pink). The intersections would be the centers of the circles to be constructed.


问:
find $f \in L^2(0,1)$ such that $f = g + K(f)$
I'm preparing for my qualifying exam and the question bank had the following question under functional analysis section and I'm stuck with this problem.

Let $L^2(0,1)$ be the Hilbert space of square integrable functions on the interval $[0,1]$ with Lebesgue measure. Define $K : L^ 2 (0,1) \rightarrow L^ 2 (0,1)$ by $K(f)(t) =\int_0^t (t − s)f(s)\,ds$. Show that
$||K|| < 1$. Given $g \in L^ 2 (0,1)$ find $f \in L^ 2 (0,1)$ such that $f = g + K(f)$.

My idea for the second part of the problem: I feel like it is the kernel of some map and that the question is asking to find an element in the coset of $g$ (the coset that you get after you quotient out by the kernel).

I really have no idea how to proceed. Any help would be appreciated.

Body:
Here's the proof for the first part , I'll try for the second part.
$||Kf||_{L^2(0,1)}=(\int_{0}^1 |Kf(t)|^2 dt)^{\frac{1}{2}}=(\int_{0}^1 |\int_{0}^t (t-s)f(s)ds|^2 dt)^{\frac{1}{2}} \le \int_{0}^t (\int_{0}^1 |(t-s)f(s)|^2dt)^{\frac{1}{2}}ds$ The inequality appears by Minkowski's integral inequality.
Now the last term $=\int_{0}^t (\int_{0}^1 |(t-s)|^2dt)^{\frac{1}{2}}|f(s)|ds$ As translations give isometry on $L^2(0,1)$  and letting $id :(0,1) \to (0,1)$ be $x \mapsto x$

we have, $$\int_{0}^t \Big(\int_{0}^1 |(t-s)|^2dt \Big)^{\frac{1}{2}}|f(s)|ds$$ $$=\int_{0}^t ||id||_{L^2(0,1)}|f(s)|ds$$ $$=\frac{1}{\sqrt{3}}\int_{0}^t|f(s)|ds$$ $$\le \frac{1}{\sqrt{3}} ||f||_{L^1(0,1)}$$ $$\le \frac{1}{\sqrt{3}} ||f||_{L^2(0,1)}$$ Thus $$||Kf||_{L^2(0,1)} \le \frac{1}{\sqrt{3}} ||f||_{L^2(0,1)} \implies ||K|| <1$$


问:
Proving an absolute inequality
No answers please, just hints!

I'm asked to prove that, on $ 0\leq x < \pi$, where $n\in\mathbb N$ and $|\sin(nx)|\leq n\sin x$, that $|\sin((n+1)x)|\leq(n+1)\sin x$ suing the fact that $\sin(x+y)=\sin x \cos y+\cos x \sin y$.

What I've done so far is $\begin{align*} |\sin(nx+x)|&=|\sin(nx)\cos x + \cos(nx)\sin x| \\ &\leq |\sin(nx)\cos x|+|\cos(nx)\sin x| \\ &\leq |\sin(nx)*1|+|\cos(nx)*1| \\ &\leq n\sin x+|\cos(nx)|\end{align*}$

and here I'm stuck

答: Use induction. Use your first step to get $|\sin ((n+1)x)| \leq |\sin (nx)|+|\sin x|$ and the apply induction hypothesis for the first term. You can start the induction argument with the case $n=0$.

Title: Proving an absolute inequality


问:
How do I find the limit $\lim_{x \to 1} \frac{1}{1 - x} - \frac{3}{1 - x^3}$
How do I calculate the following limit? $$\lim_{x \to 1} \frac{1}{1 - x} - \frac{3}{1 - x^3}$$

I already transformed this into

$$\lim_{x \to 1} \frac{(1-x)(1-x)(x+2)}{(1-x)(1-x^3)} = \frac{(1-x)(x+2)}{(1-x^3)}$$

but this still has zeroes up- and downstairs. L'Hospitals rule is not allowed. Any advice?

EDIT: Sorry, I was doing confused. There is a 3 in the second numerator.

EDIT 2: Wolfram Alpha says the limit is $-1$

Body:
$$\frac{1}{1 - x} - \frac{1}{1 - x^3}=\frac1{1 - x} - \frac1{(1-x)(1+x+x^2)}$$
$$=\frac{1+x+x^2-1}{1-x^3}=\frac{x(1+x)}{(1-x)(1+x+x^2)}$$

$$\implies \lim_{ x\to1^+}\frac{1}{1 - x} - \frac{1}{1 - x^3}=+\infty$$

$$\implies \lim_{ x\to1^-}\frac{1}{1 - x} - \frac{1}{1 - x^3}=-\infty$$


问:
Finding supremum & infimum
$B=a+(2a)^{-1}: a\in \mathbb Q, 0.1≤a≤5$.

What I tried to do is work on the limits,  $0.1≤a≤5 \implies 1/10≤1/2a≤5$
than to find the supremum and infimum. I tried to find the limit for $a+(2a)^{-1}$
which was $1/10≤a+1/2a≤5+1/10$. Making the supremum and maximum to be $5+1/10$ and infimum and min to be $1/10$.

Is this correct?

答: Plotting $y = f(x) := x + \frac{1}{2x}$ for $0.1\leq x\leq5$ reveals that your answer is incorrect. Although the supremum does appear to occur at the boundary of the interval $[0.1, 5]$, the infimum can be clearly seen to occur in the interior of the interval. Since $\mathbb{Q}$ is dense in $\mathbb{R}$ you can find the infimum by applying the first derivative test for closed intervals to the function $f$.

Title: Finding supremum & infimum


问:
If $PP^*=P^*P$ and $P^2=P$ then show that $P^*=P$
To show an idempotent normal matrix is an $\textbf{orthogonal}$ projection.

Proof: If we use, spectral decomposition of normal matrices we can write, $P=U\Lambda U^* \implies P^2=U\Lambda^2U^*$. Idempotence gives $\lambda(P)\in \{0,1\}$, so that  $P^*=U\Lambda^* U^*=U\Lambda U^*=P$.

Can we do this proof without using spectral demcomposition?

答:
Here is an elementary proof (which works in infinte dimesional spaces too): $\|P^{*}x-Px\|^{2}= \langle P^{*}x, P^{*}x \rangle-2\Re \langle P^{*}x, Px\rangle+\|Px\|^{2}=2\|Px\|^{2}-2\Re \langle P^{*}x, Px\rangle$ since the $P^{*}P=PP^{*}$. Now consider two cases: $x=Py$ for some $y$ and $x=(I-P)y$ for some $y$. In both of these cases it follows easily that $\|P^{*}x-Px\|^{2}=0$ so $P^{*}x=Px$. Now use the fact that $x =Px+(x-Px)$ so the $P^{*}$ and $P$ coincide at every point $x$.

[I have used the following calculation: $\langle P^{*}x, P^{*}x \rangle =\langle PP^{*}x, x \rangle = \langle P^{*}Px, x \rangle=\langle Px, Px \rangle=\|Px\|^{2}$].


问:
Confusion about derived functors and inverse image sheaves
I am confused about the following (probably obvious) assertion.

Let $X, Y$ be topological spaces and let $f: X \to Y$ be a continuous map. Let $\mathcal{F}$ be a sheaf of abelian groups on $Y$ and let $f^{-1}Y$ be the inverse image.

In a book I am reading, it is claimed without proof that there is a natural map $R \Gamma(F) \to R\Gamma(f^{-1} F)$, where $\Gamma(-)$ is the global sections functor.

Why does this map exist? (I agree that there is a natural map $\Gamma(F) \to \lim_{V \supset f(f^{-1}(U)} F(V) \to \Gamma(f^{-1} F)$, but I don't see why we have induced maps at the level of derived functors.)

答: Let $\mathcal{F} \to \mathcal{J}^{\bullet}$ be an injective resolution in $\mathbf{Ab}(Y)$; then $f^{-1}$ gives a resolution $f^{-1}\mathcal{F} \to f^{-1}\mathcal{J}^{\bullet}$ (the $f^{-1}\mathcal{J}^{n}$ may not be injective). Let $f^{-1}\mathcal{F} \to \mathcal{I}^{\bullet}$ be an injective resolution in $\mathbf{Ab}(X)$. By "Comparison Theorem 2.3.7" of Weibel, Homological Algebra, there exists a morphism of complexes $f^{-1}\mathcal{J}^{\bullet} \to \mathcal{I}^{\bullet}$ extending the identity $f^{-1}\mathcal{F} \to f^{-1}\mathcal{F}$. Taking global sections $\Gamma(X,-)$ gives a morphism of complexes of abelian groups $\Gamma(X,f^{-1}\mathcal{J}^{\bullet}) \to \Gamma(X,\mathcal{I}^{\bullet})$. We precompose by the natural morphisms $\Gamma(Y,\mathcal{J}^{\bullet}) \to \Gamma(X,f^{-1}\mathcal{J}^{\bullet})$ in your post.


问:
Integral of Gaussian Distribution
I'm trying to solve the below integral and my idea is to substitute $x$ by $\sqrt[]{t}x$. And my final answer is $\sqrt[]{t}$. But my textbook shows that the result should be $t$. Could someone explain where I did wrong? Thank you very much.

$$\int_{-\infty}^{\infty}\frac{x^2}{\sqrt[]{2\pi}*t}exp(-\frac{x^2}{2t})dx$$

I think the original statement is like this: For a random variable with probability density for x at time t give that $f(x,t) = $
$$\frac{1}{\sqrt[]{2\pi}*t}exp(-\frac{x^2}{2t})$$
And it says that $E(x(t)^2) = t$. I'm not sure whether I misunderstand something.

答:
Your result looks correct to me

$$\frac{1}{\sqrt{t}}\int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi t}} x^2 e^{-x^2/(2t)}dx =\frac{1}{\sqrt{t}}t=\sqrt{t}   $$

Title: Integral of Gaussian Distribution


问:
Direct proof to continuity in both variables implying continuity

Say I have a real valued function $f$ defined on $\mathbb{R}^2$ such that it is continuous in each variable. Also, assume that for any compact subset $K$ of $\mathbb{R}^2$, it's image under $f$ is also compact. Then $f$ is continuous.

So I've seen an argument by contradiction, but I was hoping someone could provide a direct argument. Also, can any of the assumptions be dropped or loosened? What if it is only continuous in one variable?

Can the last assumption be reduced down to "closed sets get sent to closed sets"? Why is $f$ sending compact sets to compact sets so important?

答:
I don’t at the moment see a direct proof, but it’s not necessary to prove the result by contradiction: one can simply prove the contrapositive. Specifically, here’s a proof that if $f:\Bbb R^2\to\Bbb R$ is separately continuous in each variable but not continuous, then there is a compact (hence closed) subset $K$ of $\Bbb R^2$ such that $f[K]$ is not closed in $\Bbb R$.

Proof. By translating $f$ if necessary, we may assume that $f$ is not continuous at the origin, and that $f(0,0)=0$. For $n\in\Bbb N$ let

$$B_n=(-2^{-n},2^{-n})\times(-2^{-n},2^{-n})\,.$$

Since $f$ is not continuous at $\langle 0,0\rangle$, there is an $r>0$ such that $f[B_n]\nsubseteq(-r,r)$ for each $n\in\Bbb N$. Let

$$\mathscr{B}=\{B_n:f[(-2^{-n},2^{-n})\times\{0\}]\subseteq(-r,r)\}\,;$$

since $f$ is continuous in the first variable, $\mathscr{B}$ is a local base at the origin.

Suppose that $B_n\in\mathscr{B}$. On the one hand, $f[B_n]\nsubseteq(-r,r)$, so there is a point $\langle x_n,v_n\rangle\in B_n$ such that $f(x_n,v_n)\notin(-r,r)$, i.e., such that $|f(x_n,v_n)|\ge r$. On the other hand, $f[(-2^{-n},2^{-n})\times\{0\}]\subseteq(-r,r)$, so $f(x_n,0)\in(-r,r)$, i.e., $|f(x_n,0)|<r$. And $f$ is continuous in the second variable, so there is a $y_n$ in the closed interval with endpoints $0$ and $v_n$ such that $r(1-2^{-n})<|f(x_n,y_n)|<r$.

By passing to a subsequence and if necessary replacing $f$ by $-f$, we may assume that we have an increasing sequence $\langle n_k:k\in\Bbb N\rangle$ in $\Bbb N$ such that $B_{n_k}\in\mathscr{B}$ for each $k\in\Bbb N$, and $r(1-2^{-n_k})<f(x_{n_k},y_{n_k})<r$ for each $k\in\Bbb N$. Let

$$K=\{\langle 0,0\rangle\}\cup\{\langle x_{n_k},y_{n_k}\rangle:k\in\Bbb N\}\,;$$

$K$ is closed (and indeed compact) in $\Bbb R^2$. However, $f[K]$ is not closed in $\Bbb R$, because the sequence $\langle f(x_{n_k},y_{n_k}):k\in\Bbb N\rangle$ is a sequence in $(-r,r)$ converging to $r$, and $r\notin f[K]$. $\dashv$

You definitely cannot relax the requirement that $f$ be separately continuous in each variable. Let $f(x,y)=\lfloor x\rfloor$; $f$ is continuous in $y$, and if $K\subseteq\Bbb R^2$ is compact, then $K$ is bounded, and $f[K]$ is finite and hence compact. However, $f$ is clearly not even continuous in $x$, let alone continuous.

You may find a result of Z. Piotrowski and E. Wingler in this paper of interest. They show that if $f:X\times Y\to Z$ is separately continuous in each variable, $Y$ is locally connected, $Z$ is locally compact, and $f$ has a closed graph, then $f$ is continuous. In particular, any separately continuous function from $\Bbb R^2$ to $\Bbb R$ that has a closed graph is continuous.