问:
The closes point to a curve in space.
I am working on the following problem.

Find the point closest to the origin, of the curve of intersection of the plane $2y+4z =5$ and the cone $z^2 = 4(x^2+y^2)$

I was able to see that the intersection will be a tilted ellipse with coordinates $(x,y,4(x^2+y^2))$.

So I was thinking that I could try putting $F(x,y) = \sqrt{x^2+y^2+16(x^2+y^2)^2}$ and optimize it. But this seems to be too long a calculation and I want to somehow use vectors, such as dot products and gradients because that seems to be easier.

Can anyone help me get started ?

答:
This is a typical exercise for Lagrange multipliers theorem (see http://en.wikipedia.org/wiki/Lagrange_multiplier). You want to refolmulate the problem to the following:

Find the minimum of function $\sqrt{x^2 + y^2 + z^2}$ (that is the distance from origin, you can even use easier function x^2 +  y^2 + z^2, because the solution will be the same) on the set which is intersection of the plane $2y+4z=5$ and the cone $z^2=4(x^2+y^2)$.

I guess you know a little about Lagrange multipliers, this shouldn't be hard to compute.

Title: The closes point to a curve in space.


问:
Significance of zero people out of 80 choosing something as their favourite from a selection of 42 things
80 people were asked to choose their favourite object from 42 objects. Not one person chose a particular object.

The criteria for people choosing objects might be random or there might be a pattern so that people like and dislike the same things.

I know that the chances that zero people choose a particular object are 14% - so this to me suggests there is something unfavourable about the object. That doesn't mean though that there is an 86% chance that this hasn't happened by random and there is a bias/pattern in the selections?

How do I determine the existence of whether there is a pattern or not and quantify the faith in which the fact that no person chose an object suggest it is unfavourable?

To me it seems pretty obvious that if no people out of 80 chose something as their favourite then we can say with reasonable certainty that it is not favoured. But somebody else is telling me that no it is not enough of a sample size.

I also have the results of all the 80 peoples selections if that helps.

答:
Maybe you have in mind to test a hypothesis:

If your null hypothesis is that each of the objects has an equal chance of being chosen, then the number of people out of 80 choosing your specific object, identified in advance is $X \sim \mathsf{Binom}(80, 1/42)$ and $P(X = 0) = 0.1455,$ as you say. In R:

```<code>dbinom(0, 80, 1/42)
[1] 0.1454673
</code>```

Then you don't have evidence at the 5% level of significance that
choices are other than random.

Reasoning similarly: if you had 130 people all avoiding this particular object, then you would have significant evidence to reject the null hypothesis at the 5% level:

```<code>dbinom(0,130,1/42)
[1] 0.04360072
</code>```

There is nothing sacred about the 5% level, but it is often used.
However, few statisticians would judge something that can occur with probability 0.14 as being especially rare.

If, after all 80 have made their choices, you discover
that one or more items were not chosen by anyone, that is not at all surprising. The probability is almost 1 that
fewer than 42 items are chosen under random sampling. On average, only about 36 of the 42 items will be chosen.
(Illustrated by simulation of 100,000 iterations in R below.)

```<code>set.seed(1012)
pop = 1:42
nr.ch = replicate(10^5, length(unique(sample(pop, 80, rep=T))))
mean(nr.ch < 42)
[1] 0.99968
mean(nr.ch)
[1] 35.88572
</code>```

Even with 130 people there is still probability about 0.87 that fewer than 42 items are chosen under random sampling. (Variant of above simulation not shown.)

ContentLicense: CC BY-SA 4.0

Body:
Maybe you have in mind to test a hypothesis:

If your null hypothesis is that each of the objects has an equal chance of being chosen, then the number of people out of 80 choosing your specific object, identified in advance is $X \sim \mathsf{Binom}(80, 1/42)$ and $P(X = 0) = 0.1455,$ as you say. In R:

dbinom(0, 80, 1/42)
[1] 0.1454673

Then you don't have evidence at the 5% level of significance that
choices are other than random.

Reasoning similarly: if you had 130 people all avoiding this particular object, then you would have significant evidence to reject the null hypothesis at the 5% level:

dbinom(0,130,1/42)
[1] 0.04360072

There is nothing sacred about the 5% level, but it is often used.
However, few statisticians would judge something that can occur with probability 0.14 as being especially rare.

If, after all 80 have made their choices, you discover
that one or more items were not chosen by anyone, that is not at all surprising. The probability is almost 1 that
fewer than 42 items are chosen under random sampling. On average, only about 36 of the 42 items will be chosen.
(Illustrated by simulation of 100,000 iterations in R below.)

set.seed(1012)
pop = 1:42
nr.ch = replicate(10^5, length(unique(sample(pop, 80, rep=T))))
mean(nr.ch < 42)
[1] 0.99968
mean(nr.ch)
[1] 35.88572

Even with 130 people there is still probability about 0.87 that fewer than 42 items are chosen under random sampling. (Variant of above simulation not shown.)


问:
P-test for Improper integral
I am trying to see if this integral converges or not, so we have $$\int_{1}^{\infty}\frac{1}{x^{1+\frac{1}{x}}}$$
Using P-Test I know if $1+\frac{1}{x}>1$, then this integral converges, but could the $x$ be negative so $1+\frac{1}{x}<1$?? then the integral diverges?
Thanks

答: The integral has form $\int_{1}^{\infty} f(x) dx$, where $\frac{1}{f(x)} = x \cdot x^{\frac{1}{x}} = x \cdot e^{\frac{\ln x}{x}} = x \cdot e^{\bar{o}(1)} \sim x$. Hence $f(x) \sim \frac1{x}$. It's well known that $\int_{1}^{\infty} \frac1{x} dx$ diverges. So the intergal $\int_{1}^{\infty} f(x) dx$ diverges also.

Title: P-test for Improper integral


问:
Show that $X$ and $Y$ are homeomorphic
Let $X \subseteq \mathbb{R^n}$ and $Y \subseteq \mathbb{R^n}$ be compact such that $a \in X$ and $b \in Y$. If $X - $ {$a$} is homeomorphic to $Y - $ {$b$}, show that $X$ and $Y$ are homeomorphic.

Things I've thought of using:

$(*)$ $X \subseteq \mathbb{R^n}$. We say that $X$ is compact if it is closed and bounded.

$(*)$ Let $f:X \subseteq \mathbb{R^n} \rightarrow Y \subseteq \mathbb{R^n}$ bijective and continuous. If $f^{-1}: Y \rightarrow X$ is continuous, then it is called a homeomorphism of $X$ over $Y$.

$(*)$ Let $K \subseteq \mathbb{R^n}$ be compact. Then every continuous injective function $f: K \rightarrow \mathbb{R^n}$ is a homeomorphism on $f(K)$

Could someone help me that I don't have much idea how to proceed

答:
I guess the way to go about this is pretty straightforward.
Suppose $g: X\setminus\{a\} \longrightarrow Y\setminus\{b\}$ is a homeomorphism, and define $f:X\longrightarrow Y$ by $f = g$ on $X\setminus\{a\}$ and $f(a)= b$.

Clearly, $f$ is injective and $f(X) = Y$.
Can you show that $f$ is continuous?
If you can, then you may apply your third point to conclude that $f$ is a homeomorphism.

Title: Show that $X$ and $Y$ are homeomorphic


问:
Suppose $\mu(E\cap I)\geq C\mu(I)$ for every interval $I\subseteq [0,1]$. Show that $\mu(E)=1$.
Suppose $E\subseteq [0,1].$

$C$ is some positive number and $\mu$ is Lebesgue measure.

My approach:

Assume $\mu(E)<\frac{q}{p}<1$.

We know $E$ can be approximated by finite summation of intervals. Intervals can be approximated by summation of intervals like $[\frac{i}{2^N},\frac{i+1}{2^N}]$. Therefore, $E$ can be approximated by finite summation of intervals like $[\frac{i}{2^N},\frac{i+1}{2^N}]$.

Choose a $N$ such that there is a $\delta$-approximation.

Suppose there are $n$ intervals used in the approximation. Since $n\cdot 2^{-N}\leq \frac{q}{p}+\delta$, we have $2^N-n\geq \frac{p-q-p\delta}{p}2^N$, which means there are at least about $\frac{p-q-p\delta}{p}2^N$ intervals remained blank.

Then, since it's a $\delta$-approximation, there is some interval on which $\mu(E\cap I)\leq \frac{\delta p}{p-q-p\delta}2^{-N}$.

Then, we have $\mu(E\cap I)/\mu(I)\leq \frac{\delta p}{p-q-p\delta}\to 0$ as $\delta\to 0$, which is a contradiction.

I think this is a correct proof, but is too ugly. Is there an elegant one?

答:
There's this nice proof using Lebesgue's differentiation theorem, at least assuming $E$ is measurable:

Let $I_{\varepsilon, x} := [x - \varepsilon, x + \varepsilon]$.  We know that for almost every $x \not\in E$ we have $$\lim_{\varepsilon \rightarrow 0} \frac{\mu(E \cap I_{\varepsilon, x})}{\mu(I_{\varepsilon, x})} = 0,$$

but

$$\lim_{\varepsilon \rightarrow 0} \frac{\mu(E \cap I_{\varepsilon, x})}{\mu(I_{\varepsilon, x})} \ge C > 0$$

so every $x \in E^c$ is contained in the exceptional set (which has measure $0$).  Hence we must have that $\mu(E^c) = 0$ and therefore $\mu(E) = 1$.


问:
A polynomial has the same remainder when divided by $x+k$ or $x-k$; what is $k$?

Question

Given that $y = 3x^3 + 7x^2 - 48x + 49$ and that $y$ has the same remainder when it is divided by $x + k$ or $x - k$, find the possible values of $k$.

My attempt

Let $f(x) = 3x^3 + 7x^2 - 48x + 49$

$\text{Using Remainder Theorem,}$

\begin{align}
f(-k) &= f(k) \\
3(-k)^3  + 7(-k)^2 - 48(-k) + 49 &= 3(k)^3 + 7(k)^2 - 48(k) + 49 \\
-3k^3 - 7k^2 + 48k + 49 &= 3k^3 + 7k^2 - 48k + 49 \\
-3k^3 --3k^3 - 7k^2 - 7k^2 + 48k + 48k + 49 - 49 &= 0 \\
-6k^3 + 14k^2 + 96k &= 0 \\
\frac{-6k^3}{k} + \frac{14k^2}{k} + \frac{96k}{k} &= \frac{0}{k} \\
6k^2 + 14k - 96 &= 0
\end{align}

$\text{Comparing } 6k^2 + 14k - 96 = 0 \text{ with } ak^2 + bk + c = 0, a = 6, b = 14, c = -96$

\begin{align}
k = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a } &= \frac{ -(14) \pm \sqrt{(14)^2 - 4(6)(-96)} }{ 2(6) } \\
k &= 3 \text{ and } -5\frac{1}{3}
\end{align}

$\therefore k = 3, -5\frac{1}{3} \text{ or } 0 $

My answer is incorrect. The correct answer is $k = 0, 4 \text{ or } -4$

Body:
\begin{align}
f(-k) &= f(k) \\
3(-k)^3  + 7(-k)^2 - 48(-k) + 49 &= 3(k)^3 + 7(k)^2 - 48(k) + 49 \\
\bf-3k^3 - 7k^2 + 48k + 49 &= 3k^3 + 7k^2 - 48k + 49 \\
-3k^3 --3k^3 - 7k^2 - 7k^2 + 48k + 48k + 49 - 49 &= 0 \\
-6k^3 + 14k^2 + 96k &= 0 \\
\frac{-6k^3}{k} + \frac{14k^2}{k} + \frac{96k}{k} &= \frac{0}{k} \\
6k^2 + 14k - 96 &= 0
\end{align}

Right there. $7(-k)^2 = 7k^2$. Looking through it, there are more sign errors throughout. $-7k^2 - 7k^2$ would be $-14k^2$, not $14k^2$.

Another one: $-3k^3 - -3k^3 = 0$.

Dividing by $k$ isn't exactly a kosher operation. What if $k = 0$?


问:
Series expansion with remaining $\ln n$
I'm studying the asymptotic behavior $(n \rightarrow \infty)$ of the following formula, where $k$ is a given constant.
$$ \frac{1}{n^{k(k+1)/(2n)}(2kn−k(1+k) \ln n)^2}$$

I'm trying to do a series expansion on this equation to give the denominator a simpler form so that it is easier to make an asymptotic analysis.

I used mathematica/wolframalpha to expand the formula (the documents say Taylor expansion is used).
http://www.wolframalpha.com/input/?i=1%2F%28n%5E%28k+%28k+%2B+1%29%2F%282+n%29%29+%282+k+n+-+k+%281+%2B+k%29+Log%5Bn%5D%29%5E2%29

However in series expansion at $n \rightarrow \infty$, the result still has $\ln n$. This is actually a form I prefer, compared to the form
$$a_0 + a_1x + a_2 x^2+...$$
Does anyone see how the result may be produced? Any help is much appreciated. Thanks.

答:
Hint: $n^{k(k+1)/(2n)}=\mathrm e^{k(k+1)\log(n)/(2n)}=1+u+u^2/2+\ldots$ with $u=k(k+1)\log(n)/(2n)$.

This yields the expansion
$$
\frac1{4k^2n^2}\left(1-\frac{(k+1)(k-2)}2\frac{\log n}n+O\left(\left(\frac{\log n}n\right)^2\right)\right).
$$

Title: Series expansion with remaining $\ln n$

ContentLicense: CC BY-SA 3.0


问:
How would I prove that f is onto then gcd(m,n) = 1?
I'm having trouble getting started on this question:

Define f: ℤn → ℤn by f(x) = mx mod n. Use the Bézout’s theorem to prove that if f is onto, then gcd(m,n)=1.

答: If $f$ is onto then there is some $a$ for which $f(a) = 1$ which is to say $ma = 1 \mod n$. This means that there is some $k$ such that $ma = kn + 1$. Let $d = \gcd(m, n)$. Then $(d|m \land d|n) \implies (d|ma \land d|kn) \implies d|(ma - kn) \implies d|1 \implies d = 1$.


问:
show that $\intop_{0}^{1}\frac{\sin^{2}\left(x\right)}{x^{2}}$ diverges
I want to show that this integral diverges
$$\intop_{0}^{1}\frac{\sin^{2}\left(x\right)}{x^{2}}$$
i tried to do
$$\intop_{0}^{1}\frac{\sin^{2}\left(x\right)}{x^{2}}=\intop_{0}^{1}\frac{\sin^{2}\left(x\right)}{x^{2}}=\intop_{0}^{1}\frac{1-\cos^{2}\left(x\right)}{x^{2}}=\intop_{0}^{1}\frac{1}{x^{2}}-\intop_{0}^{1}\frac{\cos^{2}\left(x\right)}{x^{2}}$$

and thus
$$\intop_{0}^{1}\frac{1}{x^{2}}=\lim_{a\rightarrow0^{+}}\intop_{a}^{1}\frac{1}{x^{2}}=\lim_{a\rightarrow0^{+}}-\frac{1}{x}=\infty$$

but i don't sure that if $$\intop_{0}^{1}\frac{1}{x^{2}}$$
diverges so the whole expression diverges, and if it does then why?

答:
It should not diverge. $\frac{\sin^2x}{x^2}$ is bounded over $[0,1]$, which means that its integral over the same range is bounded.

When you split the integral into $\int\frac1{x^2}-\int\frac{\cos^2x}{x^2}$, both of those integrals do diverge, so you created an $\infty-\infty$ scenario, which is nonsense.


答: It doesn't diverge, in fact $\displaystyle\int_0^1 \frac{\sin^2(x)}{x^2} \approx .89734$

Title: show that $\intop_{0}^{1}\frac{\sin^{2}\left(x\right)}{x^{2}}$ diverges

Body:
I want to show that this integral diverges
$$\intop_{0}^{1}\frac{\sin^{2}\left(x\right)}{x^{2}}$$
i tried to do
$$\intop_{0}^{1}\frac{\sin^{2}\left(x\right)}{x^{2}}=\intop_{0}^{1}\frac{\sin^{2}\left(x\right)}{x^{2}}=\intop_{0}^{1}\frac{1-\cos^{2}\left(x\right)}{x^{2}}=\intop_{0}^{1}\frac{1}{x^{2}}-\intop_{0}^{1}\frac{\cos^{2}\left(x\right)}{x^{2}}$$

and thus
$$\intop_{0}^{1}\frac{1}{x^{2}}=\lim_{a\rightarrow0^{+}}\intop_{a}^{1}\frac{1}{x^{2}}=\lim_{a\rightarrow0^{+}}-\frac{1}{x}=\infty$$

but i don't sure that if $$\intop_{0}^{1}\frac{1}{x^{2}}$$
diverges so the whole expression diverges, and if it does then why?


问:
Is there a clever general way of disconnecting those sets?
Let $X$ be a normal space. For the sake of brevity, I say that set $A$ "connects" $B$ with $C$ iff there is a connected set $\ell \subseteq A$ such that $\ell \cap B \neq \emptyset \neq \ell \cap C$.

Let $A_1,A_2,B_1,B_2$ be closed sets. Let $C_1,C_2$ be open sets. Suppose that:

$A_1 \cap B_1 = \emptyset$

$A_2 \cap B_2 = \emptyset$

$C_1 \cup C_2 = X$

$C_2$ doesn't connect $A_1$ with $B_1$

$C_1$ doesn't connect $A_2$ with $B_2$

I wonder if there is some general way of using $C_1,C_2$ to construct open sets $D_1,D_2$ such that:

$D_1 \cup D_2 = X$

Neither $D_1$ nor $D_2$ connects $A_1$ with $B_1$

Neither $D_1$ nor $D_2$ connects $A_2$ with $B_2$

It's quite intuitive to me, and it looks like it can be done with some clever application of Urysohn's lemma. But loooks and intuition are often deceiving, and so far I don't feel like I have mady any real progress and my mind is completely blank.

答:
Here is a proof from stronger versions of your hypotheses (4) and (5).  Specifically, let's assume that actually $A_1\cup B_1\subseteq C_2$ and that $A_1$ and $B_1$ can be separated by clopen sets in $C_2$, and similarly for $C_1$.

Let $E_1=X\setminus C_2$ and $E_2=X\setminus C_1$, so $E_1$ and $E_2$ are disjoint closed sets that each $E_i$ is disjoint from $A_i$ and $B_i$, and $A_i$ and $B_i$ are separated by clopen sets in $X\setminus E_i$.  We may enlarge the $E_i$ to assume they are in fact zero sets (take a continuous function that separates $E_1$ from $E_2\cup A_1\cup B_1$ to enlarge $E_1$ to a $G_\delta$ set with the same properties, and then do the same to enlarge $E_2$).  Let $U_i$ and $V_i$ be complementary clopen subsets of $X\setminus E_i$ which contain $A_i$ and $B_i$, respectively.  Now we can find a continuous function $f_i:X\setminus V_i\to[0,1]$ such that $f_i$ is $1$ on $A_i$ and $f_i^{-1}(\{0\})=E_i$ (the latter being possible since $E_i$ is a zero set).  Similarly, we can find $g_i:X\setminus U_i\to[-1,0]$ such that $g_i$ is $-1$ on $B_i$ and $g_i^{-1}(\{0\})=E_i$.  By the pasting lemma, $f_i$ and $g_i$ combine to give a continuous $h_i:X\to[-1,1]$ such that $h_i$ is $1$ on $A_i$, $-1$ on $B_i$, and $h_i^{-1}(\{0\})=E_i$.

Now define $h:X\to[-1,1]^2$ by $h(x)=(h_1(x),h_2(x))$.  This $h$ maps $A_1$ and $B_1$ to the vertical sides of the square and $A_2$ and $B_2$ to the horizontal sides of the square.  Moreover, $h^{-1}(\{(0,0)\})=E_1\cap E_2=\emptyset$.  Composing with a retraction from $[-1,1]^2\setminus\{(0,0)\}$ to the boundary of the square, we get a map $X\to \partial[-1,1]^2$ which maps $A_1$ and $B_1$ to the vertical sides of the square and $A_2$ and $B_2$ to the horizontal sides.  It thus suffices to solve the problem in the special case where $X=\partial[-1,1]^2$ and the $A_i$ and $B_i$ are the sides of the square, in which case it is easy.

More generally, the final paragraph above shows that it suffices to find a non-surjective map $h:X\to[-1,1]^2$ that maps the $A_i$ and $B_i$ to the sides of the square.  However, your hypotheses (even with the additional requirement that $A_1\cup B_1\subseteq C_2$ and $A_2\cup B_2\subseteq C_1$) do not suffice to prove the existence of such an $h$, so this method cannot solve your problem in full generality.  For instance, let $$X=[-1,1]^2\times\{1,1/2,1/3,\dots\}\cup\partial[-1,1]^2\times\{0\}\subset\mathbb{R}^3,$$ and let the $A_i$ and $B_i$ be the sides of the square $\partial[-1,1]^2\times\{0\}$.  If you let $C_1$ consist of the complement of a non-vertex point from each of $A_1$ and $B_1$ and similarly for $C_2$, then this satisfies all your hypotheses, and additionally $A_1\cup B_1\subseteq C_2$ and $A_2\cup B_2\subseteq C_1$.  However, no non-surjective map $h:X\to [-1,1]^2$ mapping the $A_i$ and $B_i$ to the sides of the square exists.  Indeed, if $h$ maps the $A_i$ and $B_i$ to the sides of the square, then it must have degree $1$ when restricted to a map $\partial[-1,1]^2\times\{0\}\to[-1,1]^2\setminus\{(0,0)\}$.  It then follows that $h$ is also has degree $1$ when restricted to $\partial[-1,1]^2\times\{1/n\}\to[-1,1]^2\setminus\{(0,0)\}$ for all sufficiently large $n$.  But then any extension of this degree $1$ map to all of $[-1,1]^2\times\{1/n\}$ will have to surject onto $[-1,1]^2$, so $h$ must be surjective.

(Note though that this example is not a counterexample to your question, since the desired $D_i$ exist--you can just take them to be complements of appropriate finite subsets of $\partial[-1,1]^2\times\{0\}$.  At this point my intuition is that the answer to your question is no in general, but I haven't been able to come up with a counterexample.  Mainly, the "not connecting" hypotheses seem rather hard to use in cases like the example above where components and quasicomponents are not the same.)


问:
consequence of Integral limits same like integrating variable
I am thinking about an intuitive explanantion (e.g. suitable for people not having insights in higher mathematics) for what would happen if the limits of the integral were really the same as the integrating variable (not just a writing error or bad notation style):

Look at this thing: $g(x):= \int_{0}^x f(x)\mathrm{d}x$. Some people might say this is wrong since the limits of the integral shouldn't correspond to the integrating variable.
But can 't we say that $g(x)$ with this definition is $0$ for every concrete $x$? E.g.
$g(1)=\int_0^1 f(1) \mathrm d1 = 0$ since $\mathrm d 1= 0$.
This is because in general
$$\mathrm d \ \text{constant}=0$$ because $$\frac{\mathrm d}{\mathrm dt} \text{constant} = 0 \Rightarrow \mathrm d\ \text{constant} = 0\cdot t = 0$$

$\Rightarrow g(x)= 0$ for every concrete $x$.

Does this make sense to explain why we have to be careful about notation?

答:
No, you are misinterpreting $\int_0^x f(x)dx.$

The variable $x$ in the integrand is a dummy variable that has absolutely no relation to the $x$ that occurs as one of the limits of integration.  This means that

$$\int_0^x f(x)dx = \int_0^x f(t)dt.$$

In fact, there are two reasons why the first expression above can be construed as bad or misleading or invalid.

In your formulation, attempting to relate the $x$ variable that occurs in the limit of integration to the $x$ variable that occurs as a dummy variable in the integrand is wrong.  Therefore, a case can be made that your formulation is invalid because you are overloading (or overusing) the $x$ variable.

Even ignoring whether your formulation is invalid, it is certainly confusing because it masks that the $x$ variable that occurs in the integrand is a dummy variable.


问:
Use Lagrange multipliers to find the exact minimum value
The given function is $$f(x,y)= 5x +7y$$
and the constraint is $$g(x,y)=x^2y-10=0$$
First I took the gradient of $f$ and $g$ and got
$$f(x,y)=\left<5,7 \right>$$
$$\lambda g(x,y)=\left< 2xy, x^2\right>$$
Then I have two equations
$$5=2xy\lambda$$
$$7=x^2 \lambda$$
Solving for $\lambda$ I get
$$\lambda = \frac{5xy}{2}$$
Plugging this into one of the two equations I get
$$7 =x^2(\frac{5xy}{2})$$
Which then works out to
$$x=\sqrt{\frac{7}{\lambda}}$$
Then solving for $y$ I get
$$y= \frac{5\sqrt{7}\sqrt{\lambda}}{14\lambda}$$
And then $\lambda$ then works out to be
$$\lambda = \frac{\sqrt[3]{7}*2^{2/3}}{4}$$
Which is approximately $0.7591$ This doesn't feel right to me what with all the weird terms and powers. Did I do it right?

答:
$x^2λ = 7 \implies \lambda = \frac{7}{x^2}$ ...(i)

$2xyλ = 5$ ...(ii)

Plugging value of $\lambda$ from (i) into (ii)

$\displaystyle \frac{14y}{x} = 5 \implies x = \frac{14y}{5}$ ...(iii)

Now plugging $x$ into our constraint,

$x^2y = 10 \implies y^3 = \frac{250}{196}$

$\displaystyle y = (\frac{250}{196})^{1/3} \, \approx 1.0845$

Using (iii), $x \displaystyle = \frac{14}{5} (\frac{250}{196})^{1/3} = (28)^{1/3} \, \approx 3.0366$


问:
How do I solve inequalities where both sides are not equal to zero, and that use absolute values and fractions?
For example something like

$$\dfrac{|x-2|+3}{4-|2x+8|}\geq-5.$$

I understand how to solve inequalities where there is an absolute value, however in fraction form just confuses me.

答:
The answer by Especially Lime shows a simpler way that works in your particular case, but a general method that always works (but in certain specific cases, including this one, this method will be longer to carry out) is to solve separately in intervals where the absolute value expressions can be written without the use of absolute values.

Note that $|x-2|$ is equal to $-(x-2)$ when $x-2 \leq 0$ and $|x-2|$ is equal to $(x-2)$ when $x-2 \geq 0.$ This is just the definition of absolute value (i.e. $|u|$ equals $-u$ when $u \leq 0$ and $|u|$ equals $u$ when $u \geq 0$; now rewrite this replacing $u$ with $x-2).$ Thus, simplifying a bit, we have $|x-2|$ equal to $-x+2$ when $x \leq 2$ and equal to $x-2$ when $x \geq 2.$ That is, the non-absolute value version of $|x-2|$ changes its form at $x=2.$

In the same way, the non-absolute value version of $|2x + 8|$ changes its form where $2x + 8 = 0,$ or $x = -4.$ For $x \leq -4$ we have $|2x + 8| = -2x - 8$ and for $x \geq -4$ we have $|2x + 8| = 2x + 8.$

Combining the results in the last two paragraphs, we see that the entire fractional expression you're dealing with has three different forms of expression in three intervals, which are determined by the points $x = 2$ and $x = -4,$ the places where the insides of at least one of the absolute value expressions is equal to zero.

Interval 1: $\;x \leq -4$

Replace $|x-2|$ with $(-x+2)$ and replace $|2x + 8|$ with $(-2x - 8).$

$$\frac{(-x + 2) + 3}{4 - (-2x - 8)} \; \geq \; -5$$

$$\frac{-x + 5}{2x + 12} \; \geq \; -5$$

Solve this the usual way, and then include ONLY those values you get that ALSO satisfy $x \leq -4.$

Interval 2: $\; -4 \leq x \leq 2$

Replace $|x-2|$ with $(-x+2)$ and replace $|2x + 8|$ with $(2x + 8).$

$$\frac{(-x + 2) +3}{4 - (2x+8)} \; \geq \; -5$$

$$\frac{-x + 5}{-2x - 4} \; \geq \; -5$$

Solve this the usual way, and then include ONLY those values you get that ALSO satisfy $-4 \leq x \leq 2.$

Interval 3: $\;x \geq 2$

Replace $|x-2|$ with $(x-2)$ and replace $|2x + 8|$ with $(2x + 8).$

$$\frac{(x - 2) + 3}{4 - (2x + 8)} \; \geq \; -5$$

$$\frac{x + 1}{-2x - 4} \; \geq \; -5$$

Solve this the usual way, and then include ONLY those values you get that ALSO satisfy $x \geq 2.$

Final Solution

To obtain the final solution, combine together (i.e. take the union) the solutions in the three intervals.

Although this seems rather lengthy, the same method works for equations as well as inequalities, and it also works when exponential and/or logarithmic and/or trig functions are involved (and other possibilities). Just divide up the number line into intervals in such a way that, within each interval, you can replace ALL of the absolute value expressions with the inside of the absolute value expression or the negative of the inside of the absolute value expression, as appropriate.

A Harder Example

To solve

$$ 2\left | x^2 \; - \; |x^2 - 1|\, \right| \;\; \geq \;\; 1,$$

first observe that the inside of the innermost absolute value, namely $x^2 - 1,$ is equal to $0$ when $x = -1,\,1.$ Now note that $x^2 - 1$ is negative when $-1 < x < 1$ and $x^2 - 1$ is positive when $x < -1$ or $x > 1.$ (Sometimes you might have to work a bit to find this part out. Here, however, it's easy to see if you think about the graph of $y = x^2 - 1,$ which is the graph of $y = x^2$ shifted down by $1$ unit.) Thus, we can remove the inner absolute value by breaking this down into two cases.

Case 1: $\; -1 \leq x \leq 1$

$$ 2\left | x^2 \; - \; (-x^2 + 1)\, \right| \;\; \geq \;\; 1$$

$$ 2\left | 2x^2 - 1 \, \right| \;\; \geq \;\; 1$$

To solve this, note that the inside of the absolute value is equal to $0$ when $2x^2 - 1 = 0,$ or $x = \pm \frac{1}{\sqrt 2}.$ Note that $2x^2 - 1$ is negative when $-\frac{1}{\sqrt 2} < x < \frac{1}{\sqrt 2}$ and $x^2 - 1$ is positive when $x < -\frac{1}{\sqrt 2}$ OR $x > \frac{1}{\sqrt 2}.$ Thus, we can remove the last absolute value by breaking this down into two sub-cases.

Sub-case A: $\;-\frac{1}{\sqrt 2} \leq x \leq \frac{1}{\sqrt 2}$

$$ 2(-2x^2 + 1 ) \;\; \geq \;\; 1$$

$$ x^2 \leq \frac{1}{4}$$

$$-\frac{1}{2} \leq x \leq \frac{1}{2}$$

Since we're in sub-case A, we can only use those values above that also satisfy $-\frac{1}{\sqrt 2} < x < \frac{1}{\sqrt 2}.$ Since $\frac{1}{\sqrt 2}$ is approximately $0.707$ (all we need to know is that it's greater $\frac{1}{2}$ to deduce what follows, and this can be seen by comparing the squares of $\frac{1}{\sqrt 2}$ and $\frac{1}{2},$ or by observing which of the denominators $2$ and $\sqrt 2$ is greater), this means that all of these values qualify for inclusion in Sub-case A. Therefore, the values of $x$ we get from Sub-case A satisfy $-\frac{1}{2} \leq x \leq \frac{1}{2}.$

Sub-case B: $\;x \leq -\frac{1}{\sqrt 2}$ OR $x \geq \frac{1}{\sqrt 2}$

$$ 2(2x^2 - 1 ) \;\; \geq \;\; 1$$

$$ x^2 \geq \frac{3}{4}$$

$$x \leq -\frac{\sqrt 3}{2} \;\; \text{OR} \;\; x \geq \frac{\sqrt 3}{2}$$

Since we're in sub-case B, we can only use those values above that also satisfy $x \leq -\frac{1}{\sqrt 2}$ OR $x \geq \frac{1}{\sqrt 2}.$ Since $\frac{1}{\sqrt 2} < \frac{\sqrt 3}{2}$ (you can see this by squaring each of these positive fractions), and hence $-\frac{\sqrt 3}{2} < -\frac{1}{\sqrt 2},$ it follows that all of the values of $x$ we got from Sub-case B qualify for inclusion in Sub-case B. To see this, it will help to draw a number line, labeling the following points in order from left to right: $-\frac{\sqrt 3}{2},\;$ $-\frac{1}{\sqrt 2},\;$ $0,\;$ $\frac{1}{\sqrt 2},\;$ $\frac{\sqrt 3}{2}.$

Therefore, the values of $x$ we get from combining Sub-case A and Sub-case B are $-\frac{1}{2} \leq x \leq \frac{1}{2}$ along with $x \leq -\frac{\sqrt 3}{2}$ along with $x \geq \frac{\sqrt 3}{2}.$

Therefore, the values of $x$ we get in Case 1 are the combined values of $x$ (just listed) that ALSO satisfy $-1 \leq x \leq 1$ (the original condition we started with for Case 1). This gives $-\frac{1}{2} \leq x \leq \frac{1}{2}$ along with $-1 \leq x \leq -\frac{\sqrt 3}{2}$ along with $\frac{\sqrt 3}{2} \leq x \leq 1.$

Case 2: $\;x \leq -1$ OR $x \geq 1$

$$ \left | x^2 \; - \; (x^2 - 1)\, \right| \;\; \geq \;\; 1$$

$$ \left | 1 \, \right| \; \geq \;\; 1$$

Since this is true for all values of $x,$ we obtain no restriction on $x$ within Case 2, and thus the values of $x$ from Case 2 are just the Case 2 restriction, namely $x \leq -1$ OR $x \geq 1.$

Final Solution

Combining all values of $x$ we got in Case 1 with those we got in Case 2 gives $x \leq -\frac{\sqrt 3}{2}$ OR $-\frac{1}{2} \leq x \leq \frac{1}{2}$ OR $x \geq \frac{\sqrt 3}{2}.$ See this graph for a visualization of what's going on.

Final Thoughts

If you know how to quickly sketch the graph of $y = |f(x)|$ when the graph of $y = f(x)$ is known (leave stuff above $x$-axis alone, reflect about $x$-axis the stuff that's below the $x$-axis), then the solution above can be shortened quite a bit. In particular, it is not difficult to hand-sketch the graph of $y = |2x^2 - 1|$ (graph $y = 2x^2,$ then shift down by $1$ unit, then graph the absolute value of the result). Having done this, you can then solve $|2x^2 - 1| \geq \frac{1}{2}$ after figuring out the coordinates of some appropriate points on your graph.


答: The function $f(x)=\dfrac{\sin^2x}{x^2}$ is defined on $\mathbf R^*$, but it a continuation $\tilde f$to $\mathbf R$, setting $\tilde f(0)=1$ which is continuous, and the improper integral of $f$ exists if and only if the integral of $\tilde f$ exists.


问:
What properties $f$ must have so that $\overline{f(z)} =f(\,\overline {z} \,)$
What properties function $f$ must have so that $\overline{f(z)}  =f(\,\overline {z} \,)$ where $z\in\mathbb{C}$

My working : I can prove it for f being a polynomial in $z$ with real coefficient using conjugates distributing properties.
I know how to prove  $\overline{\tan z}=\tan \overline z.$

I am searching for a necessary and sufficient condition on $ f$ for which it is true.

Please help me with sufficient but may not be necessary conditions also.

答: Assuming that $f$ is an entire function, then we have$$(\forall z\in\Bbb C):\overline{f(z)}=f\left(\overline z\right)\tag1$$if and only if $f(\Bbb R)\subset\Bbb R$. In fact, if we $(1)$, then, if $x\in\Bbb R$, $\overline{f(x)}=f\left(\overline x\right)=f(x)$, and so $f(x)\in\Bbb R$. And, if $f(\Bbb R)\subset\Bbb R$, then the function $g\colon\Bbb C\longrightarrow\Bbb C$ defined by $g(z)=\overline{f\left(\overline z\right)}$ is also analytic. But $g(x)=f(x)$ when $x\in\Bbb R$ and therefore, by the identity theorem, $f=g$. But this means that we have $(1)$.


问:
A matrix differentiation with trace and kronecker product
I'm new to matrix calculus and want to differentiate the following function w.r.t $X$, $Y$
$$\phi(X,Y) = Y^TA^T(L\otimes X):Y^TA^T(L\otimes X)
=tr((L\otimes X)^TAYY^TA^T(L\otimes X)) $$
I know the derivative w.r.t $Y$, but have no idea how to start it w.r.t $X$

I have checked some related questions and solutions, but I still have no idea how the 'differential' $d\phi$ are derived.

Is there any reference about calculating the differential of a matrix?

Any help will be appreciated!!

答:
The differential in this case will be
$$
d\phi = 2 Y^TA^T(L \otimes X):Y^TA^T(L \otimes dX).
$$


问:
What does the graph of $N(\mu, +\infty)$ density look like?
I was wondering what the plot of an $N(\mu, +\infty)$  density look like.

Per my understanding, a variance of zero means a horitonzal line (i.e. constant density).
As variance increases, so does the slope of the peak you see on the graph.

If we take this reasoning to the limit, my conjecture is that the graph of a Gaussian variable with infinite variance would be the x-axis, except for a single point where the value of the function is strictly positive.

Is this correct? If so, what would that point be and what would the value of the function be in that point? My guess would be that it doesn't really matter, as far as the density being integrable into a cumulative distribution function, as the definite integral of a single-point interval is $0$ no matter the value of the function in that point.

答:
You have it the wrong way round. $N(\mu,\infty)$ would be "infinitely spread out", the $x$-axis with no spikes, a distribution where every real number has the same chance of appearing – which is impossible.

What you have in mind is the limit $N(\mu,0)$. It is not a function in the usual sense, but it is still a distribution. It is the Dirac delta dunction $\delta(\mu)$.

The Wikipedia page on it has this cute little graphic showing how the normal distribution becomes $\delta$ in the zero-variance limit:


问:
Newton's method and rearranging
I am new to root-finding and using numerical methods.

Using Newton's method below: $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_0)}$$ using this chord formula where the chord length $c$ is 1 cm: $$c=2r\sin\frac{\theta}{2}$$ supposing the radius is 1.1 cm and the angle θ is unknown, show the iterative Newton's Method equation you would use to find an approximate value for $\theta$ in the context of this problem (using the appropriate function and derivative).

I have tried rearranging it to $\theta$ but Symbolab gives me two solutions which add an $n$ to the equation. I found this equation but not sure if it right:$$\theta = 2\arcsin\frac{c}{2r}$$

But when I derive this, it becomes zero so I'm unsure if my equation is right either. I would just like help deriving this equation and then applying it to the problem.

答:
You are expected to work with
$$
f(x)=c-2r\sin\frac x2\implies f'(x)=-r\cos\frac x2
$$
and indeed this equation has multiple solutions. You need to pick the one in the correct interval, what that is comes from additional knowledge about the geometrical situation. One could for instance ask for the smallest positive solution, which will be inside the interval $[0,\pi]$ if $0<c<2r$.

Title: Newton's method and rearranging


问:
Need Help with the definition of product of two ideals
I know this are not ideals but for the purpose of being clear imagine $I=\{1,2,3\}$ $J=\{4,5,6\}$ where I and J are ideals

$IJ = \{ a_1b_1 + a_2b_2 + ...+a_nb_n \mid a_i\in I b_i\in J \space i=1,2...,n \space \land \space n\in \mathbb{N}^*\}$

Then $IJ= \{1*4,1*4+2*5,1*4+2*5+3*6\} =\{4,14,32\}$

Would this be right or Am I missing something about the definition?

答:
In the definition of the product of two ideals, we're not saying take the first element of one ideal and multiply it by the first element of the next ideal, and so on. What the definition is saying is that you need to allow for arbitrary sums of any product of an element of $I$ with an element of $J$. Maybe a definition you would be more comfortable with is to define $IJ$ to be the ideal generated by the set $\{ab:a\in I\, b\in J\}$.

As an example, let's show that the ideal of even integers $I$, and the ideal of integers divisible by $3$ which we'll call $J$, has as its product the ideal of all integers divisible by $6$ which we'll call $K$.

Every multiple of $6$ can be written as $6z$ for some integer $z$, which is the same as $2\cdot 3\cdot z$ so that $6z$ is a multiple of $2$ times a multiple of $3$. Thus every multiple of $6$ is an element of $IJ$, so that $K\subseteq IJ$.

Conversely, every element of $IJ$ has the form $\sum_{i=1}^{n}a_ib_i$ where $a_i$ is in $I$ and $b_i \in J$.
But then every summand is divisible by $6$, so that the sum is divisible by $6$. This means that every element of $IJ$ is a multiple of $6$ so that $IJ\subseteq K$.

Hope this helps!


问:
Borceux. Handbook of Categorical Algebra I. Proposition 3.4.2.
I'm trying to understand proposition 3.' HoCA (vol. I).

Proposition 3.4.2 Consider a functor $F\colon \mathcal A \to \mathcal B$ with both a left adjoint functor $G$ and a right adjoint functor $H$. If functor is full and faithful, si is the other adjoint functor.

There are some minor typos in the proof, but they're not a big deal.

Borceux calls $\epsilon \colon GF \Rightarrow 1_{\mathcal A}$, $\eta\colon 1_{\mathcal B} \Rightarrow FG$ and $\alpha \colon FH \Rightarrow 1_{\mathcal B}$, $\beta\colon 1_{\mathcal A} \Rightarrow HF$ the counits and units of the adjuctions.

By duality, it is enough to prove, wlog, that if H is fully faithful then the same holds for G. Further, this statement is equivalent (by proposition 3.4.1) to say that $\eta$ is iso whenever $\alpha$ is.

The claim is that

$\alpha \circ (1_F *\epsilon * 1_H) \circ (1_{FG}*\alpha^{-1}) = \eta^{-1}$.

By the naturality of $eta$,

$\alpha \circ (1_F *\epsilon * 1_H) \circ (1_{FG}*\alpha^{-1}) \circ \eta =
\alpha \circ (1_F *\epsilon * 1_H) \circ (\eta*1_{FH})\circ \alpha^{-1}.$

The trangular equality, i.e., $(1_F *\epsilon) \circ (\eta*1_{F}) = 1_{\mathcal F}$ (thus, in particular $(1_F *\epsilon *1_H) \circ (\eta*1_{FH}) = 1_{\mathcal FH}$), gives us

$alpha \circ \alpha^{-1} = 1_{\mathcal B}$.

On the other hand, first note that the triangular equality

$(\alpha *1_F)\circ (1_F * \beta) = 1_F$

implies that $1_F* \beta = (\alpha *1_F)^{-1}\quad(\heartsuit)$.

Now consider

$\star = \eta\circ \alpha \circ (1_F *\epsilon * 1_H) \circ (1_{FG}*\alpha^{-1})$

$\star = (\text{by naturality of }\alpha) = (\alpha*1_{FG})\circ (1_{FH}*\eta) \circ (1_F *\epsilon * 1_H) \circ (1_{FG}*\alpha^{-1})$

$\star = (\text{by naturality of }1_F *\epsilon * 1_H) = (\alpha*1_{FG})\circ (1_F *\epsilon * 1_{HFG})\circ (1_{FGFH}*\eta) \circ (1_{FG}*\alpha^{-1})$

$\star = (\text{by naturality of }1_{FG}*\alpha^{-1}) = (\alpha*1_{FG})\circ (1_F *\epsilon * 1_{HFG})\circ (1_{FG}*\alpha^{-1}*1_{FG}) \circ (1_{FG} *\eta)$

Using $(\heartsuit)$, $\star$ is thus equal to

$(\alpha *1_{FG}) \circ (1_F*\epsilon *1_{HFG}) \circ (1_{FGF}*\beta *1_G) \circ (1_{FG}*\eta)$.

(Borceux writes

$(1_{FG}*\alpha) \circ (1_F*\epsilon *1_{HFG}) \circ (1_{FGF}*\beta *1_G) \circ (1_{FG}*\eta)$,

but I think there's a small typo).

The next (and last) equality is

$(\alpha *1_{FG}) \circ (1_F*\epsilon *1_{HFG}) \circ (1_{FGF}*\beta *1_G) \circ (1_{FG}*\eta) = (1_F*1_G)\circ (1_F*1_G)$.

Where does this equality come from?

Aside question. Proving all these last equalities I've realized that, if $A$ is a category, $F, G\colon A \to A$ are endofunctors, $\alpha\colon F\to 1_A$, $\beta\colon 1_A \to G$ are natural transformations, then

$\beta \circ \alpha = (\alpha *1_G)\circ (1_F *\beta)$.

I suppose that, maybe using the interchange law for 2-cells and doing some checks, lemma general enough to prove immediately the equalities in the last part of the proof.

答:
I will use the same notation. By naturality,
$$(1_F * \epsilon * 1_{H F G}) \circ (1_{F G F} * \beta * 1_G) = (1_F * \beta * 1_G) \circ (1_F * \epsilon * 1_G)$$
and so
$$(\alpha * 1_{F G}) \circ (1_F * \epsilon * 1_{H F G}) \circ (1_{F G F} * \beta * 1_G) \circ (1_{F G} * \eta)$$
is equal to
$$(\alpha * 1_{F G}) \circ  (1_F * \beta * 1_G) \circ (1_F * \epsilon * 1_G) \circ (1_{F G} * \eta)$$
which by the triangle identities is just $1_{F G}$.</p>


问:
Evaluating $\int_0^\infty\sqrt{\frac{x-1}{x^n-1}}\,dx$
I had a lot of fun in this answer where I worked out
$$\int_0^\infty\frac1{\sqrt{x^4+x^3+x^2+x+1}}\,dx=\frac4{\sqrt{4\varphi+3}}F\left(\frac{3\pi}{10},m=8\varphi-12\right)$$
But what happens if the largest exponent in the denominator polynomial is not $4$ but some other integer? In other words, is there a general closed form or single series expression for
$$\int_0^\infty\sqrt{\frac{x-1}{x^n-1}}\,dx\ ?$$
For $n=5$ the answer is as above and for $n=4$
$$\int_0^\infty\frac1{\sqrt{x^3+x^2+x+1}}\,dx=2^{-1/4}F\left(\cos^{-1}\frac{1-\sqrt2}{1+\sqrt2},\frac12+\frac1{2\sqrt2}\right)$$
The integrals for $n=1,2,3$ diverge. Evaluating the integral for $n\ge6$, however, appears to be infeasible even with series; while the gamma product sum in Jack d'Aurizio's answer here looks quite appealing, it only works for $n=5$ – only then can it be shown that the integral over $[0,\infty]$ is twice the integral over $[0,1]$, at which point you bring in beta functions. The other result in Jack's answer is a double sum, which can be generalised to other $n$ but is not very elegant (partly because of the double sum and partly because one bound of that sum uses a floor function).

If an approach that solves the task also gives integrals for the same integrand but with other bounds (e.g. $[0,1]$), that would be appreciated.

答:
I'll offer a "single series expression"; hopefully someone eagle-eyed can spot what that is in hypergeometric terms, thereby achieving a closed form.

For $x\in[0,\,1]$, apply $x=\sin^{2/n}t$; for $x\ge1$, apply $x=\csc^{2/n}t$. In terms of falling Pochhammer symbols, the integral is$$\begin{align}&\frac2n\int_0^{\pi/2}(\sin^{2/n-1}t+\sin^{-3/n}t)\sqrt{1-\sin^{2/n}t}dt\\&=\frac2n\sum_{k\ge0}\frac{(\tfrac12)_k(-1)^k}{k!}\int_0^{\pi/2}(\sin^{2(k+1)/n-1}t+\sin^{2(k-3/2)/n}t)dt\\&=\frac1n\sum_{k\ge0}\frac{(\tfrac12)_k(-1)^k}{k!}(\operatorname{B}(\tfrac{k+1}{n},\,\tfrac12)+\operatorname{B}(\tfrac{k-3/2}{n}+\tfrac12,\,\tfrac12))\\&=\frac{\sqrt{\pi}}{n}\sum_{k\ge0}\frac{(\tfrac12)_k(-1)^k}{k!}\left(\tfrac{\Gamma\left(\tfrac{k+1}{n}\right)}{\Gamma\left(\tfrac{k+1}{n}+\tfrac12\right)}+\tfrac{\Gamma\left(\tfrac{k-3/2}{n}+\tfrac12\right)}{\Gamma\left(\tfrac{k-3/2}{n}+1\right)}\right).\end{align}$$


问:
Question regarding number of polynomials over $\mathbb F_3$
I have a question regarding polynomials over $\mathbb F_3$ ($\mathbb Z/3\mathbb Z$).
In a homework problem we have to check how many third degree polynomials there are over $\mathbb F_3$.

The question is whether there are exactly $54$ or not. I'm confused as to how to go about this.

I can see how there should be a finite number of possibilities, but I don't know how to prove that there is a certain number of them.

答:
$$a x^3+b x^2+c x+d$$
Modulus $3$ coefficient $a$ can be $1$ or $2$, while $b,c,d$ have a range $(0..3)$

Thus they are $2\times 3\times 3\times 3=54$


答:
If $f(x) = mx\equiv k\pmod n$ then $n|mx -k$.  As $\gcd(n,m)|n$ then $\gcd(n,m)|mx-k$.  And as $\gcd(n,m)|m$ then $\gcd(n,m)|k$.  So $f(x)$ can only be equivalent to multiples of $\gcd(n,m)$.

If any $k$ exist that are not multiples of $\gcd(m,n)$ then $f(x) \equiv k \pmod n$ will be impossible and $f$ can not be onto. Now $k\equiv 1$ or $k\equiv n-1$ will be relatively prime to $n$ so $\gcd(m,n)|k$ can only happen if $\gcd(m,n) =1$.

.......

So that was the brainstorm.

Here's the prooof.

If $\gcd(m,n) =1$ then there exist $a,b\in \mathbb Z$ so that $am + bn =1$. (By Bezout.)  Thus for any $k$ we have $kam + kbn = k$ and $kam = k-kbn$ so $f(ka)\equiv k \pmod n$ and $f$ is onto.

If $f$ is onto then there exists an $a$ so that $f(a)=ma \equiv 1 \pmod n$.  That is there exist a $b \in \mathbb Z$ so that $am = 1 + bn$ so $am -bn =1$.  So by Bezout's Lemma $\gcd(m,n) =1$.

Title: How would I prove that f is onto then gcd(m,n) = 1?

Body:
I'm having trouble getting started on this question:

Define f: ℤn → ℤn by f(x) = mx mod n. Use the Bézout’s theorem to prove that if f is onto, then gcd(m,n)=1.


问:
Convergence of random subsequence of random variables
Let $Y_1, Y_2, \dots$ be independent random variables such that
$$
P(Y_i = 0) = 1-\frac{1}{i}, \ P(Y_i = 1) = \frac{1}{i}.
$$
Show that there is a sequence of random variables $Z_m$ taking positive integer values such that $Z_m \to \infty$ a.s. and
$$
Y_{Z_m} \to 1 \ a.s.
$$
This is easy if we don't have the $Z_m \to \infty$ requirement since we can just let $Z_m \equiv 1$. But how do I approach this if $Z_m \to \infty$? I have no idea where to start. Can anyone provide some hint? Thank you!

答:
This is a proof obtained by the suggested comment.

Since it is easy to see that
$$
P(Y_i=1 \ i.o.) = 1
$$
by Borel-Cantelli lemma, let
$$
A = \{\omega: Y_i(\omega) = 1 \ i.o.\}.
$$
Now pick $\omega \in A$, then we can find a subsequence $i_{m,\omega}$ such that $Y_{i_{m,\omega}}(\omega) = 1$ for all $m$. Then define
$$
Z_m(\omega) = i_{m,\omega}.
$$
It is easy to see that $Z_m \to \infty$ a.s. since $i_{m,\omega} \to\infty$ and $P(A) =1$. Therefore,
$$
P(Y_{Z_m} = 1 \text{ for all }m)=1.
$$

Edit: The above proof has a problem in that randomly choosing a subsequence $i_{m,\omega}$ does not gurantee that $Z_m$ is measurable. Hence, a slight modfication is required:

Let $Z_m(\omega) = \inf\{t \in \mathbb{N}^+: \sum_{i=1}^tY_i(\omega) = m\}$. Then it is still the case that $Z_m \to \infty$ a.s. since for any $\omega\in A$, $Z_m(\omega) \geq m$. Hence, $Z_m \to \infty$ a.s. To show that $Z_m$ is measurable, note that for any positive integer $k$, if $k < m$, then
$$
\{\omega: Z_m(\omega) = k\} = \emptyset \in \mathcal{F}.
$$
If $k\geq m$, then
$$
\{\omega: Z_m(\omega) = k\} = \bigg\{\omega: Y_k(\omega) = 1\bigg\} \cap \bigg\{\omega: \sum_{i=1}^{k-1}Y_i(\omega) = m-1\bigg\},
$$
which is an intersection of measurable sets as the sum of $Y_i$ are measurable. Therefore, $Z_m$ is measurable.


问:
Prove that $\sqrt{3} + \sqrt{7}$ is a primitive element of $\mathbb{Q}(\sqrt{3},\sqrt{7})$
I want to find a primitive element of $\mathbb{Q}(\sqrt{3},\sqrt{7})$ over $\mathbb{Q}$. I try with $\sqrt{3} + \sqrt{7}$.

Let $X = \sqrt{3} + \sqrt{7}$ we get

$$ X^4 - 20X^2 + 16 = 0 $$

We can prove that $[\mathbb{Q}(\sqrt{3},\sqrt{7}) : \mathbb{Q}] = 4$ since $\sqrt{3}$ and $\sqrt{7}$ are the roots of $X^4 - 10X^2 + 21$ and it is irreducible over $\mathbb{Q}$ by the Eisenstein criterion with $p = 3$.

Let $P(X) = X^4 - 20X^2 + 16$. Since $\sqrt{3} + \sqrt{7}$ is a root of $P$ then $[\mathbb{Q}(\sqrt{3}+\sqrt{7}) : \mathbb{Q}] \leq 4$.
If we prove that $P$ is irreducible over $\mathbb{Q}$ then the degree is equal to 4. But the Eisenstein criterion does not seem to work...

So, I try another thing. Let $\alpha = \sqrt{3} + \sqrt{7}$, we get $\alpha^2  = 10 + 2 \sqrt{21}$ and $\mathbb{Q} \subset \mathbb{Q}(10 + 2 \sqrt{21}) \subset \mathbb{Q}(\sqrt{3} + \sqrt{7})$. Now

$$ [\mathbb{Q}(10 + 2 \sqrt{21}) : \mathbb{Q}] = 2 \Rightarrow 2 \mid [ \mathbb{Q}(\sqrt{3} + \sqrt{7}) : \mathbb{Q} ] $$

The degree must be 1, 2 or 4.

But  $\mathbb{Q}(10 + 2 \sqrt{21}) \subset \mathbb{Q}(\sqrt{3} + \sqrt{7})$ and $ [ \mathbb{Q}(\sqrt{3} + \sqrt{7}) : \mathbb{Q}(10 + 2 \sqrt{21}) ] > 1$ then the degree must be 2 or 4. I don't know how to conclude with the formula

$$ [ \mathbb{Q}(\sqrt{3} + \sqrt{7}) : \mathbb{Q} ] = [\mathbb{Q}(\sqrt{3} + \sqrt{7})  :\mathbb{Q}(10 + 2 \sqrt{21}) ] [\mathbb{Q}(10 + 2 \sqrt{21}) : \mathbb{Q}]  $$

Thank you. I think I did a mistake somewhere.

答:
Cubing $\alpha=\sqrt 3+\sqrt 7$, you obtain the linear system
\begin{cases}
\alpha=\sqrt 3+\sqrt 7, \\[1ex]
\alpha^3=24\sqrt 3+16\sqrt7,
\end{cases}
from which can easily deduce $\sqrt 3$ and $\sqrt 7$ as a linear combination of $\alpha$ and $\alpha^3$.


问:
How is the SVD of an item document matrix - $A$'s component equal to the eigenvector matrix of $AA^T$?
I'm trying to get an intuitive grip on SVDs to explore LSI as part of NLP. I understand that it involves the usage of SVDs. As I understand in SVD

$A = SΣU^T$

I understand that $S$ is a matrix of eigenvectors of $A^TA$ and $\Sigma$ is the square root of its eigenvalues as a diagonal matrix.

I don't understand how $U$ is the matrix of eigenvectors of $AA^T$

In LSI, we decompose this SVD matrix to get a shorter vector for each term and document to find out the relative distance between those vectors and the centroid

I don't understand how doing an SVD converts words to their own vectors, by which we can estimate the distance between documents

I'm very new to NLP, I'm finding it hard to get an intuitive sense of the math, I'd sincerely appreciate any help. Is there any books that I can refer to, thanks!

答:
The item-document matrix $A$ is a $m \times n$ matrix where there are $m$ items and $n$ documents. Each entry shows the counts for the word in that document. The SVD is used to decompose the matrix in the product of several matrices. In the process of finding the SVD you're using Gram-Schmidt to form the left and right singular vectors which are the eigenvectors of $AA^T$ and $A^TA$. You can check this as

$$ AA^T = (U \Sigma U^T) (U \Sigma U^T)^T $$
$$  = U \Sigma U^T U^T \Sigma^T U^T  $$
then since $U^TU = UU^T =I$ we get
$$  = U \Sigma\Sigma^T U^T  $$
and $\Sigma^T = \Sigma \implies \Sigma \Sigma^{T} = \Sigma^{2} =\Lambda$

$$  = U \Lambda U^T  $$

You can do the same thing for $A^TA$ to see it equals $V \Lambda V^T$. The purpose is the Gram-Schmidt takes the columns (which are incredibly sparse here) and forms orthogonal vectors where the entries are describing the weights for the terms in this new factor. By doing this you can truncate it. This is called the truncated SVD. If $A$ is $ m \times n$ then we  have

$$ A_{m \times n} = U_{m \times m} \Sigma_{m \times n} V_{n \times n}^{T}$$

as the full SVD. Now if we take a threshold value $\epsilon$ such that we make all the singular values $\sigma < \epsilon$  equal to zero we get a new matrix

$$ A_{m \times n} = U_{m \times k} \Sigma_{k \times k} V_{k \times n}^{T}$$

where $k$ is the number singular values greater than $\epsilon$. By doing this you can then use those vectors to compute the similarity between documents.

Another way to view what the SVD is doing is that we can rewrite it

$$ A = \sum_{i=1}^{k} \sigma_{i}u_{i}v_{i}^{T}$$

This says the SVD decomposes a matrix (here the term document matrix) into the sum of $k$ rank-one matrices $u_{i}v_{i}^T$ which are weighted or scaled by the non-negative constants $\sigma_{1} \geq \sigma_{2} \geq \cdots \sigma_{k} \geq 0$ called the singular values. These are the square roots of the eigenvalues of the covariance matrices $AA^T$ and $A^TA$.

The actual distance between the original matrix and low-rank approximation is given by a theorem called the Eckart-Young-Mirsky theorem which says using the euclidean norm we have

$$ \|A - A_{k} \|_{2} =  \bigg\| \sum_{i=k+1}^{n}  \sigma_{i}u_{i}v_{i}^{T} \bigg\|  = \sigma_{k+1}$$

which means for a rank $k$ approximation using the $2$ norm the actual distance between them is given by the $\sigma_{k+1}$ which is the $k+1$th singular value.


问:
A system of 2 equations
During a physics problem, I have encountered a set of 2 equations and after trying to solve them for quite a while, I start to thing there may be no solution. Am I missing something really easy?

$$ (1) A_{n}\cos\left(\sqrt{\lambda}_{n}R\right)+B_{n}\sin\left(\sqrt{\lambda}_{n}R\right)=0$$

$$(2)A_{n}\cos\left(\sqrt{\lambda}_{n}\left(R+\pi\right)\right)+B_{n}\sin\left(\sqrt{\lambda}_{n}\left(R+\pi\right)\right)=0$$

I'm searching for $\lambda_n$ that solves both equations, but nothing seems to work. I can choose any values I want for $A_n$ and $B_n$, I tried taking one of them to be zero but it doesn't help.

答:
HINT.-$$A_n\sin(X)+B_n\cos(X)=0\Rightarrow \frac{A_n\sin(X)+B_n\cos(X)}{\sqrt{A_n^2+B_n^2}}=\sin(\theta+X)=0$$ where $\cos(\theta)=\dfrac{A_n}{\sqrt{n^2+B_n^2}}$ and $\sin(\theta)=\dfrac{B_n}{\sqrt{A_n^2+B_n^2}}$.

Consecuently $$\theta +X=?$$ Similarly you can get $$\theta +Y=?$$

Title: A system of 2 equations


问:
How do you solve $y'=\sin(y)$?

How do you solve $y'=\sin(y)$?

I ended up with
$$-\ln(\csc(y)+\cot(y)) = x+C$$

But then I simplified it to
$$e^{-x+C} = \frac1{\sin(y)} + \frac{\cos(y)}{\sin(y)}$$

Where do I go from here?

答:
$$y'=\sin y\iff\frac{dy}{\sin y}=dx\iff \int\frac{dy}{\sin y}=\int dx\implies$$

$$\implies\ln\tan\frac y2+c\implies$$

$$\implies\tan\frac y2=e^{c+x}\implies \frac y2=\arctan\left(e^ce^x\right)\implies\ldots$$

Title: How do you solve $y'=\sin(y)$?


问:
A polynomial has the same remainder when divided by $x+k$ or $x-k$; what is $k$?

Question

Given that $y = 3x^3 + 7x^2 - 48x + 49$ and that $y$ has the same remainder when it is divided by $x + k$ or $x - k$, find the possible values of $k$.

My attempt

Let $f(x) = 3x^3 + 7x^2 - 48x + 49$

$\text{Using Remainder Theorem,}$

\begin{align}
f(-k) &= f(k) \\
3(-k)^3  + 7(-k)^2 - 48(-k) + 49 &= 3(k)^3 + 7(k)^2 - 48(k) + 49 \\
-3k^3 - 7k^2 + 48k + 49 &= 3k^3 + 7k^2 - 48k + 49 \\
-3k^3 --3k^3 - 7k^2 - 7k^2 + 48k + 48k + 49 - 49 &= 0 \\
-6k^3 + 14k^2 + 96k &= 0 \\
\frac{-6k^3}{k} + \frac{14k^2}{k} + \frac{96k}{k} &= \frac{0}{k} \\
6k^2 + 14k - 96 &= 0
\end{align}

$\text{Comparing } 6k^2 + 14k - 96 = 0 \text{ with } ak^2 + bk + c = 0, a = 6, b = 14, c = -96$

\begin{align}
k = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a } &= \frac{ -(14) \pm \sqrt{(14)^2 - 4(6)(-96)} }{ 2(6) } \\
k &= 3 \text{ and } -5\frac{1}{3}
\end{align}

$\therefore k = 3, -5\frac{1}{3} \text{ or } 0 $

My answer is incorrect. The correct answer is $k = 0, 4 \text{ or } -4$

答:
Duncan Ramage has patiently showed you where you went wrong. I would like to provide some tips on how to avoid the errors you made and simplify your proof.

There is no need to write out every term of $f(k)$ and $f(-k)$. Instead you take the difference:

$$0=f(k)-f(-k)=3(k^3-(-k)^3) + 7(k^2-(-k)^2)-48(k-(-k))+(49-49)\\
=6k^3-96k=6k(k^2-16) \implies k=0, \pm 4.$$

When you get experienced you can skip the even-order terms altogether:

$$0=f(k)-f(-k)=3(k^3-(-k)^3) -48(k-(-k))\\
=6k^3-96k=6k(k^2-16) \implies k=0, \pm 4.$$


答:
It is better to write
$$-\log(\csc y+\cot y)=\log\tan\frac y2$$
Then we have
$$\log\tan\frac y2=x+K$$
$$\tan\frac y2=Ae^x$$
$$y=2\tan^{-1}Ae^x$$

Title: How do you solve $y'=\sin(y)$?

Body:
How do you solve $y'=\sin(y)$?

I ended up with
$$-\ln(\csc(y)+\cot(y)) = x+C$$

But then I simplified it to
$$e^{-x+C} = \frac1{\sin(y)} + \frac{\cos(y)}{\sin(y)}$$

Where do I go from here?


问:
The semifinite portion of a measure $\mu$
Let $\mu$ be a measure and define $\mu_1$ such that $\mu(E)=\mu_1(E)$ for $\mu(E)$ finite.  And for $\mu(E)$ infinite definite $\mu_1$ such that:

(i) if $E$ contains finite subsets of arbitrarily large measure then $\mu_1(E)=\infty.$

(ii) and if not then $\mu_1(E)=0$.

Prove $\mu_1$ is a measure.$$$$

So I'm really having trouble seeing how countable (even finite) additivity holds for sets $E$ such that $\mu(E)=\infty$ but which don't contain finite subsets of arbitrarily large measure.  For instance if $E$ can be partitioned into $A\cup B$ with $\mu(A)=1$ and $\mu(B)=\infty$, then $$0=\mu_1(E)=\mu_1(A\cup B)\neq\mu_1(A)+\mu_1(B)=1 + 0 = 1.$$

答:
I agree with you that $\mu_1$ as defined in Royden-Fitzpatrick does not yield a measure. An example that shows this is obtained by considering the measure on $\mathbb{N}$ defined on all subsets $A \subseteq \mathbb N$ by
$$
\mu(A) = \begin{cases} 0, & \text{if } A = \emptyset, \\ 1, & \text{if } A = \{0\} \\ \infty, & \text{otherwise.} \end{cases}
$$
For this measure $\mu$ we have $\mu_1(\emptyset) = 0$, $\mu_1(\{0\}) = 1$ and $\mu_1(A) = 0$ for all other $A \subseteq \mathbb{N}$, so $\mu_1$ is obviously not a measure.

The usual definition of the semi-finite version $\mu_{\rm sf}$ of a measure $\mu$ on $(X,\Sigma)$ is given by
$$
\mu_{\rm sf}(E) = \sup\{\mu(A) \mid A \subseteq E \text{ measurable, } \mu(A) \lt \infty\} \quad \text{for }E \in \Sigma
$$
In this definition it is clear that $\mu_{\rm sf}$ is non-negative, that $\mu_{\rm sf}(\emptyset) = 0$ and that $\mu(E) =\mu_{\rm sf}(E)$ whenever $E \in \Sigma$ has finite $\mu$-measure.  Using this, it is not hard to check that $\mu_{\rm sf}$ is $\sigma$-additive on $\Sigma$, hence it is a measure on $(X,\Sigma)$. Moreover, $\mu_{\rm sf}$ is semi-finite and we have $\mu_{\rm sf} = \mu$ if and only if $\mu$ is semi-finite.

One can also show that $\mu_{\rm sf}$ and $\mu$ have the same integrable functions (up to $\mu_{\rm sf}$-null sets) and that $\mu_{\rm sf}$ is complete only if $\mu$ is complete.


问:
Integral over product of functions is $0$
Suppose a measurable function $|g(x)|\leq M$ is bounded above by $M>0$.

If for every interval $[a,b]\subset\mathbb{R}$ $$\int_a^b g(x)\,dx=0$$
Is the following statement true?

$$\forall f\in L^1,\int_{\mathbb{R}}f(x)g(x)dx = 0$$

I am not sure how to proceed, and thanks in advance for any ideas.

答:
I don't see how this can work in general. Take for instance $f(x)=g(x)=\sin(x)$ on $[0,2\pi]$, then
$$\int_0^{2\pi} g(x)\,dx=0$$
however
$$\int_0^{2\pi} f(x)g(x)\,dx=\int_0^{2\pi}\sin^2(x)\,dx>0$$


问:
Math question. My professor and me have different answer. Please help.
A squad of 4 soldiers is selected  from a 15 man platoon to keep watch each night.

How many different squad can be selected if either Bob or Roy ( but never both) must be on the squad watch all the time?

My professor said it's 2* 13C2

But I think it's 2* 13C3

Please correct me.

I really don't know about professor answer. But mine Explanation is Bob or roy must always be present in any team thus forming a team is: 13C3 x 2 = 572 Combination is used because order is not considered. and x 2 since the 13 men would be interchangeable but the 2 must be always there thus. Multiply 2 to the combination of the squad formed.

答:
There is only one right answer - in this case, $572$. But as is often the case in combinatorics, there are many perfectly valid ways to arrive at the answer.

One way is to divide the two platoon into two groups:

$A = \{\mbox{Bob}, \mbox{Roy}\}$

$B = \{\mbox{everyone else}\}$

Choose one member from the first group, and three members from the second group. There are $2{13 \choose 3} = 572$ ways to do this, as you concluded.

Here's a totally different way:

Count all the possible squads. ${15 \choose 4} = 1365$.

Count all the squads that contain neither Bob nor Roy. ${13 \choose 4} = 715$. We reject these.

Count all the squads that contain both Bob and Roy. ${13 \choose 2} = 78$. We reject these too.

So the answer is ${15 \choose 4} - ({13 \choose 4} + {13 \choose 2}) = 1365 - (715 + 78) = 572$. This agrees with the earlier answer.

This generalizes to a combinatorial identity: $2 {n - 2 \choose k - 1} = {n \choose k} - \left( {n - 2 \choose k} + {n - 2 \choose k - 2} \right)$. It is possible to prove this by moving the negative terms on the right side over to the left side and then combining using the identity ${n \choose k} + {n \choose k + 1} = {n + 1 \choose k + 1}$ twice.


问:
Problem with Discrete Parseval's Theorem
I think I must be missing something obvious, but I can't for the life of me see what it is. The discrete version of Parseval's theorem can be written like this:

$\sum_{n=0}^{N-1} |x[n]|^2 = \frac{1}{N}\sum_{n=0}^{N-1} |X[k]|^2 $

Now, say you've got some function in time, like $x = \sin(\omega t)$. Depending on how large your value of $N$, the LHS could be arbitrarily large.

The FT of this function is a delta function at $\omega$. Everywhere else is zero, so your RHS is given by $1/N$. So how on earth does one side equal the other?! It seems to me like you've got a summation one one side and an average value on the other...

I must be missing something obvious!

答:
I completely understand your confusion. It is indeed not so obvious. First of all, you have to be careful which version of Parseval's theorem you are considering. For continuous-time signals or for discrete-time signals, and in the second case for infinitely long signals or for the DFT relation, where only signals of length $N$ are considered. The way you've stated it is the DFT version. This means that the transform of a sampled sinusoid is usually NOT a delta impulse unless the frequency is $2\pi k/N$ (i.e. an integer number of periods fit in the DFT interval of length $N$). If you consider the discrete-time case for infinitely long signals, then neither of the two sums converges in the case of a sinusoidal signal.

Example:
$x(n)=\cos\frac{2\pi m}{N}n$ with $0\le m\le N-1$. In this case the DFT
$$X(k)=\sum_{n=0}^{N-1}x(n)e^{-jn\frac{2\pi k}{N}}$$ is indeed given by two discrete deltas:

$$X(k)=\frac{N}{2}\left [ \delta(k-m) + \delta(k+m-N)\right ]$$

Applying Parseval's theorem gives

$$\sum_{n=0}^{N-1}|x(n)|^2=\frac{1}{N}\sum_{k=0}^{N-1}|X(k)|^2=\frac{N}{2}$$


问:
Prove that sequence of real numbers $(d(x_n,p))_{n \in \mathbb{N}}$ converges to $0$ in $\mathbb{R}$
be $M$ a metric space. if a sequence $(x_n)_{n\in\mathbb{N}}$ is such that $\lim_{n\rightarrow +\infty}x_n=p$. Prove that sequence of real numbers $(d(x_n,p))_{n \in \mathbb{N}}$ converges to $0$ in $\mathbb{R}$

I have tried to use the following the way, however I do not know if I am in the way direction. Could someone help me?

If $x_n\rightarrow p, y$ then $d(p, y) ≤ d(p, x_n) + d(x_n , y)\rightarrow 0.$
So $d(p, y) = 0$, implying that $p = y$.

答: Asserting that $\lim_{n\to\infty}x_n=p$ means that$$(\forall\varepsilon>0)(\exists N\in\Bbb N):n\geqslant N\implies d(x_n,p)<\varepsilon$$and asserting that $\lim_{n\to\infty}d(x_n,p)=0$ in $\Bbb R$ means that$$(\forall\varepsilon>0)(\exists N\in\Bbb N):n\geqslant N\implies|d(x_n,p)|<\varepsilon.$$It's not that much different, right?!


问:
Find a Noether normalization

Let $I \subseteq k[X_{1}, X_{2}, X_{3}, X_{4}]$ be the ideal generated by the maximal minors of the $2 \times 3$ matrix
$$\begin{pmatrix}
X_1 & X_2 & X_3\\
X_2 & X_3 & X_4
\end{pmatrix}.$$

Find a Noether normalization $k[Y_1, Y_2, Y_3, Y_4] \subseteq k[X_1, X_2, X_3, X_4]$ such that $I \cap k[Y_1, Y_2, Y_3, Y_4] = (Y_1, \ldots, Y_r)$ for a suitable $r$.

What I've done: I never encountered what a maximal minor is, so after some searching I suppose it means the determinant(s) of the maximal submatrices, i.e. in this case all $2 \times 2$ submatrices which then are (by deleting a column): $\begin{pmatrix}
X_1 & X_2\\
X_2 & X_3
\end{pmatrix}, \begin{pmatrix}
X_1 & X_3\\
X_2 & X_4
\end{pmatrix}, \begin{pmatrix}
X_2 & X_3\\
X_3 & X_4
\end{pmatrix}.$ Then, taking determinants we get $I = (X_1 X_3 - X_{2}^{2}, X_1 X_4 - X_3 X_2, X_2 X_4 - X_{3}^{2})$. Is this correct?

The next step would be to use the constructive proof of Noether's Normalization Lemma. However, I can't seem to understand the entire procedure of that proof and how to apply it to this problem. Perhaps if someone can illustrate this process, then I will better understand it after seeing it done.

答:
The function $\mathrm{noethnorml}$ receives

A polynomial ring $R = A[x_1,\ldots,x_m]$ over a coefficient field $A$.

An ideal $I \subseteq R$ given by its generators.

It returns:

A map of polynomial rings

$$\theta: T_2 = A[w_1,\ldots,w_m] \to R$$

such that

$\theta^{-1}(I) = J = (w_p,\ldots,w_m) \subseteq T_2$ is an ideal of $T_2$. It can be zero (then $p > m$).

$R/I$ is an integral ring extension of $T_2/J$.

$\mathrm{noethnorml}$ does

A. If $I=0$ then return $\theta = \mathrm{id}_R:R \to R$.

Otherwise do

Select $f \in I$. We want to put $f$ in Noether-Position, that is make a coordinate change, such that $f$ in the new coordinates $y_i$ is of the form

$$(*) y_m^N + a_1 y_m^{N-1} + \cdots + a_m$$

where the $a_i$ are polynomials in $y_1,\ldots,y_{m-1}$.

So define $S=A[y_1,\ldots,y_m]$ and maps

$$\phi: S \to R, \quad y_i \mapsto x_i + \alpha_i x_m \text{ for } i < m \text{ and } y_m \to x_m$$

$$\psi: R \to S \quad x_i \mapsto y_i - \alpha_i y_m \text{ for } i < m \text{ and } x_m \to y_m$$

which are obviously inverse to each other.

The $\alpha_1,\ldots,\alpha_{m-1} \in \mathbb{Z}$ are random integers.

Now for generic $\alpha_i$ the polynomial

$$f_1 = \psi(f)$$

is of the above form $(*)$.

B. Now compute $I_1 = \phi^{-1}(I) \subseteq S$.

C. Define a ring $T=A[u_1,\ldots,u_{m-1}]$ and a map

$$\gamma:T \to S, \quad u_i \mapsto y_i$$

C1. Set $I_2 = \gamma^{-1}(I_1) \subseteq T$.

D. This is the recursion:

Call $\mathrm{noethnorml}$ with arguments $(T, I_2)$.
The result is

$$\theta: T_1 = A[v_1,\ldots,v_{m-1}] \to T$$

E. Call $\rho = \phi \circ \gamma \circ \theta:T_1 \to R$

F. Introduce a ring $T_2 = A[w_1,\ldots,w_m]$ and a ring map

$$\theta':T_2 \to R \quad \text{ with } w_i \mapsto \rho(v_i) \text{ for } i < m \text{ and } w_m \mapsto f$$

G. Return $\theta'$ as the result.

Title: Find a Noether normalization


问:
Is this a permissible way of proving/showing this statement?
The task is to show that $$\sin\phi=\frac{1}{2i}(e^{i\phi}-e^{-i\phi}) $$

with an emphasis on show. The task does not specifically state, prove this statement.

So, what I did, is to transform the right side until I got what is stated on the left side. $$ \sin\phi=\frac{1}{2i}(e^{i\phi}-e^{-i\phi}) = ... = ...=\sin\phi  $$

Was this a valid way to show the given statement, or should I have transformed the left side until I get the statement on the right side, like this: $$ \sin\phi=...=...=\frac{1}{2i}(e^{i\phi}-e^{-i\phi}) $$

Or, is there perhaps even no difference, and I can choose either way?

I am sorry for this seemingly trivial question, as I am still a Calc 1 newbie and would love to know the difference between showing and actually proving a statement. Because it seems to me like the first way is merely showing what is given, thus not a full proof, while the nature of the second one, namely taking the left side as the springboard and transforming it until we have what was to be shown, looks more reasonable. Maybe I'm also just overthinking. Any help would be appreciated.

答:
It seems you have a complicated definition $F(\phi)$ of $\sin\phi$ in terms of simpler things defined before, and in the same way you have a complicated definition $G(\phi)$ of $e^{i\phi}$ in terms of simpler things defined before. You are told to prove
$$F(\phi)=G(\phi)\qquad\forall\,\phi\ ,$$
using the facts valid in the "simpler" world. As "$=$" is a symmetric relation such a proof can go in the direction
$$F(\phi)= \ldots =\quad\ldots\quad=\ldots=G(\phi)$$
as well as in the direction
$$G(\phi)= \ldots =\quad\ldots\quad=\ldots=F(\phi)\ .$$


问:
What do you call (and how do you solve) this kind of optimization problem?
I have a system of equations like:

\begin{cases}
X_1 = a_{11} x_1 + a_{12} x_2 + a_{13} x_3\\\\
X_2 = a_{21} x_1 + a_{22} x_2 + a_{23} x_3\\\\
X_3 = a_{31} x_1 + a_{32} x_2 + a_{33} x_3
\end{cases}

Then I have a "target" such as

$$T = t_1 x_1 + t_2 x_2 + t_3 x_3 $$

I want to choose coefficients $c_1, c_2, c_3$ such that $c_1 X_1 + c_2 X_2 + c_3 X_3$ is as close to $T$ as possible. By "as close to" I mean a least squares distance, so minimizing

$$ ((c_1 a_{11} + c_2 a_{21} + c_3 a_{31}) - t_1)^2 +  ((c_1 a_{12} + c_2 a_{22} + c_3 a_{32}) - t_2)^2 + ((c_1 a_{13} + c_2 a_{23} + c_3 a_{33}) - t_3)^2 $$

My end goal is actually to do this kind of problem in R--if you're able to show me that, amazing, but I'm really just trying to start by figuring out what this kind of problem is called! Right now I don't know what terms to look up or google....

Addendum: in case my notation is unclear, an example with numbers

\begin{cases}
X_1 = 2x_1 + 3x_2 + 4x_3\\\\
X_2 = 3x_1 - 2x_2 + 9x_3\\\\
X_3 = 5x_1 + 4x_2 + 4x_3
\end{cases}

The target is $T = 24x_1 + 12.5x_2 + 36x_3$.

In this case there is a perfect solution $c_1 = 1.5$, $c_2 = 2$, $c3=3$, but in general there won't be a perfect solution and I want to find the closest solution.

答: This is a quadratic programming problem, and the Wikipedia page lists a bunch of solvers, many of which are accessible via R.


问:
How to show that a stochastic process is predictable?
Suppose $(X_n, \mathbb{F}_n)$ is a submartingale and $\tau$ is a stopping time. Now look at the process $H_n$ given by:
$$H_n=   \left\{
\begin{array}{ll}
      1 & \text{if }n\leq\tau \\
      0 & \text{if }n>\tau \\
\end{array}
\right. $$
How can I show that $H_n$ is predictable? Intuitively it makes really good sense. I tried working out how to show that $H_n$ is either $\mathbb{F}_{n-1}$- or $\mathbb{F_{\tau}}$-measurable, but got nowhere. In addition a hint is given: what is the relation between $(H\cdot X)_n$ and $X_{\tau\wedge n}$.

答:
I think that hint massively over-complicates it; we can just show $H_n$ is $\mathcal F_{n-1}$ measurable directly.  Notice that

\begin{align*}
\{H_n = 0 \} &= \{ \tau < n \} = \{\tau \le n-1\} \in \mathcal F_{n-1}
\end{align*}

by the definition of a stopping time.  Since $H_n$ can only take on two distinct values, this proves $H_n$ is $\mathcal F_{n-1}$ measurable.


问:
How may I integrate $I = \int_{0}^{a} f(x) x^2 dx $?
How may I integrate $I = \int_{0}^{10} f(x) x^2 dx $ where $f(x)$ is:

$$ \begin{equation}
  f(x)=\begin{cases}
   f_1, & x \in (0,8) ∪ (9,10)\\
    f_2, & x \in (8,9)
  \end{cases}
\end{equation}  $$

EDIT: upper interval is 10, not $a$

答:
Assuming that $f_1, f_2$ are constants, just consider all possible values for $a$.

If $a < 8$,
$$\int_0^a f(x) x^2 dx = \int_0^a f_1 x^2 dx = \frac{f_1 a^3}{3}.$$

If $8 \leq a \leq 9$,
$$
\int_0^2 f(x) x^2 dx = \int_0^8 f_1 x^2 dx + \int_8^a f_2 x^2 dx = \frac{f_1 8^3}{3} + \left(\frac{f_2 a^3}{3}-\frac{f_2 8^3}{3}\right) = \frac 13 f_2 a^3+\left(f_1-f_2\right)\frac{512}{3}
$$

I trust you can now deal with the case $a>9$.


问:
A matrix differentiation with trace and kronecker product
I'm new to matrix calculus and want to differentiate the following function w.r.t $X$, $Y$
$$\phi(X,Y) = Y^TA^T(L\otimes X):Y^TA^T(L\otimes X)
=tr((L\otimes X)^TAYY^TA^T(L\otimes X)) $$
I know the derivative w.r.t $Y$, but have no idea how to start it w.r.t $X$

I have checked some related questions and solutions, but I still have no idea how the 'differential' $d\phi$ are derived.

Is there any reference about calculating the differential of a matrix?

Any help will be appreciated!!

答:
$\def\p#1#2{\frac{\partial #1}{\partial #2}}$
Define $B=(L\otimes X)^TAY$ and assume the following sizes for the matrices
$$\eqalign{
 m,n &= {\rm size}(X) \\
 p,q &= {\rm size}(L) \\
 mp,r &= {\rm size}(A) \\
 r,s &= {\rm size}(Y) \\
 nq,s &= {\rm size}(B) \\
}$$
Kronecker products can be vectorized with the aid of a
Commutation matrix $\,(K_{np})$
$$\eqalign{
{\rm vec}(L\otimes X)
 &= \left(I_q\otimes K_{np}\otimes I_m\right)\cdot
\left({\rm vec}(L)\otimes I_m\otimes I_n\right)\cdot {\rm vec}(X) \\
 &\doteq  M\,{\rm vec}(X) \\
}$$
Write the function in terms of $B$.
Then calculate its differential and gradient.
$$\eqalign{
\phi &= B^T:B^T \\
d\phi
 &= 2B^T:dB^T \\
 &= 2B^T:(AY)^T(L\otimes dX) \\
 &= 2AYB^T:(L\otimes dX) \\
 &= 2\,{\rm vec}(AYB^T):M\;{\rm vec}(dX) \\
 &= 2\,M^T{\rm vec}(AYB^T):{\rm vec}(dX) \\
 &= 2\;{\rm devec}\Big(M^T{\rm vec}(AYB^T)\Big):dX \\
\p{\phi}{X} &= 2\;{\rm devec}\Big(M^T{\rm vec}(AYB^T)\Big) \\
}$$
The hardest part of the process is freeing $dX$ from the Kronecker term.

Vectorization/devectorization was used to handle the Kronecker product
in this example, but one might also employ the Singular value decomposition
or the Pitsianis decomposition of the matrix $(AYB^T)$,
both of which were demonstrated in your
linked answer.

Another possibility is a
block Kronecker
decomposition of the $(AYB^T)$ matrix.

The key idea is that the Kronecker and Frobenius products have a nice distributive property
$$(A\otimes B):(C\otimes dX) \;=\; (A:C)\,B:dX$$
for compatibly dimensioned matrices.


问:
Evaluating $\int_0^\infty\sqrt{\frac{x-1}{x^n-1}}\,dx$
I had a lot of fun in this answer where I worked out
$$\int_0^\infty\frac1{\sqrt{x^4+x^3+x^2+x+1}}\,dx=\frac4{\sqrt{4\varphi+3}}F\left(\frac{3\pi}{10},m=8\varphi-12\right)$$
But what happens if the largest exponent in the denominator polynomial is not $4$ but some other integer? In other words, is there a general closed form or single series expression for
$$\int_0^\infty\sqrt{\frac{x-1}{x^n-1}}\,dx\ ?$$
For $n=5$ the answer is as above and for $n=4$
$$\int_0^\infty\frac1{\sqrt{x^3+x^2+x+1}}\,dx=2^{-1/4}F\left(\cos^{-1}\frac{1-\sqrt2}{1+\sqrt2},\frac12+\frac1{2\sqrt2}\right)$$
The integrals for $n=1,2,3$ diverge. Evaluating the integral for $n\ge6$, however, appears to be infeasible even with series; while the gamma product sum in Jack d'Aurizio's answer here looks quite appealing, it only works for $n=5$ – only then can it be shown that the integral over $[0,\infty]$ is twice the integral over $[0,1]$, at which point you bring in beta functions. The other result in Jack's answer is a double sum, which can be generalised to other $n$ but is not very elegant (partly because of the double sum and partly because one bound of that sum uses a floor function).

If an approach that solves the task also gives integrals for the same integrand but with other bounds (e.g. $[0,1]$), that would be appreciated.

答:
Let $n\ge5$.
$$J_n=\int_0^\infty\sqrt{\frac{x-1}{x^n-1}}dx=\int_0^1\sqrt{\frac{x-1}{x^n-1}}dx+\int_1^\infty\sqrt{\frac{x-1}{x^n-1}}dx.$$
Then do $x\mapsto 1/x$ in the second integral, and add the two together:
$$J_n=\int_0^1\left(1+x^{(n-5)/2}\right)\sqrt{\frac{x-1}{x^n-1}}dx.$$
Using $$(1-q)^{-\alpha}=\,_1F_0(\alpha;;q),$$ we have
$$\sqrt{\frac{x-1}{x^n-1}}=\sum_{k\ge0}\beta_k^{(n)}x^k,$$
where $$\beta_k^{(n)}=\sum_{r=0}^{k}[n|r]\frac{(\tfrac12)_{r/n}(-\tfrac12)_{k-r}}{(r/n)!(k-r)!},$$
with $[a|b]$ being the Iverson Bracket for $b/a\in\Bbb Z$:
$$[a|b]=\left\lfloor \exp\left(a\left\lfloor \frac{b}{a}\right\rfloor-b\right)\right\rfloor.$$
Thus
$$J_n=\sum_{k\ge0}\beta_k^{(n)}\int_0^1(1+x^{(n-5)/2})x^kdx=\sum_{k\ge0}\beta_k^{(n)}\left(\frac1{k+1}+\frac{2}{2k+n-3}\right).$$
We have, additionally, that
$$\beta_k^{(n)}=\sum_{r=0}^{\lfloor k/n \rfloor}\frac{(\tfrac12)_{r}(-\tfrac12)_{k-nr}}{r!(k-nr)!}.$$
I am doubtful that there is a general closed form.


问:
Let $a$ be an algebraic number, prove that $ia$ is also algebraic
Of course this is a particular case of the closure under addition and multiplication of algebraic numbers (which proof would also be appreciated), but I would prefer a simple transformation on the already known polynomial giving $a$ as a root in order to obtain a polynomial giving $ia$ as a root. Thank you in advance for your help.

答:
Lemma: for any $p\in\mathbb{C}[x]$, we have $p\overline p\in\mathbb{R}[x]$, where $\overline p$ is obtained by complex conjugating every coefficient of $p$.

Proof: By separating out the real and imaginary parts of its coefficients, we can write $p$ as $p_{\text{Re}}+ip_{\text{Im}}$, where $p_{\text{Re}},p_{\text{Im}}\in\mathbb{R}[x]$. Then $\overline p=p_{\text{Re}}-ip_{\text{Im}}$, whence \begin{align}p\overline{p}&=(p_{\text{Re}}+ip_{\text{Im}})(p_{\text{Re}}-ip_{\text{Im}})\\&=p_\text{Re}^2+p_{\text{Im}}^2\in\mathbb{R}[x],\end{align} as desired.

Now, since $a$ is algebraic, let $p\in\mathbb{Q}[x]$ be an annihilating polynomial for $a$. By the lemma above, the polynomial $q(x):=p(-ix)\overline{p(-ix)}\in\mathbb{R}[x]$. Also, we have $p(-ix),\overline{p(-ix)}\in\mathbb{Q}[i][x]$, so in fact $q\in\mathbb{R}[x]\cap\mathbb{Q}[i][x]=\mathbb{Q}[x]$. We further have $$q(ia)=p(-i^2a)\overline{p}(-i^2a)=p(a)\overline{p}(a)=0,$$ so that $q$ is a polynomial with rational coefficients, annihilating $ia$, as desired.


问:
Inverse laplace transform $1/(s^2+9)^2$

Find inverse laplace transform of $$\frac{1}{(s^2+9)^2}$$

I've tried to decompose the fraction using

$$\frac{As+B}{s^2+9}+\frac{Cs+D}{(s^2+9)^2}$$

$$1=(As+B)(s^2+9)+Cs+D$$

yet D=1, still giving me the same exact equation

$$\frac{1}{(s^2+9)^2}$$

any help?

答:
One could make an ansatz with "known" expressions from a simple Laplace Transform Table:

$$\frac1{(s^2+a^2)^2}=\frac{Aa}{s^2+a^2}+\frac{Bs}{s^2+a^2}+\frac{2Cs}{(s^2+a^2)^2}+\frac{D(s^2-a^2)}{(s^2+a^2)^2},$$

given the Laplace transforms of $\sin at$, $\cos at$, $t\sin at$, and $t\cos at$, respectively. Hence, we get $A=1/(2a^3)$, $B=C=0$, and $D=-1/(2a^2)$. Therefore

$$\frac1{(s^2+a^2)^2}=\frac1{2a^3}\cdot\frac a{s^2+a^2}-\frac1{2a^2}\cdot\frac{s^2-a^2}{(s^2+a^2)^2}.$$

Title: Inverse laplace transform $1/(s^2+9)^2$


答:
This is not an answer « per se ». But should be useful for you to think about the topic.

How do you define $\lim\limits_{n \to \infty} x_n=p$?

Body:
be $M$ a metric space. if a sequence $(x_n)_{n\in\mathbb{N}}$ is such that $\lim_{n\rightarrow +\infty}x_n=p$. Prove that sequence of real numbers $(d(x_n,p))_{n \in \mathbb{N}}$ converges to $0$ in $\mathbb{R}$

I have tried to use the following the way, however I do not know if I am in the way direction. Could someone help me?

If $x_n\rightarrow p, y$ then $d(p, y) ≤ d(p, x_n) + d(x_n , y)\rightarrow 0.$
So $d(p, y) = 0$, implying that $p = y$.


问:
Study the uniform convergence of the integral $\int_0^\infty\frac{\sin \alpha x}{\alpha^2+x^2}\mathrm{d}x, \alpha\in(0,+\infty)$

Study the uniform convergence of the integral
$$\int_0^\infty\frac{\sin \alpha x}{\alpha^2+x^2}\mathrm{d}x,\quad \alpha\in(0,+\infty)$$

I have tried the Weierstrass criterion, but $\int_0^\infty\frac{1}{\alpha^2+x^2}\mathrm{d}x$ does not converge. Then I tried the Dirichlet criterion, but $\int_0^A \sin \alpha x$ are not bounded uniformly. So I turned to prove it does not converge uniformly. My attempt are as follows

$\forall A>1$, choose $\alpha=\frac{\pi}{4A}$ then $\forall x\in[A, 2A]$
$$
\left|\int_A^{2A}\frac{\sin \alpha x}{\alpha^2+x^2}\mathrm{d}x\right|>\frac{\sqrt{2}}{2}\frac{1}{\alpha}(\arctan\frac{2A}{\alpha}-\arctan\frac{A}{\alpha})
$$
but I cannot find a lower bound of RHS, so the method seems do not work.

Is my way wrong? You can also suggest your way. Appreciate any help!

答:
The improper integral

$$I(\alpha) = \int_0^\infty \frac{\sin \alpha x}{\alpha^2 + x^2} \, dx$$

converges uniformly by the Weierstrass-M test on any interval $[A,\infty)$ where $A > 0$ since for all $\alpha > A$,

$$\left|\frac{\sin \alpha x}{\alpha^2 + x^2} \right| \leqslant\frac{1}{A^2 + x^2} \,\,\text{and }\,\, \int_0^\infty\frac{dx}{A^2 + x^2 }< \infty$$

We can also show that convergence is uniform on $(0,A]$ and, hence, on $(0,\infty]$ using the Dirichlet test.

We can rewrite the integral as

$$I(\alpha) = \int_0^\infty \frac{x}{\alpha^2+x^2} \frac{\sin \alpha x}{x} \, dx$$

With the change of variables $t = \alpha x$ we have
$$\int_0^c \frac{\sin \alpha x}{x} \, dx = \int_0^{\alpha c} \frac{\sin t}{t} \, dt = \mathrm{Si}(\alpha c) $$

and the integral (see  sine integral ) is uniformly bounded for all $\alpha,c \in [0,\infty)$.

We also have the uniform convergence $f(x,\alpha) = \frac{x}{\alpha^2+x^2} \leqslant \frac{1}{x} \to 0$ as $x \to \infty$. Furthermore,  $f(x, \alpha)$ is monotonically decreasing with respect to $x$ for all $x \geqslant A$ independent of $\alpha \in (0,A]$. Note that $f(0,\alpha) = 0$, there is a maximum for $x = \alpha$, and then $f(x,\alpha)$ is decreasing to $0$ for all $x > \alpha$. Therefore, the improper integral is uniformly convergent for $\alpha \in (0,A]$.

Note that we only need eventual monotonicity of $f$ (independent of the parameter) to retain applicability of the Dirichlet test.  To see this, suppose $f(x,\alpha) \searrow 0$ uniformly and $f(x,\alpha)$ is decreasing with respect $x$ when $x \geqslant A$.

For all $c_2 > c_1 >A$, by the second mean value theorem for integrals (which requires monotonicity of $f$), there exist $\xi \in (c_1,c_2)$ such that

$$\int_{c_1}^{c_2} f(x,\alpha) g(x,\alpha) \, dx = f(c_1,\alpha) \int_{c_1}^\xi g(x, \alpha) \, dx +  f(c_2,\alpha) \int_{\xi}^{c_2} g(x, \alpha) \, dx, $$

and

$$\left|\int_{c_1}^{c_2} f(x,\alpha) g(x,\alpha) \, dx\right| \leqslant |f(c_1,\alpha)| \left|\int_{c_1}^\xi g(x, \alpha) \, dx \right|+  |f(c_2,\alpha)| \left|\int_{\xi}^{c_2} g(x, \alpha) \, dx\right|$$

Using the uniform boundedness of the integral of $g$ and the uniform convergence of $f(x,\alpha)$ to zero, we can show that for all $c_2 > c_1$ sufficiently large (and independent of $\alpha \in (0,A]$), the Cauchy criterion for uniform convergence is satisfied.


问:
A system of 2 equations
During a physics problem, I have encountered a set of 2 equations and after trying to solve them for quite a while, I start to thing there may be no solution. Am I missing something really easy?

$$ (1) A_{n}\cos\left(\sqrt{\lambda}_{n}R\right)+B_{n}\sin\left(\sqrt{\lambda}_{n}R\right)=0$$

$$(2)A_{n}\cos\left(\sqrt{\lambda}_{n}\left(R+\pi\right)\right)+B_{n}\sin\left(\sqrt{\lambda}_{n}\left(R+\pi\right)\right)=0$$

I'm searching for $\lambda_n$ that solves both equations, but nothing seems to work. I can choose any values I want for $A_n$ and $B_n$, I tried taking one of them to be zero but it doesn't help.

答:
Rewrite your set of equations in terms of matrices, it has the form:
$$\begin{bmatrix}
\cos(\sqrt{\lambda_n}R) & \sin(\sqrt{\lambda_n}R)\\
\cos(\sqrt{\lambda_n}(R+\pi)) & \sin(\sqrt{\lambda_n}(R+\pi))\end{bmatrix}
\begin{bmatrix}A_n\\ B_n\end{bmatrix} = \begin{bmatrix}0\\ 0\end{bmatrix}
$$
In order to have non-trivial solutions (i.e. $A_n, B_n$ not both zero), a necessary and sufficient condition is the $2\times 2$ matrix is degenerate and has zero determinant. i.e.

$$\cos(\sqrt{\lambda_n}R)\sin(\sqrt{\lambda_n}(R+\pi)) -
\sin(\sqrt{\lambda_n}R)\cos(\sqrt{\lambda_n}(R+\pi)) = 0\tag{*1}$$
Compare this with the trigonometry identity:

$$\sin(a - b) = \sin(a)\cos(b) - \sin(b)\cos(a)$$

One find $(*1)$ is equivalent to
$$\sin(\sqrt{\lambda_n}(R + \pi) - \sqrt{\lambda_n}R) = \sin(\sqrt{\lambda_n}\pi) = 0$$
Since $\sin(x) = 0$ if and only if $x$ is an integer multiple of $\pi$, to have non-trival solutions, you need
$$\sqrt{\lambda_n} = k \iff \lambda_n = k^2$$
for some non-negative integer $k$.

Title: A system of 2 equations


问:
Understanding the proof for "non empty intersection of closed sets with F.I.P" imply compactness
I was trying to understand the proof of the following theorem :

" A space X is compact if and only if every collection of closed subsets of X satisfying the finite intersection property has non-empty intersection."

The standard proofs of this theorem that I am seeing are same as the one covered in the question below :

Finite Intersection Property implies compactness?

As in the question above I could understand the proof for compactness implying non empty intersection of collection of closed sets having finite intersection property however I am not clear about the proof for the other direction .

Coming to the second proof as provided in the first answer above which goes as follows :

Suppose K has the finite intersection property. To prove that K is compact let {Ui}i∈I be a collection of open sets that cover K. We claim that this collection contains a finite subcollection of sets that also cover K.

Suppose that $K \neq\bigcup_{j \in J} U_j$ where $J\subset I$ is finite. Taking compliments gives $K^c \neq \bigcap U_j^c$, which by hypothesis is non-empty - since $U_i$ is open, $U_i^c$ is closed. Since $K$ has the fip we thus have that

$ \emptyset \neq \bigcap_{i \in I} U_i^c = \left( \bigcup_{i \in I} U_i \right)^c$. This contradicts $U_i$ being an open cover for $K$.

Here  I am unable to see the significance of the part of this proof with the finite set $ J $  which is contained in $ I $ . Could we not have directly started with  the following part ?

$ \emptyset \neq \bigcap_{i \in I} U_i^c = \left( \bigcup_{i \in I} U_i \right)^c$

Most importantly we obtained a contradiction to {Ui}i∈I being an open cover of the set in question . I can't see how does that lead to compactness ?  How does it mandate the existence of an open subcover for all such open covers of this set ?

I saw a similar proof in the following blog :

https://dantopology.wordpress.com/2009/11/30/the-finite-intersection-property-in-compact-spaces-and-countably-compact-spaces/

but here as well I could not see how the contradiction leads to existence of a subcover .

答:
A family of sets with the finite intersection property is said to be centred; for convenience I will use that term.

Dan Ma’s proof is not by contradiction. He wants to prove that if every centred family of closed sets in $X$ has non-empty intersection, then $X$ is compact. To do this, he proves the contrapositive: if $X$ is not compact, then $X$ has a centred family of closed sets whose intersection is empty. This is logically equivalent to the desired implication.

The argument itself is straightforward. Suppose that $X$ is not compact; then it has an open cover $\mathscr{U}$ with no finite subcover. For each $U\in\mathscr{U}$ let $F_U=X\setminus U$, and let $\mathscr{F}=\{F_U:U\in\mathscr{U}\}$; clearly $\mathscr{F}$ is a family of closed sets. Let $\mathscr{F}_0$ be any finite subset of $\mathscr{F}$. There is a finite $\mathscr{U}_0\subseteq\mathscr{U}$ such that $\mathscr{F}_0=\{F_U:U\in\mathscr{U}_0\}$. Then

$$\bigcap\mathscr{F}_0=\bigcap_{U\in\mathscr{U}_0}F_U=\bigcap_{U\in\mathscr{U}_0}(X\setminus U)=X\setminus\bigcup\mathscr{U}_0\,.$$

$\mathscr{U}$ has no finite subcover, so $\bigcup\mathscr{U}_0\ne X$, and therefore

$$\bigcap\mathscr{F}_0=X\setminus\bigcup\mathscr{U}_0\ne\varnothing\,.$$

Thus, $\mathscr{F}$ is centred: every finite subset of $\mathscr{F}$ has non-empty intersection. But

$$\bigcap\mathscr{F}=\bigcap_{U\in\mathscr{U}}(X\setminus U)=X\setminus\bigcup\mathscr{U}=\varnothing\,,$$

since $\mathscr{U}$ is a cover of $X$, so $\mathscr{F}$ is a centred family of closed sets in $X$ whose intersection is empty.

The proof that you copied into your question uses essentially the same idea but does organize it as a proof by contradiction. I’ll try to present it a bit more clearly. We start with an arbitrary open cover $\mathscr{U}=\{U_i:i\in I\}$ of a compact space $K$, and we suppose, to get a contradiction, that it has no finite subcover. Then for each finite $J\subseteq I$ we know that $\bigcup_{j\in J}U_j\ne K$. Now for each $i\in I$ let $F_i=K\setminus U_i$; then $\mathscr{F}=\{F_i:i\in I\}$ is a family of closed sets in $K$, and for each finite $J\subseteq I$ we have

$$\bigcap_{j\in J}F_j=\bigcap_{j\in J}(K\setminus U_j)=K\setminus\bigcup_{j\in J}U_j\ne\varnothing\,,$$

so $\mathscr{F}$ is centred. We are assuming that every centred family of closed sets in $K$ has non-empty intersection, so we conclude that $\bigcap\mathscr{F}=\bigcap_{i\in I}F_i\ne\varnothing$. But then

$$\bigcup\mathscr{U}=\bigcup_{i\in I}U_i=\bigcup_{i\in I}(K\setminus F_i)=K\setminus\bigcap_{i\in I}F_i\ne K\,,$$

contradicting the fact that $\mathscr{U}$ is a cover of $K$. This contradiction shows that there must in fact be a finite $J\subseteq I$ such that $\bigcup\{U_j:j\in J\}=K$, i.e., such that $\{U_j:j\in J\}$ is a finite subcover.


问:
Prove by induction that $1*3+2*3^2+3*3^3 + \cdots + n*3^n = \dfrac{3}{4}(3^n(2n-1)+1)$
I'm trying to prove this using induction
$1*3+2*3^2+3*3^3 + \cdots + n*3^n = \dfrac{3}{4}(3^n(2n-1)+1)$

So far I have:

Base case: true

Induction step:
$\dfrac{3}{4}(3^n(2n-1)+1)+(n+1)*3^{n+1}=\dfrac{3}{4}(3^{n+1}(2(n+1)-1)+1)$

here is where I get stuck on simplification of both left and right side so that they are equal
$\dfrac{3}{4}(3^n(2n-1)+1)+3^{n+1}+3^{n+1}n=\dfrac{3}{4}(3^{n+1}(2(n+1)-1)+1)$

Thanks for your help!

答:
Notice that
$$3^{n+1}(n+1) = 3 \cdot 3^n (n+1) = \frac{3}{4} \cdot 3^n \cdot 4(n+1)$$
Because of the 3/4, you can move this term into the parentheses and combine it with the $3^n(2n-1)+1$ part. And because of the $3^n$, you can move the $4(n+1)$ into the parentheses and combine with the $2n-1$ part, to get

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

Notice that $6n+3 = 3(2n+1)$, so in the parentheses you get $3^{n+1}(2n+1)+1$.


问:
Proving $\exp (T + S) = \exp (T) \exp (S) = \exp (S) \exp (T)$ for $T,S\in\mathcal{L}(E)$

Let $E$ be a Banach space and $T,S \in \mathcal L (E)$ be two commuting maps i.e., $TS = ST.$ Then, $$\exp (T + S) = \exp (T) \exp (S) = \exp (S) \exp (T).$$

I can able to prove it using the formula for exponential but the proof is very clumsy. Is there any easier way to prove the above equality? Any help in this regard will be appreciated.

Thanks in advance.

答:
Here's a different proof than the one linked (it also works in any Banach algebra). First, you can check that the power series $$e^{tS}=\sum\limits_{n=0}^\infty\frac{t^n}{n!}S^n$$ makes sense for all $t\in\mathbb{R}, S\in\mathcal{L}(E).$ Differentiating in $t$ gives $$\frac{d}{dt}e^{tS}=Se^{tS}=e^{tS}S.$$

Next, compute that $$\frac{d}{dt}(e^{t(S+T)}e^{-tT}e^{-tS})=e^{t(S+T)}Se^{-tT}e^{-tS}-e^{t(S+T)}e^{-tT}Se^{-tS}.$$ We claim this equals zero. Indeed, the fact that $ST=TS$ implies that $$e^{-tT}S=\sum\limits_{n=0}^\infty \frac{(-t)^n}{n!}T^nS=S\sum\limits_{n=0}^\infty \frac{(-t)^n}{n!}T^n=Se^{-tT},$$ from which it follows that $Se^{-tT}=e^{-tT}S$.

Hence, $$e^{t(S+T)}e^{-tT}e^{-tS}$$ is constant in $t$. Evaluating at $t=0$ gives that $$e^{t(S+T)}e^{-tT}e^{-tS}=I.$$ All that we must show is that the inverse of $e^{tS}$ is $e^{-tS},$ as the result will follow from inversion (multiply on the right by $e^{tS}$, then by $e^{tT}$). Indeed, differentiate $e^{(s+t)S}e^{-tS}$ in $t$. You'll get that it's zero, so it's constant in $t$, and evaluation at $t=0$ gives $$e^{(s+t)S}e^{-tS}=e^{sS}.$$ Evaluate this at $s=0$ to get the inversion property.

This shows that $$e^{t(S+T)}=e^{tS}e^{tT}.$$ Evaluating at $t=1$ gives the result that you wanted (well, the first equality; to get the other, simply follow the same argument with $e^{t(S+T)}e^{-tT}e^{-tS}$).

Alternatively, this results directly from e.g. the holomorphic functional calculus. This also works in any Banach algebra.


问:
Open, closed and continuous
I have some troubles to understanding something:

We were asked to find a function that is open and continuous but not closed and actually I found such a function, but our tutor gave us this example

$ e^x$ since this function is continuous as a mapping $$\exp: \mathbb{R} \rightarrow \mathbb{R}_{>0}$$ and the inverse function is continuous too, so this function is also open.

and the counterexpample was, that $\mathbb{R}$ is closed and is mapped to $\mathbb{R}_{>0}$ which is open.

then i thought, hey if the inverse function is continuous that means that the fiber of closed sets have to be closed, but $\ln^{-1}(\mathbb{R})$ is not closed at all. where am i wrong?

答: Your tutor is wrong. Every homeomorphism (such as $\exp:\mathbb R\to\mathbb R_{>0}$) is closed. The set $\mathbb R_{>0}$ is a closed set in the standard topology on $\mathbb R_{>0}$.</p>

Title: Open, closed and continuous


答:
In the induction step you want to prove the increment from $n$ to $n+1$ on both sides are the same, namely,

$$(n+1)3^{n+1} = \dfrac{3}{4}(3^{n+1}(2n+1)-3^n(2n-1))\tag1$$

By looking at the increment you got rid of the annoying constant $1$ from the RHS. Then you notice immediately that you can divide $3^{n+1}$ from both sides, and $(1)$ becomes

$$n+1 = \dfrac{1}{4}(3(2n+1)-(2n-1))=\frac 14 (6n+3-2n+1)=\frac 14 (4n+4)=n+1$$

then you are done.

Title: Prove by induction that $1*3+2*3^2+3*3^3 + \cdots + n*3^n = \dfrac{3}{4}(3^n(2n-1)+1)$

Body:
I'm trying to prove this using induction
$1*3+2*3^2+3*3^3 + \cdots + n*3^n = \dfrac{3}{4}(3^n(2n-1)+1)$

So far I have:

Base case: true

Induction step:
$\dfrac{3}{4}(3^n(2n-1)+1)+(n+1)*3^{n+1}=\dfrac{3}{4}(3^{n+1}(2(n+1)-1)+1)$

here is where I get stuck on simplification of both left and right side so that they are equal
$\dfrac{3}{4}(3^n(2n-1)+1)+3^{n+1}+3^{n+1}n=\dfrac{3}{4}(3^{n+1}(2(n+1)-1)+1)$

Thanks for your help!


问:
Semi-Complicated recurrence relation to be solved via generating functions
If $(a_n)_{n\ge 1}$ is a sequence such that $a_0=3$ and for all $n \ge 1$,
\begin{equation}
\sum_{k=1}^{\lceil \frac{n}{2} \rceil}a_{n-2k+1}=na_n
\end{equation}
, use generating functions to find an explicit formula for $a_n$.

So far I've been swimming around in the algebra a bit, and I've gotten

$$A'(x)=\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=1}^\infty\left(\sum_{k=1}^{\lceil n/2 \rceil} a_{n-2k+1}\right)x^{n-1}$$

But I'm not sure how to get a nice expression in terms of $A(x)$ for the RHS of this equation.

答:
Here are the steps to obtain the differential equation:
\begin{align}
A'(x)
&= \sum_{n=1}^\infty\left(\sum_{k=1}^{\lceil n/2 \rceil} a_{n-2k+1}\right)x^{n-1} \\
&= \sum_{k=1}^\infty x^{2k-2} \sum_{n=2k-1}^\infty a_{n-2k+1} x^{n-2k+1} \\
&= \sum_{k=1}^\infty x^{2k-2} A(x) \\
&= A(x) \sum_{k=1}^\infty (x^2)^{k-1} \\
&= \frac{A(x)}{1-x^2} \\
\end{align}


问:
If $p \in\mathbb{R}^n$, $r > 0$, and $U = \{x \in\mathbb{R}^n \mid\|p-x\| > r\}$, then $U$ is an open subset of $\mathbb{R}^n$
Prove or give a counterexample: if $p \in \mathbb{R}^n$, $r > 0$, and $U = \{x \in \mathbb{R}^n \; | \; ||p − x|| > r\}$, then $U$ is an open subset of $\mathbb{R}^n$.

I think that it should be open but I can't figure out how to prove it.

The definition of open I am  using is for every $x \in U$, there exists a $\delta > 0$ such that  $B(x, \delta)$ is a subset of $U$.

答:
Let $W = \mathbb{R}^n \backslash U$ then $W = \mathcal{B}(0,r) = \{x \in \mathbb{R}^n \; | \; || p-x||\leq r\}$ is by definition a closed ball, and you can show that a closed ball is always a closed subset of $\mathbb{R}^n$. Therefore its complement that is $U$ is an open set of $\mathbb{R}^n$.

If you are not at ease with my first explanation, try to draw it in two dimensions : here $U$ is the portion of the plan the outside of the circle centered in $p $ with radius $r$. Take a point $q = (x,y) \in U $ and start to draw a circle that would stay in $U$, i.e. that will not intersect with your first circle.

Hint : you can take for the radius of your second circle $ \delta = \frac{1}{2}(||p-q||-r)$, you can choose also $\delta = ||p-q||-r$ as @Kevin John suggests.


问:
How do you solve $y'=\sin(y)$?

How do you solve $y'=\sin(y)$?

I ended up with
$$-\ln(\csc(y)+\cot(y)) = x+C$$

But then I simplified it to
$$e^{-x+C} = \frac1{\sin(y)} + \frac{\cos(y)}{\sin(y)}$$

Where do I go from here?

答:
It's better to use the Weierstrass substitution :
$$I=\int \dfrac {dy}{\sin y}=\int \dfrac {1+t^2}{2t}\dfrac {2}{1+t^2}dt$$
$$I=\int \dfrac {dt}{t}=\ln t +C= \ln \left|\tan \dfrac y 2 \right|+C$$
Where $t=\tan (y/2)$

Title: How do you solve $y'=\sin(y)$?


问:
Partial fraction expansion two variables
How to expand

$$\frac{y}{(x-y)(y-1)}$$

by partial fraction expansion.

Body:
HINT

$$\frac{y}{(x-y)(y-1)} = \frac{Ax+By+C}{x-y} + \frac{Dx+Ey+F}{y-1}$$


问:
Find the distribution of a sum of Poisson random variable
Let $X_1,X_2,..,X_N$ be independent identical samples from a Poission distribution with unknown parameter $\theta$

How would I find the sum of the distribution $X_1+X_2+X_3+..+X_N$ when n is finite. I looked up convolution but that only works for two random variables and I have multiple random variables and they are all poisson.

答:
This what I did not sure if it is right.
$W=X_1+X_2+..+X_N$
we have $M_w(t)=E(e^{X_1+X_2+..+X_N})=E(e^{x_1t}*e^{x_2*t}*..e^{x_n*t})$
And this equals

$M_{X_1}(T)*M_{X_2}(T)*..*M_{X_N}(t)=[M_X(t)]^n=[e^{\theta(e^t-1)}]^n$

So now you $W~Poisson(n*\theta))$ which is the same as pmf $(e^{-n*\theta}(n*\theta)^x)/x!$


问:
find $\lim_{x \to \infty} \frac{\ln(x+1)}{x}$
I need to find: $$\lim_{x \to \infty} \frac{\ln(x+1)}{x}$$
without using L'Hospital, integrals or derivative!

I would like your help in order to know i'm not doing something invalid.

I was thinking of the following way:

$$
\lim_{x \to \infty } \frac{\ln(x+1)}{x}  =  \lim_{x \to \infty } \frac{1}{x} \ln(x+1)  =  \lim_{x \to \infty } \ln(x+1)^\frac{1}{x}
$$

and because $\frac{1}{x}$ tend to $0$ as $x$ approaches $\infty$, than $(x+1)^\frac{1}{x}$ tends to $1$ and $\ln(1) = 0$
and the limit is $0$.

But this way sounds off. I was trying also to express $x$ with $t$ and use $e$ (euler) somehow, but it didn't work.

Is my way valid? If not, do you have a better way without using L'Hospital's rule?

Thanks!

Edit:

While reading answers I was thinking of another way,

$$\lim_{x \to \infty} \frac{\ln(x+1)}{x}  =  \lim_{x \to \infty} \frac{\ln(x(1+\frac{1}{x}))}{x} = \lim_{x \to \infty} \frac{\ln(x)}{x} + \frac{\ln(1+\frac{1}{x})}{x} = $$ $$\lim_{x \to \infty} \ln(\sqrt[x]{x}) + \frac{\ln(1+\frac{1}{x})}{x}$$

now $\sqrt[x]{x}$ tends to $1$ as x approaches to $\infty$, hence $\ln(\sqrt[x]{x})$ tends to $0$ (is it valid to say so?)

and $\frac{\ln(1+\frac{1}{x})}{x}$ tends to $0$ is pretty obvious.

Am on missing something in this one?

答:
It is easy to show (See Here) that $\log(x)<\sqrt{x}$ for all $x>0$.  Therefore, we have

$$\frac{\log(1+x)}{x}\le \frac{\sqrt{x+1}}x$$

And you can finish now.


答: Without L'Hospital Rule (or series for that matter!) you are left with using some basic properties of functions. For example set $\ln(x+1)=t$ so that $x=e^t-1$ and now you have an expression $\frac{t}{e^t-1}$ with $t $ going to infinity. To what extend does your teacher want you to prove that an exponential functions grows much faster than a linear term?

Title: find $\lim_{x \to \infty} \frac{\ln(x+1)}{x}$

Body:
I need to find: $$\lim_{x \to \infty} \frac{\ln(x+1)}{x}$$
without using L'Hospital, integrals or derivative!

I would like your help in order to know i'm not doing something invalid.

I was thinking of the following way:

$$
\lim_{x \to \infty } \frac{\ln(x+1)}{x}  =  \lim_{x \to \infty } \frac{1}{x} \ln(x+1)  =  \lim_{x \to \infty } \ln(x+1)^\frac{1}{x}
$$

and because $\frac{1}{x}$ tend to $0$ as $x$ approaches $\infty$, than $(x+1)^\frac{1}{x}$ tends to $1$ and $\ln(1) = 0$
and the limit is $0$.

But this way sounds off. I was trying also to express $x$ with $t$ and use $e$ (euler) somehow, but it didn't work.

Is my way valid? If not, do you have a better way without using L'Hospital's rule?

Thanks!

Edit:

While reading answers I was thinking of another way,

$$\lim_{x \to \infty} \frac{\ln(x+1)}{x}  =  \lim_{x \to \infty} \frac{\ln(x(1+\frac{1}{x}))}{x} = \lim_{x \to \infty} \frac{\ln(x)}{x} + \frac{\ln(1+\frac{1}{x})}{x} = $$ $$\lim_{x \to \infty} \ln(\sqrt[x]{x}) + \frac{\ln(1+\frac{1}{x})}{x}$$

now $\sqrt[x]{x}$ tends to $1$ as x approaches to $\infty$, hence $\ln(\sqrt[x]{x})$ tends to $0$ (is it valid to say so?)

and $\frac{\ln(1+\frac{1}{x})}{x}$ tends to $0$ is pretty obvious.

Am on missing something in this one?


答: Let $c\in\mathbb{R}$ and $f(x)=\log(x)-\sqrt{x}+c$, then $f'(x)=\frac{1}{x}-\frac{1}{2\sqrt{x}}=\frac{2-\sqrt{x}}{2x}$, therefore the maximum of $f$ is $f(4)=\log 4-2+c$. We take $c=2-\log 4$, we thus have $f(x)\leqslant 0$ for all $x>0$, which leads to $\frac{\log(x)}{x}\leqslant\frac{1}{\sqrt{x}}+\frac{\log 4-2}{x}\underset{x\rightarrow +\infty}{\longrightarrow}0$.


答:
As @Axel said, you can show $B(0,r)$ is closed. You can also show the openness directly. Suppose $y\in U$ and let $\delta = ||y-p||-r$. You may show this $\delta$ works.

Edit: Let $y \in U = \{x \in \mathbb{R}^n \; | \; ||p − x|| > r\}$ and set $\delta = ||y-p||-r$. Then for any $z\in B(y,\delta)$, we have$||z-p|| = ||(z-y)+(y-p)||\\\quad\quad\quad\,\geq ||\,||y-p||-||z-y||\,||
\\\quad\quad\quad\,=||y-p||-||z-y||>||y-p||-\delta = r$

so we have $B(y,\delta) \subset U$. And therefore $U$ is open.

Title: If $p \in\mathbb{R}^n$, $r > 0$, and $U = \{x \in\mathbb{R}^n \mid\|p-x\| > r\}$, then $U$ is an open subset of $\mathbb{R}^n$

Body:
Prove or give a counterexample: if $p \in \mathbb{R}^n$, $r > 0$, and $U = \{x \in \mathbb{R}^n \; | \; ||p − x|| > r\}$, then $U$ is an open subset of $\mathbb{R}^n$.

I think that it should be open but I can't figure out how to prove it.

The definition of open I am  using is for every $x \in U$, there exists a $\delta > 0$ such that  $B(x, \delta)$ is a subset of $U$.


答:
I will use the substitution $y=\ln(1+x)$.
Let $n$ be $n=[y]$
Then $x=e^y-1$. We can simply estimate for $x>10$ (say):
$$
0\le \frac{\ln(x+1)}x=\frac y{e^y-1}<\frac y{e^y/2}<\frac y{2^y/2}
<
\frac {n+1}{2^{n-1}}
=
\frac {n+1}{(1+1)^{n-1}}
<
\frac {n+1}{\binom{n-1}2}
<\frac {2n}{\frac n2\cdot\frac n2\cdot \frac 12}
=\frac {16}n\to 0
\ .
$$