问:
Why doesn't my derivation of the arcsec derivative formula not work?
We would like to find the derivative of $\operatorname{arcsec}(x) = y$. Rearranging this, we get $x = \sec(y)$. Taking the derivative of both sides, we get $1 = \sec(y)\tan(y)y^\prime$. Thus, $$y^\prime = \frac{1}{\sec(y)\tan(y)}.$$ Now, draw a right triangle with an acute angle $y$, hypotenuse $x$, and a side adjacent to angle $y$ with length $1$. By the Pythagorean theorem, the side opposite angle $y$ is equal to $\sqrt{x^2-1}$. Thus, we have $x = \sec(y)$, which we already knew and got from the triangle, and $$\tan(y) = \frac{1}{\sqrt{x^2-1}}$$ from the triangle we drew. We substitute these values in to our derivative expression to get $$y' = \frac{1}{x\sqrt{x^2-1}}.$$

However, $y^\prime$ should equal $$\frac1{|x|\sqrt{x^2-1}}$$ (note the absolute value). Since the difference occurred in the $|x|$ part, that means I must've done something wrong when I said $\sec(y) = x$, but I do not know why this assumption is wrong. Please correct my proof.

答:
The sloppy reasoning occurs with the right triangle you wrote. You must be careful about the range (values) of the arcsec function. $y=\text{arcsec}(x)$ can lie either in $[0,\pi/2)$ or in $(\pi/2,\pi]$. (These correspond, respectively, to $x>0$ and $x<0$.)

$\tan(y)<0$ when $y\in (\pi/2,\pi]$, and so $\tan(y)=-\sqrt{x^2-1}$ in that event. Working with your formula for $y'$, we note that when $x<0$,
$$\sec(y)\tan(y)= x(-\sqrt{x^2-1})= (-x)\sqrt{x^2-1} =|x|\sqrt{x^2-1},$$
and this explains the formula. To be honest, it's a bit sneaky to move the negative to the other term, but it allows us to write a single formula, rather than writing down cases. (There's no problem, of course, when $x>0$.)


问:
Solving exponential equations for maxent with opensource applications or coding
I am working on maximum entropy. And I have a problem solving some equations.

Look at these 2 equations:

$\frac{(2*e^{-a*2-b*2^2}) +(6*e^{-a*6-b*6^2}) +(10*e^{-a*10-b*10^2}) +(14*e^{-a*14-b*14^2})}{(e^{-a*2-b*2^2})+(e^{-a*6-b*6^2})+(e^{-a*10-b*10^2})+(e^{-a*14-b*14^2})}  = 5.77 $

$\frac{(4*e^{-a*2-b*2^2}) +(36*e^{-a*6-b*6^2}) +(100*e^{-a*10-b*10^2}) +(196*e^{-a*14-b*14^2})}{ (e^{-a*2-b*2^2})+(e^{-a*6-b*6^2})+(e^{-a*10-b*10^2})+(e^{-a*14-b*14^2})} = 45.48 $

I need advice for solving for a and b. There are 2 equation and 2 unknown variables $\{a,b\}$

These equations can be 3 or 4 in number, and unknown variables also be 3 or 4 in number.

And that coefficient $(2,6,10,14)$ can increase to $2,6,10,14,...,n$ or $1,3,5,7,...,n$

My real problem is solving equations and I am in need of ideas to solve them. I tried Matlab and solved them but I need to create that algorithm or using some open source solutions.

You can look here ; http://mathbin.net/175372

OwnerDisplayName: mary blasso

ContentLicense: CC BY-SA 3.0

Body:
This can be solved because there is 2 unknown var and 2 equations. It is exponantial but it can be solved.

I tried to use octave (http://www.gnu.org/software/octave/)
octave -q filename.math

and filename.math seems like;

function y = f (x)

       y = zeros (2, 1);

    y(1) = ((0.1*exp(-0.1*x(1)-0.1^2*x(2)))  +(0.3*exp(-0.3*x(1)-0.3^2*x(2)))  +(0.5*exp(-0.5*x(1)-0.5^2*x(2)))  +(0.7*exp(-0.7*x(1)-0.7^2*x(2)))  +(0.9*exp(-0.9*x(1)-0.9^2*x(2)))  )/((exp(-0.1*x(1)-0.1^2*x(2)))+(exp(-0.3*x(1)-0.3^2*x(2)))+(exp(-0.5*x(1)-0.5^2*x(2)))+(exp(-0.7*x(1)-0.7^2*x(2)))+(exp(-0.9*x(1)-0.9^2*x(2))))-0.471287128713;

    y(2) = ((0.01*exp(-0.1*x(1)-0.1^2*x(2)))  +(0.09*exp(-0.3*x(1)-0.3^2*x(2)))  +(0.25*exp(-0.5*x(1)-0.5^2*x(2)))  +(0.49*exp(-0.7*x(1)-0.7^2*x(2)))  +(0.81*exp(-0.9*x(1)-0.9^2*x(2)))  )/((exp(-0.1*x(1)-0.1^2*x(2)))+(exp(-0.3*x(1)-0.3^2*x(2)))+(exp(-0.5*x(1)-0.5^2*x(2)))+(exp(-0.7*x(1)-0.7^2*x(2)))+(exp(-0.9*x(1)-0.9^2*x(2))))-0.307722772277;

endfunction

  [x] = fsolve (@f, [0; 1])

I am not sure about selecting root for solving.
I am not sure fsolve has a good solver but i have one solution at least.


问:
Beta Reduction Constraints
The definition of $β$ reduction is the following :

$$(λx.M)N \rightarrow_{β} Μ[x∶=N] $$

So basically we stop treating $x$ as a bound variable and we perform substitution of the now free variable $x$ in the $M$ term under the constraint that the free variables of $N$ are still free. I was browsing the Internet for some examples to solve and I came across the following Stack Overflow Question

I mostly care about the first 2 terms of the $λ$ expression which are :
$$λz.(λm.λn.m)(z)$$
When it comes to the evaluation of this expression we start from the $λm$ term and the result according to the answer to the posted question will be :
$$λz.λn.z$$

Before the evaluation $N$ was $(z)$ and $FV(N)={z}$.

However after the evaluation the constraint is broken and now $z$ is bound in the expression  $λz.λn.z$.

Maybe there is some correlation between the parenthesis before the $λm$ and the scope of $λz$.

Any help would be appreciated.

答:
The point is that the $\beta$-rule
$$
(\lambda x.M)N \to_\beta M[x:= N]
$$
can be applied in any context. This means that, in a term $P$, if there is a subterm of the form $(\lambda x.M)N$, you can apply the $\beta$-rule to that subterm, leaving the rest of $P$ unchanged.

In your example, the context is $\lambda z.\langle\,\rangle$, which means that the term $\lambda z.(\lambda m.\lambda n. m)z$ is obtained as the plugging of the subterm $(\lambda m.\lambda n. m)z$ into the hole $\langle\,\rangle$ of the context $\lambda z. \langle\,\rangle$. Often this is denoted by $\lambda z. \langle(\lambda m.\lambda n.m)z\rangle$.

The $\beta$-rule acts inside the hole $\langle\,\rangle$ of the context, and stuff outside the hole is left unchanged.
Therefore, the $\beta$-step is
\begin{align}\tag{1}
\lambda z. (\lambda m.\lambda n.m)z &= \lambda z. \langle(\lambda m. \lambda n.m)z\rangle \\
&\to_\beta \lambda z. \langle(\lambda n. m)[m := z]\rangle = \lambda z. \langle\lambda n.z\rangle = \lambda z. \lambda n.z
\end{align}

The constraint about free variables holds for the subterm where you apply the $\beta$-rule (in this case, $(\lambda m.\lambda n.m)z$ where $z$ is free and it is still free after the $\beta$-step), but not for the whole term that is evaluated (in this case $\lambda z. (\lambda m .\lambda n.m)z$). Plugging a term into a context can bind some free variables (in this case, it happens for $z$), but this is irrelevant for the $\beta$-rule, since the $\beta$-rule is carried out inside the hole. Note that in $(1)$, if you consider the whole terms, the variable $z$ is bound before and after the $\beta$-step.

ContentLicense: CC BY-SA 4.0

Title: Beta Reduction Constraints


问:
Does algebraic structure on a set has any influence on its topological structure?
In the beginning chapters of real analysis we learn the algebraic properties of the real no.s (i.e., operations defined on $\mathbb{R}$, order properties and completeness properties of $\mathbb{R}$) as an algebraic structure imposed over $\mathbb{R}$. And then suddenly we are introduced to the topology of real no.s where we learn about neighbourhood, open set, limit points etc. But doesn't the algebraic properties of real no.s like the total order property, play any role in the topology of real no.s? I feel like there's an abrupt gap between the bridging of algebra and topology of $\mathbb{R}$ and I want to know if there's any source which can explain to me how topology of real no.s (or in that sense general topology) flows out naturally from it's algebraic structure.

答:
If the algebraic structure is unrelated to the topological structure no, but if they are linked, then absolutely yes.

For example, a topological vector spaces become completely regular when they are only Hausdorff because the vector space structure with continuous operations makes it a uniform space.

This isn't quite what you asked for but I think it is also interesting: when you have a smooth manifold, the differentiable structure inherited by the cotangent bundle gives rise to a special algebraic operation on the algebra of smooth functions on the cotangent bundle: a Poisson bracket.


问:
Constructing the submodule in Maschke's Theorem
Given a finite group $G$ and a field $K$ such that the characteristic of $K$ does not divide $|G|$, Maschke's Theorem states that if $V$ is a $KG$-module and $V_1 \subseteq V$ is a submodule, then there exists a complementary submodule $V_2 = \ker q \subseteq V$, where $q \in \text{Hom}_{KG}(V,V)$ is given as $q(v) := \frac{1}{|G|} \sum_{g\in G} g(\pi(g^{-1}v)) $, and $\pi \in \text{Hom}_K (V,V_1)$ is a $K$-linear projection. My question is related to actually writing down (a $K$-basis for) $V_2$. Is it fair to say that so long as the hypotheses are satisfied I can think of the problem as if it were just a $K$-vector space? In other words, if I can write down $V_1, V_2$ as $K$-vector spaces satisfying $V_1 \oplus V_2 = V$ as a $K$-vector space, then $V_2$ as a $KG$-module has the same basis as $V_2$ as a $K$-vector space?

For context, the concrete problem is to have $\mathbb{Q}S_3$ acting on $\mathbb{Q}\{x_1,x_2,x_3\}$ in the natural way. Then I have a known submodule $V_1$, and I want to write down the complementary submodule given by Maschke's Theorem. My thought process is to just forget the $\mathbb{Q}S_3$ structure of $V_1$ and deal with it as just as a three-dimensional $\mathbb{Q}$-vector space. Then I can find a basis for $V_2$ satisfying $V_1 \cap V_2 = \emptyset$ and $V_1 \oplus V_2 = \mathbb{Q}^3$. Then I checked if that basis generates a $\mathbb{Q}S_3$-submodule with the necessary properties.

答:
Yes. You're right that this is Maschke's Theorem. No. You're not quite right that we can take $V_2$ itself to be invariant.

Consider $G = \mathbb{Z} / 2$ acting on $k^2$ by reflection, for instance. Then obviously the axis of reflection is a sub-$kG$-module, and any other (one dimensional) subspace is a complement of vector spaces. However, it's clear that the only choice of complement that is itself a sub-$kG$-module is the line orthogonal to the axis of reflection (do you see why?).

What Maschke's Theorem buys us is the existence of a complement that's also a $kG$-module, but it might not be the one we started with. When we perform the "averaging" maneuver, we will almost certainly get back a different complement.

So, to answer your original question: How do we get the basis of a $kG$-complement? It may be easy to guess one, but it would be nice to know how to do it in general.

Let $U \subseteq V$ be a sub-$kG$-module. All the usual caveats apply ($G$ should be finite and its order should be coprime to the characteristic of $k$). Then if $\pi : V \to U$ is any old projection of vector spaces, we can average it to get

$$\pi_G = \frac{1}{|G|} \sum_{g \in G} g \pi g^{-1}$$

Then (for a certain definition of "easy to see") it's easy to see that $\pi_G$ is still a $k$-linear map from $V$ onto $U$ (indeed, it still splits the inclusion map), but now it has the bonus property of being a $kG$-linear map (do you see why?). So $\text{ker}(\pi_G)$ (which is itself a $kG$-module) is the desired complement.

Now we're in the home stretch, though, because we can regard $\pi_G$ as an ordinary $k$-linear map. So finding a basis for its kernel is "just linear algebra" in the sense that any of us could do it (or at least, we should be able to... I, for one, am probably rusty). But a computer algebra system like sage can absolutely slaughter this problem. As a (fun?) exercise, you might write up this algorithm in sage. It's not so bad, and it'll convince you that you understand it.

Of course, for many problems there's an "obvious" complement to your subspace. In practice, the guess-and-check method that you employed with the obvious complement is a good way to go. If worse comes to worse, though, your only recourse might be some honest computation:

Fix any old $k$-linear projection $\pi : V \to U$

Average it to $\pi_G : V \to U$

Compute a basis for $\text{ker}(\pi_G)$

I hope this helps ^_^


问:
Torus: Circle cut
Given a Torus $T$ with major and minor radius $R$ and $r$, respectively, I can obtain a circle lying in $T$ by cutting $T$ with a bi-tangential plane.

Now I don't want circles, but Tori with major radius $R$ and minor radius $r' \ll r$.

Given $R,r,r'$ and assuming that $T$ lies in the $X-Y$ plane:

What is the minimum angle I have to rotate the cutting plane around the $Z$-axis, such that I can place two $(R,r')$ tori as close together as possible?

Thanks.

答:
The Villarceau circles can be parametrized by

$$\vec{x}=\left(\begin{array}{ccc}
\cos\phi & \sin\phi & \\
-\sin\phi & \cos\phi & \\
&&1
\end{array}
\right)
\left(
\begin{array}{c}
\rho\sin\theta\\
r+R\cos\theta\\
r\sin\theta
\end{array}
\right)\;,
$$

where $\theta$ is the angle along the circle, $\phi$ is the orientation of the circle around the $z$-axis and $\rho=\sqrt{R^2-r^2}$. At $\phi=0$, the derivatives with respect to $\theta$ and $\phi$ are

$$\frac{\partial\vec{x}}{\partial\theta}=
\left(
\begin{array}{c}
\rho\cos\theta\\
-R\sin\theta\\
r\cos\theta
\end{array}
\right)\;,
\;\;\;
\frac{\partial\vec{x}}{\partial\phi}=
\left(
\begin{array}{c}
r+R\cos\theta\\
-\rho\sin\theta\\
0
\end{array}
\right)\;.
$$

We want to know where changing $\phi$ will be least effective in getting us away from the circle. This is determined on the one hand by the magnitude of the derivative with respect to $\phi$ and on the other hand by the angle between the two derivatives; the rate at which we move away from the circle as we change $\phi$ is the product of that magnitude and the sine of that angle:

$$
\begin{eqnarray}
\frac{\mathrm ds}{\mathrm d\phi}
&=&
\left|\frac{\partial\vec{x}}{\partial\phi}\right|
\sqrt{1-\left(\frac{\frac{\partial\vec{x}}{\partial\phi}\cdot\frac{\partial\vec{x}}{\partial\theta}}{\left|\frac{\partial\vec{x}}{\partial\phi}\right|\left|\frac{\partial\vec{x}}{\partial\theta}\right|}\right)^2}
\\
&=&
\sqrt{\left|\frac{\partial\vec{x}}{\partial\phi}\right|^2-\left(\frac{\frac{\partial\vec{x}}{\partial\phi}\cdot\frac{\partial\vec{x}}{\partial\theta}}{\left|\frac{\partial\vec{x}}{\partial\theta}\right|}\right)^2}
\\
&=&
\sqrt{(r+R\cos\theta)^2+\rho^2\sin^2\theta-\left(\rho(r\cos\theta+R)/R\right)^2}\;.
\end{eqnarray}
$$

We need to find the minimum of this expression with respect to $\theta$. Setting the derivative of the radicand with respect to $\cos\theta$ to zero yields

$$R(r+R\cos\theta)-\rho^2\cos\theta-r\rho^2(r\cos\theta+R)/R^2=0\;,$$

which simplifies to

$$\cos\theta=-\frac{R}{r}\;.$$

Since $R> r$, this has no solution, so the extrema occur on the boundary at $\cos\theta=\pm1$. At these points, $|\partial\vec{x}/\partial\theta|$ is parallel to the projection of $|\partial\vec{x}/\partial\phi|$ into the cutting plane, so the result is the same as would have been obtained by using the angle between $|\partial\vec{x}/\partial\phi|$ and the plane instead:

$$
\frac{\mathrm ds}{\mathrm d\phi}=(R\pm r)\frac{r}{R}\;\;\text{for}\;\;\cos\theta=\pm1\;.
$$

The factor $R\pm r$ is $|\partial\vec{x}/\partial\phi|$, and the factor $r/R$ is the sine of the angle between $\partial\vec{x}/\partial\phi$ and the plane. The lesser of these is the inner one at $\cos\theta=-1$, and the angle by which you need to rotate to make two tori with minor radii $r'$ fit is thus (to first order for $r'\ll r$)

$$\Delta\phi=\frac{2r'}{\mathrm ds/\mathrm d\phi}=\frac{2r'R}{(R-r)r}\;.$$

Title: Torus: Circle cut


问:
Closed form for n-th anti-derivative of $\log x$
Is it possible to write a closed-form expression with free variables $x, n$ representing the n-th anti-derivative of $\log x$?</p>

答:
Related problems: (I), (II). Here is a unified formula for the $n$th derivative and the $n$th anti-derivative of real order of $\ln(x)$ in terms of the Meijer G function

$$ G^{1, 2}_{2, 2}\left(x-1\, \Big\vert\,^{1-n, 1-n}_{1-n, 0}\right). $$

The above formula gives

a) derivatives of real order if $n>0$,

b) anti-derivatives of real order if $n<0$.

One can have the above formula in terms of the hypergeometric formula

$$ {\frac { \left( x-1 \right) ^{1-n}{\mbox{$_2$F$_1$}(1,1;\,2-n;\,1-x)}
}{\Gamma  \left( 2-n \right) }},$$

with some restrictions on $n$.


问:
Behaviour of the curve $(y^4 − 4y^2 + 3)+i(-y^3+2y)$ in the complex plane
I was reading an example of the application of the Argument Principle and I encountered the following reasoning, which I do not understand.

At a certain point in the example, we consider $f(iy)=(y^4 − 4y^2 + 3)+i(-y^3+2y)$ and the following asseveration is made: since $\lim_{y\to \infty}\frac{-y^3+2y}{y^4 − 4y^2 + 3}=0$, then $f(iy)$ reaches the positive real axis as $y\to\infty$.
(You can see the context of this reasoning on page 6 of this PDF http://legacy-www.math.harvard.edu/archive/115_fall_06/argument_principle.pdf).

I do not understand why this is true. I fact, I have ploted the curve (in this context $y$ takes values from $0$ to $\infty$), and it does not seem like it reaches the positive real axis as $y\to\infty$.

If anyone could clarify why this reasoning is valid it would be very helpful.

Thanks.

答:
Not an answer

$(p=\text{Re }z=t^4-4 t^2+3,q=\text{Im }z=2 t-t^3)$

eliminating the parameter $t$ I got the cartesian equation of the curve
$$-p^3+p^2-4 p q^2+5 p+q^4-4 q^2+3=0$$
that I used to get the plot in the image below.

If you manage to plot the curve for huge values of $q$ you will see that $\arg z\to 0$ as the limit says
$$\lim_{y\to \infty}\frac{-y^3+2y}{y^4 − 4y^2 + 3}=0\to \arg z\to 0$$

$$...$$


问:
Nested equilateral triangles
Let triangle $ABC$ is an equilateral triangle. Triangle $DEF$ is also an equilateral triangle and it is inscribed in triangle $ABC \left(D\in BC,E\in AC,F\in AB\right)$. Find $\cos\measuredangle DEC$ if $AB:DF=8:5$.

Firstly, I would be very grateful if someone can explain to me how I am supposed to draw the diagram. Obviously I have made it by sight.

Let $\measuredangle DEC=\alpha$. We can note that $\triangle AEF \cong \triangle BFD \cong CDE$. This is something we can always use in such configuration. So $$AE=BF=CD, $$ $$AF=BD=CE.$$ Let $AB=BC=AC=8x$ and $DF=DE=EF=5x$. If we denote $CD=y,$ then $CE=AC-AE=AC-CD=8x-y$. Cosine rule on $CED$ gives $$25x^2=(8x-y)^2+y^2-2y\cos60^\circ(8x-y)$$ which is a homogenous equation. I got that $\dfrac{y}{x}=4\pm\sqrt{3}.$ Now using the sine rule on $CED$ $$\dfrac{CD}{DE}=\dfrac{\sin\alpha}{\sin60^\circ}\Rightarrow \sin\alpha=\dfrac{\sqrt{3}}{10}\cdot\dfrac{y}{x}=\dfrac{4\sqrt3\pm3}{10}.$$ Now we can use the trig identity $\sin^2x+\cos^2x=1$ but it doesn't seem very rational. Can you give me a hint? I was able to find $\sin\measuredangle DEC$ in acceptable way, but I can't find $\cos\measuredangle DEC$...

答:
Note that there are two distinct solutions given the conditions in the problem.  This is because there are two admissible orientations of $\triangle DEF$:  one in which $AF > BF$, and one in which $AF < BF$.  We can understand this by noting that since $\triangle DEF$ and $\triangle ABC$ share the same center, the circumcircle of $\triangle DEF$ will intersect $\triangle ABC$ at six points, unless the side length of $\triangle DEF$ is less than or equal to half the side length of $\triangle ABC$ (in which case there will be either exactly three intersection points in the tangent case, or zero), or greater than or equal to the side length of $\triangle ABC$.

A simple way to solve for the required angle is to place the triangle on a coordinate system, say with $A = (0,0)$, $B = (8,0)$, $C = (4, 4 \sqrt{3})$, and let $F = (x,0)$ for some $0 \le x \le 8$.  Then $$E = \left(\frac{8-x}{2}, \frac{(8-x)\sqrt{3}}{2}\right).$$  We then require $EF = 5$, or $$25 = \left(x - \frac{8-x}{2}\right)^2 + \left(\frac{(8-x)\sqrt{3}}{2}\right)^2 = 3x^2 - 24x + 64.$$  Consequently $$x = 4 \pm \sqrt{3}$$ and depending on which root is chosen, $\angle DEC$ can be determined through a straightforward application of the Law of Sines.

Title: Nested equilateral triangles


问:
Knights and Knaves - Three People

Knights and Knaves Problem: There are three people A, B, and C. One of them is a cop. They say the following:

A: I am not a cop

B: The cop is a knave

C: All three of us are knaves

Is the cop a knight or a knave?

My answer: the cop is a knave because that person is person A. If Person C is a knight, and everyone is a knave, then person A is lying. I am not sure if my reasoning or answer is correct and would really like some help please. Thank you.

答:
$C$ cannot be a knight - he would be calling himself a knave.

(from $C$): $C$ is a knave, and either $A$ or $B$ (or both) is a knight.

(from $B$): $B$ is not a cop.

(from $A$): $A$ is either a knave cop, or he is a knight non-cop meaning $C$ is a knave cop.

Which also incidentally shows that $B$ is a knight.

Title: Knights and Knaves - Three People


问:
$I+J$ is the smallest ideal containing $I$ and $J$
I'm reading Dummit and Foote and it says:

It is easy to see that the sum $I+J$ of the ideals $I$ and $J$ is the smallest ideal of $R$ containing both $I$ and $J$.

There is a similar question here but the link alluding to a proof is no longer valid. Any help?

答: This is trivial: Let $L$ be an ideal containing $I$ and $J$. Then since an ideal is additively closed, we must have $I + J \subseteq L$. Hence, $I+J$ is the smallest such ideal.


问:
Computing $xyz$ divided by $xy$, where $x = 4, y = 2, z = 5$.
I'm struggling on writing the expression. Since they're small number I don't think I'll struggle with the multiplying but I'm fairly new to expressions so I don't know how to write the expression.</p>

答: We have $\frac{xyz}{xy}$. Since we are multiplying these fractions together, we can rewrite this as $\frac{x}{x}\times\frac{y}{y}\times z$. Since any nonzero number divided by itself is equal to $1$, we have $1\times1\times z = z$. Thus we have $\frac{xyz}{xy} = z$, thus we can evaluate your expression by plugging in the value of $z$, which is $5$.


问:
A paradox? Or a wrong definition?
Let $A$ be a commutative ring with $1 \neq 0$. Then writing $V(1) = V((1))$, we have $\bigcap_{\mathfrak{p} \in V(1)}\mathfrak{p} = \sqrt{(1)} = (1)$.

But then $\bigcap_{\mathfrak{p} \in V(1)}\mathfrak{p} = \{x : x \in \mathfrak{p} \text{ for all prime ideals } \mathfrak{p} \supseteq (1)\} \supseteq \text{Set}$, where the last containment is justified since there is no prime ideal that contains $(1) = A$, so $x$ can be anything by vacuous truth.

I had a similar confusion like this in my topology course, but the answer that I heard from my teacher was that "of course, $A$ is the universe that we consider," so $\bigcap_{\mathfrak{p} \in V(1)}\mathfrak{p} = \{x \in A : x \in \mathfrak{p} \text{ for all prime ideals } \mathfrak{p} \supseteq (1)\} = (1)$.

But then why "pure" definition from (naive) set theory fails and forces us to consider $A$ as the ambient space?

答:
I think the following is what Asaf Karagila means (of course, I am not going to select my answer).

Definition. Let $\{X_{\alpha}\}_{\alpha \in I}$ be a collection of subsets of $A$. We define $\bigcap_{\alpha \in I}X_{\alpha} := \{x \in A : x \in X_{\beta} \text{ for all } \beta \in I\}$.

Now, the problem I had can be resolved by reading "$\mathfrak{p} \in V(1)$" as "prime ideal $\mathfrak{p}$ of $A$ that contains $(1)$," which again can be paraphrased as "a subset $\mathfrak{p}$ of $A$ that is a prime ideal and contains $(1)$."

Title: A paradox? Or a wrong definition?


问:
showing $\sqrt{n}$ is not bounded
I want to show that $a_{n} = \sqrt{n}$ is not a bounded sequence.

Definition: We say that a sequence is bounded if it is bounded above and below. A sequence $a_{n}$ is bounded above if there exists $C$ such that, for all $n$, $a_{n} \leq C$. A sequence $a_{n}$ is bounded below if there exists $C$ such that, for all $n$, $a_{n} \geq C$.

Well clearly, our sequence is bounded below by $0$. So we need to show it is not bounded above.

Proof:

We need to show that for every $C$, there exists an $n$ such that $a_{n} > C$. Observing that for every $C \geq  0, \sqrt{C^{2}} = C$. Suppose that $ n > C^{2}$ then we have $\sqrt{n} > C$

答:
hint

$ (a_n) $ is bounded $ \iff $

$$(\exists m,M\in \Bbb R)\;: $$
$$ \;(\forall n\in\Bbb N)\; m\le a_n\le M$$

$$\iff (\exists M\ge 0)\; : $$
$$(\forall n\in \Bbb N)\; 0\le |a_n|\le M$$

$$\implies (\forall n\in \Bbb N)\; 0\le a_n^2\le M^2$$

$$\implies (a_n^2) \text{ is bounded}$$

If $ a_n=\sqrt{n} $ then $ a_n^2=n $ and $ (a_n^2)$ is not bounded because for any $ C\ge 0 $, $ n=\lfloor C \rfloor +2>C $.

Title: showing $\sqrt{n}$ is not bounded


答:
WLOG, $AB = 8, DF = 5$. Say $\angle DEC = \theta$

If $CD = x$ then $CE = 8 - x$

Applying sine law in $\triangle CDE$,

$\displaystyle \frac{\sin 60^0}{5} = \frac{\sin \theta}{x} = \frac{\sin(60^0+\theta)}{8-x}$

From first two, we have $\sin \theta = \frac{x \sqrt3}{10}$

From first and third, $\sin 60^0 \cos \theta + \cos 60^0 \sin\theta = \frac{4\sqrt3}{5} - \frac{x \sqrt3}{10}$

$\implies \cos\theta = \frac{8}{5} - \frac{3x}{10}$

Applying $sin^2\theta + \cos^2\theta = 1$, we get

$x^2 - 8x+\frac{39}{3} = 0 \implies x = 4 \pm \sqrt3$

So, $\cos\theta = \frac{4 \mp 3\sqrt3}{10}$ (please note $\pm$ for $x$ and corresponding $\mp$ for $\cos \theta$)

Title: Nested equilateral triangles

Body:
Let triangle $ABC$ is an equilateral triangle. Triangle $DEF$ is also an equilateral triangle and it is inscribed in triangle $ABC \left(D\in BC,E\in AC,F\in AB\right)$. Find $\cos\measuredangle DEC$ if $AB:DF=8:5$.

Firstly, I would be very grateful if someone can explain to me how I am supposed to draw the diagram. Obviously I have made it by sight.

Let $\measuredangle DEC=\alpha$. We can note that $\triangle AEF \cong \triangle BFD \cong CDE$. This is something we can always use in such configuration. So $$AE=BF=CD, $$ $$AF=BD=CE.$$ Let $AB=BC=AC=8x$ and $DF=DE=EF=5x$. If we denote $CD=y,$ then $CE=AC-AE=AC-CD=8x-y$. Cosine rule on $CED$ gives $$25x^2=(8x-y)^2+y^2-2y\cos60^\circ(8x-y)$$ which is a homogenous equation. I got that $\dfrac{y}{x}=4\pm\sqrt{3}.$ Now using the sine rule on $CED$ $$\dfrac{CD}{DE}=\dfrac{\sin\alpha}{\sin60^\circ}\Rightarrow \sin\alpha=\dfrac{\sqrt{3}}{10}\cdot\dfrac{y}{x}=\dfrac{4\sqrt3\pm3}{10}.$$ Now we can use the trig identity $\sin^2x+\cos^2x=1$ but it doesn't seem very rational. Can you give me a hint? I was able to find $\sin\measuredangle DEC$ in acceptable way, but I can't find $\cos\measuredangle DEC$...


问:
Examples of non-compact 3-dim submanifolds with non-trivial normal bundle
Let $\Sigma^3 \subset \mathbb{R}^7$ be a non-compact, orientable, smooth $3$-dimensional submanifold.  Its normal bundle $N\Sigma \to \Sigma$ has rank $4$, and therefore (I think) has a non-vanishing section.

I would be happy to learn that $N\Sigma$ must, in fact, be trivial, but I'm guessing that's not true.  So, are there simple examples -- the simpler the better, really -- of non-compact, orientable, smooth $\Sigma^3 \subset \mathbb{R}^7$ with non-trivial normal bundle?

(I'm guessing there are easy examples, but frankly, even if I were to start listing simple embeddings $\Sigma^3 \hookrightarrow \mathbb{R}^7$, I wouldn't know how to quickly compute the relevant characteristic classes (in fact, what would those even be here?).)

答:
Every orientable 3-manifold is parallelizable, i.e. its tangent bundle is trivial. This implies that any normal bundle is stably trivial. So $N\Sigma$ is a stably trivial bundle of rank 4 over a 3-dimensional manifold.

Milnor proved that if a vector bundle has rank higher than the dimension of your CW complex, then the vector bundle is trivial, if and only if, the vector bundle is stably trivial. So we conclude that the normal bundle of any embedding of a 3-manifold into $\mathbb{R}^7$ (or any greater dimension), is trivial.


答: Assume that $\sqrt n\le u$. But then $n=(\lfloor u\rfloor+1)^2$ is such that $\sqrt n>u$, a contradiction.

Title: showing $\sqrt{n}$ is not bounded

Body:
I want to show that $a_{n} = \sqrt{n}$ is not a bounded sequence.

Definition: We say that a sequence is bounded if it is bounded above and below. A sequence $a_{n}$ is bounded above if there exists $C$ such that, for all $n$, $a_{n} \leq C$. A sequence $a_{n}$ is bounded below if there exists $C$ such that, for all $n$, $a_{n} \geq C$.

Well clearly, our sequence is bounded below by $0$. So we need to show it is not bounded above.

Proof:

We need to show that for every $C$, there exists an $n$ such that $a_{n} > C$. Observing that for every $C \geq  0, \sqrt{C^{2}} = C$. Suppose that $ n > C^{2}$ then we have $\sqrt{n} > C$


问:
Explaining why choose from a set without choose consecutive numbers is equal to the number of solutions of a equation
I need to solve this question:

We want to count the number of ways to choose five elements from the set $\left\{1,2,\dots,20 \right\}$ with the restriction that we may not choose consecutive integers.

$\bullet$ Why is this number equal to the number of positive integer solutions to the equation $$l_1+l_2+l_3+l_4+l_5+l_6 = 21$$ with the additional restriction that $l_2,l_3,l_4,l_5 > 1$?

My solution:

To choose elements from a set without get consecutive numbers, we have a stars-and-bars problem. With this, we get the number of ways for choose from the set $\left\{1,2,\dots,20\right\}$.$$\left(\begin{array}{c} n-k+1 \\ k\end{array}\right) = \left(\begin{array}{c} 16 \\ 5\end{array}\right)$$.

Now, we need to get the number of solutions for the equation.

$l_1,l_6 \subseteq \left\{0,1,\dots,13\right\}$

$l_2,l_3,l_4, l_5 \subseteq \left\{2,3,\dots,15\right\}$

With this, we have the number of solutions: $$\left(\begin{array}{c} 14 \\ 2\end{array}\right) + \left(\begin{array}{c} 13 \\ 4\end{array}\right)$$

Here is where I got stuck. I don't know how to explain why this occurs.

答:
Hint to your problem: Consider the 5 nonconsecutive numbers you picked, and look at the gaps between them (think of the chosen numbers as the "bars" in stars-and-bars). What do you observe about these gaps?

This is a bijective approach, namely we will show for each way to pick 5 nonconsecutive numbers from $1,\ldots,20$, we can find a corresponding sum for 21 satisfying the constraints you described.

Indeed, say you picked five nonconsecutive numbers and order them  $a_1 < a_2 < a_3 < a_4 < a_5$ from $\{1,\ldots,20\}$. Then consider the six numbers
$$a_1 ,\quad a_2 - a_1 ,\quad a_3-a_2 ,\quad a_4-a_3 ,\quad a_5 - a_4,\quad 21-a_5.$$ Note their sum is 21, and every number is positive with the middle four strictly greater than 1, as $a_i$'s are nonconsecutive. You will have to convince yourself why if you pick different five nonconsecutive numbers we will get a different set of six in this way.

Conversely, we can reverse this process. If we pick six positive numbers $l_1,\ldots,l_6$ such that $l_1 + \cdots + l_6 = 21$ with $l_2,\ldots,l_4>1$ , then consider five numbers:
$$
l_1,\quad l_1+l_2,\quad l_1+l_2 +l_3,\quad l_1+l_2+l_3+l_4,\quad l_1+l_2+l_3+l_4+l_5
$$
Then you can check indeed these five numbers are nonconsecutive, and all from $\{1,\ldots,20\}$, and that if you pick a different set of $l_i$'s, you will get a different set of five nonconsecutive numbers.


问:
Determining $\sin(15)$, $\sin(32)$, $\cos(49)$, etc.
How do you in general find the trigonometric function values? I know how to find them for 30 45, and 60 using the 60-60-60 and 45-45-90 triangle but don't know for, say $\sin(15)$ or $\tan(75)$ or $\csc(50)$, etc.. I tried looking for how to do it but neither my textbook or any other place has a tutorial for it. I want to know how to find the exact values for all the trigonometric functions like $\sin x$, $\csc x$, ... opposed to looking it up or using calculator. According to my textbook, $\sin(15)=0.26$, $\tan(75)=3.73$, and $\csc(50)=1.31$ but doesn't show where those numbers came from, as if it was dropped from the Math heaven!</p>

答:
For some angles (such as 15 or 75 degrees), you can apply the formulas for doubling and halving the angles and those for sums and differences of angles. For example, if you know that $\sin x=0.4$, these formulas allow you to calculate $\sin 2x$ or $\sin 17x$ or $\sin \frac{11}{32}x$, ...

Unfortunately, such formulas are not powerful enough to calculate the values for arbitrary angle (e.g. $\sin 43$). For those, one needs to employ the real-analysis definition of $\sin x$ using the Taylor series which might be beyond the scope of what you have learned so far.


问:
Prove greedy algorithm coin change lemma
Lemma:If $n$ is a positive integer, then $n$ cents in change using quarters, dimes, nickels, and pennies using the fewest coins possible has at most two dimes, at most one nickel, at most four pennies, and cannot have two dimes and a nickel. The amount of change in dimes, nickels, and pennies cannot exceed 24 cents.

I am trying to understand this proof and tried to prove it by contradiction following the book's proof , but I am unable to fully understand the gist behind the proof.

Proof by Contradiction: the negation of conditional statement is: if $p \wedge \sim q$. We need to show that $\sim(p\rightarrow q) \equiv (p \wedge \sim q)$ is true otherwise original assumption is true.

In this proof:

$p$: in this proof is statement "$n$ is a positive integer".

$q$: $n$ cents in change using quarters, dimes, nickels, and pennies using the fewest coins possible has at most two dimes, at most one nickel, at most four pennies, and cannot have two dimes and a nickel. The amount of change in dimes, nickels, and pennies cannot exceed 24 cents.

Now, the proof proceeds as follow: We note that if we had three dimes we could replace them with a quarter and a nickel, if we had two nickels we could replace them with a dime, if we had five pennies we could replace them with a nickel, and if we had two dimes and a nickel we could replace them with a quarter. Because we can have at most two dimes, one nickel, and four pennies, but we cannot have two
dimes and a nickel, it follows that 24 cents is the most money we can have in dimes, nickels,
and pennies when we make change using the fewest number of coins for $n$ cents.

We assume in the proof that we can have more coins of each denomination (quarter, nickel, etc.), then we can get change with fewer coins. So the proof proceeds and then reached to conclusion that, "because we can have at most ...", which I don't understand given that we assumed we have more of each denomination.

Can you please rephrase the proof or correct my understanding of it in simple words?

答: If you change $p$ to "$n$ cents has been changed into quarters, dimes, nickels, and pennies using the fewest coins possible," and $q$ to "at most two dimes, at most one nickel, at most four pennies have been used, and two dime and a nickel have not been used," I think you'll see it.


问:
how to solve a random selection problem?
I need help with a statistics problem, it is for a test but the subject is not mentioned, I have tried to search for information, but I have not found anything clear

A brand of chocolates offers a promotion by giving away chocolates in packages that have “You won. Claim your chocolate bar at the nearest store ”. A quarter of the production distributed by the region is marked with a prize.

If Juan buys two chocolates, the probability that he has at least one free chocolate is?

答:
The probability of getting at least $1$ free chocolate is opposite the probability of getting no free chocolates. That means that if the probability of getting at least $1$ free chocolate is $1$, the probability of getting no free chocolates is $0$. Thus, the sum of these probabilities is always $1$.

We know that the probability of getting a free chocolate prize is $1/4$. That means the probability of NOT getting a free chocolate prize is $3/4$. So, the probability of getting no free chocolates at all is $3/4$ * $3/4$ $=$ $9/16$. Thus, since the probability of getting no chocolates free and the probability of getting at least one free chocolate is equal to $1$, that means the probability of getting at least $1$ free chocolate is $7/16$ because $7/16$ + $9/16$ = $1$.


问:
Two questions about the infinitude of prime Euclid numbers: $E_{n}=p_{1}\cdot p_{2}\cdot \ldots \cdot p_{n} + 1$
First of all I was wondering what results there are out there about the prime Euclid numbers? I would like to read up on any results that are out there.

Secondly I wanted someone to look at the following argument about the infinitude of the prime Euclid numbers.

According to Dirichlets theorem: If $a,b\in \mathbb{Z^{+}}$ such that $(a,b)=1$, then there are infinitely many prime numbers of the form $am+b$ where $m$ is a non-negative integer. Now, if we denote the $n$th primorial (i.e the product of the first $n$ primes) as $\displaystyle \prod_{i=1}^{n} p_{i}$ we define the Euclid numbers as $$E_{n}={\displaystyle \prod_{i=1}^{n} p_{i}}+1.$$ Note that $$E_{n}={\displaystyle \prod_{i=1}^{n} p_{i}}+1=2\cdot {\displaystyle \prod_{i=2}^{n} p_{i}}+1$$ where the first term $$2\cdot {\displaystyle \prod_{i=2}^{n} p_{i}}\equiv 2\pmod 4.$$ Thus $$E_{n}\equiv 3\pmod 4.$$ Meaning in particular that $$\boxed{E_{n}={\displaystyle \prod_{i=1}^{n} p_{i}}+1=4m+3}$$ for some non-negative integer $m$.

Since $(4,3)=1$, then by Dirichlets theorem we have that $4m+3$, where $m$ is a non-negative integer is a prime number infinitely often. And since the $RHS$ of the boxed equation above has this form, then does this imply that whatever is on the $LHS$ must also be a prime number infinitely often? In particular, does this mean that $E_{n}$ is prime infinitely often?

答:
No, you can't arguee that. You only know that at least one prime factor of $E_n$ is of the form $4m+3$. But it is not clear at all how many of the $E_n$ are prime.

The set
$$\{ E_n : n \in \mathbb N\}$$
is a proper subset of
$$\{ 4m+1 : m \in \mathbb N_0\}.$$

We only know that in the latter are infinitely many primes.


问:
Ultrapower construction of ${}^{\ast \ast}\Bbb R$
Suppose, we want to add positive elements that are smaller than all positive elements in ${}^{\ast}{\Bbb R}$. One way to do this, as shown in this  master’s thesis, is to construct sequences of elements in ${}^{\ast}{\Bbb R}$ that range over ${}^{\ast}{\Bbb N}$ and find an ultrafilter on ${}^{\ast}{\Bbb N}$, ${}^{\ast}{\mathcal U}$.

Can we find such ultrafilter  ${}^{\ast}{\mathcal U}$? Do we need to require all sets in ${}^{\ast}{\mathcal U}$ to be internal sets? Or we have to live with a cheap version, ${}^{\ast}\mathcal {P}{(\Bbb N)} \cap {}^{\ast}{\mathcal U}$?

The author want to eliminate the possibility of constructing an positive element in ${}^{\ast}\Bbb R$ that is strictly smaller than all positive elements in ${}^{\ast\ast}\Bbb R$ by excluding $\Bbb N$ from ${}^{\ast} \mathcal U$. But it seems to me it's far from enough.  ${}^{\ast} \mathcal U$ shouldn't contain any element with the cardinality of $\Bbb N$. I think what he need is a uniform ultrafilter in which all elements have the cardinality of $2^{\aleph_0}$. Is it right?

答: $^\star\mathbb{R}$ only satisfies countable saturation, so to defeat it it suffices to take an index set of cardinality the continuum in the ultraproduct construction, so using $^\star\mathbb{N}$ is fine.  It is not entirely clear to me why one needs such a $^{\star\star}\mathbb{R}$.</p>


问:
find the locus of points M
find the locus of points M, the difference between the squares of the distances from which to two given points A and B is equal to a given value C, At which C does the problem have a solution?

Anyone know how to solve this?

答:
This locus is a line perpendicular to $AB$. For proving it we'll prove the following:

In Figure1, $AB$ is perpendicular to $CD$ iff $AC^2-AD^2=BC^2-BD^2$

Proof: Assume that $H_a,H_b$ are the feet of altitudes from $A,B$ to $CD$, respectively. By pythagorean theorem,

$$AC^2-AD^2=AH_a^2+CH_a^2-AH_a^2-DH_a^2=CH_a^2-DH_a^2$$

Similarly, $BC^2-BD^2=CH_b^2-DH_b^2$. Therefore it suffices to prove that $H_a\equiv H_b$ iff $CH_a^2-DH_a^2=CH_b^2-DH_b^2$ and this is trivial since:

$$CH_a^2-DH_a^2=CH_b^2-DH_b^2 \implies (CH_a+DH_a)(CH_a-DH_a)=(CH_b+DH_b)(CH_b-DH_b)$$

But since $CH_a+DH_a=CH_b+DH_b$,

$$CH_a-DH_a=CH_b-DH_b$$

But since sum of them are also equal, $CH_a=CH_b$ and hence the result.

Title: find the locus of points M


问:
existence of an inner product making an endomorphism symmetrical
Let $E$ a $\mathbb R$- space of finite dimension. Let $u\in L(E)$ diagonalizable and $F$ a sub-space of E stable by $u$.
Show that it exists an inner product on $E$ so that $u$ is a symmetrical endomorphism.

Any hint to begin the resolution ?

答: Note that an $u \in L(E)$ is symmetric with respect to an inner product, iff the matrix of $u$ with respect to an orthonormal base is symmetric. If $u$ is diagonalizable, there is a basis $B$ of $E$ such that the matrix of $u$ with respect to $B$ is diagonal (hence: symmetric). Now define the inner product such that $B$ is an orthonormal base. Hint: Use coordinates with respect to $B$.


问:
Intuitive explanation of integration
An integral gives the area of the curve between $f(x)$ and the $x$-axis over some domain. Why is the lower bound of this area the $x$-axis? I've always accepted this at face value, and I know the pictorial explanation, but I don't see why it works in the integration itself, in which we explicitly state the function, the two boundaries for $x$, but not the lower horizontal boundary i.e. the $x$-axis. I hope this question makes sense. Thank you!

答: For any given value of $x$, e.g. $x_0$, the value $f(x_0)$ is equal to the length between the $x$-axis point $(x_0,0)$ and the curve point $(x_0, f(x_0))$. So, $f(x) dx$ is the area of an infinitesimal rectangle between the $x$-axis and the curve.

Title: Intuitive explanation of integration


问:
A simple geometry problem with points
Given the points $M(3,4)$ and $N(1,2)$, find $x$ in the point $P(x,0)$ so that $PM + PN$ is a minimum.</p>

答:
Here's an approach that doesn't use calculus. Consider the line segments $PM$ and $PN$. Now reflect the line segment $PM$ about the $x$-axis. Observe that the value of $PM+PN$ does not change, regardless of the point $P$ chosen. Thus, we have legally reduced your problem to this simpler one:

Given the points $M(3,-4)$ and $N(1,2)$, find $x$ in the point $P(x,0)$ such that $PM+PN$ is a minimum.

Now that $M$ and $N$ are on opposite sides of the $x$-axis, it should be clear that the shortest distance between them is a straight line! That is, we must choose $x$ such that $P(x,0)$ is a point on the line $MN$. To do this, it suffices to choose $x$ such that the slope of $PM$ is the same as the slope of $PN$:
$$ \begin{align*}
\frac{x-3}{0-(-4)} &= \frac{x-1}{0-2} \\
\frac{x-3}{4} &= \frac{x-1}{-2} \\
-2(x-3) &= 4(x-1) \\
(x-3) &= -2(x-1) \\
x-3 &= -2x+2 \\
3x &= 5 \\
x&=5/3
\end{align*}$$

Title: A simple geometry problem with points


问:
Existence of neighborhood monotonic increasing
I was thinking about the following statement:

Supose $f\in C^1$ a function s.t. $f(b)>f(a) \forall b\in (a,a+\epsilon)$. So, there's exist a neighborhood $(a,a+\delta)$ s.t. $f$ is monotonic increasing on that.

Is it true? Thank you in advance.

答:
Not necessarily. Think of the function $f(x)=x^3\left(2+\sin\frac{1}{x}\right)$. You can define the function in 0 as $f(x)=0$. In the photo, the function is in green (it´s obviously $>0$ in $(0,\infty)$), but its derivative, in red, has negative values at any neighborhood of 0.

Also, the function is $C^1$. That is obvious at any point different than 0, and using the definition of derivative you can check that the derivative of the function at 0 is 0. Moreover, as can be seen in the picture (you can also check it yourself), the derivative of the function tends to 0 in 0, so the function has a continuous derivative.


问:
Find the Maximum Trigonometric polynomial coefficient $A_{k}$

Let $n,k$ be given positive integers and $n\ge k$. Let $A_i, i=1, 2, \cdots, n$ be given real numbers. If for all real numbers $x$ we have $$A_{1}\cos{x}+A_{2}\cos{(2x)}+\cdots+A_{n}\cos{(nx)}\le 1$$
Find the maximum value of $A_{k}$.

I don't know if this question has been studied

If  $n=2$ it is easy to solve it.

答:
I'll just do a subcase of the $n=3$ case.

$A_{k}=1$ is the maximum for k=2, 3.

First notice that letting $A_{k}=1$ and taking the other values of $A_{j}=0$ we have

$$cos(kx)\leq 1$$

which holds for all x. So each $A_{k}$ is at least 1. Now we will show that each $A_{k}$ is at most 1 for $k=2,3$.

Plugging in $x=0$ gives us $A_{1}+A_{2}+A_{3} \leq 1$.

If $A_{3}>1$ then this forces $A_{1}+A_{2} \leq 1-A_{3} < 0$. However, plugging in $x=\frac{2\pi}{3}$ we have $A_{3} + \frac{-A_{1}-A_{2}}{2} \leq 1$

So $A_{3}\leq 1+\frac{A_{1}+A_{2}}{2} < 1$

Therefore $A_{3}\leq 1$ thus $A_{3}=1$.

If $A_{2}>1$ then $A_{1}+A_{3} \leq 1-A_{2} < 0$.

Plugging in $x=\pi$ we have $A_{2} + -A_{1}-A_{3} \leq 1$.

So $A_{2}\leq 1+ A_{1} + A_{3} < 1$.

Therefore $A_{2}\leq 1$ thus $A_{2}=1$.


问:
Proving that the curvature of the induced connection satisfies $k_{\nabla^{*}}(X,Y) = k_{\nabla}(X, Y)^{*}$.
I am trying to make Exercise 19 of these lecture notes.
I'll briefly summarize the question below.

Exercise Let $V$ be a dinite dimensional vector space and $A \in \mathrm{End}(V)$. We consider
$$
A^* \in \mathrm{End}(V^*), \quad \text{given by } A^*(\xi)(v) = \xi(A(v)).
$$
We keep the same notation for endomorphisms of vector bundles.
Show that the curvature of $\nabla^*$ is given by
$$
k_{\nabla^*}(X,Y) = k_{\nabla}(X,Y)^*.
$$

The lecture notes use the following definition of curvature:

Definition For any connection $\nabla$ on a vector bundle $E \to M$, the curvature is given by
$$
k_{\nabla}(X,Y)s = \nabla_X \nabla_Y(s) - \nabla_Y \nabla_X(s) - \nabla_{[X,Y]},
$$
where $X,Y \in \mathfrak{X}(M)$, $s \in \Gamma(E)$ and $[\cdot, \cdot]$ denotes the Lie bracket.

So it is viewed as a map $\mathfrak{X}(M) \times \mathfrak{X}(M) \times \Gamma(E) \to \Gamma(E)$ (or $k_{\nabla} \in \Gamma(\Lambda^2 T^*M \otimes \mathrm{End}(E)) = \Omega^2(M; \mathrm{End}(E)).)$.

Also, the induced connection $\nabla^*$ on $E^*$ is characterized by the equation
$$
L_X(\xi(s)) = \nabla_X^*(\xi)(s) + \xi(\nabla_X(s)).
$$

I have attempted this exercise by writing both sides out, but it gets very messy very quickly. Is there a smarter way to tackle this question, or is it a matter of really diligently writing everything out and hoping for the best? If that's the case, I am afraid I will have more questions, but that's a worry for later ;)

Thanks in advance for the advice or help! :)

答:
Here's the core step of the computation:

$$ \nabla_X^\ast \nabla_Y^\ast (\xi)(s) = \nabla_X^\ast (\nabla_Y^\ast (\xi))(s) = L_X (( \nabla_y^\ast \xi)(s)) - (\nabla_Y^\ast)(\nabla_Xs) \\
= L_X(L_Y\xi(s) - \xi(\nabla_Ys))-L_Y\xi(\nabla_X s)+\xi(\nabla_Y\nabla_Xs) \\= L_X L_Y\xi(s) - L_X \xi(\nabla_Ys)-L_Y\xi(\nabla_X s)+\xi(\nabla_Y\nabla_Xs) .$$

Can you take it from there?


问:
Prove that there is a number such that elements of a vector space are linearly dependent
I am inexperienced with math and linear algebra and need help with the following question.

Let $v, w$ be elements of a vector space and assume that $v\neq0$. If $v, w$ are linearly
dependent, show that there is a number $a$ such that $w = av$.

My professor said that to prove $w=av$ I need to prove an existence of $a$. How do I do that?

I know that if vectors $x,y$ are linearly dependent then there are scalars $c,d\in\mathbb{F}$, not all zero, such that $cx+dy=0$. $($Where $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$.$)$

答: As you have said, we can find $a,b\in \Bbb{F}$ not both zero so that $av+bw=0$. Suppose that $a\ne 0$, then $av=-bw$ and hence dividing we get $v=-\frac{b}{a}w$. The case of $b\ne 0$ is analogous.


答:
HINT:

$av+bw=0 \Rightarrow av=-bw \Rightarrow a^{-1}av=a^{-1}(-b)w \Rightarrow v=a^{-1}(-b)w$

Note i've used some properties of fields here (which one?)

Title: Prove that there is a number such that elements of a vector space are linearly dependent

Body:
I am inexperienced with math and linear algebra and need help with the following question.

Let $v, w$ be elements of a vector space and assume that $v\neq0$. If $v, w$ are linearly
dependent, show that there is a number $a$ such that $w = av$.

My professor said that to prove $w=av$ I need to prove an existence of $a$. How do I do that?

I know that if vectors $x,y$ are linearly dependent then there are scalars $c,d\in\mathbb{F}$, not all zero, such that $cx+dy=0$. $($Where $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$.$)$


问:
Help proving a field property
I need to prove this part of a theorem: given a field $K$ such that $|K| = p^n$, a subfield $H \subset K$, and $\xi$ a primitive element of $K$; i need to say that $H(\xi) \subseteq K$.

Of course $K$ contains all the polynomial in $H[ \xi ]$.

$\xi$ is a root of the polynomial $x^{|K| - 1} - 1 \in H[x]$, so $\xi$ is a root of a monic factor of $x^{|K| - 1} - 1 \in H[x]$ irreducible in $H[x]$.

Now, i think that i must require also that this monic factor has degree $n$. Is it right? If so, how to prove it?

答:
Here we actually have $H(\xi)=K$.

Let $F=\mathbb{F}_p$ be the common prime field of both $H$ and $K$. $K$ is a finite extension of $F$, so all of its elements, $\xi$ in particular, are algebraic over $F$. Therefore $H(\xi)=H[\xi]$.

Because $\xi$ is a primitive element of $K$, we have
$K=F(\xi)=F[\xi]$. Therefore we have the following inclusions
$$
K=F[\xi]\subseteq H[\xi]\subseteq K[\xi]=K.
$$
From this chain of inclusions we can infer the claimed equality.

Title: Help proving a field property


问:
Highly Recommended References for Hyperbolic Planes and Modular Surfaces
I would be very grateful if one could suggest highly recommended references for Hyperbolic Planes and Modular Surfaces that provide a readable self-contained introduction to these concepts.

答:
For references, you can use, for instance:

The book "Fuchsian groups" by S.Katok. It covers both hyperbolic geometry and the modular group.

Or the book Anderson, James W., Hyperbolic geometry, Springer Undergraduate Mathematics Series. London: Springer (ISBN 1-85233-934-9/pbk). xi, 276 p. (2005). ZBL1077.51008.

(but it only covers hyperbolic geometry and general discrete groups)

Or

Bedford, Tim (ed.); Keane, Michael (ed.); Series, Caroline (ed.), Ergodic theory, symbolic dynamics, and hyperbolic spaces. Lectures given at the workshop ”Hyperbolic geometry and ergodic theory”, held at the International Centre for Theoretical Physics in Trieste, Italy, 17-28 April, 1989, Oxford Science Publications. Oxford etc.: Oxford University Press,. xv, 369 p. (1991). ZBL0743.00040.

Chapters 1 and 5: Chapter 1 covers basics of hyerbolic geometry and Fuchsian groups while chapter 5 will deal with the modular group, symbolic dynamics and continuous fractions part of the story.

Or, freely available Pete Clark's lecture notes on Shimura curves
here, lectures 0, 2, 3, 4.


问:
Asymptotic Winding of the Geodesic Flow on Modular Surfaces and Continuous Fractions
Asymptotic Winding of the Geodesic Flow on Modular Surfaces and Continuous Fractions. Y. Guivarc'h and Y. Le Jan. Page 24.

How to make sure that $K$ and $D$ are normal subgroups of the group $G$ to take the quotients?

And, in general, I look for a good reference for such concepts.

答: Of course these are not normal subgroups, but these are closed subgroups; the quotients are understood as smooth manifolds.


问:
sequences and bounds proof
I am solving the following problem:

Let $S \subset R$ be a nonempty subset which is bounded above. Let $z=\sup{S}$. Prove that there exists a sequence $(a_n)$ such that $a_n \in S$ for all $n$, and $\lim_{n \to \infty}{a_n}=z.$

I know $z-\epsilon<z$, and if I can find a an inside then I will have found the Sup. What I mean by that is that we can find an element in $S$ that lies inside this epsilon interval.

From here I'm kind of lost. I would appreciate some help. Please and thank you.

答: Take $\epsilon=\frac{1}{n}>0$ $\forall n$ in natural numbers, assings that $n$ to diferent $a_{n}$ elements on $S$ using the $\epsilon$ definition of supreme. Then the fact that {$\dfrac{1}{k}$} $\longmapsto 0$.

Title: sequences and bounds proof


问:
$A=\{a_{\lambda_1}+a_{\lambda_2}+...+a_{\lambda_k}| k\in\mathbb N, \lambda_1,...,\lambda_k\in\mathbb R \;\text{are all distinct}\}$ is not bounded
For any real number $\lambda\in\mathbb R$ choose a positive real number $a_\lambda>0$.
Show that the subset $A$ below is unbounded.
$$A=\{a_{\lambda_1}+a_{\lambda_2}+...+a_{\lambda_k}| k\in\mathbb N, \lambda_1,...,\lambda_k\in\mathbb R \;\text{are all distinct}\}$$

And consider the case when $\lambda$s are from integers.

Here is an answer A boundedness question.

By hypothesis,

$$\begin{array}{l|rcl}
b : & \mathbb R & \longrightarrow & \mathbb R \\
    & \lambda & \longmapsto & b_\lambda \end{array}$$ is supposed to be a one-to-one map. Without loss of generality by changing $b$ into $-b$, we can suppose that $b(\mathbb R) \cap (0,\infty)$ is uncountable. Therefore, one of the sets $B_n = b(\mathbb R) \cap (1/n, \infty)$ is uncountable  (a union of countably many countable sets is countable). Taking $k$ distinct elements of $B_n$ we have $\sum_{i=1}^k b_{\lambda_i} \ge k/n$ proving that $B$ in unbounded.

The result doesn't hold if the $\lambda$s belong to $\mathbb N$. Take for example $b_n = 1/n^2$. As $\sum 1/n^2$ converges, every partial sum is also finite.

but I am not satisfied.

Why we write -b into b? What is the point?

It is obvious that $b(\mathbb R)\cap (1/n,\infty)$ is uncountable but $a_{\lambda_i}s$ ($a_{\lambda_1}+a_{\lambda_2}+...+a_{\lambda_k}$) are not supposed to be inside $(1/n,\infty)$
I understand why we take $(1/n,\infty)$ because we want an lower bound that can be go to the infinity. But it is not good with me.

And for the second consideration with $\lambda$s from $\mathbb Z$ I didnot quite understand why the $1/n^2,1/n^3$ sums are working?

答:
For each $n\in\mathbb{N}$, let $B_{n}=\{\lambda\in\mathbb{R}\mid a_{\lambda}>\frac{1}{n}\}$.
Clearly $\cup_{n}B_{n}=\mathbb{R}.$ Since $\mathbb{R}$ is uncountable,
there exists $n$ such that $B_{n}$ is uncountable. For any $k\in\mathbb{N}$,
we can choose pairwisely distinct $\lambda_{1},\lambda_{2},\ldots,\lambda_{k}\in B_{n}$.
Then $\sum_{i=1}^{k}a_{\lambda_{i}}>\frac{k}{n}.$ Note that $\sum_{i=1}^{k}a_{\lambda_{i}}\in A.$
This shows that the set $A$ is unbounded above.

If $\lambda$ is only allowed to take values in $\mathbb{Z},$ the
set $A$ may be bounded or unbounded, depending on $a_{\lambda}$.
For example, if $a_{n}=\frac{1}{2^{|n|}},$ then $A$ is bounded. If
$a_{n}=|n|$, then $A$ is unbounded.


问:
Extended Linear Diophantine Equation
While solving some linear equation I Stopped on the following:
How to rearrange coefficients and constants to find integer solutions for:
$2xy+3x+3y=106$. If we take $x$ as a common factor from the first two terms then we get:
$x(2y+3)+3y=106$. Any suggestion how to go farther..

Title: Extended Linear Diophantine Equation

Body:
$$4xy+6x+6y= 212$$
$$4xy+6x+6y +9= 221$$
$$ (2x+3)(2y+3) = 221 = 17 \cdot 13$$


问:
Hölder inequality and interpolation
I was seeking the other day for intuition about the Hölder inequality and more precisely the geometric intuition behind the relation $\frac{1}{p} + \frac{1}{q} = 1$.

I found this interpretation of the Hölder inequality.

It says that we are trying to find an estimate of :

$$ \int f^a g^b$$ with the knowledge of $\int f^p$ and $\int g^q$.

For $(a, b) = (1,1)$, we need the relationship $\frac{1}{p} + \frac{1}{q} = 1$ so that the points : $(p, 0), (1,1), (0, q)$ lie on the same line. We can know use interpolation to have the Hölder inequality which give an estimate at the point $(1, 1)$.

But I feel like an argument is missing here. Why the estimate is using the geometric mean of the  function $f$ and $g$ ?  Why we don't have something like (using arithmetic mean) :
$$
\int \mid fg \mid \leq \frac{1}{p} \int f^p + \frac{1}{q}\int g^q \;\;?
$$
Moreover the article is saying we can generalize this approach to approximate $\int f^ag^b$ but in this case we need three points such that $(a,b)$ is in the convex hull of these three points. Why for $(1, 1)$ two points are needed while for general $(a, b)$ we need three points ?

答:
In my opinion, the correct answer to the question has been already given by John Dawkins: thus the following one is simply an illustration of his answer based on the proof of Hölder's inequality given in reference [1], §4.1, pp. 100-101, the best I've seen in print.

The proof given by Vol'pert and Hudjiaev starts by giving a geometric proof of Young's product inequality based on the following picture ([1], p. 100, Figure 1),

where
$$
\eta=\xi^{p-1} \iff \xi = \eta^{q-1}.
$$
As correctly stated by Dawkins, $\eta=\eta(\xi)$ and $\xi=\xi(\eta)$ are inverse functions and this is the main reason for which we have the relation $\frac{1}{p}+\frac{1}{q}=1$: indeed
$$
(u^{p-1})^{q-1}= (u^{q-1})^{p-1}= u \iff (p-1)(q-1)=1\iff \frac{1}{p}+\frac{1}{q}=1.
$$
The proof then goes as follows: calling respectively $A_1$ and $A_2$ the areas of the green and pink plane regions shown, we have
$$
\begin{split}
A_1 &=\int\limits_0^a \xi^{p-1} \mathrm{d}\xi = \frac{a^p}{p}\\
A_2 &=\int\limits_0^b \eta^{q-1} \mathrm{d}\eta = \frac{b^q}{q}\\
\end{split}\implies A_1+A_2 \ge ab \iff  \frac{a^p}{p} + \frac{b^q}{q} \ge ab \label{1}\tag{Y}
$$
since the rectangle whose sides are $[0, a]$ and $[0, b]$ is evidently contained in the union of the two shown plane regions.

Now, in order to estimate the product of the two functions $f(x)$ and $g(x)$ by using \eqref{1}, it is only needed a wise choice of $a$ and $b$:
$$
\begin{split}
a &= \frac{|f(x)|}{\|f\|_p}\\
b &= \frac{|g(x)|}{\|g\|_q}
\end{split}
$$
Then we have
$$
|f(x)g(x)|\le \frac{\|f\|_p^{1-p} \|g\|_q}{p} |f(x)|^p  + \frac{\|g\|_q^{1-q} \|f\|_p}{q} |g(x)|^q
$$
and integrating both members of this inequality, one gets the sought for result.

Reference

[1] Aizik Isaakovich Vol'pert and Sergei Ivanovich Hudjaev (1985), Analysis in classes of discontinuous functions and equations of mathematical physics, Mechanics: analysis, 8, Dordrecht–Boston–Lancaster: Martinus Nijhoff Publishers, pp. xviii+678, ISBN 90-247-3109-7, MR 0785938, Zbl 0564.46025.

Title: Hölder inequality and interpolation


问:
Weierstrass theorem for holomorphic function.
I want to show that function $f(z)=\sum_{n=1}^\infty \frac{(z-n)^2}{n^4}$ is holomorphic on $\mathbb{C}$. I think, that I should to apply Weierstrass Theorem. Unfortunately I have no idea. I was thinking about creating a sequence $f_k=\sum_{n=1}^k \frac{(z-n)^2}{n^4}$.

答:
Here is an outline: choose $R>0$ and for $\lvert z\rvert\le R$ apply the bound $\lvert z-n\rvert^2\le (R+n)^2$ to get
$$
\lvert f(z)\rvert\le \sum_{n=1}^\infty \frac{(R+n)^2}{n^4}<\infty.
$$
You can prove that the above expression is finite by expanding the numerator and comparing with $\sum \frac{1}{n^2}$. Then you have shown convergence for $\lvert z\rvert\le R$ by the $M-$test. This will hold for any fixed $R$ and so $f(z)$ converges for all $z\in \Bbb{C}$.


问:
Nested equilateral triangles
Let triangle $ABC$ is an equilateral triangle. Triangle $DEF$ is also an equilateral triangle and it is inscribed in triangle $ABC \left(D\in BC,E\in AC,F\in AB\right)$. Find $\cos\measuredangle DEC$ if $AB:DF=8:5$.

Firstly, I would be very grateful if someone can explain to me how I am supposed to draw the diagram. Obviously I have made it by sight.

Let $\measuredangle DEC=\alpha$. We can note that $\triangle AEF \cong \triangle BFD \cong CDE$. This is something we can always use in such configuration. So $$AE=BF=CD, $$ $$AF=BD=CE.$$ Let $AB=BC=AC=8x$ and $DF=DE=EF=5x$. If we denote $CD=y,$ then $CE=AC-AE=AC-CD=8x-y$. Cosine rule on $CED$ gives $$25x^2=(8x-y)^2+y^2-2y\cos60^\circ(8x-y)$$ which is a homogenous equation. I got that $\dfrac{y}{x}=4\pm\sqrt{3}.$ Now using the sine rule on $CED$ $$\dfrac{CD}{DE}=\dfrac{\sin\alpha}{\sin60^\circ}\Rightarrow \sin\alpha=\dfrac{\sqrt{3}}{10}\cdot\dfrac{y}{x}=\dfrac{4\sqrt3\pm3}{10}.$$ Now we can use the trig identity $\sin^2x+\cos^2x=1$ but it doesn't seem very rational. Can you give me a hint? I was able to find $\sin\measuredangle DEC$ in acceptable way, but I can't find $\cos\measuredangle DEC$...

答:
You may calculate $\cos\alpha $ directly. Per the sine rule for $\triangle CDE$
$$\frac{\sin \alpha }{CD}= \frac{\sin (120-\alpha) }{AB-CD}= \frac{\sin 60 }{DE}
$$
Eliminate $CD$ to get
$$ \frac{AB}{DE}\sin 60 = \sin\alpha + \sin(120-\alpha)=2\sin 60\cos(60-\alpha)$$
which leads to $\cos(60-\alpha)=\frac45$. Then,
$$\cos\alpha = \cos(60\pm\cos^{-1}\frac45)
=\frac12\cdot \frac45\mp\frac{\sqrt3}2\cdot \frac35
=\frac{4\mp 3\sqrt3}{10}
$$

Title: Nested equilateral triangles


问:
Information lost in going from n-dice rolls to the sum of their faces?
I'm trying to understand how to figure out how much information is lost in going from n dice to the total score.

If I rolled n-dice I could add up their face values to find the total sum, however if I just take the total sum I most likely can't tell which dice had which face value (except for all 1's or 6's). That is, such as the sum of 3 dice to be 11, just knowing 11, I can't tell exactly how much each die contributed to the sum. So it seems as if there is a loss of information in going from knowing the faces of the die to just knowing the total score.

Now, from A Webb's answer here it is simple to find the amount of information in n dice rolls:

$$- \log_2{((\frac{1}{6})^{n})} =  - n * \log_2{(\frac{1}{6})} = n * 2.58$$ bits of information. As the probabilities follow a uniform distribution and all face probabilities are the same we'd expect the average information of each roll to be the same as each of the individual events. So the entropy (average information) for each die roll 2.58 bits. We can then sum this entropy $n$ times for $n$ dice.

Next, as the expected value of the sum of random variables is the sum of their expected values and the expected value for each die is $\frac{1+2+3+4+5+6}{6} = \frac{21}{6}$ we have that the expected value of the sum of n dice/rolls to be $n * 3.5$. I however am not sure where this gets me (if anywhere) in finding the information entropy of the sum of scores so as to compare that to that of the entropy of the n dice.

答:
As noted in the comments, you need to compute the entropy of the sum numerically. By analogy with the result for a binomial, you might assume that for large $n$

$$H(S) \approx \frac1 2 \log_2 \big( 2\pi e\, \sigma^2 \big) = \frac1 2 \log_2 \big( 2\pi e\, \frac{35}{12} \big) + \frac12\log_2(n) = 2.819256 +\frac12\log_2(n) \tag 1$$

For example, for $n=5$ we get $H(S) \approx  3.9802$ while the true value is $3.9774$

For $n=9$ the approximation gives $4.4042$ , vs the exact $4.4035$

Hence, $(1)$ seems a good approximation, and the information loss is big: the entropy of the die is linear in $n$, while the entropy of the sum grows logarithmically.


答: For every $n \in \mathbb{N},\ $ the interval $\left(z-\frac{1}{n},\ z\right]\ $ contains a member of $S,\ $ for otherwise $z$ would not be the supremum of $S$.

Title: sequences and bounds proof

Body:
I am solving the following problem:

Let $S \subset R$ be a nonempty subset which is bounded above. Let $z=\sup{S}$. Prove that there exists a sequence $(a_n)$ such that $a_n \in S$ for all $n$, and $\lim_{n \to \infty}{a_n}=z.$

I know $z-\epsilon<z$, and if I can find a an inside then I will have found the Sup. What I mean by that is that we can find an element in $S$ that lies inside this epsilon interval.

From here I'm kind of lost. I would appreciate some help. Please and thank you.


问:
$\partial_i (dx^j/ds)=0?$
Consider the curve $x^i=x^i(s)$, where $i=1,2,...,N$, $x^i$ are coordinates, not power indices.

Is $$\frac{\partial}{\partial x^i}\left(\frac{dx^j}{ds}\right)=0$$?

It wouldn't make any sense to use the chain rule to convert $s$ back to $x^i$, no?

答: For a counterexample let $s$ be $\theta$ on the circle $x^2+y^2=1$ so$$x=\cos s,\,y=\sin s\implies\frac{\partial}{\partial x}\frac{dy}{ds}=\frac{\partial}{\partial x}x=1.$$

Title: $\partial_i (dx^j/ds)=0?$


问:
On the definition of Liouville number

Definition: (from Wikipedia)

In number theory, a Liouville number is a real number $x$ with the property that, for every positive integer $n$, there exist integers $p$ and $q$ with $q > 1$ and such that

$$
  {\displaystyle 0<\left|x-{\frac {p}{q}}\right|<{\frac {1}{q^{n}}}.}
$$

A Liouville number can thus be approximated "quite closely" by a sequence of rational numbers. [....]

My question: How can I convince myself that the above definition is not arbitrary.  In other words, how nice is to know that a given number $\alpha$ is a Liouville number?

答:
$P$ is hypoelliptic iff $u \in C^k$ and $Pu$ smooth imply $u$ is smooth.

If you find this (not rigorous) definition of hypoellipticity of an operator $P$ on a function space containing functions $u$ to be useful for facilitating proofs on the smoothness of solutions of differential equations such as $Pu = f$, where $f$ is smooth, then the definition of Liouville numbers is far from arbitrary, for if we let $P = \partial/\partial_x - \alpha \partial/\partial_y$, with $\alpha$ irrational, then P is hypoelliptic when (and only when) $\alpha$ is not a Liouville number!

See 'Global Hypoellipticity and Liouville Numbers' (Greenfield and Wallach) for more details.

Title: On the definition of Liouville number


问:
How to evaluate transition rates of a Markov Jump Process
I am trying to revise for my exam, the following question is 'unseen' material from the course hence we haven't directly studied it and as a result i'm struggling to make a good start on the question.

https://i.gyazo.com/0eb37facbb88a95f22c460e4ce6b9f27
And here's the context we are given:

Bob is twice as likely to be admitted to Hospital A than to Hospital B.

The average waiting time from the time of Bob’s accident until the time of his surgery is 1 month.

The average waiting time from admission to Hospital A until full recovery is 4 months.

The average waiting time from admission to Hospital B until full recovery is 5 months. Based on this information determine the values of the parameters α, β, μ and ν.

So to determine the values of the parameters would I have to set up a system of equations? Would it be in terms of the transition rates eg, 2α = β2α=β? or is the problem more complex than that

答:
Hints:

The average waiting time from the time of Bob’s accident until the time of his surgery is just the average length of time he spends in state $1$.  The length of time he spends in state $1$ has cumulative distribution function $\ 1-e^{-(\alpha+\beta)t}\ $ (for $\ t\ge0\ $)—that is, probability density function $\ (\alpha+\beta)e^{-(\alpha+\beta)t}\ $.

The probability that Bob gets admitted to hospital A is  $\ \frac{\alpha}{\alpha+\beta}\ $, and the probability that he gets admitted to hospital B is $\ \frac{\beta}{\alpha+\beta}\ $, so you're correct that if the former is twice the latter, then $\ \alpha=2\beta\ $.

The average waiting time from admission to Hospital A until full recovery is just the average length of time Bob will spend in state $2$, given that he enters that state.  The length of time he will spend in that state after entering it has cumulative distribution function $\ 1-e^{-\mu t}\ $ (for $\ t\ge0\ $)—that is, probability density function $\ \mu e^{-\mu t}\ $.

Similarly, the waiting time from admission to Hospital B until full recovery has cumulative distribution function $\ 1-e^{-\nu t}\ $ (for $\ t\ge0\ $)—that is, probability density function $\ \nu e^{-\nu t}\ $.


问:
Find the affine transformation between 2 triangles in 3D
I have a triangle T and its transformed version T' - i.e, I know the correspondence of vertices between the two triangles.

Is it possible to find the affine transformation A, that transformed T to T'?

I know v1,v2,v3 of T and v1',v2',v3' of T', where each v has a 3D coordinate (x,y,z).

The affine transformation I believe has 12 parameters, so ideally I'd need 4 points to find A. But is there a way to do it with 3 known points (even if approximately)?

Thank you for any ideas..

答:
We assume that the two triangles are non-degenerate.

In fact, in the general case, a linear transform  fulfilling the task. Here is how.

As $A$ must send $V_k$ onto $V'_k$ for $k=1,2,3$, we can write:

$$A \times \underbrace{\begin{pmatrix}|&|&|\\V_1&V_2&V_3\\|&|&|\end{pmatrix}}_{B} = \underbrace{\begin{pmatrix}|&|&|\\V'_1&V'_2&V'_3\\|&|&|\end{pmatrix}}_{B'}$$

Two cases :

The general case : If $\{V_1,V_2,V_3\}$ are independent, matrix $B^{-1}$ exists, and we can take

$$A=B'B^{-1}$$

for the matrix of the transformation.

If ${V_1,V_2,V_3}$ are dependent (it means that the plane defined by the initial triangle passes through the origin), corresponding matrix $B$ has rank $2$. Then, taking a translation $t$ along the normal vector of this plane, replacing  $\{V_1,V_2,V_3\}$ by $t(\{V_1,V_2,V_3\})$, we are back to the previous case with the existence of a linear transform $\ell$ such that :

$$\ell(t(\{V_1,V_2,V_3\}))=\{V'_1,V'_2,V'_3\}$$

Therefore, composition $a=\ell \circ t$ is a solution.


问:
Values of $a_1$ and $a$ for which $a_{n+1}=a_n^2+(1+2a)a_n+a^2$ is convergent
I need to find the values of $a_1$ and $a \in \Re$ that make convergent  the sequence $(a_n)$ defined as:

$a_1 \in \Re $

$a_{n+1} = a_n^2+(1-2a)a_n+a^2$

I can state easily that the sequence is increasing

$a_{n+1} = a_n + (a_n-a)^2 \ge a_n$ obviously, because $(a_n-a)^2 \ge 0$

Now I know that the sequence is increasing, if it happens to be convergent, the limit has to be $a$ because

$$\lim_{n\to \infty} a_n = \lim_{n\to \infty} a_{n+1} = \lim_{n\to \infty} a_n + (a_n-a)^2$$

$$0 = \lim_{n\to \infty} (a_n-a)^2$$

$$\lim_{n\to \infty} a_n = a$$

Now, the sequence

is increasing

if it converges, it converges at $a$

As a consequence, if $a_1 > a$ then the sequence diverges.

Now I'm wondering how to prove that the sequence is

convergent for $a-1\le a_1 \le a$

divergent for $a_1 \lt a-1$

For the second request, i know it could be something related to $a_2 \gt a$, but I don't know how to proceed from $a_1 \lt a-1$ to $a_2 \gt a$.

Instead, for the first one, I thought it could be convenient to prove by induction that

$$a-1\le a_n \le a \quad \forall n \quad if \quad a-1\le a_1 \le a$$

I can easily get to state that $a_{n+1} \gt a-1$, but I got a few problem with the other hand side.

If you guys could give me some tip or hint, it would be much appreciated.

答:
hint

As you said $ (a_n) $ is increasing because

$$a_{n+1}-a_n=(a_n+a)^2\ge 0$$

If it converges to $ L $, we should have

$$L-L=(L+a)^2\implies L=-a$$


问:
Show the convergence of a sequence of integrals to a given integral
I would like to show that :

$$\int_{0}^1(3t^2 -2t^3)^n \mathrm{d}t \underset{n \rightarrow+\infty}{\sim} \frac{1}{\sqrt{n}} \int_0^{+\infty}e^{-3x^2}\mathrm{d}x$$

I thought of showing the equivalent result:

$$\sqrt{n}\int_{0}^1(3t^2 -2t^3)^n \mathrm{d}t \underset{n \rightarrow+\infty}{\longrightarrow}  \int_0^{+\infty}e^{-3x^2}\mathrm{d}x$$

I tried using the binomial expansion and that for all $x\in\mathbb{R},$ $e^{-3x^2} = \displaystyle \sum_{k=0}^{+\infty}\frac{(-3x^2)^k}{k!}$, I also thought of changing the bounds of the integrals. But I don't really know yet how to proceed.

答:
Edited based on comment by Travor Liu

This is going to be a bit messy, and I'm not going to be super precise with justifying why everything works, but we're essentially going to use Laplace's method for integral asymptotics.

We begin with rewriting the integral as follows:
$$\int_{0}^1(3t^2 -2t^3)^n \, dt = \int_{0}^{1}e^{n\ln(3t^{2} - 2t^{3})}\,dt,$$ and we note that $\ln(3t^2 - 2t^3) < 0$ on $(0,1)$, and that as $n\to \infty$ the area on $(0,\varepsilon)$, for $\varepsilon > 0$ becomes negligible.  With that in mind, we expand $\ln(3t^{2} - 2t^{3})$ around $t = 1$ (via Taylor series) to get $$\ln(3t^{2} - 2t^{3}) = -3(t-1)^{2} + O((t-1)^3),$$ and so $$\int_{0}^{1}e^{n\ln(3t^{2} - 2t^{3})}\,dt \sim \int_{0}^{1}e^{-3n(t-1)^{2}}\,dt, \quad\text{as }n\to\infty.$$  Now, we do a change of variables
\begin{align}
u &= 3n(t-1)^2 \implies t- 1 = \pm\sqrt{\frac{u}{3n}}\\
du &= 6n(t-1)\,dt = -6n\sqrt{\frac{u}{3n}}\,dt = -2\sqrt{3n}\sqrt{u}\,dt.
\end{align}
Note that, as we will see momentarily, we choose the negative root so that the integral remains positive.  With this substitution, we get
\begin{align}
\int_{0}^{1}e^{-3n(t-1)^{2}}\,dt = -\frac{1}{2\sqrt{3n}}\int_{3n}^{0}u^{-1/2}e^{-u}\,du = \frac{1}{2\sqrt{3n}}\int_{0}^{3n}u^{-1/2}e^{-u}\,du.
\end{align}
Now, the because we have the factor of $e^{-u}$, the integrand becomes exponentially small as $u\to \infty$, and so we only introduce exponentially small error by extending the upper limit of the integral to infinity.  This gives us
$$
\int_{0}^1(3t^2 -2t^3)^n \, dt \sim \frac{1}{2\sqrt{3n}}\int_{0}^{\infty}u^{-1/2}e^{-u}\,du\quad\text{as }n\to\infty.$$

At this point we make us of the change of variables $u = 3x^2$, which gives us

$$\frac{1}{2\sqrt{3n}}\int_{0}^{\infty}u^{-1/2}e^{-u}\,du =\frac{1}{2\sqrt{3n}}\int_0^\infty(3x^2)^{-1/2}e^{-3x^2}(6x)\,dx = \frac{1}{\sqrt{n}}\int_{0}^{\infty} e^{-3x^2}\, dx.$$

So, we obtain

$$\boxed{\int_{0}^1(3t^2 -2t^3)^n \, dt\sim \frac{1}{\sqrt{n}}\int_{0}^{\infty} e^{-3x^2}\, dx\quad\text{as }n\to\infty}$$

Now, if we wanted to be a bit more precise, we could write
\begin{align}
\frac{1}{2\sqrt{3n}}\int_{0}^{\infty}u^{-1/2}e^{-u}\,du&= \frac{1}{2\sqrt{3n}}\Gamma\left(\frac{1}{2}\right)\\[5pt]
&= \frac{1}{\sqrt{n}}\frac{1}{2}\sqrt{\frac{\pi}{3}},
\end{align}

where here $\Gamma$ is the gamma function, and so in reality we have
$$\int_{0}^1(3t^2 -2t^3)^n \, dt\sim \frac{1}{\sqrt{n}}\frac{1}{2}\sqrt{\frac{\pi}{3}},\qquad\text{as }n\to\infty.$$


问:
Proving that $\tan{x}-x$ has an inverse and than determining it
Consider this problem;

Show that the function $\tan{x}-x$ can be inverted and find the corresponding inverse for which applies

$$ f^{-1} (y) = (f^{-1}(y) + y)^{-2}$$

So to prove that a function can be inverted it has to be bijective, which means it has to be surjective and injective

Now to prove for injective I derived the function $\tan{x}$ and got that it is constantly growing so it is injective. To prove that it is surjective we are already given a solution. Now what I did is derived the function $\tan{x}-x$ and got that its simply $\tan^2{x}$ since the $1$ and $-1$ cancel out. Then out of the starting equation I found a way to explicitly write $\tan{x}$;

$$\tan{x}=f(x)+x$$ and with that we can also determine this;

$$\tan^2{x}=(f(x)+x)^2$$

Now I've used the formula $$ (f^{-1})'(y) = \frac {1}{f'(x)} $$ and this is where I am stuck, the only progress I could make is if instead of $f'$ I insert $\tan^2{x}$ but that is it.

Looking forward to your answers !

EDIT: The domain of the function was also given (my bad)

$$\left(\mbox{-}\frac{\pi}{2}, \frac{\pi}{2}\right)$$

答:
Okay so as I've just mentioned in the comments I've been unable to solve this so I looked in the solutions.The solution is very unclear to me,so I'd like if someone could try to explain or give me a step in the right direction.This is what the solution in its full detail looks.

$$ f(x) = tan x - x => tan(x) = f(x) + x$$
$$ f'(x) = 1 + tan^2(x) - 1 = tan^2(x) $$
$$ tan^2(x) = (f(x) + x)^2 $$

$$ (f^-1)'(x) = \frac{1}{f'(f^-1)} = \frac{1}{tan^2(f^-1)} = \frac{1}{(f(f^-1)+f^-1)^2} =\frac{1}{(x+f^-1(x))^2}$$
$$ (f^-1)'(y) = \frac{1}{f'(f^-1(y))}$$

Now the only thing more standing in the solutions is that $$(f(f^-1) = id(x) = x $$
Could anyone give me insight on what happened here?

Thank you.


问:
If $p$ is an odd prime (different than $5$) prove that $10 \mid p^{2}-1$ or $10 \mid p^{2}+1$
I was reading some properties on prime numbers in a book and I’ve found this problem.
Using Modular Arithmetic isn’t allowed

Using the division algorithm I have found:

$p \in \{5m+1, 5m+2, 5m+3, 5m+4\}$

I’ve tried to plug those values of $p$ to :$\frac{p^{2}\pm1}{10}$, but I didn’t get an integer.

Thank you for your help

答:
$p$ is odd so $p^{2}$ is odd so $p^{2}\pm 1$ is even. That takes care of the factor of 2. Now we just need the factor of 5. It will be a little tedious since you disallow modular arithmetic but here it goes.

If p=5m+a then

$$p^{2}=25m^{2}+5am+a^2$$

So for $a=1,2,3,4$

$$p^{2}=25m^{2}+5m+1$$

$$p^{2}=25m^{2}+10m+4$$

$$p^{2}=25m^{2}+15m+9$$

$$p^{2}=25m^{2}+20m+16$$

Respectively.

$25m^{2}+20m+16 -1 = 25m^{2}+20m+15$ is clearly divisible by 5.

$25m^{2}+15m+9 + 1 = 25m^{2}+15m+10$ is clearly divisible by 5.

$25m^{2}+10m+4 +1  = 25m^{2}+15m+5$ is clearly divisible by 5.

$25m^{2}+5m+1 -1 = 25m^{2}+5m$ is clearly divisible by 5.


问:
Proving union of closures is closure of unions
Is this okay for proving this subset direction, $\text{cl}(A) \cup \text{cl}(B) \supset \text{cl}(A \cup B)$?

Let $x \in \text{cl}(A \cup B)$. Let $U$ be any open set containing $x$. Then $U \cap (A \cup B)=(U \cap A) \cup (U \cap B) \neq \emptyset$.Then $U \cap A \neq \emptyset$ or $U \cap B \neq \emptyset$
If $U \cap A \neq \emptyset \implies x \in \text{cl}(A)$. So $x \in  \text{cl}(A) \cup \text{cl}(B)$.If $U \cap B \neq \emptyset \implies x \in \text{cl}(B)$. So $x \in  \text{cl}(A) \cup \text{cl}(B)$.

答:
No, because you deduced that $U\cap A$ for some neighborhood of $x$, but not for any neighborhood. (You can say that without loss of generality, $U\cap A\neq\varnothing$ for given $U$, but you can´t say that without loss of generality, $U\cap A\neq\varnothing$ for all $U$).

You can prove that fact by contradiction (spoilers ahead). If $x$ wasn´t in $cl(A)\cup cl(B)$, then you would have $U,V$ neighborhoods of $U$ such that $U\cap A=\varnothing$ and $V\cap B=\varnothing$. Thus, $U\cap V$ is a neighborhood of $x$ with $(U\cap V)\cap(A\cup B)=\varnothing$.

Edit: Some more clarification on why the answer doesn´t work. You can have some neighborhood $U$ with $U\cap A\neq\emptyset$ and other different neighborhood $V$ such that $V\cap B\neq\emptyset$. That is, for different open sets containing $x$, there can be some that intersect $A$ and others that intersect $B$, but you haven´t yet proved that every neighborhood intersects $A$, or that every neighborhood intersects $B$. You just proved that every neighborhood intersects either $A$ or $B$, which isn´t enough.

You know that every nhood intersects $A$ or $B$:

$\forall U$ neighborhood of $x$, ($U\cap A\neq\varnothing$ or $U\cap B\neq\varnothing$).

And to prove that $x$ is in the closure of $A$ or in the closure of $B$ you need the following:

($\forall U$ neighborhood of $x$ $U\cap A\neq\varnothing$) or ($\forall U$ neighborhood of $x$ $U\cap B\neq\varnothing$).

However, the second statement isn´t implied by the first one, so the proof is incomplete.


问:
A particle is moving in a straight line with velocity v m.s⁻¹
Question

My Attempt

I have done part A and got it right, but am struggling with part B. I have attached my attempt on the question as well as the full question.

The correct answer for part B is:
t1 = 4.07 s  or  t2 = 7.93 s (5 marks)

But I'm not sure where I went wrong and what I have to do.

答:
Your formula 1) for the position is fine, but it is valid only up to $t=4$. Hence your solution $t=4.065$ must be discarded.

Your formula 2) isn't correct, because position $s(t)$ at $t=4$ is $8/15$: a constant must then be added to account for that. The correct formula is as follows:
$$
s(t)=\int_0^t v(\tau)\,d\tau=
\int_0^4 (\tau^2-0.4\tau-4.4)\,d\tau+
\int_4^t (30-5\tau)\,d\tau=30t-{5\over2}t^2-80+{8\over15}.
$$


问:
A person of mass 80kg is standing in a lift of mass 600kg.
The lift is supported by a cable which is modelled as light and inextensible. The lift is moving upwards and decelerating at 1.8m.s⁻²

a) Find the tension in the lift cable.
b) Find the magnitude of the normal reaction exerted by the floor of
the lift on the person.

I have attempted this question, and got very close answers to the mark scheme, but I can't figure out why my answer is different.

The correct answers are:

(a)  5576 N (5 marks)
(b)  656 N (4 marks)

My attempt

答: The so-called correct answers seem to be obtained by taking the acceleration of gravity to be 10 m/sec$^2$ instead of 9.8 m/sec$^2$.


问:
Is it possible to determine f(7)if f(x) = 3 for all x < 7 and f is right-continuous at x = 7? What if f is left-continuous?

I am having a hard time really understanding the question, but what I am gathering is if the question had said "f(x)=3 for all x GREATER THAN 7" then we would be able to determine f(7) because it is right continuous until it gets to 3? Which 3 would be our value for f(7) in the new condition I created?

Maybe I just need a rewording of the question.

答:
If $ f $ is right-continuous at $ x=7$ then
$$f(7)=\lim_{x\to 7^+}f(x)=L$$

But if we do not have a way to compute $ L $, we cannot get $ f(7)$.

But, if $ f $ is left-continuous at $ x=7 $ then

$$f(7)=\lim_{x\to 7^-}f(x)=3$$
because
$$(\forall x<7)\;\;f(x)=3$$


问:
Inverse Semigroups, Partial Bijections, and Semilattice of Idempotents
I have a question about a passage from this paper. First, some definitions

A Semigroup $S$ is said to be an inverse semigroup provided that for every $x \in X$, there exists a unique element $x^{-1}$ such that $x^{-1}xx^{-1} = x^{-1}$ and $xx^{-1}x=x$.

The semilattice $E$ of idempotents in an inverse semigroup $S$ is given by $E = \{x^{-1}x \mid x \in S\} = \{xx^{-1} \mid x \in S\} = \{e \in S \mid e = e^2 \}$.

On page 18 of the above linked paper, the author writes:

In the case of partial bijections, the semilattice of idempotents is given by all domains and images. Multiplication in this semilattice is intersection of sets, and $\le$ is $\subseteq$ for sets, i.e., containment.

Given a set $X$, the collection of all partial bijections on $X$ is denoted as $I(X)$. The semilattice of idempotents $I(X)$ is a subset of $I(X)$, so it itself should consist of partial bijections. But how can the semilattice of idempotents consist of sets? Shouldn't it consist of all partial bijections $f$ such that $f^2 = f$? Are we identifying $f$ with its domain and image? I don't think this makes sense because there are distinct functions with the same domain and image.

答:
The idempotents in this semigroup are those partial functions $f$ such that the range is a subset of the domain and $f$ maps every element of the range to itself.

CORRECTION: I  misread "partial bijections" as "partial functions". So the requirements on $f$ in my answer together with the "bijection" requirement imply that $f$ is simply the identity function on some subset of $X$. Apparently the author of the quoted text identifies "the identity function on a subset $Y$ of $X$" with simply $Y$ and thus gets that the idempotents amount to subsets of $X$.


问:
$(n+1)!>2^{(n+1)}$ for $n$, where $n \geq 4$. (Mathematical Induction)
By using Mathematical Induction prove that $(n+1)!>2^{(n+1)}$ for $n$, where $n \geq 4$.

Body:
First, check the base case:
$$((4)+1)!=120>32=2^{(4)+1}$$
Next we want to show that $(k+1)!>2^{k+1}\implies (k+2)!>2^{k+2}$. Since $(k+2)>2$ for all $k \geq 4$ and by hypothesis $(k+1)!>2^{k+1}$, we get
$$(k+2)!=(k+2)(k+1)!>2\cdot 2^{k+1}=2^{k+2}$$


问:
True/False test: $A\in\mathbb C^{2\times 2}$ is nilpotent. Then $A^2=0.$

True/False test: $A\in\mathbb C^{2\times 2}$ is nilpotent. Then $A^2=0.$

My Attempt:

True: $A^k=0$ for some $k\in \mathbb Z^+\\\implies m_A(x)|x^k\\\implies x\text{ is the only irreducible factor of }m_A\\\implies x\text{ is the only irreducible factor of }\chi_A(\text{since }\chi_A\text{ and }m_A\text{ have same irreducible factors})\\\implies\chi_A=x^2\\\implies A^2=0$

Am I correct?

Body:
Your answer is:

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,▕▔╲ \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,▏▕\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,▏▕▂▂▂\\
▂▂▂▂▂▂╱.┈▕▂▂▂▏\\
▉▉▉▉▉▉.┈┈▕▂▂▂▏\\
▉▉▉▉▉▉.┈┈▕▂▂▂▏\\
▔▔▔▔▔▔╲▂▕▂▂▂$


问:
find a function $f(t)$ such that $\mathscr{L}\{f(t)\}(s)=f(s)/s$
find a function $f(t)$ such that $\mathscr{L}\{f(t)\}(s)=f(s)/s$

Does such a function $f(t)$ exist?

This came up while I was playing around with p-adic Norms.

答:
As shown in almost any book on the Laplace transform, we have that
$$
\mathscr{L}\left[\int\limits_0^t f(r)\mathrm{d}r\right](s)=\frac{F(s)}{s} + \frac{1}{s} \int\limits_{-\infty}^{0^+} f(r)\mathrm{d}r \iff \frac{F(s)}{s} = \mathscr{L}\left[\int\limits_0^t f(r)\mathrm{d}r\right](s) - \frac{1}{s} \int\limits_{-\infty}^{0^+} f(r)\mathrm{d}r
$$
and then using the fact that in this case you have $F(s)= f(s)$ we have
$$
\begin{split}
f(t)& =\mathscr{L}^{-1}\big[\mathscr{L}[f(t)](s)\big](t)\\
& = \mathscr{L}^{-1}\left[\mathscr{L}\left[\int\limits_0^t f(r)\mathrm{d}r\right](s) - \frac{1}{s} \int\limits_{-\infty}^{0^+} f(r)\mathrm{d}r\right](t)\\
& = \int\limits_0^t f(r)\mathrm{d}r - H(t)\int\limits_{-\infty}^{0^+} f(r)\mathrm{d}r
\end{split}
$$
where $H(t)$ is the Heaviside function. Thus you have an integral equation which can be reduced by differentiation to the following first order linear ordinary differential equation:
$$
\frac{\mathrm{d}}{\mathrm{d} t}f(t)= f(t) - \delta (t)\int\limits_{-\infty}^{0^+} f(r)\mathrm{d}r\label{1}\tag{1}
$$
where $\delta(t)$ is the Dirac delta distribution.

Solution to equation \eqref{1}.

First note that \eqref{1} is linear since
$$
c_\infty=\int\limits_{-\infty}^{0^+} f(r)\mathrm{d}r\label{ic}\tag{IC}.
$$
should be considered as a datum. Said that, we start seeking the solution of \eqref{1} in the form $f(t)= f_o(t) + f_p(t)$, where $f_p(t)$ is a particular solution found as described in this answer and $f_o(t)$ is the solution of the associated homogeneous equation
$$
\frac{\mathrm{d}}{\mathrm{d} t}f_o(t)= f_o(t).\label{2}\tag{2}
$$
We have
$$
\begin{split}
f_o(t) &=c e^t\\
f_p(t) &=c_\infty H(t)e^t
\end{split}
$$
where $c$ is a constant to be determined: and in order to determine it, we use the integral condition \eqref{ic}, obtaining
$$
c_\infty = \int\limits_{-\infty}^{0^+} f(r)\mathrm{d}r = \int\limits_{-\infty}^{0^+} f_o(r)\mathrm{d}r =c \int\limits_{-\infty}^{0^+} e^r\mathrm{d}r = c\big[e^t\big]_{-\infty}^{0+} = c
$$
Finally,
$$
f(t)= c_\infty e^t\big[1+H(t)\big]
$$


问:
Anti Logarithm Formula
Is there any formula to calculate anti logarithm just using simple calculator.. I already know how to calculate logarithm digit by digit exactly just like this What is the best way to calculate log without a calculator? But how to calculate the antilog like antilog(0.456)=10^0.456=2.857590 using simple calculator..</p>

答:
I realize I'm a little late to the party, but if you need a simple way to calculate a rough approximation of antilog base 10, use the following procedure:

Separate the integer part from the fractional part.  So 1.3979 is split into 1 and 0.3979.

Use the fractional part on this formula:
5.5x^3+  x^2 + 2.5x + 1
So in this example, 5.5(0.3979^3) + 0.3979^2 + 2.5(0.3979) + 1
which equals: 2.499 or 2.5 with a little rounding.

Now write the answer and then E and then the integer part of the original number.
So we get 2.5E1 which is short for 2.5x10^1 or 25.

If you aren't looking for exact values and just need to be in the ballpark, this formula will usually do.  If you are making some kind of graphics program where speed is more important than accuracy, then this is what you want.

Title: Anti Logarithm Formula


问:
Geometric meaning of r-cycles, r-boundaries and homology groups for a geometric simplicial complex.
I just started learning about algebraic topology, and some things are already not so clear to me. If I consider a geometric simplex $K$, I kind of understand what $H_0(K)$ is. it is a set of equivalent classes [x], where x is a vertex of the simplex, and [x]=[y] means that x and y are in the same connected components. Now for $H_1(K)$ (and more generally for $H_n(K)$, it is not so clear. Elements of $H_1$ are equivalent classes of edges of the simplicial complex, like [xy] or [uv]. [xy]=[uv] means that xy-uv is in the boundary of a 2-simplex. But is geometrically the boundary of a 2-simplex? Also what is a cycle geometrically of a 1,2,3,...-simplex?

答:
Geometrically, think of an n-simplex as an n-sphere ( it makes no difference whatsoever from a topological perspective ). So the n-th homology is simply put "how many ways there are to fit an n-shpere into my topological space". That means, just like the 0-th homology tells you how many parts there are in a space, the n-th lets you know how many n-dimension holes there are in the space you are examining. A cycle is nothing more than an element in the kernel of the boundary operator, homology takes those n-cycles and those n-boundaries and tries to figure out how many of them are redundant ( meaning how many elements in the kernel don't come from higher dimension boundaries ).
Now I know that my explanation can be a bit overly simplified, but to understand topology you are heavily reliant in your intuition rather than strict and vigorous mathematical calculations and logical steps due to the nature of the subject. Try to figure things out in a more instrictive rather than a more technical manner.</p>


问:
Nested equilateral triangles
Let triangle $ABC$ is an equilateral triangle. Triangle $DEF$ is also an equilateral triangle and it is inscribed in triangle $ABC \left(D\in BC,E\in AC,F\in AB\right)$. Find $\cos\measuredangle DEC$ if $AB:DF=8:5$.

Firstly, I would be very grateful if someone can explain to me how I am supposed to draw the diagram. Obviously I have made it by sight.

Let $\measuredangle DEC=\alpha$. We can note that $\triangle AEF \cong \triangle BFD \cong CDE$. This is something we can always use in such configuration. So $$AE=BF=CD, $$ $$AF=BD=CE.$$ Let $AB=BC=AC=8x$ and $DF=DE=EF=5x$. If we denote $CD=y,$ then $CE=AC-AE=AC-CD=8x-y$. Cosine rule on $CED$ gives $$25x^2=(8x-y)^2+y^2-2y\cos60^\circ(8x-y)$$ which is a homogenous equation. I got that $\dfrac{y}{x}=4\pm\sqrt{3}.$ Now using the sine rule on $CED$ $$\dfrac{CD}{DE}=\dfrac{\sin\alpha}{\sin60^\circ}\Rightarrow \sin\alpha=\dfrac{\sqrt{3}}{10}\cdot\dfrac{y}{x}=\dfrac{4\sqrt3\pm3}{10}.$$ Now we can use the trig identity $\sin^2x+\cos^2x=1$ but it doesn't seem very rational. Can you give me a hint? I was able to find $\sin\measuredangle DEC$ in acceptable way, but I can't find $\cos\measuredangle DEC$...

答:
As others have done, I'll assume $AB=8,DF=5$ without loss of generality. Then equilateral triangles $\triangle ABC$ and $\triangle DEF$ have area $(\sqrt{3}/4)AF^2 = 16\sqrt{3}$ and $(\sqrt{3}/4)DF^2=(25/4)\sqrt{3}$ respectively. Since the triangles  $\triangle FAE,\triangle DBF, \triangle ECD$ are all congruent, they must each have area $$\frac{1}{3}\left(16\sqrt{3}-\frac{25}{4}\sqrt{3}\right)=\frac{13\sqrt{3}}{4}.$$ But each triangle has a 60-degree base angle, so we can also write the area of $\triangle FAE$ as

$$\frac12 AF\cdot AE \cdot \sin 60^\circ$$

Since $AE=AC-EC=8-AF$, this becomes $$\frac12 AF(8-AF)\frac{\sqrt{3}}{2}= \frac{13\sqrt{3}}{4}\implies AF(8-AF)=13$$
with $x=4\pm \sqrt{3}$ as solutions. As a minor variation on the other approaches, we may then use the law of cosines on triangle $\triangle CED$ to write

$$CD^2=CE^2+ED^2-2CE\cdot ED\cos\measuredangle DEC$$
and therefore

\begin{align}
\cos\measuredangle DEC
&=\frac{CE^2+ED^2-CD^2}{2CE\cdot ED}\\
&=\frac{(4\pm \sqrt{3})^2+5^2-(4\mp \sqrt{3})^2}{2(5)(4\mp \sqrt{3})}\cdot \frac{4\mp \sqrt{3}}{4\mp \sqrt{3}}\\
&=\frac{(25\pm 16\sqrt{3})(4\mp \sqrt{3})}{130}\\
&=\frac{52\pm 39\sqrt{3}}{130}\\
&=\frac{4\pm 3\sqrt{3}}{10}
\end{align}
as others have obtained.

Title: Nested equilateral triangles


问:
Finding convolution of two functions
A common engineering notational convention is: wikipedia

${\displaystyle f(x)*g(x)\,:=\underbrace {\int_{0}^{x}f(\tau )g(x-\tau )\,d\tau } _{(f*g)(x)}.}$

I want to write the following expression as the convolution of two functions. In the other words;
what is the function $g(x)$ in the definition for the integral below:

\begin{equation}
\frac{(\rho+1)^{1-\alpha}}{\Gamma(\alpha)} \int_{a}^{x}\left(x^{\rho+1}-\tau^{\rho+1}\right)^{\alpha-1} \tau^{\rho} f(\tau) \,d \tau
\end{equation}
where
$\alpha$ and $\rho \neq-1$ are real numbers and $x > a$, $\Gamma$ is gamma function.

My Try:
Let

\begin{align}
k&=\tau^{\rho+1}\\
dk&=(\rho+1)\tau^{\rho}\,d \tau \implies \frac{dk}{(\rho+1)}=\tau^{\rho}\,d \tau.
\end{align}

Substituting the last equality to the integral in the question, we have

\begin{equation}
\frac{(\rho+1)^{-\alpha}}{\Gamma(\alpha)} \int_{a^{\rho}}^{x^{\rho}}\left(x^{\rho+1}-k\right)^{\alpha-1} f(k^{1/(\rho+1)})\, dk
\end{equation}

And then?

P.S.
If the last expression was

\begin{equation}
\frac{(\rho+1)^{-\alpha}}{\Gamma(\alpha)} \int_{0}^{x}\left(x-k\right)^{\alpha-1} f(k)\, dk
\end{equation}
$g(x)$ would be as follows:

$$g(x)=\frac{(\rho+1)^{-\alpha}}{\Gamma(\alpha)} x^{\alpha-1}.$$

答:
Let $z:=x^{\rho+1}$ and $\sigma:=\tau^{\rho+1}$.

$$\int(x^{\rho+1}-\tau^{\rho+1})^{\alpha-1}\tau^\rho f(\tau)d\tau=\frac1{\rho+1}\int(z-\sigma)^{\alpha-1}f(\sigma^{1/(\rho+1)})d\sigma$$

and you could be tempted to conclude

$$g(t)\equiv t^{\alpha-1}$$ and $$f(t)\equiv f(t^{1/(\rho+1)}).$$

Unfortunately, the bounds are not those of a convolution. What you are asking is not possible.

Title: Finding convolution of two functions


问:
formula of pascal's triangle
I want to make a program for pascal's triangle,I was reading through the details and found something like this:

```<code>0:                                1
1:                             1     1
2:                           1    2    1
3:                         1   3     3   1
4:                       1   4    6    4   1
5:                      1  5   10   10   5   1
</code>```

by looking at row number 5 of the triangle, one can quickly read off that:

$$(x + y)^5 = 1\cdot x^5 + 5\cdot x^4y + 10\cdot x^3y^2 + 10\cdot x^2y^3 + 5\cdot x y^4 + 1\cdot y^5$$

but I haven't been able to understand what are these x and y and what shall be their values.and also how shall I proceed for further rows.
I shall be really thatnkful if someone explained this to me.

答: Look at binomial coefficients. the number in the rth position in the nth row is given by n choose r, calculated by $ {n \choose r} = \frac{n!}{r!(n-r)!} $. This gives you the number of ways of choosing r objects from a set of n, you can see why we want to consider this since in your expansion of $(x+y)^5$ we want to count the various ways to get the different powers of xy.</p>

Title: formula of pascal's triangle