问:
$\sqrt{\frac{x-8}{1388}}+\sqrt{\frac{x-7}{1389}}+\sqrt{\frac{x-6}{1390}}=\sqrt{\frac{x-1388}{8}}+\sqrt{\frac{x-1389}{7}}+\sqrt{\frac{x-1390}{6}}$

How many solutions the following equation has in real numbers
?$$\sqrt{\frac{x-8}{1388}}+\sqrt{\frac{x-7}{1389}}+\sqrt{\frac{x-6}{1390}}=\sqrt{\frac{x-1388}{8}}+\sqrt{\frac{x-1389}{7}}+\sqrt{\frac{x-1390}{6}}$$

$1)1\quad\quad\quad\quad\quad\quad\quad2)2\quad\quad\quad\quad\quad\quad\quad3
 )3\quad\quad\quad\quad\quad\quad\quad4)4\quad\quad\quad\quad\quad5)\text{zero}$

Obviously we have $x\ge1390$. starting from $x=1390$ we can see LHS is very close to $1+1+1=3$ and RHS is close to zero. for higher values of $x$ we can see right side of the equation grows much faster than left side. so it seems the equation has not answer in real numbers.

I'm not sure how to proceed mathematically. I can rewrite it as follow:
$$\left(\sqrt{\frac{x-8}{1388}}-\sqrt{\frac{x-1388}{8}}\right)+\left(\sqrt{\frac{x-7}{1389}}-\sqrt{\frac{x-1389}{7}}\right)+\left(\sqrt{\frac{x-6}{1390}}-\sqrt{\frac{x-1390}{6}}\right)=0$$
I putted fractions with similar numbers in the same parentheses so it looks better now, but don't know how to continue.

Body:
$$\sqrt{\frac{x-8}{1388}}+\sqrt{\frac{x-7}{1389}}+\sqrt{\frac{x-6}{1390}}=\sqrt{\frac{x-1388}{8}}+\sqrt{\frac{x-1389}{7}}+\sqrt{\frac{x-1390}{6}} \tag {1}$$

Since $x \geq 1390$, let us put $x = 1390 + k$ where $k \geq 0$

$(1) \implies$

$$\sqrt{\frac{1382+k}{1388}}+\sqrt{\frac{1383+k}{1389}}+\sqrt{\frac{1384+k}{1390}}=\sqrt{\frac{k+2}{8}}+\sqrt{\frac{k+1}{7}}+\sqrt{\frac{k}{6}}$$

Note that since $k \geq 0$,

we have
$$\left(\sqrt{\frac{1382+k}{1388}} < \sqrt{\frac{1383+k}{1389}} < \sqrt{\frac{1384+k}{1390}}  < 1 \right) \land \left(\sqrt{\frac{k}{6}} < \sqrt{\frac{k+1}{7}} < \sqrt{\frac{k+2}{8}}  < 1 \right)$$ for $k < 6$,

$$\left(\sqrt{\frac{1382+k}{1388}} > \sqrt{\frac{1383+k}{1389}} > \sqrt{\frac{1384+k}{1390}}  > 1 \right) \land \left(\sqrt{\frac{k}{6}} > \sqrt{\frac{k+1}{7}} > \sqrt{\frac{k+2}{8}}  > 1 \right)$$ for $k > 6$ and

$$\left(\sqrt{\frac{1382+k}{1388}} = \sqrt{\frac{1383+k}{1389}} = \sqrt{\frac{1384+k}{1390}} = 1 \right)  \land \left(\sqrt{\frac{k}{6}} = \sqrt{\frac{k+1}{7}} = \sqrt{\frac{k+2}{8}} = 1 \right)$$ for $k = 6$

Can you argue why the first two cases do not lead to any solution but the last case does?

In fact, you can show that

$\left(\sqrt{\frac{k}{6}} < \sqrt{\frac{k+1382}{1388}} \right) \land \left(\sqrt{\frac{k+1}{7}} < \sqrt{\frac{k+1383}{1389}} \right) \land \left(\sqrt{\frac{k+2}{8}} < \sqrt{\frac{k+1384}{1390}} \right)$ for $k < 6$

and

$\left(\sqrt{\frac{k}{6}} > \sqrt{\frac{k+1382}{1388}} \right) \land \left(\sqrt{\frac{k+1}{7}} > \sqrt{\frac{k+1383}{1389}} \right) \land \left(\sqrt{\frac{k+2}{8}} > \sqrt{\frac{k+1384}{1390}} \right)$ for $k > 6$


问:
How to get the maximum integer value for $1+\sin^2\alpha$?
The problem is as follows:

Assume: $\alpha \in \left[ \frac{\pi}{2},\, 4 \right\rangle$
(Read as alpha belonging to the interval closed in $\frac{\pi}{2}$ and open in $4$).

$B=1+\sin^2\alpha$

Find the maximum integer value that it can take $B$.

The choices given are:

$\begin{array}{ll}
1.&\textrm{4}\\
2.&\textrm{3}\\
3.&\textrm{0}\\
4.&\textrm{2}\\
\end{array}$

I'm stuck here from the very beginning. How exactly should I find the maximum integer value for that expression here?.

Does it exist a criteria for this?. I'm requesting for help here because I feel lost.

I think I could use:

$-1 < \sin \alpha < 1$

But I doubt that's the approach right for here. Can someone help me here?. This topic isn't well covered in my precalculus book. But I'm certain that such simple approach would be used. As I mentioned, it is not much that I can offer. Some other ideas which had come to my mind are perhaps the use of.

$AM\geq GM$

But I don't know if such would be used or have to be used?.

Other approach could be derivatives.

Trying this would be:

$B(\alpha)=1+\sin^2\alpha$

$B'(\alpha)=2\sin\alpha\cdot \cos\alpha$

Equating this to zero yields:

$2\sin\alpha\cdot \cos\alpha = 0$

$\sin 2\alpha = 0$

Hence this can take values in the given range:

$2\alpha \in \left[ \pi,\, 8 \right\rangle$

Gee, $8$ is somewhat close to $2.54 \pi$

Then this means I have to look for values equal to zero in that range:

Those would be:

$\pi$, $2\pi$

But since it is divided by two for \alpha would be:

$\frac{\pi}{2}, \pi$.

This means:

Inserting any of these values in the function would give me the maximum?.

But again this would make any of those values $B=1$ and none of those values appears in the choices. Thus can someone help me here?.

答: The only integer values achieved by $\sin^2\alpha$ are $0$ and $1$ (for $\alpha=k\frac\pi2$). Depending on the allowed range of $\alpha$, you can tell which holds.

ContentLicense: CC BY-SA 4.0


问:
Linear Programming Cutting Problem
I'm doing some self studying and I have come across a problem I don't quite understand. The problem only gives me the single constraint:

The first question asks me to show that the given constraint implies the following cuts:

The second questions requires me to come up with a general rule for making cuts like the second cut from the given constraint:

The last question asks me how would I use this constraint to derive such cuts and wants me to convert it to the form in the second question:

I can see the constraint requiring less than four or five "1" values at minimum but I'm not quite sure how I would go about showing it. I haven't been able to find a question of similar nature anywhere else so I turn to people here hoping they might understand what I don't.

答:
These cuts are called cover inequalities or no-good cuts.  If your original constraint is $$\sum_{i=1}^n a_i y_i \le a_0,$$
any subset $C \subseteq \{1,\dots,n\}$ with $$\sum_{i\in C} a_i > a_0$$ yields a cover inequality
$$\sum_{i\in C} y_i \le |C|-1.$$
You can derive this via conjunctive normal form as follows:
$$
\lnot \bigwedge_{i\in C} y_i \\
\bigvee_{i\in C} \lnot y_i \\
\sum_{i\in C} (1 - y_i) \ge 1 \\
\sum_{i\in C} y_i \le |C|-1
$$
Your first cut arises from $C=\{1,\dots,6\}$ with $\sum_{i\in C} a_i = 70 > 52$,
and your second cut arises from $C=\{2,\dots,6\}$ with $\sum_{i\in C} a_i = 61 > 52$.

Title: Linear Programming Cutting Problem


问:
Compute close-form of $\int_0^{\frac\pi2}\frac{dt}{\sin t+\cos t+\tan t+\cot t+\csc t+\sec t}$
I came across the improper trigonometric integral recently shared in a Chinese web forum
\begin{align}
\int_0^{\frac\pi2}\frac{dt}{\sin t+\cos t+\tan t+\cot t+\csc t+\sec t}\\
\end{align}
which amuses me because of its appearance. What is more amusing is the claim, without providing the proof, that is has the close-form result
\begin{align}
\frac{\sqrt{8\sqrt2+2\sqrt7}}{7^{3/4} }\tanh^{-1}  \frac{\sqrt{2\sqrt7(4\sqrt2-5)}}{4\sqrt2-5 +\sqrt7}
+\frac{\sqrt{8\sqrt2-2\sqrt7}}{7^{3/4} }\tan^{-1}\frac{\sqrt{2\sqrt7(4\sqrt2-5)}}{4\sqrt2-5 - \sqrt7}
\end{align}
I was able to verify it numerically; yet rather curious in how it could ever be derived. Based on my knowledge and experience, I do not assume it would be easy.

I was able to find an unsolved post here from a long time ago, which only reexpresses the corresponding  indefinite integral via the tangent half-angle substitution as
$$\int\frac{2t(1-t)}
{2 t^4-3 t^3+3 t^2+t+1}dt$$
but gives up due to the complexity in partial fractionalization. I am doubtful that this approach would lead to the explicit expression claimed above. I am interested in any suitable methods producing the close-form.

答:
This feels more like a comment to Travis Willse's excellent answer, which essentially solves the problem, but here's the rest of the calculation:

So far we have ($u = \frac{\tan \left(\frac{\pi}{8} - \frac{t}{2}\right)}{\sqrt{2}-1}$)
\begin{align}
I &\equiv \int \limits_0^{\pi/2} \frac{\mathrm{d} t}{\sin(t) + \cos(t) + \tan(t) + \cot(t) + \sec(t) + \csc(t)} \\
&= 2 \int \limits_0^{\pi/4} \frac{\mathrm{d} t}{\sin(t) + \cos(t) + \tan(t) + \cot(t) + \sec(t) + \csc(t)} \\
&= \frac{2 \sqrt{2} \beta^2}{\sqrt{7}} \int \limits_0^1 \frac{1-u^2}{u^4 + \beta^4} \, \mathrm{d} u = \frac{2 \sqrt{2} \beta^2}{\sqrt{7}} \int \limits_0^1 \frac{1-u^2}{(u^2 + \mathrm{i} \beta^2) (u^2 - \mathrm{i} \beta^2)} \, \mathrm{d} u
\end{align}
with $\beta = \sqrt{\frac{5 + 4 \sqrt{2}}{\sqrt{7}}}$. We now need $a_{1/2} \in \mathbb{C}$ such that $a_1 (u^2 + \mathrm{i} \beta^2) + a_2 (u^2 - \mathrm{i} \beta^2) = 1 - u^2$ holds. The solution is $a_{1/2} = - \frac{1}{2} \left(1 \pm \frac{\mathrm{i}}{\beta^2}\right)$, so we can write
\begin{align}
I &= - \frac{2 \sqrt{2} \beta^2}{\sqrt{7}} \operatorname{Re} \left[\left(1 + \frac{\mathrm{i}}{\beta^2}\right) \int \limits_0^1 \frac{\mathrm{d} u}{u^2 - \mathrm{i} \beta^2} \right] = - \frac{2 \sqrt{2} \beta^2}{\sqrt{7}} \operatorname{Re} \left[\left(1 + \frac{\mathrm{i}}{\beta^2}\right) \int \limits_0^1 \frac{\mathrm{d} u}{u^2 + \left(\frac{1 - \mathrm{i}}{\sqrt{2}} \beta\right)^2} \right] \\
&= - \frac{2 \sqrt{2} \beta^2}{\sqrt{7}} \operatorname{Re} \left[\left(1 + \frac{\mathrm{i}}{\beta^2}\right) \frac{1 + \mathrm{i}}{\sqrt{2} \beta} \arctan \left(\frac{1 + \mathrm{i}}{\sqrt{2} \beta}\right) \right] \\
&= - \frac{2}{\sqrt{7}} \operatorname{Re} \left[\left(\beta - \beta^{-1} + \mathrm{i} \left(\beta + \beta^{-1}\right)\right) \arctan \left(\frac{1 + \mathrm{i}}{\sqrt{2} \beta}\right) \right] \\
&= - \frac{1}{\sqrt{7}} \operatorname{Re} \left[\left(\beta - \beta^{-1} + \mathrm{i} \left(\beta + \beta^{-1}\right)\right) \left(\arctan \left(\frac{\sqrt{2}}{\beta - \beta^{-1}}\right) + \mathrm{i} \operatorname{artanh} \left(\frac{\sqrt{2}}{\beta + \beta^{-1}}\right)\right) \right] \\
&= \frac{1}{\sqrt{7}} \left[\left(\beta + \beta^{-1}\right) \operatorname{artanh} \left(\frac{\sqrt{2}}{\beta + \beta^{-1}}\right)- \left(\beta - \beta^{-1}\right) \arctan \left(\frac{\sqrt{2}}{\beta - \beta^{-1}}\right)\right] \, .
\end{align}
In the penultimate step we have used
$$ \arctan(a + \mathrm{i} b) = \frac{1}{2} \left[\arctan \left(\frac{2 a}{1 - a^2 - b^2}\right) + \mathrm{i} \operatorname{artanh} \left(\frac{2 b}{1 + a^2 + b^2}\right)\right] $$
from here. Since $\beta \pm \beta^{-1} = \sqrt{2} \sqrt{4 \sqrt{\frac{2}{7}} \pm 1}$, we obtain
$$ I = \sqrt{\frac{2}{7}} \left[\sqrt{4 \sqrt{\frac{2}{7}} + 1} \operatorname{arcoth} \left(\sqrt{4 \sqrt{\frac{2}{7}} + 1}\right) - \sqrt{4 \sqrt{\frac{2}{7}} - 1} \operatorname{arccot} \left(\sqrt{4 \sqrt{\frac{2}{7}} - 1}\right)\right] \, , $$
which is a simplified version of the final result given in the question.


问:
Integral by parts the function in the interval $(0,1)$
There is the integral of the function $(\sin x)^2\ln x$ in the interval $(0,1)$, which I am asked to find if it converges.

I tried integration by parts but I came up with the integral of $\sin(2x)/x$ and here I stopped as I found difficulty in calculating this integral according to the integration methods.

How can I proceed?

I would appreciate for your quidance.

Thank you very much in advance.

答: You need a special function to evaluate the integral by parts. In terms of the sine integral, $$\int_0^1\sin^2x\ln xdx=\left[\frac{\sin^2x}{x}\right]_0^1-\int_0^1\frac{\sin 2xdx}{x}=\sin1-\operatorname{Si}(2)$$because $\lim_{x\to0}\frac{\sin x}{x}=1\implies\lim_{x\to0}\frac{\sin^2x}{x}=0$. But just for a convergence test, we can use a comparison. Since the logarithm is negative and the sine is bounded, you could argue$$\left|\int_0^1\sin^2x\ln xdx\right|\le\left|\int_0^1\ln xdx\right|=[x-x\ln x]_0^1=1,$$using $\lim_{x\to0^+}x\ln x=0$.


问:
Change of variables in integration on manifolds: what's wrong?
The problem is this: Let $f:M^m\to\mathbb{R}$ a continuous function with compact support and $h:N\to M$ a diffeomorphism between Riemannian compact manifolds. Then
$$\int_{M}fdV^M=\int_N f\circ h|\det dh|dV^N$$
The proof using differential forms is easy, but I'm trying to prove this without using forms, just the definition.

"Proof":

Let $\Omega$ be the support of $f$, $\Theta=h^{-1}(\Omega)$ the support of $f\circ h$, $\textbf{x}_\alpha:U_\alpha\to M$ and $\textbf{y}_\alpha:V_\alpha\to N$ parametrizations and $\{\phi_\alpha\}_{\alpha\in A}$ a partition of unity subordinate to $\{U_\alpha\}_{\alpha\in A}$. Let $g^M$ the metric of $M$ and $g^N$ the metric of $N$.Then
$$\int_M fdV^M=\sum_{\alpha}\int_{\textbf{x}_\alpha^{-1}(\Omega)}\phi_\alpha f\circ\textbf{x}_\alpha\sqrt{\det g^M}dx^M$$
Denoting $\Omega_\alpha=\textbf{x}_\alpha^{-1}(\Omega)\subset U_\alpha$,$\Theta_\alpha=\textbf{y}^{-1}_\alpha(\Theta)\subset V_\alpha$ and $T_\alpha=\textbf{x}_\alpha^{-1}\circ h\circ\textbf{y}_\alpha:V_\alpha\to U_\alpha$ we will have
$$\int_M fdV^M=\sum_{\alpha}\int_{\Omega_\alpha}\phi_\alpha f\circ\textbf{x}_\alpha\sqrt{\det g^M}dx^M=\sum_{\alpha}\int_{T_\alpha(\Theta_\alpha)}\phi_\alpha f\circ\textbf{x}_\alpha\sqrt{\det g^M}dx^M$$
by the variable change theorem,
$$\int_M fdV^M=\sum_{\alpha}\int_{\Theta_\alpha}(\phi_\alpha\circ T_\alpha)(f\circ\textbf{x}_\alpha\circ T_\alpha)\sqrt{\det g^M\circ T_\alpha}|\det dT_\alpha|dy^N$$
How
\begin{eqnarray*}
\sqrt{\det g^M\circ T_\alpha}&=&\mathrm{vol}\ P\left[\frac{\partial}{\partial x_1},\cdots,\frac{\partial}{\partial x_m}\right] \\
&=&\mathrm{vol}\ P\left[dh\left(\frac{\partial}{\partial y_1}\right),\cdots,dh\left(\frac{\partial}{\partial y_m}\right)\right] \\
&=&\mathrm{vol}\ P\left[\frac{\partial}{\partial y_1},\cdots,\frac{\partial}{\partial y_m}\right]|\det dh| \\
&=&\sqrt{g^N}|\det dh|
\end{eqnarray*}
and
$$\det dT_\alpha=\det dh$$
We have
\begin{eqnarray*}
\int_M fdV^M&=&\sum_{\alpha}\int_{\Theta_\alpha}(\phi_\alpha\circ T_\alpha)(f\circ\textbf{x}_\alpha\circ T_\alpha)\sqrt{\det g^N}|\det dh|^2dy^N \\
&=&\sum_{\alpha}\int_{\textbf{y}_\alpha^{-1}(\Theta)}(\phi_\alpha\circ T_\alpha)(f\circ h\circ \textbf{y}_\alpha)\sqrt{\det g^N}|\det dh|^2dy^N \\
&=&\int_N f\circ h|\det dh|^2dV^N
\end{eqnarray*}
Question: where is the error? It must probably be something stupid but I can't see it.

答:
Suppose $(M,g)$ is a Riemannian manifold, and $h:N\to M$ a diffeomorphism. Then, we can consider the Riemannian manifold $(N,h^*g)$. Then, for any smooth function $f:M\to\Bbb{R}$, we can write the change of variables formula as
\begin{align}
\int_Mf\,dV_g &=\int_{N}f\circ h\, dV_{h^*g}\tag{i}
\end{align}
This should remind you very much of the differential form version where $\int_M\omega = \int_{h(N)}\omega = \int_Nh^*\omega$. Here, you can think of $dV_g$ as the Riemannian volume form on $M$ induced by the metric $g$ (provided $M$ is orientable, in which case the choice of volume form provides the orientation). Or you can also think of this as a measure (which you can do even when $M$ is not orientable).

As I mentioned in the comments, writing $\det dh$ abstractly as you've done is incorrect (at the very least very bad practice) because (pointwise) $dh$ is a linear transformations between different vector spaces. Now, you seem to be starting with two Riemannian manifolds $(M,g^M)$ and $(N,g^N)$, and then considering a diffeomorphism $h:N\to M$. In this case, one can also write down a change of variables formula:
\begin{align}
\int_Mf\,dV_{g^M} &=\int_{N}f\circ h\cdot \left(\dfrac{dV_{h^*(g^M)}}{dV_{g^N}}\right)\, dV_{g^N}\tag{ii}
\end{align}
Here, the bracketed term should be interpreted as a Radon-Nikodym derivative of the two measures $dV_{h^*(g^M)}$ and $dV_{g^N}$ (which one can show are positive $\sigma$-finite measures which are mutually absolutely continuous, hence the Radon-Nikodyn theorem can be applied). This is the correct factor which should appear (very roughly speaking, it's like writing $dy=\dfrac{dy}{dx}dx$ inside the integral sign, or writing $d^ny=\left|\det\frac{\partial y}{\partial x}\right|\,d^nx$ in the multivariable setting). As you can see from the notation itself, this Radon-Nikodym derivative involves all three things: $h,g^M$ and $g^N$, and this takes into account the "change of measure" factor (which you incorrectly just wrote as $\det dh$). Of course, if one works things out in local coordinates, one will get some sort of determinant of a coordinate representation of derivatives of $h$, but in addition, one will also get extra factors involving the determinants of $g^M, g^N$ in appropriate coordinates.

As you can see (ii) is not the nicest way of writing things down; (i) is the nice way of doing things. Now, let us forget about (ii) and try to prove (i), atleast locally. So, let's forget the excess baggage of a partition of unity and work with a single chart (because like I said, once we do this, we just put in a few extra summations and we're done).

To this end, let $(U,x)$ be a chart on $M$, and $(V,y)$ a chart on $N$ such that $h(V)\subset U$ (for my own sanity, I'm working with charts... i.e the function $x:U\to x[U]\subset\Bbb{R}^n$ goes from the manifold to the Euclidean space, rather than the other way around). Also, I shall denote the metric on $M$ simply by $g$, and I consider the pull-back metric $h^*g$ on $N$. We now want to show that
\begin{align}
\int_{h(V)}f\,dV_g =\int_{V}f\circ h\,dV_{h^*g}.
\end{align}

Ok so let's start on the right. We have
\begin{align}
\int_{V}f\circ h\,dV_{h^*g} &:= \int_{y(V)}(f\circ h)\circ y^{-1}\cdot
\sqrt{\left| \det [(h^*g)_{ij}]\right|\circ y^{-1}}\, dy\tag{$*$}
\end{align}
Let's calculate what the entries of the matrix are at a point $b\in y(V)$:
\begin{align}
(h^*g)_{ij}(y^{-1}(b))&:=(h^*g)\left(
\frac{\partial}{\partial y^i}\bigg|_{y^{-1}(b)},
\frac{\partial}{\partial y^j}\bigg|_{y^{-1}(b)}\right)\\
&=g\left[h_*\left(\frac{\partial}{\partial y^i}\bigg|_{y^{-1}(b)}\right),
h_*\left(\frac{\partial}{\partial y^j}\bigg|_{y^{-1}(b)}\right) \right]\\
&= \frac{\partial (x^{\mu}\circ h)}{\partial y^i}\bigg|_{(x\circ h\circ y^{-1})(b)}\cdot
\frac{\partial (x^{\nu}\circ h)}{\partial y^j}\bigg|_{(x\circ h\circ y^{-1})(b)}\cdot
g_{\mu\nu}(x\circ h\circ y^{-1}(b))
\end{align}
The efficient way of writing this down is that the matrices are related via a congruence (I'm omitting the points of evaluation)
\begin{align}
[h^*g] &= [D(x\circ h\circ y^{-1})]^t\cdot [g_{\mu\nu}]\cdot [D(x\circ h\circ y^{-1})]
\end{align}
(This should hopefully be familiar from linear algebra: the matrix representations of a bilinear map are related by congruence). Now, since a matrix and its transpose have the same determinant, it follows that the determinants are related as:
\begin{align}
\sqrt{\left|\det (h^*g)_{ij}(y^{-1}(b))\right|}&=
\sqrt{\left|\det g_{\mu\nu}((x\circ h\circ y^{-1})(b))\right|}\cdot
\left|\det D(x\circ h\circ y^{-1})_b\right|
\end{align}
We now have everything we need to use the standard change of variables in the RHS of $(*)$:
\begin{align}
\text{RHS of $(*)$} &=
\int_{y(V)} (f\circ x^{-1})\circ (x\circ h\circ y^{-1})\cdot
\sqrt{\left|\det g_{\mu\nu}\right|}\circ (x\circ h\circ y^{-1})\cdot
\left|\det D(x\circ h\circ y^{-1})\right|\,dy\\
&=\int_{x(h(V))}f\circ x^{-1}\cdot \sqrt{\left|\det g_{\mu\nu}\right|}\,dx\\
&:=\int_{h(V)}f\, dV_g.
\end{align}
This completes the proof.


问:
Minimum of $|z_1| + |z_2|+ \dots + |z_n|$ when $z_1 + \dots + z_n = 1$ and all the arguments of $z_m$ are prescribed.
Let $-\frac{\pi}{2} \le \phi_1 < \dots < \phi_k < 0 < \phi_{k+1} < \dots < \phi_N < \frac{\pi}{2}$ be $N$ prescribed phases. Let $a_1, \dots, a_N$ be $N$ real numbers such that
$$
a_1 e^{i \phi_1} + \dots + a_N e^{i \phi_N} = 1.
$$
What is the minimum of $|a_1|+\dots+|a_N|$?

The minimum is achieved when $a_m=0$ for $m \ne k, k+1$. The proof can be easily understood in a geometrically point of view, as kindly suggested in the following answer. Many thanks to all the answers and comments.

答:
A proof by functional analysis:

Let $a=\{a_j\}_j$ be such that $$1=\sum_j{a_je^{\phi_ji}}$$

Let $f^*=\begin{bmatrix}\cos{\phi_1}&\cdots&\cos{\phi_N}\end{bmatrix}$, $g^*=\begin{bmatrix}\sin{\phi_1}&\cdots&\sin{\phi_N}\end{bmatrix}$.  Then, taking real and imaginary parts, we find that we must solve \begin{gather*}
1=f^*(a) \\
0=g^*(a)
\end{gather*} which I will abbreviate as $(1,0)=(f,g)^*(a)$.

Note that this equation is homogenous: $(\alpha,0)=(f,g)^*(a)$.  But so is the $1$-norm $\|a\|_1=\sum_j{|a_j|}$ (well, for nonnegative $\alpha$, anyways).  Thus our minimum is $$m=\min_{(1,0)=(f,g)^*(a)}{\|a\|_1}=\min_{(\alpha,0)=(f,g)^*(\alpha a)}{\frac{\|a\|_1}{|\alpha|}}$$ where $\alpha$ is any number, possibly dependent on $a$.  Changing variables $\alpha a\mapsto a$, we have $$m=\min_{\substack{0=g^*(a)\\0\neq f^*(a)}}{\frac{\|a\|_1}{|f^*(a)|}}=\left(\max_{0=g^*(a)}{\frac{|f^*(a)|}{\|a\|_1}}\right)^{-1}$$

That maximum is a well-understood quantity in functional analysis.  To see this, first recall that any normed vector space $(V,\|\cdot\|_V)$ has a dual space $V^*$ consisting of the (continuous) linear functionals $V\to\mathbb{R}$.  On this dual space, the quantity $\sup_{v\neq0}{\frac{|f(v)|}{\|v\|_V}}$ is a norm; moreover, for many different normed vector spaces it has already been computed.  For example, the dual space of $(\mathbb{R}^n,\|\cdot\|_1)$ is $(\mathbb{R}^n,\|\cdot\|_{\infty})$, where $\|x\|_{\infty}=\max_j{|x_j|}$.

So what is our vector space?  Well, $\ker{(g^*)}$ inherits a norm from the inclusion $\ker{(g^*)}\subseteq(\mathbb{R}^n,\|\cdot\|_1)$.  We want to compute the dual norm $\|\cdot\|_d$ and then $\|f^*|_{\ker{(g^*)}}\|_d$.

Well, $g^*$ defines the short exact sequence $$0\to\ker{(g^*)}\to(\mathbb{R}^n,\|\cdot\|_1)\to\mathbb{R}1\to0$$  Taking duals reverses the arrows, so that $$0\to\mathbb{R}g^*\to(\mathbb{R}^n,\|\cdot\|_{\infty})\to\ker{(g^*)}^*\to0$$  By the First Isomorphism Theorem, as Banach spaces, $\ker{(g^*)}^*=(\mathbb{R}^n,\|\cdot\|_{\infty})/(\mathbb{R}g^*)$.  What is the norm of this quotient?  For any $x^*\in(\mathbb{R}^n,\|\cdot\|_{\infty})$, $$\|x^*\|_d=\min_{\lambda}{\|x^*+\lambda g^*\|_{\infty}}$$

So now we know: we need to minimize $$\|f^*+\lambda g^*\|_{\infty}=\max_j{\left|\cos(\phi_j)+\lambda\sin(\phi_j)\right|}=\max_j{\left|\cos(\phi_j)(1+\lambda\tan(\phi_j))\right|}$$  Call the $j$th term in this maximum $l_j(\lambda)$, and the maximum as a whole $M(\lambda)$.  Each $l_j$ is the absolute value of a line making angle $\phi_j$ with the $x$-axis and intersecting said axis at $-\cot{\phi_j}$.

$M$ is convex, so it has a unique minimum.  That minimum must be the intersection of an increasing line (to the right) and a decreasing one (to the left).  We claim that it is the intersection of $l_k$ and $l_{k+1}$.

To see this, first calculate: $$M(0)=\max_j{|\cos(\phi_j)|}\in\{\cos{\phi_k},\cos{\phi_{k+1}}\}$$  Now, if we swap $\phi_j\mapsto-\phi_j$, we send $M(\lambda)\mapsto M(-\lambda)$.  Thus we may assume $M(0)=\cos{\phi_{k+1}}$ without loss of generality and $M=l_{k+1}$ on a neighborhood of $0$.

Now, $l_k$ and $l_{k+1}$ intersect at $$\lambda_0=-\frac{\cos(\phi_{k+1})-\cos(\phi_k)}{\sin(\phi_{k+1})-\sin(\phi_k)}$$  Since $l_k$ is decreasing, the minimum of $M$ is attained somewhere in $[\lambda_0,0]$.

Suppose for contradiction the minimum were attained not at $\lambda_0$, but rather at some $\lambda_1\in(\lambda_0,0]$.  Then some other, decreasing, $l_j$ intersects $l_{k+1}$ at $\lambda_1$.  Since $l_j$ and $l_k$ are decreasing but $l_{k+1}$ increasing, the relative order is then preserved as we increase $\lambda$: for $\lambda\in[\lambda_1,0]$, we have $$l_k<l_j<l_{k+1}$$  But: how do I know that $l_j$ and $l_k$ do not "bounce" off the $x$-axis on this region?  First, suppose $j<k$.  Then $l_j$ and $l_k$ both intersect the $x$-axis at $-\cot{(\phi_j)}$ and $-\cot{(\phi_k)}$, respectively...which are both positive numbers.  Second, suppose $j>k+1$.  Then $-\cot{\phi_j}$ is negative; it must be a small negative number.  But then $\cos{\phi_j}=\cot(\phi_j)\sin(\phi_j)$ is even smaller, and so $l_j$ cannot dominate $l_k$ either.

So $\lambda\to0^+$, $$l_k<l_j<l_{k+1}$$  But this is impossible: the $\cos{\phi_j}$ terms dominate the $\lambda\sin{\phi_j}$ terms, so that we must have $l_j<l_k$.  This is a contradiction, and $\lambda_0$ is the minimum.

So $M$ attains its minimum at $\lambda_0$; by algebra, $$M(\lambda_0)=\frac{\sin(\phi_{k+1}-\phi_k)}{\sin(\phi_{k+1})-\sin(\phi_k)}$$  Thus $$m=\frac{\sin(\phi_{k+1})-\sin(\phi_k)}{\sin(\phi_{k+1}-\phi_k)}$$ and it is easy to verify that this is attained when $a_j=0$ for $j\notin\{k,k+1\}$.


问:
Prove that $\{1, 11, 1001,\dots\}$ is an irregular language
Let $L:=\{1, 11, 1001,\dots\}$ be the language with alphabet $\{0,1\}$ which is formed by all powers $3^n, n=0,1,\dots$ written in binary notation. How to prove that $L$ is not regular?</p>

答:
The pumping lemma is your friend as it states that if $L$ were regular, there'd be a length $n$ such that all $w\in L$ with $|w|>n$ can be split into $w=xyz$ with $y\ne\epsilon$ such that $xy^kz\in L$ for all $k\in\mathbb N_0$. We do not know $n$, but consider one such word $w$ and let $a,b,c$ be the numbers represented by the binary strings $x,y,z$ and let $N_k$ be the number rpresented by $xy^kz$.
Then $$N_{k+1}=(N_k-c)\cdot 2^{|y|}+b\cdot 2^{|z|}+c$$
and as $N_k\to\infty$ (because $x$ begins with a $1$), we have $\frac{N_{k+1}}{N_k}\to 2^{|y|}$ as $k\to\infty$. On the other hand, $\frac{N_{k+1}}{N_k}$ is always a power of $3$. Thus for sufficiently big $k$, we find $m$ with $|3^m-2^{|y|}|<\frac12$, i.e. $3^m=2^{|y|}$. This is only possible if $m=|y|=0$, contradicting $|y|\ne 0$.</p>

ContentLicense: CC BY-SA 3.0


问:
Formula for length of diagonal in a Lambert quadrilateral
Given a Lambert quadrilateral $AOBF$ where the angles $ \angle FAO ,  \angle AOB , \angle OBF $ are right, and $F$ is opposite $O , \angle AFB$ is the acute angle , and the Gaussian curvature = -1 (so all is standard)

Then a whole lot of equations hold see https://en.wikipedia.org/wiki/Lambert_quadrilateral#Lambert_quadrilateral_in_hyperbolic_geometry

(these are the relations I found in different sources)

There is also another relation (follows from the Klein Disk model)

$$ \tanh^2 OF =\tanh^2 OA +\tanh^2 OB $$

But I was not able to deduce this formula from the others.

How can $ \tanh^2 OF =\tanh^2 OA +\tanh^2 OB $ be deduced from the other formula's?

答:
Defining $a:=|OA|$, $b:=|OB|$, $f:=|OF|$, $a':=|AF|$, $b':=|BF|$, we have

$$\begin{align}
\cosh f &= \cosh a\cosh a' &&\text{(Hyperbolic Pythagoras)}\\[4pt]
\operatorname{sech}^2f &= \operatorname{sech}^2a \operatorname{sech}^2a' \\[4pt]
\operatorname{sech}^2f&=\operatorname{sech}^2a(1-\tanh^2a') \\[4pt]
\operatorname{sech}^2f&=\operatorname{sech}^2a(1-\color{red}{\cosh^2 a\tanh^2 b}) &&\text{(Lambert quad property)}\\[4pt]
\operatorname{sech}^2f&=\operatorname{sech}^2a-\tanh^2 b \\[4pt]
1-\tanh^2f &= (1-\tanh^2a)-\tanh^2b \\[4pt]
\therefore \tanh^2f &= \tanh^2a+\tanh^2b
\end{align}$$


问:
Continuous functions do not necessarily map closed sets to closed sets
I found this comment in my lecture notes, and it struck me because up until now I simply assumed that continuous functions map closed sets to closed sets.

What are some insightful examples of continuous functions that map closed sets to non-closed sets?

答:
In addition to the excellent examples presented by Brian, I want to point out what exactly all of these have in common.

The key property that is making these examples work (i.e. closed sets whose images under a continuous function are not closed) is that they're unbounded. If instead we were dealing with closed and bounded sets, then their images would always be closed (and bounded). This is a result that can be summarized by saying "the image of a compact set under a continuous function is compact".

Here's my attempt at explaining why this is, informally, by restricting ourselves to real functions (functions from $\mathbb{R}$ to $\mathbb{R}$) with the usual Euclidean metric. In this special case, by the Heine-Borel theorem, a set is compact if and only if it is closed and bounded. (In a more general setting we would need to work with the general definition of compactness, which is not quite as intuitive.)

Intuitively, the only way you can make the image of a set under a function not closed is if, on that set, the function comes arbitrarily close to a certain value but never actually reaches it. (This is because you will not be able to draw any open ball in the codomain around that value—which will be part of the complement of the image—that doesn't contain some points from the image, which implies the complement isn't open and hence the image isn't closed.) Here are some examples of functions, continuous and discontinuous, that exhibit this behaviour (apologies for the hand-drawing):

The function $f \colon \mathbb{R} \to \mathbb{R}$ defined as $f(x) = e^x$:

The image of $\mathbb{R}$ under $f$ is $(0, +\infty)$. This function gets arbitrarily close to $0$ as $x$ tends to $-\infty$, but never attains that value; this is what causes the image to not be closed. This is a continuous function on a closed but unbounded set.

The function $f \colon (0, 1) \to \mathbb{R}$ defined as $f(x) = x$:

The image here is $(-1, 1)$. The cuplrits for the non-closedness of the image are $-1$ and $1$; again, the function approaches these values but never actually reaches them. This is a continuous function on a non-closed (and in particular,  open) and bounded set.

The function $f \colon [0, 1] \to \mathbb{R}$ defined as $\displaystyle f(x) = \begin{cases} \displaystyle\frac{1}{e^{x^2}} \quad &\text{if $x \ne 0$} \\ 0 \quad &\text{if $x = 0$} \end{cases}$:

Here the image is $\{0\}\cup [\frac{1}{e}, 1)$. Once again, $f$ approaches 1 but doesn't reach it. This is a discontinuous function on a closed and bounded set.

Now, what are the ways in which a continuous function on a closed set can approach a value without reaching it? If the domain is unbounded, that's easy: just let the function approach a value asymptotically as $x$ tends to $\pm\infty$ (like in our first example, $f(x) = e^x$). Are there any other ways? (Spoiler alert: no!)

If we try something like the third example above, approaching a value from both sides, the continuity of the function will force us to "fill in" the missing image with the value we're approaching:

But this is not specific to closed sets. What about approaching a value towards a point at the boundary of our domain? A key property of a closed set is that it contains all its limit points (in fact, we can even define closed sets by this property), hence the point on the boundary we approach will also be part of the domain. So, our function will also be defined there: which value will it take? Continuity, once again, forces the image at that limit point to be the limit of the images, i.e., the value we were approaching. Therefore this value is included in the image of the domain and thus the latter remains closed.

We have thus (intuitively) shown that the only way to make a continuous function on a closed set produce a non-closed image is to make the domain unbounded. In other words, if a set is closed and bounded (we call this compact), then the image of that set under a continuous function will also be closed.

Here's a little addendum: the result I cited at the top of this answer said that the image of a closed and bounded set under a continuous function is closed and bounded, but so far we have only seen the "closed" part.

Again, an intuitive explanation runs along the same lines. For the sake of illustration let's restrict ourselves to intervals. If you have a bounded domain, i.e. "limited horizontal space to draw your graph", and you have to draw a graph from one endpoint to another without lifting your pen, you can only "grow" your function a finite amount. You can't blow off to infinity, because then you would have to "return" from infinity to connect back to the second endpoint, creating a discontinuity (an asymptote):

The fact that the interval is closed is key: this ensures that the function is defined at the endpoints and thus, by continuity, the graph must "connect" to $(a, f(a))$, $(b, f(b))$. Otherwise, if the endpoints weren't included, the function could grow to infinity asymptotically as it approaches $a$ or $b$, as it doesn't need to be defined there. For example, $x \mapsto \frac{1}{x}$ defined on $(0, 1]$ has a vertical asymptote at $0$.


答:
Write $x=1396+t$. Then
$$\sqrt{1+\frac t{1388}}\:+\sqrt{1+\frac t{1389}}\:+\sqrt{1+\frac t{1390}}$$
$$=\sqrt{1+\frac t8}\:+\sqrt{1+\frac t7}\:+\sqrt{1+\frac t6}.$$
Clearly $t=0$ is one solution. Now, $t$ cannot be positive, because that would make the terms on the LHS each smaller than the terms on the RHS. Similarly, $t$ cannot be negative, for the opposite reason. Therefore $t=0$ or $x=1396$ is the unique solution.

Body:
How many solutions the following equation has in real numbers
?$$\sqrt{\frac{x-8}{1388}}+\sqrt{\frac{x-7}{1389}}+\sqrt{\frac{x-6}{1390}}=\sqrt{\frac{x-1388}{8}}+\sqrt{\frac{x-1389}{7}}+\sqrt{\frac{x-1390}{6}}$$

$1)1\quad\quad\quad\quad\quad\quad\quad2)2\quad\quad\quad\quad\quad\quad\quad3
 )3\quad\quad\quad\quad\quad\quad\quad4)4\quad\quad\quad\quad\quad5)\text{zero}$

Obviously we have $x\ge1390$. starting from $x=1390$ we can see LHS is very close to $1+1+1=3$ and RHS is close to zero. for higher values of $x$ we can see right side of the equation grows much faster than left side. so it seems the equation has not answer in real numbers.

I'm not sure how to proceed mathematically. I can rewrite it as follow:
$$\left(\sqrt{\frac{x-8}{1388}}-\sqrt{\frac{x-1388}{8}}\right)+\left(\sqrt{\frac{x-7}{1389}}-\sqrt{\frac{x-1389}{7}}\right)+\left(\sqrt{\frac{x-6}{1390}}-\sqrt{\frac{x-1390}{6}}\right)=0$$
I putted fractions with similar numbers in the same parentheses so it looks better now, but don't know how to continue.


问:
If $X$ is compact and space $X{/}K$ the contraction of $K$ to a point.Prove that $X{/}K$ is the compactification of $X-K$
If $X$ be Hausdorff and compact and $K\subset X$ compact, and   $X{/}K$ the contraction of $K$ to a point.Prove that $X{/}K$ is the compactification of $X-K$.

I have been tried different ways of show that both are homeomorphic.

$\pi:X{/}K-\{[K]\} \longrightarrow X-K$ given by $\pi(x)=x$ and see that is a homeomorphism, unfortunelly I can´t look that the function is continous.

$\pi:X{/}K  \longrightarrow (X-K) \cup \lbrace \infty \rbrace$ given by $\pi(x)=x$ if $x\in [x]$ or $\pi(x)=\infty$ if $x\in [k]$
but at this time is hard show that $\pi$ is a homeomorphism.
Any advice of how I should prove that?

答:
The result isn’t quite correct. Let $X=[0,1]\cup\{2\}$ with the topology that it inherits from $\Bbb R$, and let $K=[0,1]$. Then $X$ is compact and Hausdorff, and $K$ is compact. But $X\setminus K=\{2\}$ is also compact, so it does not have a one-point compactification. What is true is that $X/K$ is homeomorphic to the Alexandroff extension of $X\setminus K$, which is the one-point compactification if and only if $X\setminus K$ is not compact.

For convenience let $Y=X\setminus K$. Let $\tau$ be the topology on the Alexandroff extension $Y\cup\{\infty\}$. $X$ is Hausdorff, so $K$ is closed in $X$, and $Y$ is therefore open in $X$. Thus, a subset $U$ of $Y$ is open in $Y$ if and only if it is open in $X$. This means that $\tau$ consists of all open subsets of $Y$ together with all sets of the form $\{\infty\}\cup(Y\setminus C)$ such that $C$ is a compact subset of $Y$.

Let $\pi:X\to X/K$ be the quotient map, let $y\in K$, and define

$$h:Y\cup\{\infty\}\to X/K:x\mapsto\begin{cases}
\pi(x),&\text{if }x\in Y\\
\pi(y),&\text{if }x=\infty\,;
\end{cases}$$

we want to show that $h$ is a homeomorphism, i.e., that it is a continuous, open bijection. It’s easy to check that it’s a bijection. I’ll show that it’s open and leave it to you for now to go on and show that it is also continuous.

Let $U\in\tau$. If $\infty\notin U$, then $h[U]=\pi[U]$, which is open in $X/K$ because $U=\pi^{-1}\big[\pi[U]\big]$ is open in $X$. If $\infty\in U$, then $U=\{\infty\}\cup(Y\setminus C)$ for some compact $C\subseteq Y$, so

$$h[U]=\{\pi(y)\}\cup\pi[Y\setminus C]=\pi[K\cup(Y\setminus C)]=\pi[X\setminus C]\,,$$

and $X\setminus C=\pi^{-1}\big[\pi[X\setminus C]\big]$ is open in $X$, so again $h[U]$ is open in $X$, and we’ve now shown that $h$ is an open map.


问:
Conditional Probabilities in Bridge
The question reads as follows:

Each of 4 bridge players has been dealt 13 cards from a 52-card
deck.

(a) Given that your partner has at least one ace, what is the probability that she has at least two aces?

(b) Given that your partner has the ace of spades, what is the probability that she has at least two aces?

(c) Compare your answers. Do they make sense? Explain

For both a and b I used P(A | B) = P(intersection of A and B)/P(B)

For a) I got .37 by doing ($\binom{4}{2}$$\binom{48}{11}$+ ... $\binom{4}{4}$$\binom{48}{9}$)/($\binom{4}{1}$$\binom{48}{12}$+ ... $\binom{4}{4}$$\binom{48}{9}$) = 0.37

For b) I got .56 by doing ($\binom{1}{1}$$\binom{3}{1}$$\binom{48}{11}$+ ... $\binom{1}{1}$$\binom{3}{3}$$\binom{48}{9}$)/$\binom{1}{1}$$\binom{51}{12}$ = .56
I am not sure that I did A and B correctly, or how to explain how it makes sense in the context of part c.

答:
Your working for $(a)$ and $(b)$ is correct. Here is another way to look at it -

a) Sets of $13$ cards where at least one ace is present $ =
\displaystyle {52 \choose 13} - {48 \choose 13}$

Probability of exactly one ace, given at least one ace is present $ = \displaystyle {4 \choose 1}{48 \choose 12} \ / \ \bigg[{52 \choose 13} - {48 \choose 13}\bigg] = \frac{9139}{14498}$

Subtracting from $1$ gives probability of at least two aces, which is same as your answer.

b) Probability of only one ace, given ace of spades is present

$ = \displaystyle {48 \choose 12} / {51 \choose 12} = \frac{9139}{20825}$

Subtracting from $1$ gives you the desired probability of at least two aces, which is same as your answer.

(c) is asking you to explain why it makes sense that probability in (b) is higher than in (a), basically compare the results with explanation.

Title: Conditional Probabilities in Bridge


问:
Definition of a field in maths and physics
Is the field as defined in analysis ( i.e a set with multiplication and addition defined on it such that they obey certain field axioms like closedness and associativity .... ) related by anyway to the " field" in physics ( e.g electric field, force field …).

Additionally, if they are not equivalent, you would do me great justice if you can refer me to some rigorous definition of "field" in physics. Thank You

答:
In physics, a "scalar field" is essentially a function of position, or a number at every point. The temperature $T(x,y,z)$ at every point in a room is described by a scalar field.

A "vector field" is essentially a vector at every point.  It can have various units. So the gravitational field $\overrightarrow{g}$ has units of $m/s^2$ while the electric field $\overrightarrow E$ has units of $V/m$ and the magnetic field $\overrightarrow B$ is measured in teslas.

A gravitational field is in effect an arrow at every location in space pointing "down"--it is the definition of down, in fact.  The size of the vector indicates the strength of the gravitational field.

Mathematics is the map, and physics is the territory, in the case of physics fields. A scalar field or vector field is a mathematical object, one function or a set of functions with 3 inputs in three dimensional space. You can add these fields and so forth, do mathematical operations on them, but the physical phenomenon is the reality the model tries to describe.

As others have said, this is a very different use of the word "field" from how mathematicians use it to mean roughly "a set of things you can do arithmetic with" and more precisely follows various axioms of being a commutative group under addition with identity $0$ and nonzero elements being a commutative group under multiplication with identity $1$ and satisfying the distributive laws.


答:
There's actually no problem for the convergence, because
$$
\lim_{x\to0}\sin^2x\ln x=0
$$
so you can extend the function by continuity over $[0,1]$. (Use $x\to0^+$, if you prefer, but since the function is only defined for $x>0$, there is no issue.)

Why is the limit $0$? Because already
$$
\lim_{x\to0}x\ln x=0
$$
which is essentially the same as
$$
\lim_{t\to\infty}\frac{e^t}{t}=\infty
$$

Title: Integral by parts the function in the interval $(0,1)$

Body:
There is the integral of the function $(\sin x)^2\ln x$ in the interval $(0,1)$, which I am asked to find if it converges.

I tried integration by parts but I came up with the integral of $\sin(2x)/x$ and here I stopped as I found difficulty in calculating this integral according to the integration methods.

How can I proceed?

I would appreciate for your quidance.

Thank you very much in advance.


问:
For which values of $a$ does this system of equations $\mathbf{{not}}$ have a unique solution?
Here's my system of linear equations:

$\begin{cases}
x + 2y + 2z = 1\\x + ay + 3z = 3\\x + 11y +az = 0\\
\end{cases}$

Thus I have the augmented matrix $\left[\begin{array}{ccc|c}1&2&2&1\\1&a&3&3\\1&11&a&0\end{array}\right]$

By row reduction, I obtain:

$\left[\begin{array}
{ccc|c}1&2&2&1\\0&a-2&1&2\\0&9&a-2&-1\end{array}\right]$

Unfortunately, I am stuck at this stage. I have tried swapping rows around but I didn't have much luck.

Update: I have managed to solve this with the use of the determinant.

Matrix of minors:
$\left[\begin{array}
{ccc}a^2-33&a-3&11-a\\2a-22&a-2&9\\6-2a&1&a-2\end{array}\right]$

Matrix of cofactors:
$\left[\begin{array}
{ccc}a^2-33&3-a&11-a\\22-a&9&a-2\\6-2a&-1&a-2\end{array}\right]$

Adjugate matrix:
$\left[\begin{array}
{ccc}a^2-33&22-2a&6-2a\\3-a&a-2&-1\\11-a&-9&a-2\end{array}\right]$

Det(A) = $1(a^2 - 33) + 2(3 - a) + 2(11 - a) = a^2 - 4a - 5$

$(a - 5)(a + 1) = 0$

Thank you all for your help!

答:
Let's rearrange the augmented matrix representing the system of linear equations, by swapping $R_2$ and $R_3$ (to avoid division by $a-2$ while row-reduction), to get

$\left[\begin{array}{ccc|c}1&2&2&1\\1&11&a&0\\1&a&3&3\end{array}\right]$ and a subsequent row-reduction will give you

$\left[\begin{array}{ccc|c}1&2&2&1\\0&9&a-2&-1\\0&0&1-\frac{(a-2)^2}{9}&2+\frac{a-2}{9}\end{array}\right]$

Unique solution:
in order for the system to have a unique solution, the coefficient and the augmented matrix both must be of full rank i.e., both must have rank $n=3$. To make that happen, we must have $1\neq\frac{(a-2)^2}{9} \implies a \notin \{-1,5\}$.

No solution:
for this to happen, we need to have the augment matrix rank higher than the coefficient matrix rank. This can happen if the coefficient matrix has rank < 3, e.g., if $1=\frac{(a-2)^2}{9} \implies a \in \{-1,5\}$ and augmented matrix has rank $n=3$ simultaneously, which means $2+\frac{a-2}{9} \neq 0 \implies a \neq -16$, i.e., for $a=-1,5$ the system will have no solution.

Infinitely many solutions:
it can happen iff both the coefficient and the augmented matrix are rank-deficient, i.e. for both of them the rank is $<3$. For this the last row of the augmented matrix needs to be zero. Which implies both $1-\frac{(a-2)^2}{9}=0 \implies a \in \{-1,5\}$ and $2+\frac{a-2}{9}=0 \implies a=-16$ simultaneously. Which is clearly impossible.


问:
How to convexify a nonconvex function?
Given $b > 0$, let  $f : \Bbb R_{>0}^2 \to \Bbb R$ be

$$f(x,y) := \frac{x}{y \log_2 \left(1 + b \frac{x}{y}\right)}$$

I would like to prove that $f$ is convex.

I tried to use the definition of convexity (i.e., the Hessian matrix) but it seems to be non-convex. Is there any transformation I could do to make my function convex?

Title: How to convexify a nonconvex function?

Body:
$f(x,y)$ is not convex. For $b=1$ try points $A=[10,1]$ and $B=[30,1]$ and the value at the midpoint:

$f(A) = 2.8906482,~f(B) = 6.0554725,~f\left(\frac{A+B}{2}\right) = 4.5534049 > \frac{f(A)+f(B)}{2} = 4.4730604$


问:
choice set for equivalence classes
I'm trying to understand the proof of the Paradox of Banach-Tarski. I was reading Wagon's work about the paradox and tried to understand the proof of Theorem 1.5, only I haven't warmed up to group theory yet. Maybe someone can explain to me what exactly Wagon means by:

Let M be a choice set for equivalence classes of the relation on $S^1$ given by
calling two points equivalent if one is obtainable from the other by a
rotation about the origin through a (positive or negative) rational
multiple $2\pi$ radians.

I would appreciate it if someone could explain the statements so that I can continue to work on the proof and to understand it better. Thanks for your help!

答:
Equivalence classes are easy.  Let's say all the even numbers are equivalent to each other and all the odd numbers are equivalent to each other.  This partitions the integers into two sets:

$\{\ldots,-1,1,3,5,\ldots\}$ and $\{\ldots,-2,0,2,4,\ldots\}$

We say that all of the members of the same set are equivalent to each other.  So $-1\sim1\sim3\ldots$ and so on.

You should be able to see that every equivalence relation $\sim$ on a set partitions it.

If we map the real numbers onto the circle via the morphism $x\mapsto e^{2i\pi x}$ you should see that $0$ maps to the same point on the circle as $1$ and as $2,3,4,\ldots$.  They all go to the same place on the circle.  You can think of this as a projection in that it loses some information because if you were to pick a point on the circle and send that point back to where it came from in $\Bbb R$ you wouldn't know which point to send it back to.  You would have an infinite set $\{x+i:i\in\Bbb Z\}$.  All of those points could be thought of as equivalent.

Technically an equivalence relation is reflexive, symmetric and transitive.  This means:

every element is equivalent to itself, and

if $a\sim b$ then $b\sim a$, and

if $a\sim b$ and $b\sim c$ then $a\sim c$

But intuitively what really captures these three properties is the fact that an equivalence relation partitions a set.  A partition of a set is a disjoint family of sets that covers the original set.

Title: choice set for equivalence classes


问:
Explanation of Notation Regarding Power Set of a Set
In Grimmet and Strizaker's Probability and Random Processes it states section 1.2

The power set of $\Omega$, which is written $\{0,1\}^{\Omega}$ and contains all subsets of $\Omega$....

It goes on to use this in an example as follows:

A die is thrown once. We can take $\Omega$ = $\{1,2,3,4,5,6\}$, $F = \{0,1\}^{\Omega}$

Here $F$ is the set of all subsets of the sample space $\Omega$ which are of interest. I don't understand this notation to specify that set. How does it generate/enumerate all the subsets of interest?

答:
The Cartesian product

$$\{0,1\}\times\{0,1\}\times\{0,1\}\times\{0,1\}\times\{0,1\}\times\{0,1\}$$ denotes the set of all $6$-tuples made of zeroes and ones. There are $2^6$ of them. You could write this as

$$\{0,1\}^6.$$

Now if you interpret a $0$ as "leave" and a $1$ as "take", every tuple corresponds to a subset. E.g.$(1,1,0,1,0,0)\equiv\{1, 2, 4\}$. In this context, the notation

$$\{0,1\}^\Omega$$ makes sense, though it is purely conventional. Another common notation is

$$2^\Omega.$$


问:
Question about pure tensors
Suppose that $A$ is a unital algebra over the field $k$ and $a \in A$ satisfies

$$\forall x \in A: \exists z_x\in A: x \otimes a = z_x \otimes 1$$

Can we conclude that $a \in k1$?

It seems plausible, but strange things can happen when considering tensor products! I tried to do a functional trick but could not conclude.

答:
With $x=1$, we know that $1 \otimes a=b \otimes 1$ for some $b \in A$.

Let $e_i$ be any basis of $A$ containing $1$ (say $e_t=1$), then $(e_i \otimes e_j)$ is a basis of $A \otimes A$. Write $a=\sum_i{a_ie_i},b=\sum_i{b_ie_i}$. Then $1 \otimes a=\sum_i{a_i e_t \otimes e_i}$ and $b \otimes 1=\sum_i{b_ie_i \otimes e_t}$. By identifying coordinates, since $1 \otimes a=b \otimes 1$, it follows that $a_i=0$ if $i \neq t$, so $a=a_te_t \in k1_A$.

Title: Question about pure tensors


问:
For which $p$ does $\int_1^\infty\frac{1}{\left(x^{p^3} + x^{4p}\right)\ln^2(1+\sqrt[4]{x})}dx$ converge
The title pretty much sums up the problem. I need to examine for which values of $p\in\mathbb{R}$ the following improper integral converges:
$$I := \int_1^\infty\frac{1}{\left(x^{p^3} + x^{4p}\right)\ln^2(1+\sqrt[4]{x})}dx$$

What I initially thought about was the Abel-Dirichlet test, which only gives us a range of $p$ for which the integral converges, but I'm not sure how to find out whether that are all values and I even  suspect that they're not. So, if we define
$$f(x) := \frac{1}{\ln^2(1+x^{\frac{1}{4}})}$$
then we have that $f(x) > 0$ and is monotonically decreasing and $\displaystyle\lim_{x\to\infty}f(x) = 0$, so if we make $\displaystyle\int_1^{\infty}\frac{1}{x^{p^3}+x^{4p}}dx$ convergent, then by the Abel-Dirichlet test $I$ will also be convergent. And this happens when $max\{p^3, 4p\} > 1$, i.e. when $p > \frac{1}{4}$.

But as said, this is an artificial constraint in order to employ the test I could think of. I also figured I could somehow bound $I$ between some integrals that converge for the same values of $p$, but I could only find the upper bound in the following way by examining boundedness of $f(x)$:

Since we mentioned it was monotonically decreasing and tends to $0^+$, from some point on it will be strictly less than 1. This point is namely $a = (e-1)^4 > 1$, so we can split $I$ as sum of two integrals (one from 1 to a and another from a to $\infty$). Since the first one is a definite integral, it is a number and doesn't change convergence, so
$$I \sim \int_a^\infty\frac{1}{\left(x^{p^3} + x^{4p}\right)\ln^2(1+\sqrt[4]{x})}dx < \int_a^\infty\frac{1}{x^{p^3} + x^{4p}}dx$$
and we obtain the same results as with the test before, namely that when $p > \frac{1}{4}$ the upper bound integral converges and so does $I$. But what could be a smart lower bound? Is that even the approriate approach? Any ideas are greatly welcomed! Many thanks in advance!

答:
hint

near infinity,
$$1+x^{\frac 14}\sim x^{\frac 14}$$
and

$$\ln(1+x^{\frac 14})\sim \frac 14 \ln(x)$$

If $ p^3>4p$, then
$$x^{p^3}+x^{4p}\sim x^{p^3}$$

If $ p^3=4p$, then

$$x^{p^3}+x^{4p}\sim 2x^{4p}$$

If $ p^3<4p$, then
$$x^{p^3}+x^{4p}\sim x^{4p}$$

in the end, you will get what we call the Bertrand's integral.


问:
What is the Idea of a Relative Open Set?
My real-analysis text gave the following defintion:

Let U be a subset of E.  U is open relative to E if

for $\forall t \in U$, $\exists \epsilon$ such that $N_\epsilon(t) \cap E \subset U$.

Although the idea that U is open in $\mathbb R$ follows the definition,

I normally do not think about the intersection of $N_\epsilon(t) \cap \mathbb R$.

U is open if for $\forall t \in U$, $\exists \epsilon$ such that $N_\epsilon(t) \subset U$; every t is an interior point of U.

Intuitively, each point, t, is contained in a "bubble".

So, what is the significance of specifying the intersection?

Apparently, I can't think of a relative open set in terms like interior points and "bubbles".

答: A relative open set is essentially the restriction of an open set to a subset. For example, open sets in $\mathbb R$ are such that each point in the set is contained in an open interval or "bubble" as you say.  Given this idea of openness, how would we define an open subset of the closed interval $[0,1]$? It seems clear that something like $(\frac{1}{4}, \frac{3}{4})$ should be open but what about $[0,\frac{1}{2})$? This set is not open in $\mathbb R$ because any open interval around $0$ contains a negative number which is not in $[0,\frac{1}{2})$. However, there are no negative numbers in $[0,1]$ so this shouldn't preclude this set from being open relative to $[0,1]$. From inside $[0,1]$, $[0,\frac{1}{2})$ looks like an open neighborhood or your "bubble".</p>


问:
fitting an exponential curve with squared exponent through three points
I apologize if the format is horrible! I am not sure how to do this.

\begin{center}
I am using least squares regression to fit an exponential to my data. Thus, I need to find a good guess for initial parameters. The model function for this data is
\end{center}

\begin{equation}
y\ =\ ae^{bK^{2}} +c
\end{equation}
\begin{center}

where $\displaystyle K\ =b^{-1} \ \sqrt{2( x+3/2)}$ and $\displaystyle b\ =\ 0.15$
\end{center}
Here is the method that i have done, however it does not result in a very good fit.

\begin{gather*}
y_{1} \ =\ ae^{bK^{2}_{1}} +c\\
y_{2} \ =\ ae^{bK^{2}_{2}} +c\\
y_{3} =ae^{bK^{2}_{3}} +c
\end{gather*}
Thus,
\begin{gather*}
y_{1} '\ =\ 2abK_{1} e^{bK^{2}_{1}}\\
y_{2} '\ =\ 2abK_{2} e^{bK^{2}_{2}}
\end{gather*}

\begin{equation*}
\frac{y_{2} '}{y1'} =\frac{K_{2}}{K_{1}} e^{b\left( K^{2}_{2} -K1^{2}\right)}
\end{equation*}
Moreover, note that $\displaystyle y_{2} '\ =\ \frac{dy_{2}}{dK_{2}} =\frac{y_{3} -y_{2}}{K_{3} -K_{2}}$ and similarly, $\displaystyle y_{1} '=\ \frac{y_{2} -y_{1}}{K_{2} -K_{1}}$.

Thus $\displaystyle b$ is equal to

\begin{gather*}
b\ =\ \ln\left(\frac{y_{2} 'K_{1}}{y_{1} 'K_{2}}\right) /\left( K^{2}_{2} -K^{2}_{1}\right)\\
\end{gather*}
Now to find $\displaystyle a$, we will eliminate $\displaystyle c$ by

\begin{gather*}
y_{2} -y_{1} \ =\ a\left( e^{bK^{2}_{2}} -e^{bK^{2}_{1}}\right)\\
y_{3} -y_{2} \ =\ a\left( e^{bK^{2}_{3}} -e^{bK^{2}_{2}}\right)
\end{gather*}
so
\begin{gather*}
a_{1} \ =\ \frac{y_{2} -y_{1}}{e^{bK^{2}_{2}} -e^{bK^{2}_{1}}}\\
a_{2} \ =\ \frac{y_{3} -y_{2}}{e^{bK^{2}_{3}} -e^{bK^{2}_{2}}}
\end{gather*}
and we have $\displaystyle a\ =\frac{a_{1} +a_{2}}{2}$.

Now for $\displaystyle c$ we just plug in all the values into each equation and solve for $\displaystyle c$. i.e

\begin{gather*}
c_{1} \ =\ ae^{bK^{2}_{1}} -y_{1}\\
c_{2} \ =\ ae^{bK^{2}_{2}} -y_{2}\\
c_{3} =ae^{bK^{2}_{3}} -y_{3}
\end{gather*}
so $\displaystyle c\ =\ \frac{c_{1} +c_{2} +c_{3}}{3}$ .

When I plot this fit against the data, the curve only really fits through the first three points. I am not sure what is going on or where the problem is.

DATA:
\begin{gather*}
( 2,2.35448742279577\ \ )\\
( 4,-1.368206934161941)\\
( 6,-3.305390118335089)\\
( 8,-4.482711701457797)\\
( 10,-5.266602875206834)\\
( 12,-5.820768423448095)\\
( 14-6.229391391650249)\\
( 16,-6.540202260195814)\\
( 18-6.782285215077716)\\
( 20-6.974372715890539)
\end{gather*}
THE ATTEMPT PLOTTED IS HERE

答:
This question will probably win the contest of the most muddled wording.

The first step is to simplify it.
$$y=ae^{bK^2}+c\quad;\quad K=b^{-1}\sqrt{2(x+3/2)}$$
$$y=ae^{(2x+3)/b}+c$$
$$y=ae^{3/b}e^{(2/b)x}+c$$
Let $\quad \beta=a e^{3/b}\quad$ ; $\quad\gamma=\frac{2}{b}\quad $ and $\quad\alpha=c$ .

$$\boxed{y=\alpha +\beta e^{\gamma x}}$$
The question becomes very simple : How to fit this function to the given data. That is all the problem if I well understand the original question.

The answer is : Use any available non-linear regression solftware.

Alternatively we will use another method which doesn't require iterative calculus.

The principle is explained in https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales . Copy below of the application to the present problem.

The notations $a,b,c$ in the referenced paper are not the same as the notations $a,b,c$ in the question of the OP. This is confusing. In order to clarify the relationships are written below, in coming back to the notations of the OP.

Both equations give the same blue curve on the graph.


问:
Roulette Again Chances to Walk away with at least $1010
You go to Las Vegas with $\$1000$ and play roulette $100$ times by betting $\$10$ on the number 36
each time. Compute the probability of leaving the Casino with at least $\$1010$. Hint: Each bet
you have chance $1/38$ of winning $\$350$ and chance $37/38$ of loosing $\$10$.

So in order to have at least $\$1010$ when you leave you have to win at least $3$ times.  $$3\cdot 1050+1000=2050-(97\cdot 10)=1080$$
I am assuming that I am looking for the probability of $97$ losses or less.
I interpreted this as a Binomial Distribution, which I plugged into Excel with the following calculation:
=Binom.dist(97,100,(37/38),TRUE), however, it does not get me the right answer.  I am unsure on how to fix the program.

Thank you!

答:
I don't know what casino you've been going to but on standard Roulette there are $37$ numbers, numbered $0-36$.  So I doubt it's a maths error.

By your method, plugging

```<code>=Binom.dist(97,100,(36/37),TRUE)
</code>```

into Excel gives $\approxeq0.5093$

Well, anybody knows your expected winnings from any casino bet are less than what you took in so the answer must be $\approxeq1-0.5093=0.4907$

Body:
I don't know what casino you've been going to but on standard Roulette there are $37$ numbers, numbered $0-36$.  So I doubt it's a maths error.

By your method, plugging

=Binom.dist(97,100,(36/37),TRUE)

into Excel gives $\approxeq0.5093$

Well, anybody knows your expected winnings from any casino bet are less than what you took in so the answer must be $\approxeq1-0.5093=0.4907$


Title: Conditional Probabilities in Bridge

Body:
The question reads as follows:

Each of 4 bridge players has been dealt 13 cards from a 52-card
deck.

(a) Given that your partner has at least one ace, what is the probability that she has at least two aces?

(b) Given that your partner has the ace of spades, what is the probability that she has at least two aces?

(c) Compare your answers. Do they make sense? Explain

For both a and b I used P(A | B) = P(intersection of A and B)/P(B)

For a) I got .37 by doing ($\binom{4}{2}$$\binom{48}{11}$+ ... $\binom{4}{4}$$\binom{48}{9}$)/($\binom{4}{1}$$\binom{48}{12}$+ ... $\binom{4}{4}$$\binom{48}{9}$) = 0.37

For b) I got .56 by doing ($\binom{1}{1}$$\binom{3}{1}$$\binom{48}{11}$+ ... $\binom{1}{1}$$\binom{3}{3}$$\binom{48}{9}$)/$\binom{1}{1}$$\binom{51}{12}$ = .56
I am not sure that I did A and B correctly, or how to explain how it makes sense in the context of part c.

Body:
(a) Using hypergeomentric PDF in R:

a.2 = sum(dhyper(2:4, 4, 48, 13))
a.1 = sum(dhyper(1:4, 4, 48, 13))
a.1; a.2; a.2/a.1
[1] 0.6961825   # P(A >= 1)
[1] 0.2573349   # P(A >= 2)
[1] 0.3696372   # P(A >=2 | A >= 1)


问:
How to get the area of the triangle embedded in the unit circle when it makes an inner reflection?
The problem is as follows:

The figure represents the unit circle which has a radius of one unit. Find the value of the angle $\alpha$ such that the largest edge of the cherry shaded region is equal to $\sqrt{2+\sqrt{3}}$.

The choices given in my book are as follows:

$\begin{array}{ll}
1.&\frac{2\pi}{3}\\
2.&\frac{5\pi}{6}\\
3.&\frac{11\pi}{12}\\
4.&\frac{3\pi}{4}\\
\end{array}$

I'm not sure exactly how to solve this problem, but I belive that the given angle alpha is referred that begins on $A$ and ends in the other end from the vertex of the triangle which appears to be isosceles.

I think the approach here is that to get the value of alpha I must make a trigonometric equation and use the larger edge in terms of alpha and equate this with what I was given and this may give the desired result.

The distance from the $\textrm{x-axis}$ to any of the vertex of the base of the triangle with those parallel to $\textrm{y-axis}$ is the same, hence this is an isosceles triangle and this may help or ease the solution a bit.

Since those internal angles in the circle which make up the triangle are $\frac{\alpha}{2}$ because it is an iscribed angle, then I'm getting:

I'm naming the end point of the angle alpha as $P$:

Thus $\angle OAP=90-\alpha/2$ and the intersection of lines where is the second little square looking from the left would be point $R$.

Thus:

$\angle ROP=180-\alpha$

Therefore:

$OR=1\cdot \cos (180-\alpha)$

$AO=1$

Thus:

$AR=1-\cos\alpha$

$AR\sec \left(90-\frac{\alpha}{2}\right)= \sqrt{2+\sqrt{3}}$

From which I believe the equation to be solved would be as follows:

$(1-\cos\alpha)\cdot \sec \left(90-\frac{\alpha}{2}\right)= \sqrt{2+\sqrt{3}}$

Using the necessary algebra I'm getting this into:

$\cos\alpha=0$

$2\cos^2\alpha-(6+3)\cos \alpha=0$

$\cos\alpha=\frac{6+\sqrt{3}}{2}$

But from looking at these expressions none of these seem to help me much into the solution.

Was my strategy wrong or what?. Can someone help me here?. I honestly don't think I'm far off from the answer, so I need a helping hand which can guide me to the right path.

答:
First notice that the longest side of the cherry triangle (the hypotenuse) is congruent to its reflection above the $x$-axis: the segment connecting the initial $(1, 0)$ and terminal $(\cos \alpha, \sin \alpha)$ points of the angle $\alpha$ in standard position. Let's call this length $\ell$. Then, Pythagoras give us
$$
\ell^2 = (1 - \cos \alpha)^2 + (\sin \alpha)^2
= 1 - 2\cos \alpha + \cos^2 \alpha + \sin^2 \alpha
= 2(1 - \cos \alpha).
$$
So we need to solve the equation $2(1 - \cos \alpha) = 2 + \sqrt{3}$ for $\alpha$, i.e.
$$
\cos \alpha = -\frac{\sqrt{3}}{2}.
$$

The diagram suggest that the terminal point is above the $x$-axis (can we assume this?), in which case $0 \leq \alpha \leq \pi$. Then, there is a unique angle
$$
\alpha = \arccos \biggl(\! -\frac{\sqrt{3}\,}{2} \biggr) = \frac{5\pi}{6}.
$$


问:
When is the area of a triangle whose side lengths are consecutive integers also an integer?
Consider a triangle with side lengths 3, 4, and 5. By Heron's formula, its area is $\sqrt{6(6 - 5)(6-4)(6 - 3)} = \sqrt{6(1)(2)(3)} = \sqrt{36} = 6$. Are there any other triangles like this?</p>

答:
Because of a liking for symmetry, we let the sides be $y-1$, $y$, and $y+1$. Using the Heron Formula, we find, exactly like you did, that $\frac{3}{16}y^2(y^2-4)$ should be a perfect square. Thus $y$ must be even, say $y=2s$. So we want $3(s^2-1)$ to be a perfect square, say $(3t)^2$. We arrive at the equation
$$s^2-3t^2=1,$$
an instance of a Pell Equation.

In this case, the equation has the fundamental solution $s_1=2$, $t_1=1$. By general theory of the Pell equation, the positive solutions $(s,t)$ are given by
$$s=\frac{(2+\sqrt{3})^n+(2-\sqrt{3})^n}{2},\qquad t=\frac{(2+\sqrt{3})^n-(2-\sqrt{3})^n}{2},$$
where $n$ ranges over the positive integers.

In our case we are primarily interested in $s$. Let $s_0=1$ and $s_1=2$. Define $s_n$ by the recurrence
$$s_{n+2}=4s_{n+1}-s_n.$$
For suitable $t_n$, $(s_n,t_n)$ ranges over the positive solutions of the Pell equation. For example, $s_2=7$, giving $y=14$. This is your $x=13$. Continuing, we get $s_3=26$, and $s_4=97$. These give your computed values. There are infinitely many others.

Remark: The Pell equation is dealt with in most introductions to number theory. There is also a very nice book on the equation, by Ed Barbeau.


问:
How should I finish this by completing to square?
I am trying to calculate the integral:

$$\int_{-\infty}^{+\infty}\exp\left[-\frac{a^2(k-k_0)^2}{2} + ikx-i\frac{hk^2}{2m}t\right]dk$$

I thought on maybe mark $k-k_0=u$ but I am not see how it gonna help me

P.S the integral is only with respect to $k$. $k_0$ is constant also like the rest of paramteters

$i$ is the imagenery unit

Body:
$$
-\frac{a^2(k-k_0)^2}{2} + ikx-i\frac{hk^2}{2m}t
= -\frac12 a^2 k^2 + a^2 k_0 k - \frac12 a^2 k_0^2 + ikx - i\frac12 \frac1m ht k^2 \\
= -\frac12 (a^2+i\frac{ht}{m}) k^2 + (a^2 k_0 + ix) k - \frac12 a^2 k_0^2 \\
= -\frac12 (a^2+i\frac{ht}{m}) \left( k^2 - 2 \frac{a^2 k_0 + ix}{a^2 + i\frac{ht}{m}} k + \frac{a^2 k_0^2}{a^2+i\frac{ht}{m}} \right) \\
= -\frac12 (a^2+i\frac{ht}{m}) \left( \left[ k - \frac{a^2 k_0 + ix}{a^2 + i\frac{ht}{m}} \right]^2 + \left[ \frac{a^2 k_0^2}{a^2+i\frac{ht}{m}} - \left(\frac{a^2 k_0 + ix}{a^2 + i\frac{ht}{m}}\right)^2 \right] \right) \\
$$


问:
Ricci flow as heat flow on Riemannian manifold
I read that Ricci flow is "a nonlinear heat flow for the Riemannian metric". Can someone explain what this means?

Nonlinear heat flow has a wikipedia page but I don't understand how this differs from the traditional heat flow on Euclidean space. My guess is to replace the Laplacian with its version on a Riemannian manifold.

This might also be related, but I recently learned that heat flow is locally a mean curvature flow. I'm not sure what the relationship between MCF and Ricci flow is, but this seems like a piece of the puzzle.

答:
$\newcommand{\R}{\mathbb{R}}$
The basic heat equation on a domain in $\R^n$ is
$$ \partial_tu = \sum_{i=1}^n (\partial_i)^2u $$
The inhomogeneous version is
$$ \partial_tu = \sum_{i=1}^n (\partial_i)^2u + f, $$
where $f$ is a function of $x \in \R^n$ only.
This all can be generalized to the linear PDE
$$ \partial_tu = a^{ij}(x,t) \partial^2_{ij}u + b^k(x,t)\partial_ku + c u+ f(x), $$
where the matrix $A = [ a_{ij}]$ is positive definite.
This is the most general version of what is often called a linear heat equation on a domain in $\R^n$.

A quasilinear heat equation is simply one where the functions $a^{ij}$, $b^k$, $c$, and $f$ are functions of not only $x$, $t$, but also $u$.

On a Riemannian manifold,  one possible linear heat equation is of the form
$$
\partial_tu = \Delta_g u + b^k\partial_k u + cu + f,
$$
where $\Delta_g u = g^{ij}\nabla^2_{ij}u$.

All of the above assumes $u$ is a scalar function on its domain.

Where things get more complicated is when $u$ is either a map $u: M \rightarrow N$, as it is for the harmonic map heat flow or the Riemannian metric itself, as it is for the Ricci flow. If you write out the formulas for these PDEs in local coordinates, you will see that the coefficients of the PDE depend on the unknown map or metric and therefore the PDE is nonlinear. The fact that the harmonic map heat flow can be written simply as
$$
\partial_t u = \Delta u
$$
is misleading, because here $\Delta u$ is a nonlinear function of its first and second partial derivatives.

As for why the Ricci flow is a nonlinear heat equation is longer story, but after you use the so-called DeTurck trick, the Ricci flow looks like
$$
\partial_t g_{ij} = g^{pq}\partial^2_{pq}g_{ij} + \text{ lower order terms}
$$
which is a nonlinear heat equation for the metric $g$.

There are similar stories for other geometric heat flows, such as the mean curvature flow.


问:
Which probability formula do I use to solve the question?
An oil prospector will drill a succession of holes in a given area to find a productive well. The probability he is successful on a given trial is 0.2.

What is the probability that the tenth hole drilled yields the fourth productive well?

If ten holes are drilled, what is the probability of exactly 4 productive wells?

I am having trouble deciding what should be done on each question. I am fairly confident that I should do

$10 \choose 4$$0.2^4(1-0.2)^6$

but I am not sure which one to do it on. I think I need to use that formula to find the probability of exactly 4 wells, but then how do I solve the first question?

答:
Your expression applies to the 2nd question.

For the 1st question, you have the more restrictive situation where there were exactly 3 successes in the 1st 9 trials and a success on the 10th trial:

$\left[\binom{9}{3} \times (0.2)^3 \times (0.8)^6\right] \times (0.2).$


问:
Proof by Induction: inequalities and natural numbers.
How can I prove that

For all $n \in \Bbb N$ such that $10<n$ then $0 < n^2 -13n +24$ by induction.

This is what I have so far:

Base Case:

$n = 11$

If $n = 11$ then:

$0 < n^2 -13n +24$

$0 < (11)^2 - 13 (11) + 24$

$0 < 2$

Induction Hypothesis:

$0 < k^2 -13k +24$

Inductive Step:

$0 < (k+1)^2 -13(k+1) +24$

I’m supposed to get to this $2k-12$, but I don’t understand how, and after that I think we need to do something else.

答:
For the base case $n=11$, $f=121-143+24=1$ so the base case is positive.

Then for $n+1$, we have $f(n+1)=f(n)+2n-14$

Well, clearly $2n>14$ for all $n>11$ so $f$ is strictly increasing, QED.


答:
Given a topological space $(X,\tau_X)$ and an arbitrary set $Y \subseteq X$ we want to endow $Y$ with the structure of a topological space itself. One natural thing to do is to "restrict" the topology $\tau_X$ on $X$ to topology $\tau_Y$ on $Y$. We have to make sure that:

$Y, \emptyset \in \tau_Y$

$\forall \mathcal U \subseteq \tau_Y: \ \bigcup \mathcal U \in \tau_Y$

$\forall U,V \in \tau_Y: \ U \cap Y \in \tau_Y$

We can achieve this by taking the "relative topology" as $\tau_Y$ defined by

$\tau_Y = \{ Y \cap U \in \mathcal P(X) \mid U \in \tau_X \}$.

You might want to check that this indeed gives a topological space $(Y,\tau_Y)$ and that it coincides with your description if we take $X = \mathbb{R}$ and $Y = E$. If you know what a "metric space" is, you might want to think about this case in a more general context...

Title: What is the Idea of a Relative Open Set?

Body:
My real-analysis text gave the following defintion:

Let U be a subset of E.  U is open relative to E if

for $\forall t \in U$, $\exists \epsilon$ such that $N_\epsilon(t) \cap E \subset U$.

Although the idea that U is open in $\mathbb R$ follows the definition,

I normally do not think about the intersection of $N_\epsilon(t) \cap \mathbb R$.

U is open if for $\forall t \in U$, $\exists \epsilon$ such that $N_\epsilon(t) \subset U$; every t is an interior point of U.

Intuitively, each point, t, is contained in a "bubble".

So, what is the significance of specifying the intersection?

Apparently, I can't think of a relative open set in terms like interior points and "bubbles".


问:
If $X$ is binomial, $2X$ isn't binomial
My question is really simple. I Don't understand why if a random variable $X \sim \text{Bin}(n,p)$, then $2X$ isn't binomial. I know every value of $2X$ is even, but I can't prove $2X$ isn't binomial using this fact.

答:
It's because $X$ isn't just the value whose probability is being enumerated.  If it were, you could double those and enumerate the probabilities.

But in a binomial, the distribution is a set of pairs of frequencies and the associated probabilities, fitting a predetermined pattern. To be binomial, those frequencies must be exactly determined by the combinatorial rules that arise out of the number of different ways $TRUE$ and $FALSE$ can be picked from $m$ trials.  Doubling, or multiplying by any ratio $N$ populates the frequencies of success equivalent to $0\pmod N$ with the original probabilities rather than the ones in accordance with the new combinatorics, and populates every frequency in-between with probability zero.

Title: If $X$ is binomial, $2X$ isn't binomial


问:
Value of an integral
A function  $f$ is defined as
$$f(t)=t^{r-1}(2t-t^2)^{n-1}(2-2t)$$
where $r,n$ are natural numbers. Can we somehow find an a closed expression for the inegral $\int_0^1 f(t)\,dt$ or even an asymptotic expression as $n$ goes to infinity? One could substitution $u=2t-t^2$ but for the extra term $t^{r-1}$.

Mathematica yields the following value :
$$2^{-1 + 2 n + r} (B[1/2, -1 + n + r, n] - 2 B[1/2, n + r, n])$$
where $B(x,a,b)=\int_0^x u^{a-1}(1-u)^{b-1} du$ is the incomplete beta integral

OwnerDisplayName: sajjad veeri

Title: Value of an integral

Body:
Partial solution: asymptotics at $n\to\infty$

$$I(n)=\int_0^1t^{r-1}(2t-t^2)^{n-1}(2-2t)dt=2^n\int_0^1t^{r-1}(1-t)e^{(n-1)\log(t(1-\frac{t}{2}))}dt$$

$\log(t(1-\frac{t}{2}))$ has maximum at $t=1$.

Near this pont

$\log(t(1-\frac{t}{2}))=\log{\frac{1}{2}}-(t-1)^2-\frac{1}{2}(t-1)^4+...$

Changing the variable $\,t\to{x}=1-t$ and expanding integration to $\infty$

$$I(n)=2^n{e}^{(n-1)\log{\frac{1}{2}}}\int_0^1(1-x)^{r-1}e^{-(n-1)\bigl(x^2+\frac{1}{2}x^4+...\bigr)}xdx=$$$$=2\int_0^1\bigl(1-(r-1)x+...)\bigr)e^{-(n-1)\bigl(x^2+\frac{1}{2}x^4+...\bigr)}xdx$$$$\sim2\int_0^\infty{e}^{-(n-1)x^2}xdx-2(r-1)\int_0^\infty{e}^{-(n-1)x^2}x^2dx+O\Bigl(\frac{1}{n^2}\Bigr)$$
$$\sim\frac{1}{n}\Bigl(1-\frac{\sqrt{\pi}(r-1)}{2\sqrt{n}}\Bigr)+O\Bigl(\frac{1}{n^2}\Bigr)$$


答:
This is actually quite a profound aspect of set theory.

The powerset of $A$, or $\mathcal{P}(A)$, is defined to be $\{x : x \subseteq A\}$.

What is a subset of $A$? I suggest that, in some sense, a subset of $A$ is really just a function from $A$ to $\{0, 1\}$, where $0$ represents False and 1 represents True.

A function $f : A \to \{0, 1\}$ determines a subset of $A$ by $\{x \in A : f(x) = 1\}$.

Conversely, a subset $X \subseteq A$ determines a function $f : A \to \{0, 1\}$ by defining $f(a) = 1$ if $a \in X$, and $f(a) = 0$ otherwise.

Thus, with 0 representing False and 1 representing True, $f(a)$ tells us whether the statement $a \in X$ is true or false.

So strictly speaking, elements of $\mathcal{P}(A)$ are not functions $f : A \to \{0, 1\}$. But there is a bijective correspondence. Thus, we often write $\mathcal{P}(A) = \{0, 1\}^A$.

Title: Explanation of Notation Regarding Power Set of a Set

Body:
In Grimmet and Strizaker's Probability and Random Processes it states section 1.2

The power set of $\Omega$, which is written $\{0,1\}^{\Omega}$ and contains all subsets of $\Omega$....

It goes on to use this in an example as follows:

A die is thrown once. We can take $\Omega$ = $\{1,2,3,4,5,6\}$, $F = \{0,1\}^{\Omega}$

Here $F$ is the set of all subsets of the sample space $\Omega$ which are of interest. I don't understand this notation to specify that set. How does it generate/enumerate all the subsets of interest?


问:
Statistical test for boolean value groups

Intro

This question is somewhat of a prequel to this one - but can be seen also as a standalone!

I have two distinct populations/measurements of boolean values: $\{x_i\}_{i=1}^{m_1},x_i \in \{0,1\},|x|=m_1$ and $\{y_i\}_{i=1}^{m_2},y_i\in\{0,1\},|y|=m_2$. Each $x_i$ (and $y_i$) is independent of the other boolean values.

These numbers represent activity values ($1$ = active, $0$ = inactive) of two different group of models, namely the good ones ($x$'s) and the bad ones ($y$'s). I wanted to find a measurement of how much the (mean) activity of the good models is different than the bad models' activity. So, I started with the average values $a_1=mean(x)$ and $a_2=mean(y)$ and took the difference: $d=a_1-a_2$. Then I wanted to include a penalty related to the amount of models $m_1,m_2$ and this was solved in this question by introducing a proper scaling function for the difference $d$.

The problem

I was considering that if I had two populations of continuous measurements, then I would probably do a Wilcoxon Rank Sum test (use wilcox.test in R for example) and get also a p-value + location parameter shift estimation (a kind of $d$ as is in the respective implementation) for this (for the boolean case, ranking methods are useless right?). So, the p-value would tell how sure I am about this difference - which is kind of what I wanted to solve numerically (by let's say inserting the uncertainty in the final formula) in this other question.

All in all, I need a statistical test that would compare boolean values and get me as a result a p-value and the d difference (or similar) so to speak.

What I tried

So, since I work with two different and independent set of boolean values $\{x_i\}$ and $\{y_i\}$ the words binomial distribution and bionomial test came into my mind (from loooong ago :) But after reading about them for a while and playing with the respective R functions, it turned out not I wanted this to be.

The most significant problem for me was that I do not know the probability of success, i.e. $p=P(x_i=1)$ (or $y_i$ for that matter) - actually that's what I want to know for each model category! I tried to use the average $a_1=mean(x)=p$ and then to take a value of the cumulative distribution but it does not give results representative to what I want (the average activity). And the binomial test does not come close to what I want either.

For example: $x=\{0,0,0,0,1,0\}, m_1=n=6,p=1/6=0.1667$ and then I would calculate the $P(X\geq 1)=0.665$, which is way higher than expected.

Maybe I am overthinking it, but I believe there should be something close to what I want to do with this boolean dataset that I have. Any directions for distribution(s) or statistical tests that can accommodate my case are welcome!

Also posted on Cross Validated.

答:
For continuous measurements of two random boolean variables, in a setting in which you'd like to combine the distributions and consider them as one, I'd recommend a Poisson distribution.

It's a close approximation to a binomial and the concept of the "average frequency of event $A$" and the "average frequency of event $B$" are clearly additive since the average frequency (per unit time) of events $A$ and $B$ is the sum of the two means.


问:
Evaluating the congruence $(1+195) \pmod 7$
$$(1 + 195) \pmod 7 \equiv \quad ?$$

How would I get the answer? Because when I divide $196 / 7$ I get $28$ which is not a decimal to multiply by $7$.

答: $196$ is divisible by $7$, so $196 \equiv 0 \mod 7$</p>


问:
Continuous random walks with discrete steps?
Continuous-time random walks in continuous spaces are generally defined as follows: sample a waiting time (from a continuous pdf), wait for this time, and then jump to a random position (from another continuous pdf).

This sounds like a very discrete process to me. For instance, the walker is at a countable number of positions over time, and moves a countable number of times.

This seems different from a walker continuously moving from its current position to the one after the jump, during the waiting time (which would then be a travel time).

This is visible in drawings of random walks on the plane: the consecutive walker positions are generally linked by a (straight) line:

This does not fit the "jump" definition, where the walker stays at one place until it jumps, and is never located at intermediate positions.

Is there something I misunderstand? May someone explain why the two approaches are equivalent, if they are? Or what are their differences, if any?

I am also interested in pointers to material on random walkers that would travel from places to places, rather than jump.

I have posted a related, more formal, question on MO but it was downvoted, probably because I do not correctly understand the points above.

答:
There is indeed a difference between continuous-time random walks and the situation where a walker is moving continuously. For continuous-time random walks, the path of a walker is a series of points. For visualization purposes we can connect these to get the line segments which you mention. In contrast, a continuously moving walker's path can form any continuous line.

I'm not an expert, but an example of the latter is a Wiener process.


问:
Is $\frac{\textrm{d}y}{\textrm{d}x}$ not a ratio?
In Thomas's Calculus (11th edition), it is mentioned (Section 3.8 pg 225) that the derivative $dy/dx$ is not a ratio. Couldn't it be interpreted as a ratio, because according to the formula $dy = f'(x) \, dx$ we are able to plug in values for $dx$ and calculate a $dy$ (differential)? Then, if we rearrange we get $dy/dx=f'(x)$, and so $dy/dx$ can be seen as a ratio of $dy$ and $dx$. I wonder if the author says this because $dx$ is an independent variable, and $dy$ is a dependent variable, and for $dy/dx$ to be a ratio both variables need to be independent.

答:
Of course it is a fraction in appropriate definition.

Let me add my view of answer, which is updated text from some other question.

Accordingly, for example, Murray H. Protter, Charles B. Jr. Morrey - Intermediate Calculus-(2012) page 231 differential for function $f:\mathbb{R} \to \mathbb{R}$ is defined as function of two variables selected in special way by formula:
$$df(x)(h)=f'(x)h$$
so it is linear function with respect to $h$ approximating $f$ in point $x$. Also it can be called 1-form.

This is fully rigorous definition, which does not required anything, then definition/existence of derivative. But here is more: if we define differential as existence of  linear approximation in point $x=x_0$ for which holds
$$f(x)-f(x_0) = A(x-x_0) + o(x-x_0), x \to x_0$$
then from this we obtain, that $f$ have derivative in point $x=x_0$ and $A=f'(x_0)$. So existence of derivative and existence of differential are equivalence requirements. Rudin W. - Principles of mathematical analysis-(1976) page 213.

If we use this definition for identity function $g(x)=x$, then we obtain
$$dg(x)(h)=dx(h)=g'(x)h=h$$
This gives possibility to understand record $\frac{dy}{dx}=\frac{df}{dx}$ exactly as usual fraction of differentials and holds equality $\frac{df(x)}{dx}=f'(x)$. Exact record is $\frac{df(x)(h)}{dx(h)}=\frac{f'(x)h}{h}=f'(x)$.

Let me note, that we are speaking about single variable approach, not multivariable.

I cannot explain why someone assert, that $\frac{dy}{dx}$ cannot be understand as fraction - may be lack of knowledge about differential definition? For any case I bring, additionally to above source, list of books where is definition of differential which gives possibility understand fraction in question:

James R. Munkres - Analysis on manifolds-(1997) 252-253 p.

Vladimir A. Zorich - Mathematical Analysis I- (2016) 176 p.

Loring W. Tu (auth.) - An introduction to manifolds-(2011) 34 p.

Herbert Amann, Joachim Escher - Analysis II (v. 2)  -(2008) 38 p.

Robert Creighton Buck, Ellen F. Buck - Advanced Calculus-(1978) 343 p.

Rudin W. - Principles of mathematical analysis-(1976) 213 p.

Fichtenholz Gr. M - Course of Differential and Integral Calculus vol. 1 2003 240-241 p.

Richard Courant - Differential and Integral Calculus, Vol. I, 2nd Edition -Interscience Publishers (1937), page 107

John M.H. Olmsted - Advanced calculus-Prentice Hall (1961), page 90.

David Guichard - Single and Multivariable Calculus_ Early Transcendentals (2017), page 144

Stewart, James - Calculus-Cengage Learning (2016), page 190

Differential in Calculus

For complete justice I mention Michael Spivak - Calculus (2008) 155 p. where author is against understanding of fractions, but argument is from kind "it is not, because it cannot be". Spivak one of my most respected and favorite authors, but "Amicus Plato, sed magis amica veritas".


问:
$\sqrt{\frac{x-8}{1388}}+\sqrt{\frac{x-7}{1389}}+\sqrt{\frac{x-6}{1390}}=\sqrt{\frac{x-1388}{8}}+\sqrt{\frac{x-1389}{7}}+\sqrt{\frac{x-1390}{6}}$

How many solutions the following equation has in real numbers
?$$\sqrt{\frac{x-8}{1388}}+\sqrt{\frac{x-7}{1389}}+\sqrt{\frac{x-6}{1390}}=\sqrt{\frac{x-1388}{8}}+\sqrt{\frac{x-1389}{7}}+\sqrt{\frac{x-1390}{6}}$$

$1)1\quad\quad\quad\quad\quad\quad\quad2)2\quad\quad\quad\quad\quad\quad\quad3
 )3\quad\quad\quad\quad\quad\quad\quad4)4\quad\quad\quad\quad\quad5)\text{zero}$

Obviously we have $x\ge1390$. starting from $x=1390$ we can see LHS is very close to $1+1+1=3$ and RHS is close to zero. for higher values of $x$ we can see right side of the equation grows much faster than left side. so it seems the equation has not answer in real numbers.

I'm not sure how to proceed mathematically. I can rewrite it as follow:
$$\left(\sqrt{\frac{x-8}{1388}}-\sqrt{\frac{x-1388}{8}}\right)+\left(\sqrt{\frac{x-7}{1389}}-\sqrt{\frac{x-1389}{7}}\right)+\left(\sqrt{\frac{x-6}{1390}}-\sqrt{\frac{x-1390}{6}}\right)=0$$
I putted fractions with similar numbers in the same parentheses so it looks better now, but don't know how to continue.

Body:
The easiest but absurd way to do this is by squaring and squaring and squaring and..... To write out the equation of polynomial in $x$

$$\sqrt{\frac{x-8}{1388}}+\sqrt{\frac{x-7}{1389}}+\sqrt{\frac{x-6}{1390}}-\sqrt{\frac{x-1388}{8}}-\sqrt{\frac{x-1389}{7}}-\sqrt{\frac{x-1390}{6}} = 0$$

With the help of a cas, then we would have to observe the polynomial in $x$ to see how many real roots it has

$$( 166271366055421398492452275393603941230766862954455137380647248140804636783475714327462432993480052899307496588105274122048146156780410637021361 \cdot x^6 - 1420140036782905319437561361639895711607700448706531408943829995316272533767501640030686352288245638342331131237991828557619781758027571151457168696 \cdot x^5 + 5053362485747062046204356842284861702806221919004319273920353418806545439134535332974351841718762407835957590697696948539773086891685978542298184835440 \cdot x^4 - 9589047098047637215433435077525811816830831747049585209634085291233160531678815568240751343497601637440011806189213827409633807381969946623370920034699520 \cdot x^3 + 10233876003972506215993189306355354430848302521410622385212634701164151435860483711557613487835971006731199476091003849691449303854798610976036031554449411840 \cdot x^2 - 5824396568911083719726988788037339915692680410889570986360670278090638276801032058096056401851613752306491371113890184533047015490065483972816831803846901282816 \cdot x + 1381013855402990033245846225695823841737429380881009697772601168066108621988351698959635167325016829640478323081178371168331795718011254016938645598239579277398016 )\cdot (x-1396)= 0 $$

The polynomial is $7$th degree, and has $4$ real roots

EDIT:

But squaring an equation creates an environment for false root.... so all roots are tested, so that only $x-1396 = 0$ is a real root that satisfies the equation


问:
Question regarding r's complement
This question was asked in a test in my undergrad course

A computer, using 3-digit radix complement arithmetic with an unknown
radix r, gives the following results expressed in radix complement
form:

(m + n)r = (087)r; (m − n)r = (005)r  ; (n − m)r = (184)r

Identify the radix r and the decimal values of m and n.

This is what I did:

Adding last 2 equations,

(m-n+n-m)r = (005)r + (184)r

-> 0 = (005)r + (184)r

-> 0 = 5r0 + 1r2 + 8r1 + 4r0

Because this equation has no integer solutions, therefore no valid r exists.

Am I correct?

答:
Assuming that in $3$-digit $r$'s complement, $-x$ is the complement with respect to $r^3$. Let $x = m-n$

$x = (005)_r$

$-x = (184)_r$

$r^3 = 1\cdot r^2 + 8 \cdot r + 9$

Since $-x$ has a $8$ in it, we have $r\ge 8+1$. But then it will not satisfy our derived equation.

Perhaps there was a typo in the exam?

Title: Question regarding r's complement


问:
Relationships between $2$ mutually tangent circles in a semi-circle
In the attempt to answer this question, I was brought to another one, connected to a certain family of circles.

Consider the semicircle made of the upper half of the unit circle and the horizontal diameter with endpoints $A(-1,0)$ and $B(1,0)$.

Let us now consider the family $(F)$ of circles inscribed into this semicircle as can be seen on Fig. 1.

Fig. 1: Certain circles of family $(F).$

They can be described in a parametric way as circles

$$\text{with center} \ (x_c=-\cos a, y_c=\frac12 \sin^2a) \ \text{and radius} \ y_c, \ a \in (0,\pi) \tag{1}$$

or equivalently as circles $(C_T)$

$$\text{with center} \ \left(x_c=\frac{T^2-1}{(1+T)^2}, y_c=\frac{2T}{(1+T)^2}\right) \ \text{and radius} \ y_c, \ T \in (0,+\infty) \tag{2}$$

(Connection between the two parameterizations: $T=t^2$ where $t:=\tan \tfrac{a}{2}$).

The locus of these centers is parabola $(P)$ with cartesian equation:

$$y=\tfrac12(1-x^2),\tag{3}$$

represented on Fig. 2 as the light blue curve.

Each circle $(C_T) \in F$ is associated with a unique circle $(C_{T'}) \in F$ which is tangent to $(C)$ as depicted on Fig. 2.

Fig. 2: Two circles of family $(F)$ are mutually tangent if their parameters $T$ and $T'$ comply with relationship (4).

Let $I$ be the point of tangency of $(C_T)$ and $(C_{T'})$.

Question: how can we describe the locus of $I$?

(sketched as the magenta curve on Fig. 2.)

My work: I have obtained through tedious calculations an unexpected linear relationship between parameters $T$ and $T'$ which is:

$$T=(3-2 \sqrt{2})T'\tag{4}$$

Using (4), I have tried to obtain the locus of $I$ as the envelope of chords defined by the line segment joining the centers of circles $C_T,C_{T'}$ using homographies but to no avail.

Edit: Nice solution using inversion (a little different from the solution given in a comment by Blue).

'Refer to Fig. 3). Consider inversion with pole $S(0,-1)$ and power $2$ meaning that $M$ is exchanged with $M'$ iff $S,M,M'$ are aligned and $SM.SM'=2$. Its circle of inversion is  the black circle (center $S$, radius $\sqrt{2}$). It is easy to see that any circle of family (F) is globally invariant. This is in particular the case for the two blue circles. As the image by inversion of two tangent curves in a certain point $I$ is two tangent curves with contact point $I'$ where $M'$ is the image of $M$, point $I$ is invariant, establishing that $I$ belongs to the circle of inversion.$\square$

As a consequence, tangency points $T_1$ and $U_1$, $T_2$ and $U_2$, are exchanged, meaning that $S,T_1,U_1$ and $S,T_2,U_2$ are aligned.

Fig. 3.

Remark: Two properties of circles of family (F) deserving to be known:

http://www.gogeometry.com/problem/p311_circle_inscribed_semicircle_chord.htm

http://www.gogeometry.com/problem/p310_semicircle_circle_tangent_angle.htm

答:
Observe that the point of tangency divides the segment between the centers in the ratio of the radii, that is, in the ratio $y(T')/y(T)$. Therefore, if you call $(U,V)$ the point of tangency,
$$
U(T)=x(T)+\frac{y(T')}{y(T)}(x(T')-x(T));\quad V(T)=y(T)+\frac{y(T')}{y(T)}(y(T')-y(T))
$$
where $T'$ is expressed through $T$ using the relation you found (assuming it is correct).


问:
Classic example of a non exact form
Let $\dfrac{xdy-ydx}{x^2+y^2}$ be a 1-form defined in $\mathbb{R}^2\backslash\{0\}$.
Where can I find a detailed proof that it is not exact? I would prefer a proof that doesn't use results about conservative vector fields (I know that way). I have searched Jefree Lee, Do Carmo & some others, but they all skip this part.  Thanks a lot.</p>

答:
If you integrate it along a circle containing the origin in its interior, you get a nonzero value (try it! put $x=\cos \theta, y = \sin \theta$). If the form were exact, the integral would vanish.

I don't know if that's what you mean by a "conservative result", but this is really the reason why this form isn't exact.

Title: Classic example of a non exact form