答:
Note that we have using the $AM-GM$ inequality

$$\begin{align}
\left|\frac{d}{dx}\left(\frac{1}{x^2+n^2}\right)\right|&=\left|\frac{-2x}{(x^2+n^2)^2}\right|\\\\
&=\left|\frac{-2x}{x^2+n^2}\right|\frac1{x^2+n^2}\\\\
&\le \frac1{n^2}\frac{2|x|}{x^2+n^2}\\\\
&\le \frac{1}{n^3}
\end{align}$$

Hence, $\sum_{n=1}^\infty \frac{-2x}{(x^2+n^2)^2}$ converges uniformly for all $x$ and we conclude that the series of interest is differentiable for all $x$.

Title: Continuity and differentiability of $h=\sum_{n=1}^{\infty} \frac{1}{x^2+n^2}$

Body:
I have solved the following exercise and I would like to know if I have made any mistakes:

Let $h(x)=\sum_{n=1}^{\infty}\frac{1}{x^2+n^2}$.

(a) Show that $h$ is a continuous function defined on all of $\mathbb{R}$;

(b) Is $h$ differentiable? If so, is the derivative function $h'$ continuous?

My solution:

(a) $|h_n(x)|=\frac{1}{x^2+n^2}\leq\frac{1}{n^2}$ for all $x\in\mathbb{R}$ and $\sum_{n=1}^{\infty}\frac{1}{n^2}<\infty$ so by Weierstrass M-Test $\sum_{n=1}^{\infty}\frac{1}{x^2+n^2}$ converges uniformly on $\mathbb{R}$ and since $h_n(x)$ is continuous for all $n\geq 1$, by Term-by-term Continuity Theorem the limit function $h$ is continuous too.

(b) for $x\in [-M,M]\ (M\in\mathbb{R}^+)$, $|h'_n(x)|=|-\frac{2x}{(x^2+n^2)^2}|=\frac{2}{(x^2+n^2)^2}|x|\leq \frac{2M}{n^4}$ and $\sum_{n=1}^{\infty}\frac{2M}{n^4}<\infty$ so by Weierstrass M-Test $\sum_{n=1}^{\infty}h'_n(x)$ converges uniformly on each interval $[-M,M]$ to a function $h$ and since $0$ belongs to every such interval and $h(0)=\sum_{n=1}^{\infty}\frac{1}{n^2}<\infty$ by Term-by-Term Differentiability Theorem $h$ is differentiable; $h'$ is also continuous by Term-by-Term Continuity Theorem since each $h'_n$ is continuous and the convergence of the series of derivatives is uniform.


问:
Why is this a linear matrix inequality in $x,t$?
I'm given the following inequality

$$\begin{bmatrix}
 t & (Ax+ b)^T \\
 (Ax+b) & P_0+x_1P_1+\dots+x_nP_n
\end{bmatrix}\succcurlyeq  0$$

is said to be a linear matrix inequality in variables $x,t$. The general for of a LMI I knwo is the form $x_1A_1+\dots+x_nA_n\succcurlyeq  0$ how can the above be seen of this type in $x, t$?

答:
I assume that the matrix $A$ has size $m \times n$. Let $a_1,\dots,a_n$ denote the columns of $A$. Define
$$
M_0 = \pmatrix{0 & b^T\\b & P_0}, \quad
M_k = \pmatrix{0 & a_k^T\\ a_k & P_k} \text{ for } k = 1,\dots,n,
\\M_{n+1} = \pmatrix{1 & 0_{1 \times m}\\0_{m \times 1} & 0_{m \times m}}.
$$
We can write the inequality presented in the form
$$
M_0 + x_1 M_1 + \cdots + x_n M_n + t M_{n+1} \succeq 0.
$$

ContentLicense: CC BY-SA 4.0


问:
Existence and uniqueness of Stokes flow
What are the solution existence and uniqueness conditions for Stokes' flow?

$$\begin{gathered}
\nabla p = \mu \Delta \vec{u} + \vec{f}
\\
\nabla \cdot \vec{u} = 0
\end{gathered}$$

Maybe you could also provide some articles or books about the topic? Most physics books seem not to care about these details.

答:
There is one famous example in which there is no solution for the Stokes' flow case: Stokes flow around a cylinder, which is approriately named Stokes' paradox. In this case it is impossible to match the boundary conditions
both at infinity (uniform flow) and at the cylinder surface (no-slip) with Stokes flow dynamics. See e.g. paragraph 6.4 of the Fluid Dynamics Lecture Notes of
Jacques Lewalle. The breakdown of a solution means that there will always be some inertial effect, regardless how small the Reynolds number is.

An approximate solution to this issue was first proposed by Oseen who introduced a linearized inertial term to account for inertial contributions in the far field.

Later, Proudman and Pearson calculated a more precise solution through asymptotic expansions and matching of the far field solution and the near-cylinder solution.

So to answer your question: existence for Stokes' flow is not guaranteed even though the criterion $Re << 1$ is satisfied. A pretty good explanation for the precise reasons of non-existence is given in chapter 7 of the Fluid Dynamics I lecture notes by Prof. Childress. In the same document they also show (in paragraph 7.2) that Stokes' flow does exhibit uniqueness for non-trivial cases (i.e. $\textbf{u}\neq0$).

For more details on conditions for solvability: there is quite a lot of mathematical fluid dynamics literature on the solvability of Stokes' flow. Nazarov and Pileckas reference a  number of them in their paper with the telling title  "On the Solvability of the Stokes and Navier-Stokes Problems in the Domains That Are Layer-Like at Infinity"

OwnerDisplayName: Juris

ContentLicense: CC BY-SA 3.0

Title: Existence and uniqueness of Stokes flow

OwnerDisplayName: michielm


问:
How to solve the geodesic equations in 2D polar coordinates?
Consider the XY-plane as a surface in 3D-space, parametrized with polar coordinates $u$ an $v$ as $\vec{r}=(v\cos(u),v\sin(u),0)$. For the geodesic equations I get

Eq.1: $v~d^2u+2~du~dv=0$

Eq.2: $d^2v-v(du)^2=0$.

[EDIT]
Of course we all know that the general solution is the straight line $a=v\cos(u-\alpha)$ which passes through some point $(u,v)=(\alpha,a)$. But I would like to derive that solution, i.e. solve the system in general.

Moreover, after checking out quite a few texts (e.g. Dray) I could only find discussions where the straight line is parameterized by arc length, which I would like to avoid. Although it's obvious that using arc length simplifies such things tremendously, I like the idea of $v$ being a function of $u$, thus with $u$ as the main variable. So I would prefer to avoid using e.g. $ds$ in the derivation. (This also fits in my opinion that "the geodesic" is a geometric concept whose definition should be kept free from restrictions on the parameterization. But that's another discussion, and even my "heroes" Millman and Parker clutter their definition of a geodesic with an arc length parameterization.)

Any suggestions?

[EDIT]
Nobody seems to care? It looks like everybody thinks I'm a student doing my homework? I know it might look like a simple question but I just can't get it right. And no, I'm not a student.

Consider a straight line in the plane, e.g. $a=v*cos(u-\alpha)$ which passes through some point $(u,v)=(\alpha,a)$. After differentiating twice and rearranging things a bit I get

Eq.3: $sin(u-\alpha)*(v*d^2u+2*du*dv)+cos(u-\alpha)*(v*(du)^2-d^2v)=0$.

So, if the geodesic equations (1) and (2) hold, then (3) is fulfilled. But that's the wrong direction: in order to prove that the straight line is a geodesic we should be able to prove that (1) and (2) are fulfilled if we know that (3) holds, and not the opposite way around. But I can't really satisfactorily convince myself with my reasonings.

答:
In the mean time I worked out the following. We have $a=v\ cos(u-\alpha)$ and need to prove

Eq.1: $v\ d^2u+2\ du\ dv=0$

Eq.2: $d^2v−v\ (du)^2=0$.

We differentiate $a=v\ cos(u-\alpha)$ and obtain $0=dv\ cos(u-\alpha) - v\ sin(u-\alpha)\ du$. We square both terms of the result and get Eq.4: $(dv)^2\ cos^2(u-\alpha)=v^2\ sin^2(u-\alpha)\ (du)^2$.

We have $(dv)^2 + (v\ du)^2 = (ds)^2$ from which follows $(dv)^2 = (ds)^2 - (v\ du)^2$ which we substitute in Eq.4 and get $((ds)^2 - (v\ du)^2)\ cos^2(u-\alpha)=v^2\ sin^2(u-\alpha)\ (du)^2$. Working this out and rearranging gives $(ds)^2\ cos^2(u-\alpha)=v^2\ (du)^2$. We eliminate the cosine from this with $a=v\ cos(u-\alpha)$ and obtain $(ds)^2\ a^2 = v^4\ (du)^2$, so $a\ ds=v^2\ du$. You can stick with arc length or set $s=b\ t+c$ so the curve is not unit speed anymore and get $a\ b\ dt = v^2\ du$ . Differentiating this gives $a\ b\ d^2t = 2\ v\ du\ dv + v^2\ (d^2u)$. We set $d^2t=0$, and get $0= 2\ v\ du\ dv + v^2\ (d^2u)$. We can divide by $v$ (because $v=0$ gives the origin; no solution) and $v\ d^2u + 2\ du\ dv=0$. So far for Eq.1.

Now we differentiate $(dv)^2 + (v\ du)^2 = (ds)^2$ thus $2\ dv\ d^2v + 2\ v\ du\ (du\ dv + v\ d^2u) = 2\ ds\ d^2s = 0$. Thus $dv\ d^2v + v\ (du)^2\ dv + v^2\ du\ d^2u = 0$. We use Eq.1 to change $v\ d^2u$ in $-2\ du\ dv$ and get $dv\ d^2v + v\ (du)^2\ dv + v\ du\ (-2\ du\ dv)=0$ which simplifies to $dv\ d^2v - v\ (du)^2\ dv=0$. We can divide by $dv$ (because $dv\neq0$ for our straight line) and we get Eq.2.

There are 2 free parameters $a$ and $\alpha$ in our proposed solution $a=v\ cos(u-\alpha)$ and so we found all solutions to the system Eq.1 & Eq.2 ([EDIT] maybe this is too big a shortcut).

Pfewww... I guess that's it! Any comments still welcome, though.


问:
Evaluate $\int^{2\pi}_{0}\frac{\mathrm{d}x}{(\cos x+\sin x+\sqrt{3})^2}=2\sqrt{3}\pi.$
The problem is to evaluate the following definite integral:
$$\int^{2\pi}_{0}\frac{\mathrm{d}x}{(\cos x+\sin x+\sqrt{3})^2}=2\sqrt{3}\pi.$$

I have failed in almost every way I tried, with the exception of substitution by $\tan(\frac{x}{2})$ which leads to the integration of a rational function with annoying coefficients.

Is there any "good" way to address the problem, given that the result doesn't seem that complicated? I've also tried this website, but the result it gave was far from satisfactory. Please help.

答:
Note

$$ I(a)=\int_0^\pi \frac{dt}{(\cos t+a)^2} =
-\frac d{da}\int_0^\pi \frac{dt}{\cos t+a}
=-\frac d{da} \frac\pi{\sqrt{a^2-1}}=\frac{\pi a}{(a^2-1)^{3/2}}
$$

and apply it to

\begin{align} \int^{2\pi}_{0}\frac{\mathrm{d}x}{(\cos x+\sin x+\sqrt{3})^2}
&= \int^{2\pi}_{0}\frac{\mathrm{d}x}{(\sqrt2\cos (x-\frac\pi4)+\sqrt{3})^2}\\
&= \int^{\pi}_{0}\frac{\mathrm{d}t}{(\cos t+\sqrt{\frac32})^2}
 =I\left(\sqrt{\frac32}\right)=2\sqrt3\pi
\end{align}


问:
Demonstration Gradient Descent
The directional derivative represents the slope of a function along a direction. The minimization of the directional derivative is equal to the anti-gradient, and the proof of this is clear to me. But why is the minimization of the directional derivative at a point x, equivalent to finding the direction where the function at point x decreases faster? I can't understand why the anti-gradient gives direction and direction in which the function decreases faster. What is the relationship between the minimization of the directional derivative and the monotony of a function to be minimized?</p>

答:
The answer to your question is rooted in the relation between derivatives and linear approximations by tangents. When you take a differentiable real-valued function $f: \mathbb{R}^n \rightarrow \mathbb{R}$, a linear approximation of $f$ at a point $a = (a_1, a_2, \dots, a_n) \in \mathbb{R}^n$ could be defined by:

$$g(x) = f(a) + \sum\limits_{i=1}^{n} f_{x_i}(a)(x_i-a_i)$$
where $x = (x_1, x_2, \dots, x_n) \in \mathbb{R}^n$.

It is easy to verify that $g(a) = f(a)$ and $\nabla g = \nabla f$.

Now, take any unit vector $\hat{u}$, if we want to minimize $D_{\hat{u}}f(a)$, we have to minimize $\nabla f(a) \cdot \hat{u} = \nabla g(a) \cdot \hat{u}$. So we could look at the function $g$ instead of $f$.

For the cases of the first and second dimensions where the graph of $g$ is a line and a plane respectively, it is easy to imagine that the direction which minimizes the directional derivative is in the opposite direction of the gradient.

In higher dimensions, it becomes more difficult to see the relationship. Let's look for a unit vector $\hat{u} = (u_1, u_2, \dots, u_n)$ such that $g(a + \hat{u})$ is minimized.

$$g(a + \hat{u}) = f(a) + \sum\limits_{i=1}^{n} f_{x_i}(a)((a_i + u_i)-a_i) = f(a) + \sum\limits_{i=1}^{n} f_{x_i}(a) u_i = f(a) + \nabla g(a) \cdot \hat{u}$$

Since $f(a)$ is a constant, then we need to minimize the term $\nabla g(a) \cdot \hat{u}$. It could be easily verified that the minimum of the term occurs when $\hat{u} = - \frac{\nabla g(a)}{\| \nabla g(a)\|}$. In other words, when $\hat{u}$ is in the opposite direction of the gradient.

However, this is not true for minimizing $f(a + \hat{u})$. At the first glance, one might think that to minimize $f(a + \hat{u})$, the unit vector $\hat{u}$ must be in the opposite direction of the gradient. But the length of the vector $\hat{u}$ is 1 which is large enough for the function to have many local minima in the unit disc around $a$ so that no single direction is optimal.

This confusion in the meaning of "steepest descent" or "fast decreasing" makes one think incorrectly of gradient descent. The direction of the "steepest descent" should only be understood "infinitesimally" and not with a distance away from a point. It should be noted that the opposite direction of the gradient may not be the best direction to find an absolute minimum or even a good local minimum after going through several iterations.

Title: Demonstration Gradient Descent


问:
Calculating explicit integral with Fubini
Hey I am trying to solve following integral, but i dont have any idea how to proceed.

$\int_{0}^{1}\int_{x}^{1} \sin(y^2) dydx$

I have tried to solve it with substitution, but i am missing a term, namely sth like $ay$.

Trying to use Lebesgue integral with Fubini and transformation I got $\int_{\mathbb{R}^2}\sin(y^2)\cdot \chi_{[0,1]\times [x,1]} d\mathcal{L^2}(x,y)$, but I have still the same problem... how to proceed?

I also tried to use the series of sine $\sin(y^2) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} y^{4n+2}$, but then this series is not monotone so i can not use monotone convergence so that does not really help, or do I miss sth?

Does anyone have a hint for me?

答:
I explain my comment in more detail here:

We integrate over the set $\{(x,y)|0\leq x \leq 1, x\leq y\leq 1\}=\{(x,y)|0\leq y\leq 1,0\leq x\leq y\}$. So when interchanging the integrals we obtain $\int_0^1 \int_0^y \sin(y^2)dx dy=\int_0^1 y\sin(y^2)dy$. Now using the substitution $t=y^2, dt=2y dy$ we obtain $\int_0^1 y\sin(t)\frac{dt}{2y}=\frac{1}{2}\int_0^1 \sin(t) dt=\frac{-cos(1)+cos(0)}{2}$.


问:
Show that $\gamma:[a,b]\rightarrow \mathbb{C}$ has bounded variation iff exists $h:[0,1]\rightarrow [a,b]$ with $\gamma \circ h$ Lipschitz.
Let $~\gamma:[a,b]\rightarrow \mathbb{C}~$ be a continuous curve: show that $~\gamma~$ has bounded variation if and only if there exists $~h:[0,1]\rightarrow [a,b]~$ continuous and surjective such that $~\gamma \circ h~$ is a Lipschitz curve.

My only idea is for the proof of the reciprocal proposition: if $$
|\gamma(h(x))-\gamma(h(y))|\leq k|x-y|~\forall x,y\in [0,1],$$
then, given a partition $P=\{a=x_0,x_1,\dots, x_n=b\}$, I have that exists $~y_i~$ with $~h(y_i)=x_i~$ for each $i$ from $1$ to $n$. Then,
$$\sum_{i=1}^n|\gamma(x_i)-\gamma(x_{i-1})|\leq \sum_{i=1}^n|\gamma(h(y_i))-\gamma(h(y_{i-1}))|\leq k\sum_{i=1}^n|y_i-y_{i-1}|$$
But I don't know how to continue, because I think that $h$ is not necessarily monotonic.

Edit: For the direct implication I found a reference that helped me a lot, here goes https://mathoverflow.net/questions/253024/from-bounded-variation-to-1-lipschitz-function

答:
Your attempt works if $h$ is nondecreasing. I think my argument here shows that you can use $h$ to construct a nondecreasing function $g : [0,1] \to [a,b]$ such that $\gamma \circ g$ is also Lipschitz, after which you can apply your argument. (Please comment if you find errors in my argument.)

WLOG we may assume there exists $0 \le z_0 < z_1 \le 1$ such that $h(z_0)=a$ and $h(z_1)=b$. (If not, replace $h$ with $\tilde{h}(x)=h(1-x)$.)

Let $g:[z_0,1] \to [a,b]$ be defined by $g(x) = \sup_{t \in [z_0, x]} h(t)$. Show that $g$ is continuous, nondecreasing, and surjective onto $[a,b]$.

$\gamma$ is uniformly continuous on $[a,b]$, so there exists $\delta$ such that $|u-u'| \le \delta$ implies $|\gamma(u)-\gamma(u')| \le \epsilon$.

I claim that for $z_0 \le x < y \le 1$, there exist $x',y'$ satisfying $x \le x' \le y' \le y$ such that
$h(x') = g(x)$ and $|g(y)-h(y')| \le \delta$.

Proof of claim: If $g(y) > g(x)$, choose $y' \le y$ such that $|g(y) - h(y')| \le \min\{\delta, \frac{g(y)-g(x)}{2}\}$, by definition of $g(y)$. We must have $y' \ge x$ because $h(y') \ge g(y) - \frac{g(y)-g(x)}{2} =  \frac{g(y) + g(x)}{2} \ge g(x) = \sup_{[z_0, x]} h(t)$. Next, because $h(x) \le g(x) \le h(y')$, the intermediate value theorem implies the existence of some $x'$ between $x$ and $y'$ such that $h(x') = g(x)$.

By uniform continuity of $\gamma$, we have $$|\gamma(g(x)) - \gamma(g(y))| \le \epsilon + |\gamma(h(x'))-\gamma(h(y'))| \le \epsilon + k|x'-y'| \le \epsilon + k|x-y|.$$

Since $\epsilon$ is arbitrary, we see that $\gamma \circ g$ is also Lipschitz.

Finally, if we define the linear map $\tilde{g} : [0,1] \to [z_0, 1]$ by $\tilde{g}(t) = z_0 + t(1-z_0)$, you can note that $(g \circ \tilde{g}) : [0,1] \to [a,b]$ is your continuous nondecreasing surjection such that $\gamma \circ (g \circ \tilde{g})$ is Lipschitz.


问:
Relations and quotient groups.

We've started studying relations and I have no idea what I'm doing. Is this supposed to be the way to represent my answers?

a.

The relation is all possible relations such that the sum of a,b is equal to the sum of c,d.
So in order to avoid double counting we basically need to find a way to reach all numbers in [2,8].

$\left\{\left[1,1\right]_{R}\ ,\left[1,2\right]_{R},\left[2,2\right]_{R},\left[2,3\right]_{R},\left[3,3\right]_{R},\left[3,4\right]_{R},\left[4,4\right]_{R}\right\}$

b.

The relation is all possible relations such that the sum of a1,a2,a3 is equal to the sum of b1,b2,b3.
So in order to avoid double counting we basically need to find a way to reach all numbers in [0,3].

$\left\{\left[0,0,0\right]_{R}\ ,\left[1,0,0\right]_{R},\left[1,1,0\right]_{R},\left[1,1,1\right]_{R}\right\}$

答:
I think you're almost there. I believe they want some explanation for why these are the answers (perhaps for $a$ you can say that two pairs are equivalent if they have the same sum, so we only have to find one representative for each sum) I mean, you need to explain what you mean by "find a way to reach all [2,8]".

However I believe you have a mistake. The equivalence class of $[2,2]$ is equal to the equivalent class of $[1,3]$ because $2+2=1+3$.

Title: Relations and quotient groups.


问:
Generating a subring inside a field
Let $K:=\mathbb{Z}[u_{j}^{\pm}:j \in J]$, $J$ finite, be a ring of Laurent polynomials and let $A:=K(X_1,\dots,X_n)$ be a field of rational functions in $n$ variables. Now, let $S=\{u^{\alpha}:\alpha_i \geq 0 \}$ be a finite set of (finite) monomials with non-negative powers only. Consider the subring $\mathbb{Z}[S]$ of $A$ generated by $S$. I am not sure what $\mathbb{Z}[S]$ would look like. I think that this is (as a set) literally a polynomial ring in $S$ over $\mathbb{Z}$, that is, there aren't any Laurent polynomials in it. Is this correct?

答:
These have names! They are called semigroup algebras. In general, they are not polynomial rings. For example, if $S = \{x^2, xy, y^2 \}$, then
$$\mathbb{Z}[S] \cong \mathbb{Z}[a,b,c]/\langle ac - b^2\rangle.$$
The isomorphism just sends $a \mapsto x^2$, $b \mapsto xy$ and $c \mapsto y^2$. These sorts of rings are interesting in their own right, and they show up as coordinate rings of affine toric varieties.

Just to add a little more. These will form a polynomial ring if and only if $|S| = n$ and when you arrange the exponent vectors appearing in $S$ in a matrix, the matrix has determinant $\pm 1$.

Title: Generating a subring inside a field


问:
Is deconvolution simply division in frequency domain?
Is it correct to say that deconvolution simply division in frequency domain?
And that convolution in time domain is multiplication in frequency domain.

And is it a convention to notate a function in frequency domain with a hat above the letter?

like $\hat{f}$*$\hat{g}$=$\hat{result}$

Deconvolution in time domain: $\hat{result}$/$\hat{g}$=$\hat{f}$ for $\hat{g} \not= 0 $

And How would you pronounce $\hat{g}$? function g in frequency domain, frequency domain function g, function g transformed into frequency domain, or none of those?

答:
Under some conditions, yes - it is possible.  Say we have $h=f\star g$ where $f$ is known, $g$ is unknown, and $h$ is 'measured'.  As you state, by the convolution theorem, we have

$$
\hat{h}=\hat{f}\hat{g}
$$  Now, if $\hat{f}\neq 0$, we can claim that $\hat{g}=\hat{h}/\hat{f}$.  However this poses two issues: firstly, it is entirely too restrictive to assume that $\hat{f}\neq 0$ - take for example the simplest band-pass filter,

$$
\hat{f}(k)=\left\{\begin{array}{cc}
1 &-1\leq k\leq 1\\
0 & \text{else}\end{array}\right.
$$ This is a very common convolution kernel, and you'll be unable to perform deconvolution by dividing.  In fact, the situation is worse - if $\hat{f}(k)=0$ for any $k$, then the convolution operator $g\mapsto f\star g$ will have nontrivial null space: any function $g$ which has $\hat{g}(k)\neq 0$ will result in the zero function after applying the convolution. Hence deconvolution is ill-posed for these kernels, i.e. there will not be a unique solution!

Even if you could perform deconvolution by division, it would be a bad idea as far as numerical accuracy is concerned - division should always be avoided due to round-off error.

As for pronouncing $\hat{g}$, most people say "$g$-hat", though technically it would be more appropriate to say "the Fourier transform of $g$".

Update 1/14:

To really get in to deconvolution, one should talk more seriously about regularization methods.  The classical way to do deconvolution is to simply take a Tikhonov regularization, i.e. if $Af=f\star g$ and we want to solve $Af=h$ for $f$, we consider a sequence (taking $\gamma\rightarrow 0$) of problems of the sort

$$
\min_{f_\gamma}\|Af_\gamma-h\|_2+\gamma\|f_\gamma\|_2
$$ this is a "regularized" least squares problem which has explicit solution

$$
f_\gamma=(A^tA+\gamma I)^{-1}A^th
$$

This is essentially a filtering technique - we avoid dividing by zero by filtering out those frequencies, then hope that we recover something close to $f$ as we take $\gamma\rightarrow 0$.  A better method turns out to be "$l^1$ regularized least squares", where we replace $\|f_\gamma\|_2$ with $\|Wf_\gamma\|_1$, where $W$ is a sparsifying transform such as wavelets.  This is a broad topic - see the book "Sparse image and signal processing" by Starck et al.  For more info on the classical methods for deconvolution, check out "Introduction to Inverse Problems in Imaging" by Bertero and Boccacci.


问:
How to write the equation of a line in affine space
Write the equation (in coordinates $x_1, x_2)$ of a line in $A^2$, where A denotes Affine space: passing through the point (2, −3) and parallel to the vector (5, 2).

Can we use the equation y=mx +c to find the slope and the constant to write the equation in Affine space?

Like I can get the slope as m = 2/5 since its parallel to the vector (5,2) and can get the 'c' by substituting the value of the point(2,-3) and m in the above-mentioned equation. I get the equation as $y=(2/5)x-(19/5)$.

答:
In general, if the line passes through $P=(p_1, p_2)$ and is parallel to the vector $V = (v_1, v_2)$, one can say that $X=(x_1, x_2)$ belongs to the line if the vector $\overrightarrow{PX}$ is colinear to $V$, that is to say if the determinant
\begin{equation}
0=
\begin{vmatrix}
x_1-p_1& v_1\cr
x_2 -p_2& v_2\cr
\end{vmatrix} = v_2(x_1-p_1)-v_1(x_2-p_2)
\end{equation}
This is the equation of the line. In your case
\begin{equation}
2(x_1-2) - 5(x_2+3) = 0
\end{equation}
or $2 x_1-5 x_2=19$.


问:
$\int_{0}^{\pi} D_{n}(y)dy=\frac{1}{2}$ Dirichlet
I need to calculate that $\int_{0}^{\pi} D_{n}(y)dy=\frac{1}{2}$ with $D_{n}(y)= \frac{1}{2\pi}\frac{\sin((n+\frac{1}{2})y)}{\sin(\frac{y}{2})}$ from Dirichlet.

Now I tried to do this with the known that $D_{n}(y)=\frac{1}{2\pi}\sum_{k=-1}^{n}e^{iky}$.

I tried to things but I never become 1/2.

First I tried to solve the somation and then integrate: $\int_{0}^{\pi} D_{n}(y)dy=\int_{0}^{\pi} \frac{1}{2\pi}\sum_{k=-1}^{n}e^{iky}dy=\int_{0}^{\pi} \frac{1}{2\pi}(\frac{e^{-niy}-e^{iy(n+1)}}{1-e^{iy}})dy$. This integral doesn't converges so I did something wrong but I don't know what?

Second I tried to switch the integral and the sommetion and then solve it: $\int_{0}^{\pi} D_{n}(y)dy=\int_{0}^{\pi} \frac{1}{2\pi}\sum_{k=-1}^{n}e^{iky}dy=\int_{0}^{\pi} D_{n}(y)dy=\frac{1}{2\pi}\sum_{k=-1}^{n}\int_{0}^{\pi}e^{iky}dy=\frac{1}{2\pi}\sum_{k=-1}^{n}\frac{1}{ik}(1-e^{-\pi i k})$. But this sometion will give $i\infty -i \infty$ so this doesn't work eather.

Can someone eplain me what I do wrong and how I can fix it?

Body:
You have to do summation from -n to +n when identifying Dirichlet kernel, though your closed expression for the kernel is correct: $D_{n}(y)=\frac{1}{2\pi}\sum_{k=-n}^{n}e^{iky}=\frac{1}{2\pi}\frac{\sin((n+\frac{1}{2})y)}{\sin(\frac{y}{2})}=\frac{1}{2\pi}\frac{\sin(\frac{(2n+1)y}{2})}{\sin(\frac{y}{2})}$

$D_{n}(y)=\frac{1}{2\pi}\sum_{k=-n}^{n}e^{iky}=\frac{1}{2\pi}+\frac{1}{2\pi}\sum_{k=1}^{n}e^{iky}+\frac{1}{2\pi}\sum_{k=-1}^{-n}e^{iky}=\frac{1}{2\pi}+\frac{1}{2\pi}e^{iy}(1-e^{iyn})\frac{1}{1-e^{iy}}+\frac{1}{2\pi}e^{-iy}(1-e^{-iyn})\frac{1}{1-e^{-iy}}=\frac{1}{2\pi}\frac{e^\frac{iy}{2}-e^\frac{-iy}{2}-e^\frac{iy}{2}-e^{-iy(n+\frac{1}{2})}-e^\frac{iy}{2}+e^{iy(n+\frac{1}{2})}}{e^\frac{iy}{2}-e^\frac{-iy}{2}}=\frac{1}{2\pi}\frac{\sin((n+\frac{1}{2})y)}{\sin(\frac{y}{2})}$


问:
Number of positive odd divisors of a number.
Let $n$ be a positive integer. Then find the number of divisors of $210n$ that are odd multiples of $3$. Then answer must be in terms of $n$.

My attempt:
We have $210=2\times 3\times 5\times 7.$ The odd multiples of $3$ dividing $210$ are $$3, 15, 21, 105.$$ Let $x$ be the odd divisor of $n$. Then, the divisors of $210n$ that are odd multiples of $3$ are odd multiples of $A\times x$, where $A$ is an odd multiple of $3$ dividing $210$.

Please help how to write the answer in terms of $n$

答:
Let $X$ be a positive integer who when written in prime factorization form is $X=p_1^{x_1}p_2^{x_2}p_3^{x_3}\cdots p_k^{x_k}$

The number of positive integer divisors of $X$ is equal to $(x_1+1)(x_2+1)\cdots (x_k+1)$

This is seen by straightforward counting principles... having chosen a value for the exponent for $p_1$ to be used for our divisor from among the integers $\{0,1,\dots,x_1\}$, chosen a value for the exponent for $p_2$ and so on, noting that there are $x_1+1$ such choices for $p_1$ since we are including zero.

Letting instead $X=2^{x_1}\cdot 3^{x_2}\cdot 5^{x_3}\cdot 7^{x_4}p_5^{x_5}p_6^{x_6}\cdots p_k^{x_k}$, explicitly including $2,3,5,7$ even if that makes $x_1,x_2,x_3,x_4$ zero... we get that $$210X = 2^{x_1+1}\cdot 3^{x_2+1}\cdot 5^{x_3+1}\cdot 7^{x_4+1}p_5^{x_5}\cdots p_k^{x_k}$$

Now, counting the number of divisors of $210X$ who are odd (meaning the exponent chosen for the divisor for $2$ must be zero) who are multiples of $3$ (meaning the exponent chosen for the divisor for $3$ must not be zero)... the total will be:

$$(x_2+1)(x_3+2)(x_4+2)(x_5+1)(x_6+1)\cdots (x_k+1)$$

noting again that there is no freedom in our selection for the power of $2$ to use and that since this is $210X$ as opposed to $X$ itself, there is additional freedom for the choices of powers to be used for $3,5,7$

The answer depends on the exact prime factorization of $X$ and as such is unable to be simplified further without additional constraints on what $X$ is (e.g. $X$ being coprime to $210$ in which case you might be able to write the answer in terms of $\sigma_0(X)$, the divisor counting function)


问:
A field extension of prime degree

Suppose that $E$ is an extension of $F$ of prime degree. Show that $~~\forall~ a \in E : ~ F(a)=F$ or $F(a)=E$

Attempt: Suppose that $E$ is an extension of a field $F$ of prime degree, $p$. Therefore $p = [E :F] = [E : F(a)][F(a) : F]$. Since $p$ is a prime number, we see that either $[E : F(a)] = 1$ or $[F(a) : F] = 1$.

Now, $[E : F(a)] = 1 \implies $ there is only one element $x \in E$ which forms a basis and every element in $E$ is generated by $x$ i.e. $E = \{x~c~|~ c \in F(a)\}$

$[F(a) : F] = 1 \implies $ there is only one element $y \in F(a)$ which forms a basis and every element in $F(a)$ is generated by $y$ i.e. $F(a) = \{y~d~|~ d \in F\}$

Have I inferred it correctly? How do I move ahead?

Thank you for your help.

答: Alternative argument: If $a\in F$, clearly $F(a)=F$. Otherwise the degree of the minimal polynomial of $a$ over $F$ is $[F(a):F]$, which divides $[E:F]=p$. But $p$ is a prime, and $a\notin F$, which means that $[F(a):F]\ne 1$, so we must have $[F(a):F]=[E:F]=p$. Now, since $F(a)/F$ is a subextension of $E/F$ with the same extension degree, we get $E=F(a)$.

Title: A field extension of prime degree


问:
Prove equality in triangle inequality for complex numbers
We need to show that
$$ |z_{1}+z_{2}+\cdots+z_{n}|=|z_{1}|+|z_{2}|+\cdots+|z_{n}|$$
if and only if $z_{1},z_{2},\dots,z_{n}$ have the same argument (i.e. $z_{j}=r_{j}e^{i\theta}$ for $j=1,\dots,n$).

This way [$\Longleftarrow$] is easy, but for the other way around, I get stuck proving it for the sum of two complex numbers, even though the induction itself isn't that big a problem.

Does anyone happen to know how to prove this for two complex numbers?

答:
Let's try to solve the equation

$$
|a+b|=|a|+|b|
$$

for $a,b\in\mathbb{C}$. It's not restrictive to assume that both $a$ and $b$ are non-zero. We can square it, because both sides are non-negative:

$$
|a+b|^2=(|a|+|b|)^2
$$

so, taking into account that $|z|^2=z\bar{z}$, we get

$$
(a+b)(\bar{a}+\bar{b})=a\bar{a}+2|a|\,|b|+b\bar{b}
$$

that simplifies in

$$
\bar{a}b+a\bar{b}=2|a|\,|b|
$$

Now, write $a=|a|u$ and $b=|b|v$; the equation becomes

$$
\bar{u}v+u\bar{v}=2
$$

Recall now that $\bar{u}=u^{-1}$, since by hypothesis $|u|=1$. Then …


问:
how to solve in $\Bbb{R}$ the inequality $(x-3)(x+1)<0$

Solve the following inequality in $\Bbb{R}$:
$$(x-3)(x+1)<0$$

My answer was:
$$x^2+x-3x-3<0 \\
x^2+x-3x<+3 \\
x^2-2x<3
$$
what should i do next,is it a remarkable identity?
i dont know how to continue.

答:
Note that the product of two positive quantities or two negative quantities is positive,

but the product of a negative quantity and a positive quantity is negative.

Can you take it from here?


问:
Spivak's ambiguous popcorn function
There is a question given in Michael Spivak's Calculus chapter 6 problem 1 section iv that reads as follows.

For which of the following functions $f$ is there a continuous function $\mathrm{F}$ with domain $\mathbb{R}$ such that $\mathrm{F}(x)=f(x)$ for all $x$ in the domain of $f$?

iv) $f(x)=1/q, \;x=p/q$ rational in lowest terms.

answer book says

iv)
No $\mathrm{F}$, since $\mathrm{F}(a)$ would have to be $0$ for irrational $a$, and then $\mathrm{F}$ is not continuous at $a$ is $a$ is rational. (typo intentionally reproduced from the book)
(end section)

The function was defined for rational numbers $p/q$. The function was never defined for $\mathbb{R}$. $f(x)$ for any irrational number is therefore undefined or unknown to the reader. This seems to be a sloppy example involving the popcorn function. I believe the conclusion that $f(x) = 0$ for any irrational number $x$ is not possible with the information given or maybe there is something that mathematicians know that is implied that I do not understand.

Is the function undefined for $x$ where $x$ is an irrational number? This would also result in a discontinuous function but for different reasons then what was concluded by Spivak. I also lack the understanding that the denominator $q$ can be isolated and extracted after $\mathbb{Q}$ has been mapped to $\mathbb{R}$ but that might be off topic. The domain is given as $\mathbb{R}$ not $\mathbb{Q}$.

答:
The function $ f $ is defined only for rational numbers. The function $ F $ (capital F) is a function that we are looking for - and it must be defined on all real numbers. The argument used is that if such an $ F $ exists, then it must be true that $ F(x) = 0 $, for all $ x \in \mathbb{R} \setminus \mathbb{Q} $ (all irrational numbers). Try to see why this is true on your own. But that means that for every rational number $a$, $ F $ is not continuous at $a$ (this is easy to prove).

PS I wasn't familiar with the name "popcorn function" so I did a little search... I found the late J.H. Conway's name far superior, the "Stars over Babylon" function (link)

Title: Spivak's ambiguous popcorn function


问:
What is the CDF, pdf and expectation of $\theta^*$ of $\theta$?
I have that $x_1, x_2,...,x_n$ are from a rv $X$ that has the density function $f_X(x)=\frac{2x}{\theta^2} \quad$ for $0 \le x \le \theta  \quad$ and $f_X(x)=0 \quad$ otherwise. I have determined that $\theta^* = max(x_i)$. How Can I now calculate the CDF $F_{\theta^*}$, the pdf $f_{\theta^*}$ and the expectation $E[\theta^*]$ of the maximum likelihood estimator $\theta^*$?

答:
Hints:

If the $x_1,\ldots, x_n$ are independent, then $P(\theta^* \le t) = P(x_1 \le t, x_2 \le t, \ldots, x_n \le t) = \prod_{i=1}^n P(x_i \le t)$

$f_{\theta^*}(t) = \frac{d}{dt} F_{\theta^*}(t)$

Use the usual formula for computing expectation when you have the density function


问:
find the interest rate in the form of compound interest with recurring contribution
I have the following formula to find the final value of compound interest with recurring contribution:

```<code>Where:
S: is final amount
T: my recurring value
i: interest rate
n: recurrence period
</code>```

what I want is to isolate i in this formula

答:
Comment: If you mean i in term of S, T and n then  let $(1+i)^n=A$, you may write :

$1+i=A^{\frac 1n}$

$$S=T\cdot \frac {A-1}{A^{\frac 1n}-1}$$

$$S\cdot A^{\frac 1n}-T\cdot A=S-T$$

That is if interest rate is 0 then $A^{\frac 1n}=A=1$

You may rewrite this as:

$$S (1+i)-T(1+i)^n=S-T$$


问:
What is the moment generating function of a function of a wiener process?
If I have that $B(t)$ is a wiener process (brownian motion), what is the moment generating function of

$X(t) = e^{B(t)-t/2}$?

答:
The moment generating function of $X_t$ will be defined by

$$M_X(\lambda)=\mathbb E( e^{X_t\lambda})$$
In this case your random variable $X_t:= e^{B_t-t/2}$ is a functional of a Gaussian r.v., the the above expectation can be calculated as follows

$$M_X(\lambda)=\int_{\mathbb R} e^{\lambda e^{x-t/2}}\frac{1}{\sqrt{2\pi t}}e^{-x^2/2t}dx$$

The integrand is clearly not in $L^1$ so you can claim that no m.g.f exists for this particular random variable.


答:
There is an important distinction between $f(x)$ and $F(x)$. Here $f(x)$ has the rationals $\mathbb Q$ as its domain so is meaningless for the irrationals.

Any continuous $F(x)$ on $\mathbb R$, where $F(\frac pq)=f(\frac pq) =\frac1q$ on $\mathbb Q$, will have $\lim\limits_{x \to y} F(x)=0$ for all $y \in \mathbb R$, which clearly fails to be continuous.

Title: Spivak's ambiguous popcorn function

Body:
There is a question given in Michael Spivak's Calculus chapter 6 problem 1 section iv that reads as follows.

For which of the following functions $f$ is there a continuous function $\mathrm{F}$ with domain $\mathbb{R}$ such that $\mathrm{F}(x)=f(x)$ for all $x$ in the domain of $f$?

iv) $f(x)=1/q, \;x=p/q$ rational in lowest terms.

answer book says

iv)
No $\mathrm{F}$, since $\mathrm{F}(a)$ would have to be $0$ for irrational $a$, and then $\mathrm{F}$ is not continuous at $a$ is $a$ is rational. (typo intentionally reproduced from the book)
(end section)

The function was defined for rational numbers $p/q$. The function was never defined for $\mathbb{R}$. $f(x)$ for any irrational number is therefore undefined or unknown to the reader. This seems to be a sloppy example involving the popcorn function. I believe the conclusion that $f(x) = 0$ for any irrational number $x$ is not possible with the information given or maybe there is something that mathematicians know that is implied that I do not understand.

Is the function undefined for $x$ where $x$ is an irrational number? This would also result in a discontinuous function but for different reasons then what was concluded by Spivak. I also lack the understanding that the denominator $q$ can be isolated and extracted after $\mathbb{Q}$ has been mapped to $\mathbb{R}$ but that might be off topic. The domain is given as $\mathbb{R}$ not $\mathbb{Q}$.


问:
Question about lagrange's theorem (group theory)
Lagrange's theorem tells us:
$$\frac{\left|G\right|}{\left|H\right|}=\left[G:H\right]$$
(in group theory)

But it is not clear to me what is  the number represents,does it represent the size of subgroup $H$?

答: $|G|$ and $|H|$ are the orders of the groups $G$ and $H$, respectively, i.e. the number of elements in each. $[G:H]$ is the index of $H$ in $G$, the number of cosets of $H$ in $G$, i.e. the number of unique elements in the set $\{gH:g\in G\}$.


问:
Equivalence of norms (Frobenius and L1)
I've been trying to prove that the Frobenius and 1 norm are equivalent, satisfying the following inequalities ($A\in \mathbb F^{m\times n}$, with $m\geq n$):
$$\frac{1}{\sqrt m}||A||_1 \leq ||A||_F \leq \sqrt n ||A||_1 $$

While proving the second inequality has been quite straightforward (I have proved that $||A||_F ^2 \leq  n ||A||_1 ^2$, which is equivalent), I haven't been able to figure out the first one. I suppose that I should aim for the inequality $||A||_1 ^2 \leq  m ||A||_F ^2$, but I don't know how to continue.

(The definition of the 1 norm I'm using is $||A||_1 = \max_{j=1,...,n} \sum_{i=1}^m |a_{ij}| $)

答:
Without loss of generality, suppose the maximum of $\sum_{i=1}^m|a_{ij}|$ occurs on the first column of $A$, which for convenience we denote by $v$. Then
$$
\frac{1}{\sqrt{m}}\|A\|_1
=\frac{1}{\sqrt{m}}\|v\|_1
=\frac{1}{\sqrt{m}}\langle |v|,\mathbf 1\rangle
\le\frac{1}{\sqrt{m}}\|v\|_2\|\mathbf 1\|_2
=\|v\|_2\le\|A\|_F.
$$

Title: Equivalence of norms (Frobenius and L1)


问:
Why is ellipse convex?
Well it is, by drawing one and looking at it. But how about starting from the definition: the sum of distances to two points is constant. A more general question is to start with an $n$-ellipse (some examples here: What are curves (generalized ellipses) with more than two focal points called and how do they look like?).

Suppose $n$ points $\mathbf{p}_i$ in a two dimensional plane, define the general ellipse $E$ by
$$
\sum_i^n d_i = D_0 \quad \mathrm{with} \quad d_i = |\mathbf{r}-\mathbf{p}_i|, \quad \mathbf{r} = (x,y)
$$
Questions

is $E$ convex?

does $E$ enclose all points?

if "no" to question 2, what's the value $D_0$ such that $E$ cross a point? And which point?

To higher dimenions?

I know that's lots of questions, and I doubt there is existing knowledge on those questions. Could someone explain or point out some references?

Update: clarification on question 2. For the standard ellipse, when $D_0$ is very small, it does not exist. It come to existence firstly as a line segment connecting the two points---I still count this as "enclose". A point is not enclosed only if it totally falls outside of the shape.

答:
This is an attempt at proving that the set of all points on the interior and the boundary of the general ellipse $E$ is indeed a convex set.

Define the set $S$ as follows:
$$S= \left\{\mathbf{r} :\sum_i^n d_i(\mathbf{r}) \leq D\right \},$$
where $d_i(\mathbf{r})=\mid \mathbf r - \mathbf{p_i} \ \mid$, $D$ is some constant, and $\{\mathbf{p_1}, \mathbf{p_2} , ..., \mathbf{p_n}\}$ are the $n$ foci of the general ellipse $E$.

Now, we wish to prove that the set $S$ is convex. Using the definition of convexity, for any pair of elements $\mathbf{r_1}, \mathbf{r_2} \in S$, we have to show that $\lambda \mathbf{r_1}+(1-\lambda)\mathbf{r_2}$ is also in $S, 0 \leq \lambda \leq 1$. This can be done by a simple application of the Triangle-Inequality, as follows:

$$\sum_i^n d_i\left(\lambda \mathbf{r_1}+(1-\lambda)\mathbf{r_2}\right)$$
$$=\sum_i^n \mid \lambda \mathbf{r_1}+(1-\lambda)\mathbf{r_2} - \mathbf{p_i}  \ \mid$$
$$=\sum_i^n \mid   \lambda \mathbf{r_1} - \lambda \mathbf{p_i} + (1-\lambda) \mathbf{r_2} \ -(1-\lambda) \mathbf{p_i} \  \mid$$
$$\leq \sum_i^n \lambda \mid \mathbf{r_1} - \mathbf{p_i} \ \mid  + \  (1-\lambda) \mid \mathbf{r_2} - \mathbf{p_i} \ \mid$$
$$= \lambda \sum_i^n  \mid \mathbf{r_1} - \mathbf{p_i} \ \mid  + \ (1-\lambda) \sum_i^n \mid \mathbf{r_2} - \mathbf{p_i} \ \mid$$
$$\leq \lambda D + (1-\lambda) D$$
$$= D.$$

Where the second last inequality follows since $\mathbf{r_1}, \mathbf{r_2}$ are elements of $S$.

Title: Why is ellipse convex?


Title: Equivalence of norms (Frobenius and L1)

Body:
I've been trying to prove that the Frobenius and 1 norm are equivalent, satisfying the following inequalities ($A\in \mathbb F^{m\times n}$, with $m\geq n$):
$$\frac{1}{\sqrt m}||A||_1 \leq ||A||_F \leq \sqrt n ||A||_1 $$

While proving the second inequality has been quite straightforward (I have proved that $||A||_F ^2 \leq  n ||A||_1 ^2$, which is equivalent), I haven't been able to figure out the first one. I suppose that I should aim for the inequality $||A||_1 ^2 \leq  m ||A||_F ^2$, but I don't know how to continue.

(The definition of the 1 norm I'm using is $||A||_1 = \max_{j=1,...,n} \sum_{i=1}^m |a_{ij}| $)

Body:
$(a+b)^2=a^2+2ab+b^2\le2(a^2+b^2)$.

More generally, $$\sum_{i=1}^m|a_i|\le \sqrt{m}\sqrt{\sum_{i=1}^m|a_i|^2}$$
$$\therefore\ \|A\|_1=\max_j\sum_{i=1}^m |a_{ij}|\le\sqrt{m}\max_j\sqrt{\sum_{i=1}^m|a_{ij}|^2}\le\sqrt{m}\sqrt{\sum_{ij}|a_i|^2}=\sqrt{m}\|A\|_F$$


问:
GRE quant question
If $a > 0$ and $b < 0$, which of the following statements are true about the values of $x$ that solve the equation $x^2-ax+ b =0$?

Indicate all such statements.

A) They have opposite signs

B) Their sum is greater than zero.

C) Their product equals $-b$.

While solving some GRE practice problems, I came across the problem above. the question seemed easy enough at first sight, but after solving it and looking at the solution. the solution states that only choices A and B are correct. why is this true? why is choice C incorrect? I googled the question but I was not convinced by any of the answers. while glancing at the question again I also realized that the first statement seems to contradict the second. a>0 but x^2 - ax + b =0. how can "a" be positive but at the same time be negative in the quad. equation? "a" can be produced by the sum of the two roots of the equation. and how is choice C incorrect? the product of a positive and negative integer will produce a negative integer?  I've been looking at this question for about 30 mins and i cannot help but feel that this is a poorly written gottca question. A good explanation is appreciated thanks.

答: The roots have sum $a>0$ (so B) is true) and product $b<0$ (so A) is true and C) is false).

Title: GRE quant question


问:
How to do a 3 constant reciprocal (multiplicative inverse) regression
I am trying to fit a set of data to a curve such as:

$y=\frac{m}{x-a}+b$

Without the constant $a$, it is easy to define $z=\frac{1}{x}$ and convert it to a linear model. But I have not been able to find how to solve it with the additional constant.

Ideally I would like to use the Theil-Sen estimator (https://en.wikipedia.org/wiki/Theil%E2%80%93Sen_estimator) to solve as I expect to have significant outliers and this method in principal should be more robust. If not, I can any other method such as least squares.

The goal of this exercise is ultimately to automatically identify the outliers. Table below has an example of data points that I am trying to fit:

x
y

2.86
210.021

9.08
108.977

0.63
199.2

125.5
110.723

68.61
112.154

64.62
87.464

233.03
43.727

16.61
277.019

13.41
201.378

11.01
51.5

0.28
455.625

32.65
160.806

145.55
115.917

2.33
160.668

22.07
158.409

67.3
102.105

18.9
143.599

30.45
314.55

115.86
210.744

111.46
126.739

5.24
53

2.71
220.523

1.57
238.647

4.94
272.322

165.75
100.999

20.6
95.482

112.59
114.391

2.06
203.475

89.82
125.323

39.2
95.994

568.93
38.768

69.12
20.15

51.91
80.275

答:
$$y=\frac{m}{x-a}+b \tag 1$$
Instead of fitting the above function $(1)$ one can fit the equation $(2)$ below to which $y$ is solution.
$$(y-b)(x-a)=m$$
$$-xy+ay+bx-ab+m=0$$
$$-xy+ay+bx+c=0 \quad;\quad c=m-ab \tag 2$$
This is a linear equation wrt $a,b,c$ which can be solved thanks to linear regression.
Section 5 in https://fr.scribd.com/doc/14819165/Regressions-coniques-quadriques-circulaire-spherique .

A numetical example (with reasonable scatter) is given below :

The result of fitting is exact if the data isn't scattered.

If the scatter of data is small the result of fitting is quite the same what ever the chosen criteria of fitting is.

If the data is highly scattered the "best fit" is different depending a lot on the criteria of fitting chosen (LMSE, LMAE, LMSRE, etc.). Thus one have to use an iterative method of fitting in which the criteria of fitting is implemented.

In the OP question we observe that the given data is highly scattered. So, the question is not well posed without specifying a criteria of fitting. No definitive answer can be given because they are as many "best fit" as they are different criteria of fitting.

With the data provided by the OP, the result appears on the next figure. It is clear that with such an high scatter this result is more than doubtful. Even with a specific griteria of fitting I am affaid that the result would be doubtful as well.


答:
$$(x-3)(x+1)<0$$
The product of two factors is negative when both factors have opposite signs. So that you have to solve these inequalities:

$$ x-3 >0 \text { and } x+1 <0$$
Or:

$$x-3 <0 \text { and  } x+1>0$$

Title: how to solve in $\Bbb{R}$ the inequality $(x-3)(x+1)<0$

Body:
Solve the following inequality in $\Bbb{R}$:
$$(x-3)(x+1)<0$$

My answer was:
$$x^2+x-3x-3<0 \\
x^2+x-3x<+3 \\
x^2-2x<3
$$
what should i do next,is it a remarkable identity?
i dont know how to continue.


问:
1st derivative of $\frac{2x}{\sqrt{x^2 + 1}}$
Another simple question that I can't work out today, yet I would work it out two weeks ago!

I need to find the 1st derivative of $$\frac{2x}{\sqrt{x^2 + 1}}$$.

So I use the Quotient rule and I get: $$\frac{(x^2 + 1)^.5 (2) - (2x)(0.5x^2 + 0.5)^-5}{x^2+1}$$

Am I heading in the correct direction and do I just need to multiply and try to get rid of the exponents somehow?

Thanks

答:
The numerator is not quite right. We want $(x^2+1)^{1/2}(2)-2x g'(x)$ where $g(x)=(x^2+1)^{1/2}$. That derivative will be $(2x)(1/2)(x^2+1)^{-1/2}$.

The denominator is right. Now you can if it is useful do some manipulation ("simplification").

Remark: For square roots, I think you will find $\sqrt{x^2+1}$ easier and safer to work with than the exponent notation. For other roots, the advantage goes over to the exponent notation.


问:
How to prove that $m(f\geq a)\leq \frac 1{a^n}\int f^n$?
Let $f:[0,1]\to\mathbf R_+$ be a measurable function and $a> 0$ such that $m(f\geq a)>0$.

I know that Markov's inequality states that $m(f\geq a)\leq\frac1a\int f$

How can it be deduced that $m(f\geq a)\leq \frac 1{a^n}\int f^n$?

答:
Hint:

Apply Markov's inequality (which you already mentioned) to $f^n$ and $a^n$ instead of $f$ and $a$.


答: If you want to think geometrically, the graph of $y=(x-3)(x+1)$ is an upward opening parabola that intersects the $x$-axis at $x=-1$ and at $x=3.$ We have $y$ negative for $x$ between the two roots, that is, for $-1<x<3,$ which provides the solution to the original inequality.


答:
I'm not sure if I read your notation correctly, but it looks like there's a mistake in the numerator:

$$f(x)=2x,\quad f^{\prime}(x)=2,\quad g(x)=\sqrt{x^2+1},\quad g^{\prime}(x)=
\frac{2x}{2\sqrt{x^2+1}}$$

And the numerator should then be $f^{\prime}(x)g(x)-f(x)g^{\prime}(x)$.

Title: 1st derivative of $\frac{2x}{\sqrt{x^2 + 1}}$

Body:
Another simple question that I can't work out today, yet I would work it out two weeks ago!

I need to find the 1st derivative of $$\frac{2x}{\sqrt{x^2 + 1}}$$.

So I use the Quotient rule and I get: $$\frac{(x^2 + 1)^.5 (2) - (2x)(0.5x^2 + 0.5)^-5}{x^2+1}$$

Am I heading in the correct direction and do I just need to multiply and try to get rid of the exponents somehow?

Thanks


问:
Finding a stochastic process from Ito’s lemme on $Y_t$ such that $\int_0^1Y_s\,\mathrm{d}B_s= B_1-2B_1\mathbb{I}_{|B_1|}$
Problem:

I would like to zoom in on a particular part of a question as a follow-up on a question I have previously asked, which I feel deserves its own space. I was looking at the random variable

$$X=B_1-2B_1\mathbb{I}_{|B_1|}=\begin{cases}B_1&\text{if }|B_1|\geq1\\-B_1&\text{otherwise}\end{cases}$$

for $t\in[0,1]$, where $\{B_t\}$ is a standard BM. I also have

$$X_t=\mathbb{E}(X|\mathcal{F}_t)=\mathbb{E}(B_1-2B_1\mathbb{I}_{|B_1|<1}|\mathcal{F}_t)$$

where the filtration $\mathcal{F}_t$ satisfies the usual conditions. I need to find an adapted process $\{Y_t\}_{t\in[0,1]}$ s.t. $X=\int_0^1Y_s\,\mathrm{d}B_s$. I was thinking that $Y_t=1-2\mathbb{I}_{|B_1|<1}$ was a good choice, but apparently it is wrong, why is this so?

Furthermore, is it true that $X_t=\int_0^tY_s\,\mathrm{d}B_s$ in general? How does this lead to the result that $\int_0^tX_sY_s\,\mathrm{d}B_s$ is a martingale?

Lastly, in my original question, I was trying to figure out the distribution of $X$, and I mistakenly thought was a Brownian motion. How do we determine the distribution from the form of $X$ without looking at $Y_t$?

My attempt:

Using the Markov property and independent increments, I have established $X_t=B_t-2f(B_t,\sqrt{1-t})$ where $f(x,s)=\mathbb{E}\left[(x+sZ)\mathbb{I}_{|x+sZ|<1}\right]$ and $Z\sim\mathcal{N}(0,1)$. Letting $s=\sqrt{1-t}$, I have from Ito's formula

$$\mathrm{d}X_t=\left(1-2\frac{\partial f}{\partial B_t}\right)\mathrm{d}B_t-\left(\frac{\partial^2f}{\partial B_t^2}-\frac1s\frac{\partial f}{\partial s}\right)\mathrm{d}t.$$

Since $X_t$ is a martingale, it is also a local martingale and the drift term is zero, so the PDE satisfied by $f(x,s)$ on the interval $s\in(0,1]$ is
$$\frac{\partial^2f}{\partial x^2}=\frac1s\frac{\partial f}{\partial s}.$$

Additionally, $X_0=\mathbb{E}(X_t|\mathcal{F}_0)=\mathbb{E}(X_t)=0$ a.s., so we have initial condition $f(0,1)=0$. To my knowledge, there is no closed form solution to this PDE (I think). Then, $Y_t=1-2\frac{\partial f}{\partial B_t}$ which I got from the diffusion part of Ito's formula. If I have missed out on a possible closed form solution for $f$ or $\partial f/\partial B_t$ do let me know.

答:
Let $f(b) := b 1_{|b|< 1}$ and compute by the Markov property of Brownian motion and independent increments that $$E[f(B_1)|\mathscr{F}_t] = E[f(B_1)|B_t] = E[f((B_1-B_t) + B_t)|B_t] = E[f(Z\sqrt{1-t} + b)]|_{b=B_t},$$ where $Z \sim N(0,1)$.

Thus we need to compute for a fixed $b$ the value of the integral
$$
E[f(Z\sqrt{1-t} + b)]
= E[(Z\sqrt{1-t} + b) 1_{-1 <Z\sqrt{1-t} + b < 1}]
= E[(Z\sqrt{1-t} + b) 1_{\frac{-1-b}{\sqrt{1-t}} <Z< \frac{1-b}{\sqrt{1-t}}}]
$$
$$
=\sqrt{1-t} E[Z1_{\frac{-1-b}{\sqrt{1-t}} <Z< \frac{1-b}{\sqrt{1-t}}}] +bP(\frac{-1-b}{\sqrt{1-t}} <Z< \frac{1-b}{\sqrt{1-t}})
$$
$$
=\left(\sqrt{1-t} E\left[Z\mid{\frac{-1-b}{\sqrt{1-t}} <Z< \frac{1-b}{\sqrt{1-t}}}\right] +b\right)P(\frac{-1-b}{\sqrt{1-t}} <Z< \frac{1-b}{\sqrt{1-t}})
$$
and using the formula for the mean of a truncated normal https://en.wikipedia.org/wiki/Truncated_normal_distribution this is (using $c:= \frac{-1-b}{\sqrt{1-t}} $ and $d:=\frac{1-b}{\sqrt{1-t}}$ and $\phi, \Phi$ normal density and CDF)
$$
=\left(\sqrt{1-t}\frac{\phi(c)-\phi(d)}{\Phi(d)-\Phi(c)} +b\right)(\Phi(d)-\Phi(c))
$$
$$
= \sqrt{1-t}(\phi(c)-\phi(d)) + b(\Phi(d)-\Phi(c)).
$$

Thus $$
E[B_1-2B_1 1_{|B_1|<1}|\mathscr{F}_t] $$
$$
= B_t - 2\sqrt{1-t}\left(\phi\left( \frac{-1-B_t}{\sqrt{1-t}}\right)-\phi\left(\frac{1-B_t}{\sqrt{1-t}}\right)\right) - 2B_t\left(\Phi\left(\frac{1-B_t}{\sqrt{1-t}}\right)-\Phi\left( \frac{-1-B_t}{\sqrt{1-t}}\right)\right) = M(B_t).
$$
Thus one should have
$
E[\int_0^1 Y_s dB_s|\mathscr{F}_t] =\int_0^t Y_s dB_s
$
equal to the previous line. Thus $Y_t = d M(B_t)$ can just be computed using Ito's formula on that monstrous formula, which involves only smooth functions of $B_t$. I'm confident a dedicated person could carry out the Ito formula calculation without further instruction, hopefully leading to some great cancellation of terms using that $\Phi' = \phi$, though I cannot force myself to type it up for you.

As for your question about $\int_0^t X_s Y_s dB_s$, this is a martingale because any integral with respect to Brownian motion is a local martingale, and you can check that $E\int_0^t (X_s Y_s)^2 ds < \infty$ so it is indeed a martingale.

One has that $X$ is normally distributed, this construction is used to give a standard example of normals that are uncorrelated but not independent. Check https://en.wikipedia.org/wiki/Normally_distributed_and_uncorrelated_does_not_imply_independent. I'm not sure about $X_t$.


问:
Why is this a local isomorphism?(model theory)
In Kees Doets's Basic Model Theory, local isomorphism is defined as follows:

A local isomorphism between models A and B of the same vocabulary is a finite relation {${(a_1,b_1),...,(a_n,b_n)}$}⊆ $A×B$ such that the simple expansions $(A,a_1,...,a_n)$ and $(B,b_1,...,b_n)$ satisfy
the same atomic sentences.

And then, it says:The finite injection {${(0,0),(2,e),(5,\pi)}$} is a local isomorphism between (Z,<) and (R,<).

My question is, what about this atomic sentence: 4<c,where c is to be substituted by 5 or $\pi$?
In A, c = 5, 4<5 is a true atomic sentence,
but in B, c=$\pi$, 4<$\pi$ is false.
How comes that {${(0,0),(2,e),(5,\pi)}$} is a local isomorphism between (Z,<) and (R,<)?

答: I think the point here is that $4 < c$ cannot be expressed as an atomic sentence in the relevant expansions of $(Z, <)$ or $(R, <)$: with no constant symbols in the base language, and no ability to use quantifiers, there's no way to "name" any numbers in these expansions other than the ones specified by the local isomorphism (for which we are directly adding new constant symbols).


问:
$(f_n)$ sequence of differentiable functions on $[0,1]$ and converge pointwise to $0$.
Let $(f_n)$ sequence of differentiable functions on $[0,1]$ converging pointwise to $0$. Suppose

$|f'_n(x)| \leqslant 2015 + \cos(x)$ $\forall x \in [0,1]$ and $\forall n$. Show that $(f_n)$ converge uniformly to $0$.

''Proof''

As supposed, $|f'_n(x)| \leqslant 2015 + \cos(x)$. We can rewrite it:

$|f'_n(x)| \leqslant 2015 + \cos(x) \leqslant 2016$ as $\cos(x)\leqslant1$

Since $(f_n)$ is a sequence of differentiable functions and the derivative is bounded, $(f_n)$ is Lipschitz. Hence, $\forall x,y \in [0,1]$:

$|f_n(x)-f_n(y)| \leqslant 2016|x-y|=\epsilon/3$

Moreover, as $(f_n)$ is Lipschitz, it implies that $(f_n)$ is uniformly continious on $[0,1]$: $\forall \epsilon>0$ $\exists \delta>0$ such that $\forall x,y \in [0,1]:$

$|x-y|<\delta \Rightarrow |f_n(x)-f_n(y)| \leqslant \epsilon/3$

And we want to show that $\forall \epsilon >0 \exists$ $n_0$ such that $\forall n\geqslant n_0 \forall x \in [0,1]$:

$|f_n(x)|<\epsilon$.

Then i don't really see how to apply the pointwise convergence(i.e create kind a subdivision on $[0,1]$ or?) and conclude the proof using triangular inequality.

P.S Im stydying Analysis I right now so if you could explain with much details as possible it would be kind from your part. Thanks in advance ;)

答:
Use @Michael's idea: first, to fix your observation about the uniform continuous of $(f_n)$ on $[0,1]$, you have to break your line

$|f_n(x)-f_n(y)| \leqslant 2016|x-y|=\epsilon/3$

into two separate statements:

$(f_n)$ is $2016$-Lipschitz.

$\forall \epsilon > 0, \exists \delta > 0: \forall x,y \in [0,1], |x-y| < \delta \implies |f_n(x)-f_n(y)| < \epsilon / 3$.

But #2 is still ambiguous since there's no quantifier and domain for the variable $n$ stated, and it's not clear if $\delta$ depends on $n$.  To make things clear, you may fix $\epsilon > 0$ and define $\delta = \epsilon / 6048$, so that whenever $|x-y| < \delta$, #1 can be applied to get #2.

Then we proceed to the partition $P = \{x_0 = 0,x_1 = \delta, x_2 = 2\delta, \dots, x_{M-1} = (M-1) \delta, x_M = 1\}$, where $M = \lceil 1/\delta \rceil$ is the number of subintervals of $P$.  Use pointwise convergence on each partition point $x_i$ to yield $N_i \in \mathbb{N}$ such that for all $n \ge N_i$, $|f_n(x_i)| < \epsilon / 3$.

Observation 2: for each partition point $x_i$, $f_n(x_i)$ is $\epsilon$-small when $n$ is sufficiently large.

Now go back to the paragraph about the uniform continuity of $(f_n)$.  Apply this on a partition point $x_i$ and a (non-partition) point $x \in [x_i,x_{i+1})$, then $|x_i - x| < \delta$, so $|f_n(x_i) - f_n(x)| < \epsilon / 3$.

Observation 1: when $x$ is $\delta$-close to a partition point $x_i$, $f_n(x_i)$ approximates $f_n(x)$ $\epsilon$-closely, and the error $\epsilon/3$ is independent of $n$.

The question asks for the uniform convergence of $(f_n)$ on $[0,1]$, so we need  $N = \max\limits_{i \in \{0, 1, \dots, M\}} N_i$, so that whenever $n \ge N (\ge N_i)$ and $x \in [0,1]$, $$|f_n(x)| \le |f_n(x) - f_n(x_i)| + |f_n(x_i)| \le \epsilon /3 + \epsilon + 3 < \epsilon.$$

To recap:

bounded derivatives gives subinterval length $\delta$ and $(f_n)$ Lipschitz.

use $\delta$ to build partition $P = \{x_i\}_i$.

apply pointwise convergence on partition points $x_i$.  use $N = \max_i N_i$ to complete the logic.

use triangle inequality to conclude.

$\require{AMScd}$
\begin{CD}
f_n(x_i) @<\delta \to 0< \text{uniform convergence} < f_n(x)\\
@V n \to \infty V  \text{pointwise limit} V @.\\
0
\end{CD}


问:
Application of Cauchy Schwarz
I had a question about the inequality in the following picture. I'm working on problem 6.1.4 from Classical Fourier Analysis by Grafakos. I've been told that Cauchy-Schwarz will allow me to conclude that

However, I can't seem to find the the right application of Cauchy-Schwarz that will allow me to conclude this. So far, I've tried making $A = \{x \in \mathbb{R}^n: |x| > 2|y|\}$ and writing the outside integral using an indicator function, but that didn't work. Is there something clever I should multiply by to conclude this?

答:
Let $f(x)=\sqrt{|x|^{n+\epsilon} \int_0^\infty |\Psi_t(x-y)-\Psi_t(x)|^2 \frac{dt}{t}}$ and $g(x)=\frac{1}{\sqrt{|x|^{n+\epsilon}}}$.

Then, by C-S you have
$$\int_{|x|>2|y|} \sqrt{\int_0^\infty |\Psi_t(x-y)-\Psi_t(x)|^2 \frac{dt}{t}} dx \leq \sqrt{ \int_{|x|>2|y|}\frac{1}{|x|^{n+\epsilon}}dx } \sqrt{ \int_{|x|>2|y|} |x|^{n+\epsilon} \int_0^\infty |\Psi_t(x-y)-\Psi_t(x)|^2 \frac{dt}{t}dx}$$

Therefore, your problem boils down to showing that
$$
\int_{|x|>2|y|}\frac{1}{|x|^{n+\epsilon}}dx \leq c_n^2 |y|^{-\epsilon}
$$
which should follow immediatelly from the convergence of the integral on the LHS, with the proper choice of $c_n$.

Title: Application of Cauchy Schwarz


问:
Solving a Complex Diophantine Equation with One Known
I am struggling to find a way to implement Euclid's Algorithm in order to solve this diophantine equation. The $N$ will be known and the set of solutions I wish to find will be a set of decreasing $X$ solutions that satisfy $$x^2 - 4  y^2 = n.$$

When simplifying I have concluded that the solution could be as follows...
$$x^2 - 4 y^2 = (x - 2y)  (x + 2y)$$ when $n = 12$,
$$12 = (x - 2y) (x + 2y)$$
would it be possible to use Euclid's Algorithm to solve for the GCD in decreasing order?
The poposed solutions are a set of $n=12$, $x = 4$ and $y = 1$.

答:
The following shows that the Pythagorean Theorem can be used here.
$$x^2-4y^2=n\implies n+ 4y^2=x^2\\
\rightarrow\quad A^2+B^2=C^2:A=\sqrt{n}, B=2y, C=x\\
\implies n=A^2\quad y=\frac{B}{2}\quad x=C$$
Beginning with Euclid's formula
$ \quad A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2\quad$
we can find solutions by solving for $k$ and test a defined range of $m$-values to see which ones yield integers. Side-$A$ of a Pythagorean triple can be any odd number greater than one and so any $A=(2j+1)$ will yield a triple. Here is the process demonstrated for $A=15$

\begin{equation}
A=m^2-k^2\implies k=\sqrt{m^2-A}\\
\text{for}\qquad  \lfloor\sqrt{A+1}\rfloor \le m \le \frac{A+1}{2}
\end{equation}
The lower limit ensures $k\in\mathbb{N}$ and the upper limit ensures $m> k$.
$$A=15\implies \lfloor\sqrt{15+1}\rfloor=4\le m \le \frac{15+1}{2} =8\\\land\quad m\in\{4,8\}\implies k \in\{1,7\} $$
$$F(4,1)=(15,8,17)\qquad \qquad F(8,7)=(15,112,113) $$

Translating, we get $\quad
 (15,8,17)\rightarrow(17,4,225)\qquad (15,112,113)\rightarrow (113,56,225)\quad$
These can be obtained more directly from $(m,k)$ if we adapt Euclid's formula as follows.
$$x=m^2+k^2\qquad y=mk\qquad n=(m^2-k^2)^2$$

Notice that $A=\sqrt{n}\space $ lets us put in any odd number for $A\ge3$ and we get the other terms out automatically in translation or by applying the modified formula. Here are a $33$ [primitive] solutions from the first $25$ $n$-values in ascending order of $n.\quad$ e.g. $5^2-4(2^2)=3^2$

$$(5,2,9)\quad (13,6,25)\quad (25,12,49)\quad (41,20,81)\quad (61,30,121)\\ (85,42,169)\quad (17,4,225)\quad(113,56,225)\quad (145,72,289)\quad (181,90,361)\\
 (29,10,441)\quad  (265,132,529)\quad (221,110,441)\quad (313,156,625)\quad (45,18,729)\\
 (365,182,729)\quad  (421,210,841)\quad (481,240,961)\quad (65,28,1089)\quad (545,272,1089)\\
 (37,6,1225)\quad (613,306,1225)\quad (685,342,1369)\quad (89,40,1521)\quad (761,380,1521)\\
 (841,420,1681)\quad  (925,462,1849)\quad (53,14,2025)\quad (1013,506,2025)\quad (1105,552,2209)\\
 (1201,600,2401)\quad  (149,70,2601)\quad (1301,650,2601)\quad $$

This procedure also works for $n=(4x),x>1$ but the results are not primitive.


问:
Value that makes the piecewise function continuous
Question: Given the function $f:(0, \infty) \to \mathbb R$ defined by

$$f(x) = \left\{
        \begin{array}{ll}
            \int^{x^2}_x \frac{dt}{ln(t)} & \quad x \neq 1 \\
            l                             & \quad x = 1
        \end{array}
    \right.
$$

determine the value of $l$ that makes $f$ continuous at $1$. For this value of $l$, is the function also differentiable at $1$?

I wanted to evaluate the limit of $f(x)$ when $x$ approaches to $1^+$ and $1^-$. But I think $\frac{1}{ln(x)}$ does not have a nice indefinite integral, so I couldn't evaluate the limit. Any suggestions?

答:
Note that $\ln(x)\approx x-1$ for $x$ close to $1$. Then:

$$\lim_{x\to 1^{-}}\int_{x}^{x^{2}}\frac{dt}{t-1}=\lim_{x\to 1^{-}}\ln\vert t-1\vert\bigg\vert_{x}^{x^{2}}=\lim_{x\to 1^{-}}\ln\vert x^{2} - 1\vert -\ln\vert x - 1\vert=\lim_{x\to 1^{-}}\ln\bigg\vert\frac{x^{2}-1}{x-1}\bigg\vert$$
$$=\lim_{x\to 1^{-}}\ln\vert x + 1\vert = \ln\vert 1 + 1\vert = \ln2$$

And similarly for the limit from the positive side. Thus, $\boxed{l = \ln 2.}$

Now, we will analyze differentiability at $x=1$. Since we will be taking one-sided limits as approaching $1$, we will only need to worry about the integral piece of the function. Letting $F(x)$ be a function such that $\frac{d}{dx}F(x) = \frac{1}{\ln x}$:

$$\frac{d}{dx}\int_{x}^{x^{2}}\frac{dt}{\ln t}=\frac{d}{dx}\big(F(x^{2}) - F(x)\big)$$

By the Chain Rule:

$$=\frac{2x}{\ln x^{2}} - \frac{1}{\ln x}$$

Log properties:

$$=\frac{2x}{2\ln x}-\frac{1}{\ln x}=\frac{x - 1}{\ln(x)}$$

Now we need to find the one-sided limits of the derivative approaching $1$. We have:

$$\lim_{x\to 1^{-}}\frac{x-1}{\ln(x)}$$

Using l'Hopital:

$$\lim_{x\to 1^{-}}\frac{1}{\frac{1}{x}}=\lim_{x\to 1^{-1}}x = 1$$

Once again, we obtain the same answer for the other one-sided limit. And since we have already "patched" the function to ensure continuity, the equality and existence of these two limits is enough for us to say that $\boxed{\text{The function is differentiable at }x = 1.}$


问:
Every irreducible character of $G$ is an irreducible character of $H$?
Let $H$ be a proper subgroup of $G$ such that for all $\chi\in Irr(G)$, $\chi_H\in Irr(H)$.

That is, the restriction of every irreducible character of $G$ to $H$ is an irreducible character of $H$.

What can be said in that case ? Is there any theorem related to this case ?

答:
Theorem Let $G$ be a finite group with a proper subgroup $H$, such that all irreducible characters of $G$ restrict irreducibly to $H$. Then the following hold.

$(a)$ $H$ is a normal subgroup and $G/H$ is abelian

$(b)$ Every irreducible character of $H$ is $G$-invariant

$(c)$ $k(G)=|G:H|k(H)$ (here $k(X)$ denotes the number of conjugacy classes of a group $X$)

$(d)$ $Z(G)=C_G(H)$ and $[G,H]=H'$

$(e)$ If $H$ is a Hall-subgroup, then $H$ is a direct factor of $G$.

Proof (a) Since $H$ is proper, $(1_H)^G=1_G+\sum a_{\chi}\chi$, with $a_{\chi}$ positive integers, with at least one $\chi \in Irr(G)$. Assume $\chi \in Irr(G|1_H)$, then by Frobenius Reciprocity, $0 \lt a_{\chi}=[(1_H)^G,\chi]=[1_H,\chi_H]$. Since $\chi_H$ is irreducible we must have $\chi_H=1_H$, so $a_{\chi}=1$, $\chi$ is linear and $H \subseteq ker(\chi)$. Hence, $(1_H)^G$ is the sum of all linear characters $\lambda$ of $G$ with $H \subseteq ker(\lambda)$. This implies that $ker((1_H)^G)=core_G(H)=\bigcap \{ker(\lambda): \lambda_H=1_H\} \supseteq H$. Hence $H=core_G(H)$, that is $H$ is normal and of course $G' \subseteq \bigcap \{ker(\lambda): \lambda_H=1_H\}$, since each $\lambda$ here is linear.

(b) This follows from (a) and Clifford Theory (see for example (6.2)Theorem in Isaacs' CTFG).

(c) We can count the irreducible characters of $H$ and $G$, applying Gallagher's Correspondence Theorem (see (6.17)Corollary in CTFG): let $\chi \in Irr(G)$ and $\chi_H=\vartheta \in Irr(H)$ then $Irr(G|\vartheta)=\{\lambda\chi: \lambda \in Irr(G/H)\}$. Since $G/H$ is abelian, this yields $\#Irr(G)=|G/H|\cdot \#Irr(H)$.

(d) It is easy to prove (using Schur's Lemma) that if $\chi \in Irr(G)$ with $\chi_H \in Irr(H)$, then $C_G(H) \subseteq Z(\chi)$. Since $\bigcap_{\chi \in Irr(G)} Z(\chi)=Z(G)$ and obviously $Z(G) \subseteq C_G(H)$, the assertion follows. For the second statement remember that every linear $\lambda \in Irr(H)$ is $G$-invariant by (b). So for all $g \in G, h \in H$ we have $\lambda(ghg^{-1}h^{-1})=\lambda^g(h)\lambda(h^{-1})=\lambda(h)\lambda(h^{-1})=1$. Hence $[G,H] \subseteq ker(\lambda)$. Since $H'=\bigcap \{ker(\lambda):\lambda \in Irr(H), \lambda(1)=1\}$ and of course $H' \subseteq [G,H]$, we are done.

(e) Assume that $H$ is a Hall-subgroup. Then owing to the Schur-Zassenhaus Theorem, $H$ has a complement $A$, with $G=HA$ and $H \cap A=1$. Note that $A$ is abelian. Let $P \in Syl_p(A)$, then by (a) $P$ acts (coprimely) on $H$ by conjugation. By (b) $P$ fixes every element of $Irr(H)$, and hence we can apply Brauer's Permutation Lemma (see (6.32)Theorem in CTFG) and it follows that $P$ fixes every $H$-conjugacy class. Since for every $h \in H$ gcd$(|Cl_H(h)|,p)=1$, $Cl_H(h)$ must contain a fixed point, so an element centralized by $P$. Or otheriwse put $Cl_H(h) \cap C_H(P) \neq \emptyset$. For every conjugacy class $C$ of $H$ pick an $h_C$ fixed by $P$ and put $K=\langle h_C: C \text{ is an } H-\text{conjugacy class} \rangle \leq H$. Then obvioulsy $H=\bigcup_{h \in H}K^h$ and a classical result implies that in fact $H=K$. Hence $H$ is centralized by $P$ and since $A$ is generated by its Sylow-subgroups, it follows that $A$ centralizes $H$, so $G=H \times A$.

Note Probably more can be said on the type of group pairs $(G,H)$ above. One thing that is worth mentioning is the "dual" (restriction of characters vs. induction) of the above property. The following was proved by Eric Kuisch (see Homogeneous Character Induction, J. Alg. 149, (1992), 454-471, Theorem(1.2))

Theorem Let $G$ be a group and $H$ a proper non-trivial subgroup. Then the following are equivalent.

$(a)$ Each non-trivial irreducible character of $H$ induces irreducibly to $G$

$(b)$ $G$ is a Frobenius group with kernel $H$.

This is quite surprising: $(b) \Rightarrow (a)$ was well-known (see for example (6.24)Theorem in CTFG), but the reverse implication was unknown at that time!


问:
Solve $\begin{cases}x^2+xy-y^2+2x+y=10\\3x^2-2xy+2y^2-5x-2y=22\end{cases}$
Solve $$\begin{cases}x^2+xy-y^2+2x+y=10\\3x^2-2xy+2y^2-5x-2y=22\end{cases}.$$

I am not sure how to approach the problem. I tried mulptiplying the first equation by $2,$ then I added the equation to the first one but it seemed useless at the end. Can you give me a hint?

答:
You were right to multiply the first equation by two. Now adding them together, we get: $$5x^2-x-42=0$$
which you can solve by the quadratic equation: $$x=\frac{1\pm\sqrt {1+4\cdot5\cdot42}}{10}=\frac{1\pm 29}{10}=3, -2.8$$
Then just force through the first equation to find each $y$


Body: $$\left(\frac{2x}{\sqrt{x^2 + 1}}\right)'=\frac{(2x)'(\sqrt{x^2 + 1})-(\sqrt{x^2 + 1})'2x}{(\sqrt{x^2 + 1})^2}.$$


问:
Why can't my graphing calculator find the RREF of the transpose of a matrix?
I know this is somewhat of an odd question, but I am having trouble with my TI-84 calculator and I don't know why.

I'm trying to find the RREF of the transpose of a $4\times6$ matrix; for some reason my graphing calculator gives me an error. Something to do with the dimensions? Here is a photo of matrix $A$.

I want to find RREF$(A$ transposed$)$.

答:
To compute a unique solution for a system of equations you need the same number of equations as unknowns.  In other words if you have 4 variables (unknowns) you need only 4 equations.

Adding more equations than variables creates what is called an "over determined" system of equations.  In most "math class" type problems, all you need to do is drop 2 equations and solve using the remaining 4 equations.  It does not matter which two equations you drop (although common sense would say to drop the more complex equations).

There is a field of "applied" math in which we intentionally write more equations than variables.  This is usually done when the numerical data contains small measurement errors.  Using an "over determined" system of equations allows you to compute the most statistically likely answer for each variable.


问:
Substitution constraints in Lambda Calculus
I came across this definition of Substitution in Lambda Calculus and I am trying to wrap my head around it.$$(λy.P)[x:=N] \equiv λy.(P\space[x:=N])$$ given two constraints : $y \notin FV(N)\space$ or  $\space  x \notin FV(P)$

I fully understand the first constraint as if there is a free variable $y$ in the context of $N$ it should remain free even after the substitution but in the original expression $y$ is bound.

However I struggle with the second one as the whole point of $P\space[x:=N]$ is to substitute the free occurrences of $x$ in $P$ with the expression $N$ so $x \in FV(P)$ is something that could be occasionally true.

So I guess this is either a typo or I am missing something.

答:
The second constraint means: If there are no free occurrences of the variable $x$ in $P$ to substitute for, then nothing is going to be substituted anyway, and we don't need to worry about anything being captured that shouldn't.

If both $y \in FV(N)$ and $x \in FV(P)$, then the clause is not applicable, and bound renaming needs to be performed.


问:
Find the matrix representation $[T]_{\epsilon}^{\epsilon}$ with respect to $\epsilon$
I'm preparing for a preliminary graduate entrance exam, and i'm going over an old test. The question is worded exactly as follows, I am not leaving anything out. I am a bit confused as to what they are asking and (what is the matrix representation of $[T]_{\epsilon}^{\epsilon}$ with respect to $\epsilon$?), also, how to do it. Could somebody help me out? Thank you!

Consider the basis $\epsilon$ for $V=M_{2 \times 2}(\mathbb{R})$ with the following basis vectors:

$$ e_1 = \left( \begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix} \right) $$

$$ e_2 = \left( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right) $$

$$ e_3 = \left( \begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix} \right) $$

$$ e_4 = \left( \begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix} \right) $$

And consider the linear transformation $T: V \rightarrow V$:

$T(\left( \begin{matrix} a & b \\ c & d \end{matrix} \right)) = \left( \begin{matrix} 2a-2b & -a+3b \\ 4c-2d & 3c-d \end{matrix} \right)  $

Find the matrix representation $[T]_{\epsilon}^{\epsilon}$ with respect to $\epsilon$

答:
$[T]_\epsilon^\epsilon$ means that the input and output vectors are in $\epsilon$ basis (here the standard basis).

Represent the $2\times2$ real matrix $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ by the $4\times1$ column vector $\begin{bmatrix}a\\b\\c\\d\end{bmatrix}$.

So we are given $T\begin{bmatrix}a\\b\\c\\d\end{bmatrix}=\begin{bmatrix}2a-2b\\-a+3b\\4c-2d\\3c-d\end{bmatrix}$.

Can you figure out the matrix of the linear transformation now?

 $M_T=\begin{bmatrix}2&-2&0&0\\-1&3&0&0\\0&0&4&-2\\0&0&3&-1\end{bmatrix}$


问:
Pullback of an $n$-form by the inclusion map.
Let $\Omega = dx^1 \wedge \ldots \wedge dx^n$ an $n$-form on $\mathbb R^n$ and
$$\alpha = \sum_{j = 1}^n (-1)^{j-1}x^j dx^1 \wedge \ldots \wedge dx^{j-1} \wedge dx^{j+1}\wedge \ldots \wedge dx^n.$$
In order to prove Brouwer fixed point, I've to show that for every smooth function $F: B^n \to \mathbb S^{n-1}$, with $B^n$ the unit ball in $\mathbb R^n$, such that $F(x) = x$ for all $x \in \mathbb S^{n-1} \subset B^n$, we have $d(F^* \circ \iota^*)(\alpha) = 0$ where $\iota: \mathbb S^n \to \mathbb R^n$ is the inclusion map. Since the exterior derivative commute with the pullback, I find
$$d(F^* \circ \iota^*)(\alpha) = (F^* \circ \iota^*)(d\alpha)= n (F^* \circ \iota^*)(\Omega).$$
Now here is my problem: As $\Omega$ is an $n$-form on $\mathbb R^n$, $\iota ^* \Omega$ is an $n$-form on $\mathbb S^{n-1}$ which is zero because the dimension $\mathbb S^{n-1}$ is $n-1$.

Is this argument correct ? It seems too easy, I didn't use anywhere the fact $F|_{\mathbb S^{n-1}} = \text{Id}_{\mathbb S^{n-1}}$. More generally, if we've $N \subset M$ with $\dim N = n < \dim M = m$, does the pullback by the inclusion map of a $k$-form (for $k > n$) always vanishes ?

答:
You use the fact that $F\big|_{S^{n-1}} = \text{id}$ when you finish the proof and apply Stokes's Theorem. $\displaystyle\int_{S^{n-1}} \alpha\ne 0$ by inspection, and yet
$$\int_{S^{n-1}} \alpha = \int_{S^{n-1}} F^*\alpha = \int_{D^n} dF^*\alpha = \int_{D^n} F^*(d\alpha) = 0,$$
because of your observation. (I've omitted the $\iota^*$'s, but you can put them in.)


问:
$\lim_{ x \to 0^+ } \frac{(1+(\sin x)^2)^{\frac{1}{4}}-1}{((1+\sin x)^{(\frac{-1}{x})}-e^{-1}) \ln (1+ \sqrt{1-e^{-x^2}})}$
I have to evaluate this limit:
$$\lim_{ x \to 0^+ }  \frac{(1+(\sin x)^2)^{\frac{1}{4}}-1}{((1+\sin x)^{(\frac{-1}{x})}-e^{-1}) \ln (1+ \sqrt{1-e^{-x^2}})}$$

Applying the L'Hôpital rule. I'm really lost in the calculation and the only simplification that I managed to do is: $$ \ln (1+ \sqrt{1-e^{-x^2}}) \approx \sqrt{1-e^{-x^2}}$$
Are there any other simplifications that I can apply? In my book the suggested solution is $\frac{e}{2}$ or ~$1.359$

答:
Here, in my opinion, it is better to use Taylor's series. However, if you are going to use L'Hopital's rule, I suggest to break the problem into three more manageable chunks:
$$\begin{align*}
&\lim_{ x \to 0^+ }  \frac{(1+(\sin x)^2)^{\frac{1}{4}}-1}{x^2}=L_1,\\
&\lim_{ x \to 0^+ }  \frac{\ln (1+ \sqrt{1-e^{-x^2}})}{x}=L_2,\\
&\lim_{ x \to 0^+ }  \frac{(1+\sin x)^{(\frac{-1}{x})}-e^{-1} }{x}=\lim_{ x \to 0^+ }
\frac{\exp(-\frac{\ln(1+\sin x)}{x})-e^{-1} }{x}=L_3.
\end{align*}$$
Then the desired limit is $\displaystyle\frac{L_1}{L_2L_3}$.


答:
There's a good explanation on page 30 here: https://linear.axler.net/LinearAbridged.pdf (page 36 in the PDF, definition 3.32).

Suppose you are given finite dimensional vector spaces $V, W$, a linear map $T : V \to W$, and ordered bases $\alpha =\{v_1,\dots,v_n\}$ for $V$ and $\beta = \{w_1,\dots,w_m\}$ for $W$. Then the the matrix $[T]^\alpha_\beta$ is defined by

$$[T]^\alpha_\beta = \begin{pmatrix} [Tv_1]_\beta & \cdots & [Tv_n]_\beta \end{pmatrix}$$

where the columns are defined by

$$[x]_\beta = \begin{pmatrix} x_1 \\ \vdots \\ x_m \end{pmatrix} \iff x = \sum_{i=1}^m x_iw_i.$$

Thus to say the $(i,j)$-th entry of $[T]^\alpha_\beta$ is $A_{i,j}$ is to say
$$Tv_j = \sum_{i = 1}^m A_{i,j}w_i.$$

Body:
I'm preparing for a preliminary graduate entrance exam, and i'm going over an old test. The question is worded exactly as follows, I am not leaving anything out. I am a bit confused as to what they are asking and (what is the matrix representation of $[T]_{\epsilon}^{\epsilon}$ with respect to $\epsilon$?), also, how to do it. Could somebody help me out? Thank you!

Consider the basis $\epsilon$ for $V=M_{2 \times 2}(\mathbb{R})$ with the following basis vectors:

$$ e_1 = \left( \begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix} \right) $$

$$ e_2 = \left( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right) $$

$$ e_3 = \left( \begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix} \right) $$

$$ e_4 = \left( \begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix} \right) $$

And consider the linear transformation $T: V \rightarrow V$:

$T(\left( \begin{matrix} a & b \\ c & d \end{matrix} \right)) = \left( \begin{matrix} 2a-2b & -a+3b \\ 4c-2d & 3c-d \end{matrix} \right)  $

Find the matrix representation $[T]_{\epsilon}^{\epsilon}$ with respect to $\epsilon$


问:
Does such sequence exist?
In analysis course, my professor use a special sequence (in his notes) that I will mention below to prove a theorem.  I can't find any example of such a sequence. I will not bother to mention the full theorem (because I found another proof for it on the internet), but I only want to know does such a sequence $(a_n)$ with below properties exist in real numbers?

it should be bounded

$\limsup (a_n)=\sup_{n}a_n=1$

$\liminf (a_n)=\inf_{n}a_n=-1$

set of its sequential limit have exactly four elements.

I will appreciate if someone gives me an example.

Title: Does such sequence exist?

Body: How about $-1,0,\frac12, 1, -1,0,\frac12,1,...$?


问:
Let $G=\langle \mathbb R \times \mathbb R, +\rangle$ and $H=\{(x,y): y=2x\}$. Show that $H$ is a subgroup of $G$.

Let $G=\langle\mathbb R \times \mathbb R, +\rangle$ and $H=\{(x,y): y=2x\}$. Show that $H$ is a subgroup of $G$.

I'm trying to show that $H$ is closed and contains inverses (individually). So the ordered pairs have the form $(x,2x)$. Let $(x,y)$ and $(z,w)$ be in $H$. Then

$$\begin{align}
(x,y)+(z,w)&=(x,2x)+(z,2z)\\
&=(x+z,2x+2z)\\
&=(x+z,2(x+z)).
\end{align}$$ Thus $H$ is closed. Now, if $(x,y)=(x,2x)\in H$, then $(-x,-y)=(-x,-2x)$ implies that their sum is $(0,0)$ which is the identity element. Thus $(-x,-y)\in H$ and $H$ is a subgroup of $G$.

Is this correct? Could I also have consider ordered pairs of the form $({1\over 2}y,2x)$ or just consider two of the same elements, $(x,y)$ and done the proof that way? How do I know that I'm setting the problem up right in the first place. I feel like I don't really understand it and I'm just getting lucky sometimes. Any other solutions, hints, advice is greatly appreciated.

答:
Pairs of the form $(\frac 12 y, 2x)$ don't belong to $H$ in general; I think you meant $(\frac y2, y)$. Of course, this could have worked exactly the same way because $y=2x$ is equivalent to $x=\frac y2$.

But more about the proof technique itself: when proving that $H$ is closed under the addition operation, that was fine, but when trying to prove that $H$ is closed under taking inverses, you don't have to calculate what the inverse is if you already know how to compute inverses in the original group (in this case $\mathbb R \times \mathbb R$); you just have to show that the inverse is in $H$. So an equality as simple as
$$
-(x,2x) = (-x, -2x) = (-x, 2(-x)) \in H
$$
would suffice to show that $H$ is closed under taking inverses.

Hope that helps,


问:
1st derivative of $\frac{2x}{\sqrt{x^2 + 1}}$
Another simple question that I can't work out today, yet I would work it out two weeks ago!

I need to find the 1st derivative of $$\frac{2x}{\sqrt{x^2 + 1}}$$.

So I use the Quotient rule and I get: $$\frac{(x^2 + 1)^.5 (2) - (2x)(0.5x^2 + 0.5)^-5}{x^2+1}$$

Am I heading in the correct direction and do I just need to multiply and try to get rid of the exponents somehow?

Thanks

答:
Without messing up with nominator and denominators:

$x=\tan { u } $ then draw right triangle, see

$\\ \frac { 1 }{ \sqrt { x^{ 2 }+1 }  } =\cos { u } \\$

$\\ \frac { x }{ \sqrt { x^{ 2 }+1 }  } =\sin { u } \\$

$ \frac { du }{ dx } =\frac { 1 }{ x^{ 2 }+1 } \\$

$ f=\frac { 2x }{ \sqrt { x^{ 2 }+1 }  } =2\sin { u } \\$

$$ f'\quad =2\cos { u } \frac { du }{ dx } =\frac { 2 }{ \sqrt { x^{ 2 }+1 }  } \frac { du }{ dx } =\frac { 2 }{ \sqrt { x^{ 2 }+1 }  } \frac { 1 }{ x^{ 2 }+1 } $$


问:
Fubini's Theorem and Stochastic (Ito) Integrals
Edit: in the comments, it is pointed out by @Mag that Ito Integral can - in fact - be defined pathwise, rather than an L2 limit. A bit of googling has led me to these Oxford university notes here for example. My question: would it be after all possible to apply Fubini's theorem to Ito Integral, defined pathwise?

Original question: I wonder whether Fubini's theorem is applicable to taking the expectation of Ito integrals. It is a well known fact that Ito integral posesses the "martingale property", i.e. for any deterministic or stochastic process $H(\omega,t)$, we have:

$$\mathbb{E}\left[\int_{h=0}^{h=t}H(\omega,h)dW_h\right]=0$$

From the above, it is obvious that for many types of integrands, Fubini's theorem "doesn't work", for example, we have that $\mathbb{E}[W_t^2]=t$, but:

$$\mathbb{E}\left[\int_{h=0}^{h=t}W_h^2dW_h\right]=0\neq \int_{h=0}^{h=t}\mathbb{E}\left[W_h^2\right]dW_h =\int_{h=0}^{h=t}hdW_h$$

Question: so what is it that makes Fubini's theorem not applicable above? The usual two conditions for being able to apply Fubini are:

(i) $\mathbb{E}[|H(\omega,t)|]<\infty$

(ii) $H(\omega,t)$ is measurable

In case of $H(\omega,t):=W_t^2$, both conditions are satisfied: (i) clearly, $\mathbb{E}[|W_t^2|]<\infty \forall t \in \mathbb{R}$ and (ii) $W_t$ is adapted and measurable (with respect to the filtration it generates and if the entire probability space is generated by $W_t$, then indeed $W_t$ is "measurable" with respect to $\left(\Omega,\mathbb{P},\mathcal{F}_t\right)$).

答:
In Fubini's theorem we have two measures and two Lebesgue integrals. Here you have one Lebesgue integral (i.e the expectation) and an Ito integral. The Ito integral is not defined pathwise but as a $L^2$ limit. Also note that $dW_t$ is not a measure.

Also, it's not always true that
$$\mathbb{E}\left[\int_0 ^t H_udW_u\right]=0. \tag{1}$$

The equality (1) is true when $\mathbb{E}\left[\int_0^t H^2_u du\right]<\infty$


问:
Find the intersection of an angle bisector and the opposite side
Given a triangle with vertices $A$, $B$ and $C$ whose coordinates are $(x_1, y_1,z_1)$, $(x_2, y_2,z_2)$ and $(x_3, y_3,z_3)$ respectively, how do I determine the coordinates of $D$ such that $AD$ bisects $\angle$$BAC$?

The first idea that came to my mind is as follows:

Find the equation of the straight line for the line segment $AD$ that bisects  $\angle$$BAC$.

Find the equation of the straight line that passes through points $B$ and $C$.

Find the intersecting point of these two straight lines.

Is there an alternative method where I can find the coordinates of $D$ directly or just by using the coordinates of $A$, $B$ and $C$? If not, what is the  efficient way (computationally) to determine the coordinates of $D$?

Any suggestions or comments are kindly appreciated!

EDIT

It may be true (as suggested by jjagmath and Derive Foiler) that using barycentric instead of cartesian coordinates is elegant but I forgot to mention that what I am looking for is a cartesian solution (=

答:
Alternatively, using the Angle Bisector Theorem, we have $BA:AC = BD:DC$.

The lengths of $BA$ and $AC$ are easy to find; finally use the section formula on $BC$.


问:
Unable to reconstruct a new polynomial to find a given value
Suppose $f(x)= x^3+2x^2+3x+3$ and has roots $a , b ,c$.
Then find the value of
$\left(\frac{a}{a+1}\right)^{3}+\left(\frac{b}{b+1}\right)^{3}+\left(\frac{c}{c+1}\right)^{3}$.

My Approach :
I constructed a new polynomial $g(x) = f\left(\frac{x^{\frac{1}{3}}}{1-x^{\frac{1}{3}}}\right)$ and then used the Vieta's formula for the sum of roots taken one at a time to solve the sum.

But then I realised that I won't be able to do so as $g(x)$ is not a polynomial anymore.
Can anyone help me please.

答:
It has some boring calculations so I'll just write about the sketch of my solution:
Using Vietta formulas, find the coefficients of the polynomial that has roots ${1-\frac{1}{a+1},1-\frac{1}{b+1},1-\frac{1}{c+1}}$. Then just use the Newton formula to find the sum of cubes of that polynomial


答:
Whoops, this is an elaboration on the other answer.

Here's one way this can be done. Thinking about these like vectors, if we let $t:=\frac{BD}{CD},$ we can say
$$D=tB+(1-t)C.\tag{$*$}$$
We can compute $t$ by the angle bisector theorem, saying
$$t=\frac{AB}{AC}.$$
We can compute the lengths $AB$ and $AC$ from the coordinates, from which $(*)$ will finish.

By the way, this is a bit more elegant with barycentric coordinates instead of Cartesian ones.

Title: Find the intersection of an angle bisector and the opposite side

Body:
Given a triangle with vertices $A$, $B$ and $C$ whose coordinates are $(x_1, y_1,z_1)$, $(x_2, y_2,z_2)$ and $(x_3, y_3,z_3)$ respectively, how do I determine the coordinates of $D$ such that $AD$ bisects $\angle$$BAC$?

The first idea that came to my mind is as follows:

Find the equation of the straight line for the line segment $AD$ that bisects  $\angle$$BAC$.

Find the equation of the straight line that passes through points $B$ and $C$.

Find the intersecting point of these two straight lines.

Is there an alternative method where I can find the coordinates of $D$ directly or just by using the coordinates of $A$, $B$ and $C$? If not, what is the  efficient way (computationally) to determine the coordinates of $D$?

Any suggestions or comments are kindly appreciated!

EDIT

It may be true (as suggested by jjagmath and Derive Foiler) that using barycentric instead of cartesian coordinates is elegant but I forgot to mention that what I am looking for is a cartesian solution (=


问:
Show that exist $p\in \mathbb{R} $ such that $p^3+p=1$
Show that there exists $p\in \mathbb{R} $ such that $p^3+p=1$.

How can I prove it? I was thinking in what can I do to prove it but I really stuck .

I just see supreme axiom, Archimedean property and sequences.

Any hint of what I can do?

答:
Let $f(p) = p^{3} + p - 1$. Then $f$ is continuous because it is a polynomial.

Moreover, we also have that $f(0) = -1$ and $f(1) = 1$.

Given that $f(0) < 0 < f(1)$, we conclude (through the IVT) there exists $c\in[0,1]$ such that $f(c) = 0$.

Hopefully this helps!


问:
Probability of two random n-digit numbers dividing each other
Let $n$ be a positive integer. Suppose $a$ and $b$ are randomly (and independently) chosen two $n$-digit positive integers which consist of digits 1, 2, 3, ..., 9. (So in particular neither $a$ nor $b$ contains digit 0; I am adding this condition so that division by $b$ will be possible, and that we don't get numbers of the form $0002$ and so on). Here "randomly" means each digit of $a$ and $b$ is equally likely to be one of the 9 digits from $\{1,2,3,..., 9\}$.

My question concerns the divisibility of these integers:

1) What is the probability that $b$ divides $a$ ?

The answer, of course, will depend on $n$. Denote this probability by $p(n)$. I would be happy with rough estimates for $p(n)$ as well :)

2) Is it true that $p(n)\to 0$ as $n\to\infty$?

I think answer to question 2) is yes (just by intuition).

答:
As $a,b$ are $n$-digit numbers, we have $10^{n-1}\le a,b<10^n$. If $b|a$, this implies $\frac ab\in\{1,2,\ldots,9\}$. So at most $9$ of the $9^n$ possible $b$ are divisors of $a$. $p(n)\le \frac 9{9^n}$.
While this estimate is far from sharp, it shows $p(n)\to0$ as $n\to\infty$.

Can we calculate $p(n)$ more precisely? Your forbidding zeroes is partly making this more difficult as divisibility of $a$ by small numbers $\in\{1,2,\ldots,9\}$ becomes less easily predictable.
We need to calculate, for $d\in\{1,\ldots,9\}$, the probability that $a$ is both a multiple of $d$ and is $\ge d\cdot 111\ldots1$ (where $111\ldots1$ is the smallest $n$-digit number without zeroes) and $\frac ad$ has no zeroes.
The probability that $a\ge d\cdot 111\ldots 1$ is approximately(!) $\frac{9-d}8$.
The probability that $a$ is divisible by $d$ is approximately(!) $\frac1d$ and is almost(!) independant of whether or not $a$ is $\ge d\cdot 111\ldots1$.
The probability that, if $a$ is indeed divisible by $d$, the number $\frac ad$ has no zeroes is at most $1$; in fact for $d>1$ it looks like it decreases $\to0$ as $n\to\infty$. Therefore we estimate the expected number of zero-free $n$-digit divisors of $a$ as approximately (and likely less than)
$$\sum_{d=1}^9 \frac{9-d}{8}\cdot\frac1d=\frac{4609}{2240}\approx 2.06$$
and hence get $$\tag1p(n)\approx \frac{2.06}{9^n}.$$
For small $n$, this estimate as quite a bit off. For example, it gives $p(1)\approx0.2286$ instead of the correct value $p(1)=\frac{23}{81}=0.28395\ldots$, next $p(2)\approx 0.0254$ instead of $p(2)=\frac{163}{6561}=0.02484\ldots$, and next $p(3)\approx 0.00282$ instead of $p(3)=\frac{483}{177147}=0.00261\ldots$. However, we observe already here (though only empirically!) that the estimate $(1)$ is an upper bound for $n\ge 2$.


问:
How does $\int dz \,d\bar z$ evaluate to $\int\int dx \,dy$?
While asking us to prove this relation:

$$\int dz\,d\bar{z} e^{-\bar{z}wz} = \frac{\pi}{w}$$

our professor gave us this statement:

In the above equation the measure $\int dz\,d\bar{z}$ actually stands for $\int dR[z]dI[z] = \int\int dxdy$, the limits running from −∞ to +∞.

Given this statement, I find that proving the relation is quite easy. However, I am unable to evaluate $\int dz\,d\bar{z}$ to obtain $\int\int dxdy$. Any pointers in understanding this would be of great help.

Edit:
I am assuming that $w$ is a real positive integer.

Thanks in advance..

答:
Your professor is off by a factor of $i/2$. What's actually going on is differential forms. We have $z=x+iy$ and $\bar z=x-iy$, so
\begin{align*}
dz &= dx + i\,dy \\
d\bar z &= dx - i\,dy,
\end{align*}
from which we get
$$dz\wedge d\bar z = (dx+i\,dy)\wedge (dx-i\,dy) = -2i dx\wedge dy.$$
(Remember that $\wedge$ is anti-commutative, so $dx\wedge dx = dy\wedge dy = 0$ and $dx\wedge dy = -dy\wedge dx$.)
You can see that this "fudge factor" is the determinant of the Jacobian matrix of the change of coordinates. At any rate, the correct formula is
$$\tfrac i2 dz\, d\bar z = dx\, dy.$$


问:
Does there exist a circle with EXACTLY two rational points, no more, no less?
Topic title says it all, I think. Does there exist a circle with EXACTLY two rational points, no more, no less? I know that as soon as it has three rational points, its center must be rational too, therefore an isomorphism with the unit circle yields infintely many (dense) rational points. I have examples with only one and zero rational points, but I'm wondering about the case 2.

I already know that the center must be irrational, because else one rational point on the circle would suffice to yield infinitely many points.

答:
From cut-the-knot.org:

there are at most two rational points on a circle with irrational
center. Before proving the result, let's recollect a few fundamental
facts.

For any pair of rational points, the straight line passing through the
two may be expressed by an equation with rational (and, hence,
integer) coefficients. The midpoint of the segment joining two
rational points is itself a rational point. A line perpendicular to a
line with slope $\alpha$ has a slope $-1/\alpha$, implying that the slopes of the
two lines are either both rational or both irrational. Therefore, the
perpendicular bisector of a segment joining two rational points always
has an equation with rational coefficients. The solution to a system
of two linear equations with rational coefficients, if exists, is
necessarily rational. Which is to say, that if two straight lines that
are expressed by linear equations with rational coefficients
intersect, their point of intersection is necessarily rational.

With these preliminaries, we are ready to confront the problem. Assume
to the contrary there is a circle with irrational center on which
there are three distinct rational points, say, $P, Q, R$. The
perpendicular bisectors of $PQ$ and $QR$ would meet at the center of the
circle which, from the foregoing discussion, would be a rational point
in contradiction with the conditions of the problem.

So, choose two rational points $P,Q$ and an irrational point $C$ on their perpendicular bisector.  The circle with center $C$ that goes through $P,Q$ answers your question.


答: If $a$, $b$ and $c$ denote the lengths of the sides $BC$, $CA$ and $AB$, then $D$ has barycentric coordinates $(0:b:c)$ in the triangle $ABC$. This means that $D = \frac{b}{b+c} B + \frac{c}{b+c} C$ as vectors.


问:
$L^0( \mathbb P | \mathcal F_0 ; || \cdot ||_{\mathbb R^d} ) \subset L^0( \mathbb P | \mathcal F_1 ; || \cdot ||_{\mathbb R^d} ) $?
As stated in the title,

How to prove that $$L^0( \mathbb P | \mathcal F_0 ; || \cdot ||_{\mathbb R^d} ) \subset  L^0( \mathbb P | \mathcal F_1 ; || \cdot ||_{\mathbb R^d} ) $$

where the space $ L^0( \mathbb P | \mathcal F_0 ; || \cdot ||_{\mathbb R^d} ) $ is given by the equivalence class of all measurable random variables, given the probability measure $\mathbb P$ restricted to the filtration $\mathcal F_t $. Apparently, it should be enough to prove:

$(\Omega, \mathcal F)$, $(S, \mathcal S)$, two measurable space, $\mathcal F_t$ a filtration on $\mathcal F$,

$X \colon \Omega \to S$, an $\mathcal F, \mathcal S $ measurable function, $Y \colon \Omega \to S$, an $\mathcal F_t, \mathcal S$ measurable function, and such that $\exists A \in \mathcal F \colon \mathbb P(A) = 1$ as well as $A \subset \{ X \equiv Y \}$. Then, it implies $ X$ is a  $\mathcal F_t, \mathcal S $ measurable function.

So my question is how are these two theoremes related and how to prove the former?

My attempt.

In the proof of the later, I am stuck at the step where I have to use the probability measure. Indeed, I see that the preimage of $X$ should lie on the filtration, the problem is that I am not able to express the idea that if the preimage is on $\mathcal F \setminus F_t$ then the image has measure $0$.

答:
after discussing it with other people, the whole point comes down to using the stochastic bases.

So, first, one has that the set $A$ I have given the definition above is in the filtration $\mathcal F_t$. This is because of the stochastic bases where all sets in the kernel of the measure are in the filtration. Then, one can partition $\{ X \in B \}$ upon $A$, giving the union of two sets: one is in the filtration, and the other one is of measure zero, thus in the filtration (same argument).

Using the same method, we can show the inclusion of spaces.


答:
In case your course hasn't covered IVT yet:

Consider the set $S=\{x\in\mathbb R\mid x^3+x-1<0\}$, which is non-empty (e.g. $0\in S$). As $f(x)=x^3+x-1$ is increasing, $S$ is bounded above (e.g. by $1$). So let $s=\sup S$ (which exists by Archimedean property). We claim that $f(s)=0$.

Let $\varepsilon=|f(s)|$, and suppose $\varepsilon>0$. As $f$ is continuous, we can take $\delta$ such that $\lvert f(s)-f(t)\rvert<\varepsilon$ whenever $\lvert s-t\rvert<\delta$.

If $f(s)<0$, then $f(s+\delta/2)<0$ too, contradicting that $s$ is an upper bound. If $f(s)>0$, then $f(t)>0$ whenever $s-\delta<t<s+\delta$, so $s-\delta/2$ would be a smaller upper bound, again a contradiction.

With very little effort, this proof can be generalised to prove the Intermediate Value Theorem.

Title: Show that exist $p\in \mathbb{R} $ such that $p^3+p=1$

Body:
Show that there exists $p\in \mathbb{R} $ such that $p^3+p=1$.

How can I prove it? I was thinking in what can I do to prove it but I really stuck .

I just see supreme axiom, Archimedean property and sequences.

Any hint of what I can do?


问:
Must there be an invertible matrix $Q$ such that $Q^{-1}MQ=N$?
Let $M$ be a complex square matrix such that it has the following:

Minimal polynomial: $\mu(\lambda)= (\lambda-3)^2(\lambda+2)^3(\lambda-5)$

Characteristic polynomial: $\chi(\lambda)=(\lambda-3)^4(\lambda+2)^5(\lambda-5)^2$

$\dim(E_3)=2$ and $\dim(E_{-2})=3$, where $E_\lambda$ is the eigenspace for $\lambda \in \mathbb{R}\,$.

Now, my question is this:

If $N$ is another complex square matrix with the same characteristic polynomial, minimal polynomial, and geometric multiplicities as stated above, must there be an invertible matrix $Q$ such that $Q^{-1}MQ=N$ ?

If this is true, I would like to understand why. And if not, a counter-example would be much appreciated.

答:
All of the information that you have given can be interpreted as a description of the Jordan normal form of $M$.  In particular:

Characteristic polynomial:

$M$ has size $\deg(\chi) = 11$

The sum of the sizes of the Jordan blocks associated with $\lambda = 3$ (i.e. the number of times $3$ appears on the diagonal) is $4$. This sum is $5$ for $\lambda = -2$ and $2$ for $\lambda = 5$.

Minimal polynomial:

The largest block associated with $\lambda = 3$ has size $2$. Respectively, $3$ for $\lambda = -2$ and $1$ for $\lambda = 5$.

Algebraic multiplicity:

There are $2$ Jordan blocks associated with $\lambda = 3$ and $3$ for $\lambda = -2$.

Let $J_k(\lambda)$ denote the Jordan block of size $k$ associated with $\lambda$, and let $\oplus$ denote a direct sum.  Verify that with the information given above, the Jordan form of $M$ must be (up to a permutation of the blocks)

$$
J = [J_2(3) \oplus J_2(3)] \oplus [J_3(-2) \oplus J_1(-2) \oplus J_1(-2)] \oplus [J_1(5) \oplus J_1(5)].
$$

Two matrices are similar if and only if they have the same Jordan form. However, any matrix $N$ that has all of the listed characteristics must also have this Jordan form. It follows that for any such $N$ is similar to $M$, i.e. there is a matrix $Q$ for which $
Q^{-1}MQ = N$.


问:
Solve, $[3x + 1] = 2x - \frac{1}{2}$ , and find the sum of all roots.

Solve, $[3x + 1] = 2x - \frac{1}{2}$ , and find the sum of all roots.

What I Tried: I have :-
$$\rightarrow [3x + 1] = \frac{4x - 1}{2}$$
Now, the LHS is an integer, so $2|(4x-1)$ .
I can see that if $x$ is an integer, then $(4x - 1)$ is odd, so there are no integer solutions. I am not sure about fractions, as $x = \frac{1}{2}$ still makes it odd, but $x = \frac{1}{4}$ makes it even, although $2$ does not divided $0$.

This is where I am stuck, how should I proceed next?

Can anyone help me?

答:
Let us find  a necessary condition.

$$\lfloor 3x+1\rfloor=2x-\frac 12 \implies$$

$$\lfloor 3x\rfloor=2x-\frac 32 \implies $$

$$3x-\lfloor 3x\rfloor =x+\frac 32\implies$$

$$0\le x+\frac 32<1 \implies$$

$$-\frac 32\le x<-\frac 12\implies $$
$$-\frac 72\le 2x-\frac 12<-\frac 32\implies$$

$$2x-\frac 12\in \{-3,-2\}\implies$$
$$x\in\{-\frac 54,-\frac 34\}$$

We check this condition is sufficient.
the sum of the roots is
$$S=-\frac 54-\frac 34=-2$$