问:
Prove $ \lim_{h\rightarrow 0}\frac{1}{h}\int_a^x[f(t+h)-f(t)]\mathrm{d}t=f(x)-f(a). $

Suppose $f(x)$ is continuous on $[a, b]$. Prove that $\forall x\in (a,b)$, we have
$$
\lim_{h\rightarrow 0}\frac{1}{h}\int_a^x[f(t+h)-f(t)]\mathrm{d}t=f(x)-f(a).
$$

If $f$ id diffentiable, the conclution can be obtained by Newton-Leibniz formula easily. When $f$ is just continuous, I tried to use knowledge about integral with parameters. Write
$$
F(h,x):=\frac{1}{h}\int_a^x[f(t+h)-f(t)]\mathrm{d}t,
$$
but it seems we still need to compute $\lim_{h\rightarrow 0}\frac{f(t+h)-f(t)}{h}$, and prove uniform convergence. I have no idea how to go on.

Is my thoughts workable? If not, please suggest your way. Appreciate any help!

答:
This is a simple consequence of the Fundamental Theorem of Calculus. Also you need to have $h\to 0^{+}$ instead of $h\to 0$ (why? Explain yourself).

Clearly if $F$ is an anti-derivative of $f$ on $[a, b] $ (such an $F$ exists because of continuity of $f$ and Fundamental Theorem of Calculus) then we have $$\int_{a}^{x}f(t)\,dt=F(x)-F(a)$$ and $$\int_{a} ^{x} f(t+h) \, dt=F(x+h) - F(a+h) $$ and thus the desired limit equals the limit of expression $$\frac{F(x+h) - F(x)} {h} - \frac{F(a+h) - F(a)} {h} $$ and since $F$ is the anti-derivative of $f$ the desired limit equals $$F'(x) - F'(a) =f(x) - f(a) $$


问:
Show that the triangle $ADC$ is isosceles
Let $K$ be a circle with center $M$ and $L$ be a circle that passes through $M$ and intersects $K$ in two different points $A$ and $B$ and let $g$ be a line that goes through $B$ but not through $A$. Let $C$ be the second point of $g$ on $L$ and let $D$ be the second point of $g$ on $K$ with $C=B$ or $D=B$ if $g$ is the tangent through $B$ on the circle $K$ or $L$. Show that the triangle $ADC$ is isosceles.

For that do we have to show that that the length of each side of triangle is equal? Or how can we show that?

$$$$

EDIT:

We get the following, or not?

答:
Hint:

Let $ \angle  ADC = \alpha$.

What is $ \angle AMB$ in terms of $ \alpha$?

What is $ \angle ACB$ in terms of $ \alpha$?

What is $ \angle DAC$ in terms of $ \alpha$?

ContentLicense: CC BY-SA 4.0


问:
What's a better way to integrate this?
$$ \int \frac{1}{x^2 + z^2} dx $$

I tried substitution and also by parts. By parts is getting messy and I am not sure if I am getting the right answer. I am trying to figure out an easier way or the proper way to integrate this. Could someone please show me?

答:
You can use the substitution $u= x/z$. Then you get
$$
\frac{1}{z} \int \frac{1}{u^2 + 1} du = \frac{1}{z} \arctan\left(\frac{x}{z}\right) + C.
$$</p>

ContentLicense: CC BY-SA 3.0

Title: What's a better way to integrate this?


问:
Orthogonal endomorphisms in finite dimension
As far as I understand it, orthogonal endomorphisms in finite dimensions have an orthogonal matrix for representation only if the basis on which the matrix is created is orthonormal. Doesn't that pose a problem, as then there will exist some bases in which the orthonormal application is not orthonormal ?</p>

答:
Given a scalar product on a (finite dimensional) vector space $V$, an isometry $f:V\to V$ is a linear map preserving the distance induced by the scalar product (or, equivalently, that preserves the scalar product).

The fact that $f$ is an isometry is a base-free notion.

Now, if you fix a basis on $V$, then both the scalar product and $f$ have associated matrices. If $M$ is the matrix associated to the scalar product and $A$ is the matrix associated to $f$, then the fact that $f$ is an isometries, reads -- in COORDINATES relative to the fixed basis -- as

$$A^tMA=M$$

This holds true for ANY basis of $V$. Choosing a basis which is orthogonal w.r.t. the scalar product translates to the fact that $M$ is the identity matrix. As a consquence, the matrix of $f$ in any orthononmal basis is an orhtogonal matrix. But in general what holds true is just the above formula.


问:
An application of Weierstrass M-Test to determine sets of uniform convergence.
let $\sum_{n=1}^{\infty}fn(x) $ = $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}(\frac{x}{1+x})^n$

I am attempting to apply Weierstrass M-Test on [a,$\infty$)

I get that

$|\frac{1}{\sqrt{n}}(\frac{x}{1+x})^n|$<$|\frac{1}{\sqrt{n}}|$

but the RHS doesnt converge, so we cant use M-Test.

What other methods can I use?

答:
Let $g_m(x)=\sum_{n=1}^m \frac {1}{\sqrt n}\left(\frac {1}{1+x}\right)^n.$

If $S$ is a subset of $\Bbb R$ with no upper bound in $\Bbb R$ then $g_n$ does not converge uniformly on $S.$

Proof: We show that for any $m\in \Bbb N$ there exist $m'>m$ and $x\in S$ with $g_{m'}(x)-g_m(x)>1/(2\sqrt 2),$ as follows:

Let $m'=2m.$ Now take $x\in S$ where $x>0$ and $x$ is large enough that $(x/(1+x))^{2m}>1/2.$ Then $(x/(1+x))^n>1/2$ when $m<n\le 2m,$ so $$g_{m'}(x)-g_m(x)=g_{2m}(x)-g_n(x)=\sum_{n=m+1}^{2m}(1/\sqrt n)(x/(1+x))^n>$$ $$>\sum_{n=m+1}^{2m}(1/\sqrt n)(1/2)\ge$$ $$\ge \sum_{n=m+1}^{2m}(1/\sqrt {2m})(1/2)=$$ $$=m(1/\sqrt {2m})(1/2)=(\sqrt m)/(2\sqrt 2)\ge 1/(2\sqrt 2).$$


问:
Area between two given equations
Let $x^2 + y^2$ = 6 and $x = y^2$. What is the area between them?

I solved the two equations in terms of y and I graph it in Geogebra: https://www.geogebra.org/classic/ahnk4hfw

Solving the two equations in terms of x, the graph is:
https://www.geogebra.org/classic/nwznqk3c

When I integrate in terms of y (the graph is the first one), the bounds are -$\sqrt 2$ and $\sqrt 2$. The integrand is $ \sqrt {6-y^2} $ - $y^2$. I got 4.63 as the area.
$$\int_{-\sqrt 2}^{\sqrt 2} \sqrt {6-y^2} - y^2 \, dy = 4.63 $$
Although, I am not sure if I expressed the integral correctly because it seems that I am missing one function: the -$\sqrt {6-y^2} $.

When I follow the second graph, isn't it that the whole region is bounded by the two functions, and hence the area is the circle's area? Or should I only choose the positive square roots and not the negative ones?

My last question is, what is the best approach of finding the integral of this case? Is it by integrating with respect to y or x? Do I only need to take the smaller region?

Title: Area between two given equations

Body:
$x^2+y^2=6$ and $y^2=x$ intersect at $P_2(2,\sqrt{2}),$ and $P_1(2,-\sqrt{2})$. The area included between them is best given by
$$A=\int_{-\sqrt{2}}^{\sqrt{2}} [\sqrt{6-y^2}-y^2] dy= \left[\frac{y}{2}\sqrt{6-y^2}+3\sinh^{-1}(y/\sqrt{6})-\frac{y^3}{3}\right]_{-\sqrt{2}}^{\sqrt{2}}$$
$$A=\frac{2}{3}[\sqrt{2}+9\sin^{-1}(1/\sqrt{3})]=4.6354$$

Note:Here area projected on $y$-axis is a good idea. So $A=\int_{y_1}^{y_2}[x_1(y)-x_2(y)]dy.$


问:
Strict convexity of the norm
I'm trying to prove that Strong convexity $\Rightarrow$ strict convexity. So, given $g(x) := f(x) - \alpha\|x\|^2$ I need to verify the definition:

$$g(\lambda x + (1-\lambda)y) <\lambda g(x) + (1-\lambda)g(y).$$

So this is

$$f(\lambda x + (1-\lambda)y) -\alpha\|\lambda x + (1-\lambda)y\|^2 < \lambda f(x) + (1-\lambda)f(y) -\alpha\|\lambda x + (1-\lambda)y\|^2$$

So the first part is OK because $f$ is a convex function by definition. All I need to do now is to prove that $\|x\|^2$ is a strict convex function But in general

$$\|\lambda x + (1-\lambda)y\|^2 \le ( \lambda \|x \| + (1-\lambda)\|y\|)^2$$

But the right hand side should be $< \lambda \|x\|^2 + (1-\lambda)\|y\|^2$ and this does not hold here....

答:
To prove strict convexity of the squared norm you need to show that $\|\lambda \mathbf{x} + (1-\lambda)\mathbf{y}\|^2 < \lambda \|\mathbf{x} \|^2 + (1-\lambda)\|\mathbf{y}\|^2$. I believe this is the part that caused your confusion.

Now, rearranging the inequality above gives: $$-(\lambda(1-\lambda)\|\mathbf{x} \|^2+\lambda(1-\lambda)\|\mathbf{y} \|^2-2\lambda(1-\lambda)\mathbf{x}^T\mathbf{y})<0.$$ Last arrangement gives us $$-\lambda(1-\lambda)\|\mathbf{x-y} \|^2<0$$

which clearly holds since $0<\lambda<1$ and $\mathbf{x}\neq\mathbf{y}$.

Title: Strict convexity of the norm


问:
Find the amount of nuts and bolts produced when $P=500$
A factory simultaneously manufactures nuts and bolts. Suppose that x and y denote the amount (measured by weight in kilograms) of nuts and bolts respectively produced by the factory in a day, with $x\ge 0$ and $y \ge 0$. You may assume that both x and y are real numbers, and not necessarily whole numbers. Production is constrained so that $x^2 + y^2 = 500^2$. The daily profit $P$ is given by $P = 3x + y$.

Part A asked to show that for any fixed daily profit $P$ satisfying $500 \le P < 1500 $ there is a unique production run that produces the profit $P$.

Part B (the part I'm trying to do) asks to find the amount of nuts and bolts produced when:

(i) $P = 500$

(ii) $P = 1500$

(iii) $P = 1580$

Part C requires (iii) and Rolle's Theorem to show that the profit has a maximum value for some $x$ satisfying $468 \le x \le 480$ without the use of differentiation.

Part D asks to determine the maximum value of the profit P and the number of nuts and bolts which produce maximum profit.

答:
Part B

(i) when P = 500,

no. of nuts = 0

(ii) when P = 1500,

no. of nuts = 500

no. of bolts = 0

(iii.) When P = 1580 there are two production runs:

Production Run 1:

no. of nuts = 480

no. of bolts = 140

Production Run 2:

no. of nuts = 468

no. of bolts = 176


问:
Vague convergence - equivalence?
Are the following definitions of vague convergence equivalent?

The sequence of s.p.m $\mu_n$ vaguely converges to $\mu$ iff

(1) There exists a dense subset $D$ of the real line $\mathbb{R}$  so that $ \forall a \text{ and } b \in D \text{ with } a <b, \mu_n((a,b]) \rightarrow \mu((a,b])$.

(2) For all $f\in C_K^+ (\mathbb R)$ we have  $\int f \mu_n \rightarrow \int f \mu$.

A source connecting these two definitions is also welcome.

Proof idea: Use the fact that $L^2$ functions like indicators are dense in $C$. Also $C$ functions can be approxdimated by simple functions.

答:
It is right in fact.
The core of proof is the D-valued approximation lemma.
You can even let $f \in C_0$. And you can find the total proof in A Course in Probability Theory written by Kai Lai Chung.

Title: Vague convergence - equivalence?


答:
Whether an endomorphism is orthogonal is not basis-dependent. We say an endomorphism $T$ on a Hilbert space $V$ (e.g. a finite-dimensional inner product space) is orthogonal if $T^* T = I$, where $T^*$ is the adjoint of $T$, and $I$ is the identity map on $V$.

Now, orthogonal endomorphisms on finite-dimensional spaces do correspond to orthogonal matrices, but not as generally as some students tend to think. If $T$ is an orthogonal endomorphism and $B$ is an orthonormal basis, then the matrix $[T]_B$ of $T$, corresponding to $B$, is an orthogonal matrix. There is a converse to this too: if $B$ is orthonormal, and $[T]_B$ is orthogonal, then so is $T$.

But, the key fact here is that $B$ must be an orthonormal basis! If we take a non-orthonormal basis, we have no guarantee whatsoever that $[T]_B$ is orthogonal, even if $T$ is orthogonal. For example, take the orthogonal endomorphism:
$$T : \Bbb{R}^2 \to \Bbb{R}^2 : (x, y) \mapsto (y, -x).$$
If you're interested, this is a clockwise rotation of the plane by $\pi/2$ radians. Under the standard basis $S$, we get
$$[T]_S = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.$$
Note that the columns (and rows) form an orthonormal basis, hence the matrix is orthogonal. But, if we choose a non-orthonormal basis like
$$B = \{(1, 0), (1, 1)\},$$
then
$$[T]_B = \begin{pmatrix} 1 & 2 \\ -1 & -1 \end{pmatrix},$$
which is not orthogonal.

Title: Orthogonal endomorphisms in finite dimension

Body: As far as I understand it, orthogonal endomorphisms in finite dimensions have an orthogonal matrix for representation only if the basis on which the matrix is created is orthonormal. Doesn't that pose a problem, as then there will exist some bases in which the orthonormal application is not orthonormal ?</p>


问:
How can we work with a joint probablility density function with more than 3 variables?
This is basically more of a theoretical question and nothing I actually encountered. I have been thinking that is totally possible to need more than $3$ variables to represent a joint probability density function, in class I have learned only the two variables version where we simply do a double integral, and it is quite obvious to see that for 3 variables we can do a triple integral but what happens when we have more than three? We can't possibly draw or imagine anything in the fourth dimension fo set up the integral limits in thr first place. I know we don't need to always draw to get the correct limits but sometimes it is the case. Any help would be appreciated.

答:
In principle, "drawing" the integration boundaries is just an assist to the mind, and interchanging integration boundaries can be done formally, without need of sketches. When you reach the point in your studies that you need to integrate over multiple variables with tricky dependencies between the integration limits, you will probably have enough technical proficiency to deal with it.

If you are still looking for a method to sketch high dimension integration boundaries, you can try picking  pairs of integration variables and sketch the relationship between each pair. That is, given a complicated $\int_{f_1(x_2 \cdots x_n)} dx_1\cdots \int_{f_n(x_1\cdots x_{n-1})} dx_n$ you can try drawing the boundaries of pairs $x_a;x_b$, keeping all other variables fixed.


问:
Flipping coins problem Find E(x) and Var(x)
There is a fair coin and a biased coin that flips heads with probability 1/4. You randomly pick one of the coins and flip it until you get a heads. Let X be the number of flips you need. Compute E(X) and Var[X].

In the question, I did:

Let H denotes to get a head, $$P(H)= \frac12*\frac12+\frac12*\frac14=\frac38$$
$$E(X)=\frac1p=\frac83$$
$$Var(X)=\frac q{p^2} = \frac {40}9$$

but the TA said the answer is $$E(X)=3$$ $$Var(X)=8$$
I hava no clue where I did it wrong

答:
The expected number of tosses to get a head if the probability of a head is $p$ is $E[X_p] = {1 \over p}$.

(This can be quickly checked from the equation
$p\cdot 1 + (1-p)\cdot (E[X_p] +1) = E[X_p]$ .)

Then if $C=1$ with probability ${ 1\over 2}$ if you choose the fair coin and $C=0$ with probability ${ 1\over 2}$ if you choose the biased coin the
outcome will be
$C\cdot X_{1 \over 2} + (1-C) \cdot X_{1 \over 4}$ and so
\begin{eqnarray}
E[C\cdot X_{1 \over 2} + (1-C) \cdot X_{1 \over 4}] &=& E[C\cdot] E[ X_{1 \over 2}] + E[1-C] \cdot E[X_{1 \over 4}] \\
&=& {1 \over 2} \cdot 2 + { 1\over 2} \cdot 4 \\
&=& 3
\end{eqnarray}


问:
Plotting parametric curve with implicit parameter function
I need to plot a Smale diagram as a parametric curve $(p(x), h(x))$ where $h$ is amended potential and $p = \omega^2$ is parameter of bifurcation:
$$h:=V_\omega(x,y)=-2\cos(x)-\cos(y)-\frac{1}{2}\omega^2\bigl(\sin^2(x)+(\sin(x)+\sin(y))^2\bigr)$$
There is system of equilibrium equations (partial derivatives of amended potential are zero):
$$h_x'=2\sin(x)-\omega^2\cos(x)(2\sin(x)+\sin(y))=0$$
$$h_y'=\sin(y)-\omega^2\cos(y)(\sin(x)+\sin(y))=0$$

I am trying to get $y = y(x)$ and $p = p(x) = p(x, y(x))$ from this system. For example, isolating $\sin(y)$ and $\cos(y)$, and using $\sin^2(y)+\cos^2(y)=1$, it can be turned into:
$$4\bigl(1-\omega^2\cos(x)\bigr)^2\Bigl(\cos^2(x)+\bigl(2-\omega^2\cos(x)\bigr)^2\sin^2(x)\Bigr)=\omega^4\cos^2(x)(2-\omega^2\cos(x))^2$$
Then, substituting $\cos(y)$ and $\sin(y)$ in $h$:
$$h=-2\cos(x)-\frac{2\bigl(1-\omega^2\cos(x)\bigr)}{\omega^2\bigl(2-\omega^2\cos(x)\bigr)}-\frac{1}{2}\omega^2\sin^2(x)\biggl(1+\Bigl(1+\frac{2\bigl(1-\omega^2\cos(x)\bigr)}{\omega^2\cos(x)}\Bigr)^2\biggr)$$

So I got something like parametrization for $h$ and $p$ with the parameter $x$, but for $p(x)$ it is implicit. I had attempted to plot this in Maple 13, but unsuccessfully.

(it's described here)

What I want to know - is there a way to plot this parametric curve on $(h,\omega^2)$ coordinates with computational methods, even numerically (Wolfram Mathematica or Maple would be the best)?

Particularly, for one curve $h_-(\omega^2)$, separating from point $(x,y,\omega^2)=(0,0,2-\sqrt{2})$ with the following conditions:
$$\omega^2>2-\sqrt{2}$$
$$h(2-\sqrt{2})=-3$$
$$\sin(x)\sin(y)>0, (x,y)\in O_\varepsilon (0)$$

答:
With MATHEMATICA the leaves's graphics for $p(x) = \omega^2(x)$

```<code>ContourPlot[4 (1 - p Cos[x])^2 (Cos[x]^2 + (2 - p Cos[x])^2) Sin[x]^2 - p^2 Cos[x]^2 (2 - p Cos[x])^2 == 0, {x, -Pi, Pi}, {p, 0, 2 Pi}, ContourStyle -> {Thick, Blue}]
</code>```

Follows a script to plot the parametric $(p(x),h(x))$ for one of $p(x)$ leaves,

```<code>Clear[p, equ, h]
equ[p_, x_] := 4 (1 - p Cos[x])^2 (Cos[x]^2 + (2 - p Cos[x])^2) Sin[x]^2 - p^2 Cos[x]^2 (2 - p Cos[x])^2
h[p_, x_] := -2 Cos[x] - 2 (1 - p Cos[x])/(p (2 - p Cos[x])) - 1/2 p Sin[x]^2 (1 + (1 + 2 (1 - p Cos[x])/(p Cos[x]))^2);
x = -Pi;
Sols = {};
dx = 0.01;

While[x < Pi,
  sol = Solve[{equ[p, x] == 0, p > 0}, p, Reals];
  If[sol != {}, AppendTo[Sols, {x, Flatten[p /. sol][[1]]}]];
  x += dx
]

par = {};
For[k = 1, k <= Length[Sols], k++,
  {xx, pp} = Sols[[k]];
  If[Max[pp, Abs[h[pp, xx]]] < 10, AppendTo[par, {pp, h[pp, xx]}]]
]
ListLinePlot[par, PlotStyle -> {Thick, Blue}]
</code>```

Body:
With MATHEMATICA the leaves's graphics for $p(x) = \omega^2(x)$

ContourPlot[4 (1 - p Cos[x])^2 (Cos[x]^2 + (2 - p Cos[x])^2) Sin[x]^2 - p^2 Cos[x]^2 (2 - p Cos[x])^2 == 0, {x, -Pi, Pi}, {p, 0, 2 Pi}, ContourStyle -> {Thick, Blue}]

Follows a script to plot the parametric $(p(x),h(x))$ for one of $p(x)$ leaves,

Clear[p, equ, h]
equ[p_, x_] := 4 (1 - p Cos[x])^2 (Cos[x]^2 + (2 - p Cos[x])^2) Sin[x]^2 - p^2 Cos[x]^2 (2 - p Cos[x])^2
h[p_, x_] := -2 Cos[x] - 2 (1 - p Cos[x])/(p (2 - p Cos[x])) - 1/2 p Sin[x]^2 (1 + (1 + 2 (1 - p Cos[x])/(p Cos[x]))^2);
x = -Pi;
Sols = {};
dx = 0.01;

While[x < Pi,
  sol = Solve[{equ[p, x] == 0, p > 0}, p, Reals];
  If[sol != {}, AppendTo[Sols, {x, Flatten[p /. sol][[1]]}]];
  x += dx
]

par = {};
For[k = 1, k <= Length[Sols], k++,
  {xx, pp} = Sols[[k]];
  If[Max[pp, Abs[h[pp, xx]]] < 10, AppendTo[par, {pp, h[pp, xx]}]]
]
ListLinePlot[par, PlotStyle -> {Thick, Blue}]


问:
Finding the mle of a log normal distribution
So let $X1,X2,..,XN$ be an independent sample from log normal distribution with the pdf $f(x,\theta)=(x^2 \sigma^2*2\pi)^{(-1/2)}e^{-(log(x)-\theta)^2/{2\sigma^2}}$
and we have $\sigma^2=1$ and $\theta$ uknown

So I did the following we have the

$L(\theta,x)=(x_1^2\sigma^22\pi)^{-1/2}e^{-(log(x_1)-\theta)^2/{2\sigma^2}}*(x_2^2\sigma^22\pi)^{-1/2}e^{-(log(x_2)-\theta)^2/{2\sigma^2}}*...*(x_n^2\sigma^22\pi)^{-1/2}e^{-(log(x_n)-\theta)^2/{2\sigma^2}}$

and I get the following
$L(\theta,x)=(2\pi)^{(-n/2)}*(\sigma^2)^{-n/2}*(1)/(x_1*x_2*..*x_n)e^{(-1/2\sigma^2)\sum(log(x_i)-\theta)^2}$

So I take the $log(L(\theta,x)$ and I get

$(-n/2)log(2\pi)+log(1/x^n)-(1/2\sigma^2)\sum(log(x_i)-\theta)^2$

So now to find the mle of $\theta$
I do $d/d(\theta)log(L(\theta,x))=0$ take the derivative

and I get
$=log(x_1)-\theta+log(x_2)-\theta+...+log(x_n)-\theta$

so I get
$log(x_1)+log(x_2)+..+log(x_n)-\theta*n$

so the mle $\theta[hat]$ of $\theta$
is
$\theta[hat]=((log(x_1)+log(x_2)+..+log(x_n))/n$

I am not sure if this is right.

答:
The question is very very simple.

But the important thing you have to keep also in mind is that the likelihood is defined unless a moltipilicative constant.

Thus if you look at your density that is

$$f(x|\theta)=\underbrace{\frac{1}{x\sqrt {2\pi}}}_{\perp \!\!\!\!\!\perp \theta}e^{-\frac{1}{2}[\log x-\theta]^2}$$

So your likelihood is the following

$$L(\theta)\propto e^{-\frac{1}{2}\Sigma_i[\log x_i-\theta]^2}$$

thus immediately you get

$$l(\theta)=-\frac{1}{2}\Sigma_i[\log x_i-\theta]^2$$

$$l^*(\theta)=\Sigma_i \log x_i -n\theta=0 \rightarrow \hat{\theta}=\frac{1}{n}\Sigma_i \log x_i$$

concluding,

$$\hat{\theta}=\overline{Y}_n$$

where $Y_i=\log X_i$ and it is very natural knowing that, if $X\sim$ lognormal, then $\log X\sim$ normal


问:
Doubt in density function of scaled of random variables
The $h$ is normally distributed random variable, having PDF $f_h(x)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-x^2/2\sigma^2}$. Another random variable $\rho$ is related with $h$ by the relation $\rho=h^2\rho_t$. Then its PDF is given as $f_\rho(x)=\frac{f_h(\sqrt{x/\rho_t})}{\sqrt{\rho_tx}}$--------(1). I am having difficulty in understanding how eq (1) is derived. Because using the concept of scaling of random variable, the PDF of $\rho$ should be $\frac{1}{\sqrt{\rho_t}}f_h(\sqrt{x/\rho_t})$. Any help please as why extra $\sqrt{x}$ is coming in denominator of equation (1).

答:
Since you have the non-linear form $h^{2} \rho_t$ the fomula you used is not correct. It is applicable only for finding the density of $c h$ where $c$ is a constant.

$P(\rho \leq x)=P(h^{2} \leq \frac  x {\rho_t} )=\int_{-\sqrt {\frac x {\rho_t}}}^{\sqrt {\frac x {\rho_t} }} f_h(y)dy=2\int_{0}^{\sqrt {\frac x {\rho_t} }} f_h(y)dy$. To find the density function of $\rho$ you have to differentiate w.r.t. $x$. When you do that you have to apply Chain Rule for differentiation and the derivative of $\sqrt {\frac x {\rho_t} }$ is $\frac 1 2 (\rho_t)^{-1/2}  x^{-1/2}$. I will let you finish the computation of $f_\rho$.


问:
Deviation from average
Say I have a finite list of numbers of some statistical experiment, what is the best way to represent the typical distance from the mean of the numbers based on this data? I have found the standard deviation, but there is also average absolute deviation. How do they differ?

Thanks in advance

答:
For a data set $D=(D_j)_{j\in\Bbb N, j\leq n}$ with mean $\mu=\frac{\sum_jD_j}{n}$, I denote the standard deviation $\sigma$ and absolute deviation $\alpha$. Then:

$$\alpha=\frac {\sum_j|\mu-D_j|}{n}$$
$$\sigma=\bigg(\frac{\sum_j|\mu-D_j|^2}{n}\bigg)^\frac12$$

Title: Deviation from average


问:
Even town problem
Here's the problem.
There are $n$ citizens living in Eventown. They form clubs. In order to limit the number of citizens, they came up with following rules:

Each club has to have an even (but nonempty) number of members.

Every two clubs must have an odd number of members in common.

If $m$ is the number of clubs formed, prove that $m\le n$.

I think this is true but I am not quite sure. I'm trying to prove this induction.

I induct on the number of clubs. Let $P(m)$ be the statement:

for all $n\in \mathbb{N}$ and all $A_1, A_2, \ldots, A_m$, if $A_1, A_2, \ldots, A_m \in \mathscr{P} (\{ 1,2,\ldots , n \})$ satisfying $|A_i|$ is even for all $i$ and $|A_i \cap A_j |$ is odd for all $i\ne j$ then $m\le n$.

Base case: $m=1$. This is trivially true.

Induction hypothesis. Suppose $P(m)$ is true.

To show that $P(m+1)$ is true. Let $n\in \mathbb{N}$ and $A_1, A_2, \ldots, A_m , A_{m+1}$ be nonempty subsets of $\{ 1,2,\ldots , n \}$.

Let $J=\{ 1, 2, \ldots , n \} \setminus A_{m+1}$. Suppose $|J|=k<n$. Then $x_1< \ldots < x_k$ be members of $J$. Now consider the town with clubs $A_1 , A_2, \ldots , A_m$ with members from $J$ (thinking of them as relabelling). Then $A_1 , A_2, \ldots , A_m$ satisfy the induction hypothesis so $m\le k$. But then $m<n$ so $m+1\le n$ and hence we are done.

Is this proof correct? I hope that I conveyed by my idea properly.

答:
Note that your induction hypothesis step doesn't work.

For $A_i $ restricted to the set $J$, denoted as $A_i|J$, we do not know as yet that

$A_i|J$ has non-empty members

$A_i|J$ has an even number of members

The intersection of every 2 $A_i|J$ is odd.

$A_i|J$ are distinct on restriction (In order to conclude that$ \{ A_i | J \}$ has size $m$)

Can you prove these? If not, we can't apply the induction hypothesis.

In fact, these statements need not be true on the restricted set. As an explicit example, for $m = 2$, $n = 4$, if we have $A_1 = \{ 1, 2 \}, A_2 = \{ 2, 3 \}, A_3 = \{1, 3 \}$, then $J = \{2, 4\}$, and $A_1|J = \{2\}, A_2|J = \{2\}$. Several of the conditions are not satisfied.

Note:

Assuming these are true, then what we have is $ m \leq k \leq n-2$. This gives $ m + 2 \leq n$, which indicates that we might run into issues when $ m = n-1$.

Title: Even town problem


问:
supremum of the set T={n∈ℕ: 1/2^n ≤1}.
$T=\{n\in\mathbb ℕ: \frac1{2^n} \leq 1\}$

I worked this question and for the supremum i got $\frac{1}{2}$ because $0<\frac{1}{2^n}\leq \frac12$....Is the question even correct in the first place? May you have your analysis on this one.
The question was to state if it is true that $\sup (T)=1$.Is the question even correct in the first place?

If in the set of upper bounds the set $B=\{\frac12,1,\dots\}$ we have $\frac12$ as the smallest element how come $\sup(T)=1$?

答:
The supremum of the set $T$, as you wrote it, does not exist. This is because $T=\mathbb N$, because the inequality $\frac{1}{2^n}\leq 1$ is true for every element $n\in\mathbb N$.

Note that I answered the question you wrote, but probably not the one you wanted to write. But I can't read your mind. So if you discover that the answer is not what you needed, edit your question, and I will gladly look at it again.

Title: supremum of the set T={n∈ℕ: 1/2^n ≤1}.


问:
Find the volume of the solid enclosed by the surface $\rho=1+\cos\phi$
Find the volume of the solid enclosed by the surface $\rho=1+\cos\phi$.
How do I set up the triple integral for this? I know that $1+\cos\theta$ is a cardioid but how would it look like as a surface?

答:
The volume is found by integrating $\iiint 1\cdot dV$. The real trick is in finding the bounds. The easiest way to do this would be to set up the integral order with the angle first:

$$V = \int_0^{2\pi}\int_0^2 \int_0^{\cos^{-1}(\rho-1)} \rho^2\sin\phi \:d\phi d\rho d\theta $$


答:
There are actually two possible regions:

Or,

You see, the 'area between circle and parabola' is a bit ambiguous without saying which area specifically. But I'll take a guess that you are referring the orange (i.e: 2nd diagram). For computing the area, the y axis method and x axis method are both equivalents. Let's focus on the area in first quadrant for now because you can find the one in fourth by symmetry,

Using integration with x-axis, the area is given as:

$$ \int_0^2 y_{parabola} dx + \int_2^{\sqrt{6}} y_{circle} dx$$

If you had done integration with $y$,

$$ \int_0^{\sqrt{2} } (x_{circle} - x_{parabola})  dy$$

Perhaps the second integral is easier because you're evaluating over the same bounds.

Title: Area between two given equations

Body:
Let $x^2 + y^2$ = 6 and $x = y^2$. What is the area between them?

I solved the two equations in terms of y and I graph it in Geogebra: https://www.geogebra.org/classic/ahnk4hfw

Solving the two equations in terms of x, the graph is:
https://www.geogebra.org/classic/nwznqk3c

When I integrate in terms of y (the graph is the first one), the bounds are -$\sqrt 2$ and $\sqrt 2$. The integrand is $ \sqrt {6-y^2} $ - $y^2$. I got 4.63 as the area.
$$\int_{-\sqrt 2}^{\sqrt 2} \sqrt {6-y^2} - y^2 \, dy = 4.63 $$
Although, I am not sure if I expressed the integral correctly because it seems that I am missing one function: the -$\sqrt {6-y^2} $.

When I follow the second graph, isn't it that the whole region is bounded by the two functions, and hence the area is the circle's area? Or should I only choose the positive square roots and not the negative ones?

My last question is, what is the best approach of finding the integral of this case? Is it by integrating with respect to y or x? Do I only need to take the smaller region?


问:
$A$ is nilpotent matrix and satisfies the equation $A^5=3A^3-2A.$ Prove $A=0.$
I'm stuck at this question , I had an idea how to solve it but I think it's not good tho.

The Question :

$A$ is nilpotent and $A^5=3A^3-2A,$ prove $A=0.$

( $A^k = 0 :$ for certain  $k\in \mathbb{N}$)

My try : I tried changing the equation to the form $\Rightarrow A=\frac{A^3(3I-A^2)}{2}$ and then multiplying both sides by $A^{k-3}$ and proceed with $\Rightarrow  A^{k-2} = 0$.

But I'm not sure how to proceed with it now.

答:
Here is another approach:

$$A(\underbrace{A^4 - 3A^2}_{B:=}+2I)=O$$

Note that $B$ is nilpotent because $A$ is nilpotent.
Hence, the only eigenvalue of $B$ is $ \lambda = 0$. Hence, $B+2I$ is invertible (otherwise $\lambda = -2$ would be an eigenvalue of a nilpotent matrix which is a contradiction).

It follows by multiplying by $(B+2I)^{-1}$ from the right that $A =O$.


Title: Area between two given equations

Body:
Using $\int ydx$.

These curves intersect at $P_1(2,\sqrt{2}),$ and $P_2(2,-\sqrt{2})$

$$A=2\left[\int_{0}^{2} \sqrt{x} dx+\int_{2}^{\sqrt{6}} \sqrt{6-x^2} dx\right]$$
$$A=\frac{8\sqrt{2}}{3}+[3\pi-2\sqrt{2}-6 \sin^{-1}\sqrt{2/3}]=4.5356$$


问:
Distribution of $X(t+s)$ in Poisson proccess in terms of distribution of $X(t)$ and $X(s)$

Let $(X(t))_{t \geq0}$ be a Poisson process with intensity $\lambda$. Denote $p_t$ as distribution of $X(t)$. Write $p_{t+s}$ in terms of $p_t$ and $p_s$.

This was a homework we skipped due to insufficient time and the lecturer kinda forgot about it next time, so it is left unsolved. I appreciate any help as I wasn't really able to work it out.

答: $X(t+s)=(X(t+s)-X(s))+X(s)$. $X(t+s)-X(s)$ and $X(s)$ are independent and their distributions are $\rho_s$ and $\rho_t$. Hence the distribution of $X(t+s)$ is the convolution of $\rho_t$ and $\rho_s$.


问:
Prove that a summation is convergent

Let $\{x_n\}$ be a decreasing sequence of positive numbers and $\lim_{n \to \infty} x_n = 0$.
Prove that the following summation is convergent.(Hint: Use alternating series test.)$$S = x_1 -\frac{1}{2}(x_1+x_2) +\frac{1}{4}(x_1+x_2+x_3)-\frac{1}{8}(x_1+x_2+x_3+x_4) + \dots$$

My try: Let $$S_n = \frac{1}{2^{n-1}}\sum_{m=1}^nx_m$$So we are interested in the convergence of $$S = \sum_{n=1}^{+\infty}(-1)^{n+1}S_n$$We can prove that $\{S_n\}$ is a decreasing sequence: $$S_{n+1} - S_n = \frac{1}{2^{n}}\sum_{m=1}^{n+1}x_m - \frac{1}{2^{n-1}}\sum_{m=1}^nx_m = \frac{x_{n+1}}{2^n} + (\sum_{m=1}^nx_m)(\frac{1}{2^n} - \frac{1}{2^{n-1}}) = \frac{1}{2^{n-1}}(\frac{x_{n+1}}{2} + (\sum_{m=1}^nx_m)(\frac{1}{2} - 1))$$ $$\frac{x_{n+1}}{2} + (\sum_{m=1}^nx_m)(\frac{1}{2} - 1)\le0 \iff x_{n+1} - \sum_{m=1}^nx_m \le 0 \iff x_{n+1} \le \sum_{m=1}^nx_m$$ Which is true since $x_{n+1} \le x_n$ and $x_n\ge 0$. Therefore we have $S_{n+1}\le S_n$. Obviously $S_n \ge 0$ for all $n$ and we should prove $\lim_{n\to \infty}S_n = 0$ in order to use alternating series test. I couldn't prove that and got stuck here.

答:
$$\lim_{n\to\infty}S_n=\lim_{n\to\infty}\frac{x_1+\ldots+x_n}{2^{n-1}} =\lim_{n\to\infty}\frac{x_1+\ldots+x_n}{n}\frac{n}{2^{n-1}}.$$

Now, The first factor is the arithmetic mean of $x_n$'s, so it converges also to $0$ (see here, for instance, but it is a classical application of Stolz Cesaro).

Morteover, it is straightforward that $\lim_{n\to\infty}\frac{n}{2^{n-1}}=0$. (If you do not want to calculate this limit, you can also argue that $0\le \frac{n}{2^{n-1}}\le 1$, so bounded).

So, $\lim_{n\to\infty}S_n=0$.

Title: Prove that a summation is convergent


问:
Perimeter change when moving one point of a triangle along a circle
I am interested in the perimeter $P(\Delta')$ of a modified version of a triangle $\Delta = (A,B,C)$. I chose point $B$ to be the point I want to shift towards the inside (or on the boundary) of $\Delta$. I chose a distance $r<h$ where $h$ is the height of the triangle. $B$ can be moved to an arbitrary point $E$ which is of distance $r$ to $B$ and lays inside or on the boundary of $\Delta$. This means that $E$ must also lay on the circle of radius $r$ with the center point $B$. I denote $\Delta'=(A,E,C)$. What is the perimeter $P(\Delta')$ of the new triangle given the original side lengths, $r$ (and probably some representation of the arc length to E)? In what range does $P(\Delta)-P(\Delta')$ lie in?

I attached a small figure which visualizes both triangles.

Edit

Based on the comments, I added some notation:

Assuming $B=(0,0)$, then we have $E=(x,y)$ with $x=\lambda_A x_A + \lambda_C x_C$ and $y=\lambda_A y_A + \lambda_C y_C$ ($\lambda_A+\lambda_C = 1$).

This gives us the following set of equations:

$$r=\sqrt{(0-x)^2 + (0-y)^2} = \sqrt{(\lambda_A x_A + \lambda_C x_C)^2 + (\lambda_A y_A + \lambda_C y_C)^2}$$
$$|AE| = \sqrt{(x-x_A)^2+(y-y_A)^2}$$
$$|CE| = \sqrt{(x-x_C)^2+(y-y_C)^2}$$

Moreover, we have:
$$|AB| = \sqrt{x_A^2 + y_A^2}$$
$$|CB| = \sqrt{x_C^2 + y_C^2}$$

The perimeter change is in fact the difference $|AB|+ |CB| - |AE| - |CE|$. I would like to show that $r \leq |AB|+ |CB| - |AE| - |CE|$. I struggle with the proof because I still have six variables $x_A,x_C,y_A,y_C,\lambda_A$ and $\lambda_C$.

Background

This question arises from a computer science problem called lawn mowing/milling, where you are given a cutter (in my case a circular shape) and a Polygon to cover. I noticed that given a tour of the circular cutter one can change the cutter from a circular shape with a radius $s$ to a square with side length $2s$. Afterward, the tour can be shortened a bit. I have already proven a specific length $r$ that I can move the points of the tour "inwards" to shorten the tour. My goal is to find out how much shorter the final tour gets when I modify it. In my example, $A-B-C$ are part of the tour and $B$ is a point which can be moved inward by $r$ to a point $E$. My tour will later be $A-E-C$. I want to find out how much the tour is shortened after my transformation and I hope that it will be shortened by at least $r$.

答:
This answer deals with the minimal perimeter issue.

Using slightly different notation, we have
a known $\triangle ABC$ and given radius $r_a$
of the circle $\mathcal{A}(A,r_a)$.
The circle intersects with $AB$, $AC$ at $D,\ E$, respectively.
For the point $F$ along the interior arc $DE$,
and the point $G=AF\cap BC$,
the minimal perimeter of $\triangle FBC$
is achieved when $F$ is the tangential point,
where the ellipse focused at $B,C$ touches
the circle $\mathcal{A}$, which also means
that $\angle GFB=\angle CFG=\theta$.

Let $|BF|=u,\ |CF|=v$, $|BG|=at,\ |CG|=a(1-t)$
for some $t\in(0,1)$.

Then using the cosine rule for $\triangle ABG$,
Stewart’s theorem
for $\triangle FBC$ and the properties of
the bisector of $\angle CFB$, we can express $t$ in terms of
the side lengths $a,b,c$ of $\triangle ABC$ and the radius $r_a$
as the root of the quartic

\begin{align}
q_4\,t^4+q_3\,t^3+q_2\,t^2+q_1\,t+q_0&=0
\tag{1}\label{1}
,
\end{align}

where

\begin{align}
q_4 &= (b^2-c^2)^2-4\,r_a^2\,a^2,
\\
q_3 &= 4\,c^2\,(b^2-c^2)-4\,r_a^2\,(b^2-2\,a^2-c^2),
\\
q_2 &= r_a^2\,(4\,b^2-8\,c^2-5\,a^2)+2\,c^2\,(3\,c^2-b^2),
\\
q_1 &= r_a^2\,(a^2-b^2+5\,c^2)-4\,c^4,
\\
q_0 &= c^2\,(c^2-r_a^2)
\tag{2}\label{2}
,
\end{align}
if the vertices $A,B,C$ in counter-clockwise orientation.

The real roots of \eqref{1} need to be checked
for which one gives the actual minimum.

For the attached image, I've got all four real roots,
two of them were greater than $1$, one gives the minimal perimeter,
and, unfortunately, another one was also in a valid range,
but the corresponding perimeter was not minimal.

Also, the curved red, green and blue lines,
shown in the image,

emanating from the vertices $A,B$ and $C$, respectively,
illustrate the corresponding locus
of the points of minimal perimeter
for possible values of $r_a,\ r_b$ and $r_c$
for the circles
$\mathcal{A}(A,r_a)$,
$\mathcal{B}(B,r_b)$
and $\mathcal{C}(C,r_c)$.

Naturally, if the angles of $\triangle ABC$ are less than $120^\circ$,
all this curves intersect at the
Fermat–Torricelli point
$T$, which provides the minimum of $|TA|+|TB|+|TC|$.

Edit

On the other hand, it is much simpler to find
the corresponding radius $r_a$ given some $t\in(0,1)$:

\begin{align}
r_a
&=
\left|
\frac{t^2 b^2-(1-t)^2\,c^2}
{(1-2t)\sqrt{(c^2-a^2\,t)(1-t)+b^2\,t}}
\right|
.
\end{align}

Note that $t=\tfrac12$ is a special (and simpler) case,
which must agree only with $b=c$,
and such condition is better considered separately.

For $b=c$, the equation \eqref{1}

factors out as

\begin{align}
(1-2t)^2(b^4-r_a^2(b^2-a^2\,t\,(1-t)))
&=0
,
\end{align}

and the proper root is $t=\tfrac12$,
so, for any valid value of $r_a$,
$G$ must be the midpoint of $BC$,
to get the minimal perimeter of $\triangle FBC$,
as expected for the isosceles $b=c$ case.

Edit 2

Actually, the valid range of $t$ would be a subset of either
$(0,\tfrac12)$ or $(\tfrac12,1)$, depending of
the side lengths $b$ and $c$.


问:
A ring extension of $\mathbb Z$ in which $x^4+y^4+1$ is reducible
My first instinct was the ring of two variable polynomials modulo $a^4+b^4+1$.($\mathbb Z[a,b]/(a^4+b^4+1)$)

But I haven’t managed to reduce it to linear factors.

Can any of you help?

答:
Homogenize and reduce to: is $P(x,y,z)=x^4 + y^4 + z^4$ reducible in $K[x,y,z]$, for some ring $K\supset \mathbb{Z}$?  If it is, then some (easy to write) system of equations with coefficients in $\mathbb{Z}$ has a solution in $K$. But that means ( Hilbert nullstellensatz) that the system has a solution in $\mathbb{C}$ ( or even in
$\bar{\mathbb{Q}}$, the algebraic closure of $\mathbb{Q}$).

So assume that $P(x,y,z) = P_1(x,y,z)\cdot P_2(x,y,z)$, with $P_1$, $P_2 \in \mathbb{C}[x,y,z]$, homogenous, of positive degree, and $\deg P_1 + \deg P_2 = 4$.

The complex projective curves $P_1=0$, and $P_2=0$ have a non-void intersection ( Bezout's theorem implies this). Let $(x_0\colon y_0\colon z_0)$ a common point of these curves, that is $P_i(x_0, y_0, z_0)=0$.
But this implies with the product rule
$$\frac{\partial P}{\partial x}(x_0, y_0, z_0) = \frac{\partial P}{\partial y}(x_0, y_0, z_0)=\frac{\partial P}{\partial z}(x_0, y_0, z_0)=0$$
(a singular point of the curve $P(x,y,z) = 0$).
But this implies $x_0 = y_0 = z_0 = 0$, contradiction. ( a point in the projective plane has at least one of the projective coordinates $\ne 0$).

Note: This works for any $P$ homogenous of $3$ variables for which the curve $P=0$ is non-singular.

For comparison, the polynomial $x^4 + 2 x^2 y^2 + y^4 - z^4$ is reducible. It also has two singular points: (at infinity) $(1\colon \pm i\colon 0)$.


答: The series is absolutely convergent. Since $(x_n)$ is bounded $|x_1+x_2+..+x_n | \leq nM$ for some $M$ and $\sum \frac  n {2^{n}}$ is convergent. Use Comparison Test to finish.

Title: Prove that a summation is convergent

Body:
Let $\{x_n\}$ be a decreasing sequence of positive numbers and $\lim_{n \to \infty} x_n = 0$.
Prove that the following summation is convergent.(Hint: Use alternating series test.)$$S = x_1 -\frac{1}{2}(x_1+x_2) +\frac{1}{4}(x_1+x_2+x_3)-\frac{1}{8}(x_1+x_2+x_3+x_4) + \dots$$

My try: Let $$S_n = \frac{1}{2^{n-1}}\sum_{m=1}^nx_m$$So we are interested in the convergence of $$S = \sum_{n=1}^{+\infty}(-1)^{n+1}S_n$$We can prove that $\{S_n\}$ is a decreasing sequence: $$S_{n+1} - S_n = \frac{1}{2^{n}}\sum_{m=1}^{n+1}x_m - \frac{1}{2^{n-1}}\sum_{m=1}^nx_m = \frac{x_{n+1}}{2^n} + (\sum_{m=1}^nx_m)(\frac{1}{2^n} - \frac{1}{2^{n-1}}) = \frac{1}{2^{n-1}}(\frac{x_{n+1}}{2} + (\sum_{m=1}^nx_m)(\frac{1}{2} - 1))$$ $$\frac{x_{n+1}}{2} + (\sum_{m=1}^nx_m)(\frac{1}{2} - 1)\le0 \iff x_{n+1} - \sum_{m=1}^nx_m \le 0 \iff x_{n+1} \le \sum_{m=1}^nx_m$$ Which is true since $x_{n+1} \le x_n$ and $x_n\ge 0$. Therefore we have $S_{n+1}\le S_n$. Obviously $S_n \ge 0$ for all $n$ and we should prove $\lim_{n\to \infty}S_n = 0$ in order to use alternating series test. I couldn't prove that and got stuck here.


问:
How do you convert $(12.0251)_6$ into fractions?
How do you convert $(12.0251)_6$ (in base 6) into fractions?

I know how to convert a fraction into base $x$ by constantly multiplying the fraction by $x$ and simplifying, but I'm not sure how to go the other way?

答:
You could use continued fractions.  </p>

```<code>                       Cf     A   /  B
             12.0251           1  /  0      In the left, we produce a
    1.0000  -12.0000    8      8  /  1      continued fraction from
   -0.5420    0.0251   12     97  / 12      12.0521 and 1.0000, in base 6
    0.0140   -0.0140    1    105  / 13        There are Cf of the positive
   -0.0111    0.0111    1    202  / 25      number, which gives the negative
    0.0025   -0.0054    2    511  / 63      number.
   -0.0013    0.0013    1    713  / 88        The right is Cf(r-1) + Cf(r-2)
              &c                            starting at 1 in A and 0,1 in B.
</code>```

<p>You stop when the error gets precise enough, or you get a zero in either column.  </p>

Body:
You could use continued fractions.

                       Cf     A   /  B
             12.0251           1  /  0      In the left, we produce a
    1.0000  -12.0000    8      8  /  1      continued fraction from
   -0.5420    0.0251   12     97  / 12      12.0521 and 1.0000, in base 6
    0.0140   -0.0140    1    105  / 13        There are Cf of the positive
   -0.0111    0.0111    1    202  / 25      number, which gives the negative
    0.0025   -0.0054    2    511  / 63      number.
   -0.0013    0.0013    1    713  / 88        The right is Cf(r-1) + Cf(r-2)
              &c                            starting at 1 in A and 0,1 in B.

You stop when the error gets precise enough, or you get a zero in either column.


问:
A = [1,2,3]; Give a subset S ⊆ A x A but S must not be X x Y for any subsets X, Y ⊆ A
I did A x A and the result was [(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)]

I also tried X x Y keeping in mind that the subsets are pairs: X = [a, b]; Y = [a', b']

X x Y = [ (a,b), (a',b), (a, b'), (a',b')]

And i stated that any S which has 2 pairs as subsets must be a solution if one of those isnt (1,1), (2,2), (3,3).
So S can be for example (1,2), (1,3).

Is this correct ? Thanks a lot

答: Your example is not correct since $S=\{1\} \times \{2,3\}$. Take $S=\{(1,2),(2,3)\}$ instead. If $S=A\times B$ then $1$ and $2$ both belong to $A$ and $2$ and $3$ belong to $B$. So we must have $(1,3) \in S$ a contradiction.


问:
Show that there in no value such that f(x) = 1/x is continuous at the point 0 using the epsilon/delta-definition
Prove, that there does not exist a $c \in \mathbb{R}$, such that

\begin{equation}
f(x) = \begin{cases}\frac{1}{x}, \text{ for } x \neq 0\\
c, \text{ for } x = 0
\end{cases}
\end{equation}
is continuous for $x_0 = 0$

Using sequences I solved this the following way:

Let $(x_n)_{n \in \mathbb{N}} := (\frac{1}{n})_{n \in \mathbb{N}} \subset \mathbb{R}$\ {$0$} and $\lim_{n \rightarrow \infty} x_n = 0.$

Then, $\lim_{n \rightarrow \infty} f(x_n) = \lim_{n \rightarrow \infty} \frac{1}{1/n} = \infty$

Hence, the limit is unequal to $c = f(x_0)$ for any $c \in \mathbb{R}.$

I was wondering how to prove this using the $\epsilon$-$\delta$-method

Here is what I came up with so far:

$\exists \epsilon > 0$, $\forall \delta > 0$ , there $\exists x \in [x_0 -x] < \delta$, s.d. $|f(x_0) - f(x)| \geq \epsilon.$

Thus, in this case, we need to shwo that

$\exists x \in $ $[-x] < \delta$, s.d. $|c - \frac{1}{x}| \geq \epsilon$. Now, let $\epsilon := \frac{1}{2}$.

As $x$ becomes very small, the expression $|c - \frac{1}{x}|$ grows to infinity and is thus greater then $\frac{1}{2}$.

Is there a way to make my last sentence more mathematical? Can I somehow get rid of the $c$ or use a different argument to stick closer to the $\epsilon-\delta$-definition?

答:
Yes, just find the value of $x$ explicitly. When $c >0$ Note that for $0< x < 1/c$ we have $|c - 1/x| = 1/x - c$, so we want to show that $1/x - c > 1/2$ for some $x$ in this range. But then we just solve:
$$1/x > 1/2 +c$$
and thus any
$$x < \frac{1}{1/2 + c}$$
works, that is, choose $0<x < \min(\delta, \frac{1}{1/2 + c})$.

When $c < 0$ follow the same through for $1/c < x < 0$, and when $c = 0$ it's easier.


问:
One-sided alternative hypothesis
I am a little bit confused about definitions of null and alternative hypotheses.

My understanding is that the null and alternative hypotheses are defined based one a partition of the parameter space $\Theta$. Suppose that $\Theta$ can be partitioned into two disjoint subsets $\Theta_0$ and $\Theta_1$. Then, the null is such that $\theta \in \Theta_0$ and the alternative is such that $\theta \in \Theta_1$.

However, it is not uncommon to find examples where people test:
$$ H_0: \theta = \theta_0 \quad\text{versus}\quad H_1: \theta > \theta_0 $$
or
$$ H_0: \theta = \theta_0 \quad\text{versus}\quad H_1: \theta < \theta_0 .$$
For example, here :

https://statweb.stanford.edu/~owen/courses/200/lec07.pdf (bottom of page 3)

http://www.stat.yale.edu/Courses/1997-98/101/sigtest.htm

These hypotheses don't form a partition of the parameter space (assuming $\theta\in\mathbb{R}$). Is it still correct to do that? Do people assume (implicitly) that the equality in the null is actually an inequality?

答:
. You need some kind of monotonicity property of the power function to claim that. Right?

Yes, correct.

Remember that the I type error (test size) is defined as

$$\alpha=\sup_{\theta \in \Theta_0}\pi(\theta)$$

Here is a formal definition of UMP Test

this explains why $\mathcal{H}_0:\mu\leq\mu_0 \equiv \mathcal{H}_0:\mu=\mu_0$

Actually there are many hypothesis-testing problems for which a UMP test does not exist. In these situation usually the tests are restricted in a class of Unbiased ones. If a test is biased it cannot be used because it might give incoherent results

a test $\gamma$ is unbiased iff

$$\sup_{\theta \in \Theta_0}\pi_{\gamma}(\theta)\leq \inf_{\theta \in \Theta_1}\pi_{\gamma}(\theta)$$

Title: One-sided alternative hypothesis


问:
Calculate $I=\int_0^{2\pi}dx \int_0^{\pi} e^{\sin y(\cos x-\sin x)}\sin y\,dy$
I'm not quite sure how to calculate this integral:

$$I=\int_0^{2\pi}dx \int_0^{\pi}  e^{\sin y(\cos x-\sin x)}\sin y\,dy$$

With Mathematica I see the result is $2\pi \sqrt 2 \sinh \sqrt 2$.

Is there any special technique for calculating this integral?

Any help will be appreciated.

Thank you for these brilliant solutions using Bessel function or complex analysis.

After looking into this integral for a long time I found another "primary" solution:

Rewrite it as

$$I=\int_0^{2\pi}d\theta \int_0^{\pi}  e^{\sin \varphi(\cos \theta-\sin \theta)}\sin \varphi\,d\varphi$$

By polar coordinates it becomes

$$I=\int_{u^2+v^2+w^2=1} e^{u-v}\,dS$$

since $dS=\sin\varphi d\theta d\varphi$. Then choose an orthogonal transform such that $z=\frac{u-v}{\sqrt 2}$, we get

$$I=\int_{x^2+y^2+z^2=1} e^{\sqrt 2 z}\,dS$$

Going back to polar coordinates it becomes

$$I=\int_0^{2\pi}d\theta \int_0^{\pi}  e^{\sqrt 2 \cos\varphi}\sin \varphi\,d\varphi$$

And finally we get

$$I=2\pi \int_{-1}^1 e^{\sqrt 2 t}\,dt$$

答:
I see a solution using the modified Bessel function (not in the result, however) $$I_0(z)=\sum_{n=0}^\infty\frac{1}{n!^2}\left(\frac{z}{2}\right)^{2n}=\frac1\pi\int_0^\pi e^{z\cos t}\,dt.$$ Since $\cos x-\sin x=\sqrt2\cos(x+\pi/4)$ being integrated over its period, we may drop the $\pi/4$: $$\int_0^{2\pi}e^{r(\cos x-\sin x)}\,dx=\int_0^{2\pi}e^{r\sqrt2\cos x}\,dx=2\pi I_0(r\sqrt2),$$ and using the power series, we can compute
\begin{align*}
\int_0^\pi I_0(a\sin y)\sin y\,dy
&=\sum_{n=0}^\infty\frac{a^{2n}}{2^{2n}n!^2}\int_0^\pi\sin^{2n+1}y\,dy
\\&=\sum_{n=0}^\infty\frac{a^{2n}}{2^{2n}n!^2}\mathrm{B}\left(\frac12,n+1\right)
\\&=\sum_{n=0}^\infty\frac{a^{2n}}{2^{2n} n!^2}\frac{\sqrt\pi\cdot n!}{2^{-n-1}(2n+1)!!\sqrt\pi}
\\&=2\sum_{n=0}^\infty\frac{a^{2n}}{(2n+1)!}=\frac2a\,\sinh a.
\end{align*}
This gives the result you've got.


问:
Monotone Sequence $\frac{n}{n+1}$
I have to proof this sequence :
$(x_n) := \frac{n}{n+1}$ for $n\in \Bbb N^+$ is a monotone sequence.
I saw some examples of people using the induction method, so I tried it, but I got stuck when I was trying to prove the $A(k+1)$ is true.

Here's what I have done so far

Let $A(n) := x_n \le x_{n+1}$

$A(n)$ is true for all $n\ge1$. Assume that $A(k):=x_{k}\ge x_{k+1}$ is true, then show that $A(k+1)$ is also true.

I don't know where I should go next. I've been trying to add both side, but can't find the desired end. Should I change the proving method? For some backgrounds, before this question I had to prove that $(x_n)$ is convergent and bounded.

答: No need to use induction here. If $n\in\Bbb N$, then$$x_{n+1}-x_n=\frac{n+1}{n+2}-\frac n{n+1}=\frac{1}{(n+2)(n+1)}>0$$and therefore your sequence is strictly increasing.

Title: Monotone Sequence $\frac{n}{n+1}$


答: No need for induction. $\frac 1 n$ is decreasing so $1+\frac 1 n$ is also decreasing. This implies that $\frac 1 {1+\frac 1 n}$ is increasing. But $\frac 1 {1+\frac 1 n}=\frac n {n+1}$

Title: Monotone Sequence $\frac{n}{n+1}$

Body:
I have to proof this sequence :
$(x_n) := \frac{n}{n+1}$ for $n\in \Bbb N^+$ is a monotone sequence.
I saw some examples of people using the induction method, so I tried it, but I got stuck when I was trying to prove the $A(k+1)$ is true.

Here's what I have done so far

Let $A(n) := x_n \le x_{n+1}$

$A(n)$ is true for all $n\ge1$. Assume that $A(k):=x_{k}\ge x_{k+1}$ is true, then show that $A(k+1)$ is also true.

I don't know where I should go next. I've been trying to add both side, but can't find the desired end. Should I change the proving method? For some backgrounds, before this question I had to prove that $(x_n)$ is convergent and bounded.


问:
Injection of $E(\mathbb{Q})_{\text{tors}}$ into $\tilde{E}(\mathbb{F}_p)$?
I'm looking at Example VII.3.3.3 (p.193, 2nd ed.) of Silverman's The Arithmetic of Elliptic Curves. We have the elliptic curve $E:y^2=x^3+x$, with discriminant $\Delta=-64$, so there is good reduction for all primes $p\geq 3$. It is noted that $(0,0)$ is a point of order two in $E(\mathbb{Q})$, and that
 $$\tilde{E}(\mathbb{F}_3)=\{\mathcal{O},(0,0),(2,1),(2,2)\}\cong\mathbb{Z}/4\mathbb{Z}$$
 $$\tilde{E}(\mathbb{F}_5)=\{\mathcal{O},(0,0),(2,0),(3,0)\}\cong(\mathbb{Z}/2\mathbb{Z})^2$$
Then it is said that

Since $E(\mathbb{Q})_{\text{tors}}$ injects into both of these groups, we see that $(0,0)$ is the only nonzero torsion point in $E(\mathbb{Q})$.

Now, my understanding of Proposition VII.3.1b (p.192, 2nd ed.) is that for any discretely valued local field $K$ with residue field $k$, the reduction map from $E(K)[m]$ to $\tilde{E}(k)$ is injective for all $\gcd(m,\text{char}(k))=1$, where $E(K)[m]$ denotes the $m$-torsion subgroup of $E(K)$. So, we are looking at the compositions
$$E(\mathbb{Q})[m]\hookrightarrow E(\mathbb{Q}_3)[m]\hookrightarrow \tilde{E}(\mathbb{F}_3)\quad\text{ for all }3\nmid m$$
$$E(\mathbb{Q})[n]\hookrightarrow E(\mathbb{Q}_5)[n]\hookrightarrow \tilde{E}(\mathbb{F}_5)\quad\text{ for all }5\nmid n$$
It seems the best we can say (I think) is that the $3$-torsion-free part of $E(\mathbb{Q})_{\text{tors}}$ injects into $\tilde{E}(\mathbb{F}_3)$, and the $5$-torsion-free part of $E(\mathbb{Q})_{\text{tors}}$ injects into $\tilde{E}(\mathbb{F}_5)$. This still clearly implies that $E(\mathbb{Q})_{\text{tors}}$ must be of order 2 in this particular case, because $\tilde{E}(\mathbb{F}_5)$ is $3$-torsion-free and $\tilde{E}(\mathbb{F}_3)$ is $5$-torsion-free; but it seems to me that the reasoning in the block-quoted statement (at least without further explanation) is technically wrong.

If $p$ is a prime of good reduction for $E$, then is it true that all of $E(\mathbb{Q})_{\text{tors}}$ injects into $\tilde{E}(\mathbb{F}_p)$, or just that $E(\mathbb{Q})_{\text{tors}}[m]$ injects into $\tilde{E}(\mathbb{F}_p)$ for any $m$ relatively prime to $p$ (and hence the $p$-torsion-free part of $E(\mathbb{Q})_{\text{tors}}$ injects into $\tilde{E}(\mathbb{F}_p)$)?

答:
Let $K$ be a discretely valued field whose residue field is perfect of characteristic $p$, and let $E$ be an elliptic curve over $K$ with good reduction.  If one supposes further that the absolute ramification index $e$ of $K$ is $< p-1$,
then it follows that the reduction map is injective on the $K$-rational $p$-power torsion points of $E$.  As you already noted, the reduction map is automatically injective on the prime-to-$p$ torsion, and so we conclude that (when $e < p-1$) the reduction map is injective on all the $K$-rational torsion.

There are various ways to see the claimed injectivity.  One is via a consideration of the formal group, and another is via the theory of finite flat group schemes.  The rough idea is that any non-trivial $p$-torsion point which reduces to the identity (and hence lies in the points of the formal group) is the solution to an Eisenstein polynomial of degree divisible by $p-1$.  If $e < p-1$ then such a polynomial can
have no roots in $K$, and so the non-trivial $K$-rational $p$-torsion points cannot lie in the formal group.

If one considers the particular case when $K = \mathbb Q$ or $\mathbb Q_p$ equipped with the $p$-adic valuation, where $p$ is an odd prime, then $e = 1 < p-1$, and this result applies.  This is what Silverman is using.

I think this is discussed somewhere in Silverman, although I forget whether it is treated in the general form described above, or just in the particular case of odd primes $p$ in $\mathbb Q$.


问:
Curvature of a non-compact complete surface
Assume $\Sigma$ is a non compact, complete surface. Assume the integral $$\int_{\Sigma}K$$ is convergent, where K is the Gauss curvature of $\Sigma$. Is it always true that $$\frac{1}{2\pi}\int_{\Sigma}K$$ is an integer?</p>

答: Sure, I am going to have to go with the contrapositive on this one: $\int{\Sigma}$div$\gamma$$\nu$ = $\int{\partial\Sigma}< \nu, N> \sigma$ #. In other words Stokes theorem.</p>


Title: Monotone Sequence $\frac{n}{n+1}$

Body:
Note that for every $n$
$$
\frac{x_{n+1}}{x_n}=\frac{\frac{n+1}{n+2}}{\frac{n}{n+1}}=\frac{n^2+2n+1}{n^2+2n}>1
$$
Therefore $\frac{x_{n+1}}{x_n}>1$, so $x_{n+1}>x_n$


问:
Theorem related to the sum of harmonic functions
I am trying to solve a problem related to harmonic functions, but I do not see how to begin. The problem is

Let $Ω \subset \mathbb{R}^{n}$ be an open and connected set. Let $u :
 \Omega \to \mathbb{R}$ be a harmonic function. Suppose that for a pair of
points $(x_0, y_0) \in \Omega \times \Omega$, $u$ satisfies $u(x_0) +
 u(y_0) = M$. Then there are infinitely many pairs $(x, y) \in \Omega
 \times \Omega$ such that $u(x) + u(y) = M$.

I think that it has something to do with mean value properties. Any tip to approach this problem?.

答:
Suppose on the contrary, that there are only finitely many such pairs, then we can find $r>0$ such that the ball $B_r(x_0,y_0)\subset \Omega\times \Omega$ contains no other such point,
i.e. $\forall(x,y)\in B_r(x_0,y_0)\setminus \{(x_0,y_0)\}: u(x)+u(y)\neq M$.

Now we define the harmonic function $\tilde{u}:B_r(x_0,y_0)\to \mathbb{R}$ by $\tilde{u}(x,y)=u(x)+u(y)$. If $M=\max_{(x,y)\in B_r(x_0,y_0)} \tilde{u}(x,y)$, then we can conclude by the maximum principle that $\tilde{u}$ is constant on $B_r(x_0,y_0)$ and this contradicts the assumption that $(x_0,y_0)$ is the only point in the ball that attains the value $M$. Similarly, we can use the minimum principle to conclude that $M\neq \min_{(x,y)\in B_r(x_0,y_0)} \tilde{u}(x,y)$.

Therefore we have shown that there exists points $(x_1,y_1), (x_2,y_2)\in B_r(x_0,y_0)$ such that
\begin{align}
\tilde{u}(x_1,y_1)<M<\tilde{u}(x_2,y_2).
\end{align}
Now we can choose a contiuous curve $c:[0,1]\to B_r(x_0,y_0)$ with $c(0)=(x_1,y_1)$ and $c(1)=(x_2,y_2)$ such that $c(t)\neq (x_0,y_0)$ for all $t\in[0,1]$. If we denote the composition by $f(t)=\tilde{u}(c(t))$, then by construction we have $f(0)<M<f(1)$ and by the intermediate value theorem we can conclude that there has to be a point $t'$ such that $\tilde{u}(c(t'))=f(t')=M$.

Since $c(t')\neq (x_0,y_0)$ we have found another point in $B_r(x_0,y_0)$ that attains the value $M$, which contradicts our assumption that there are only finitely many such points in $\Omega\times\Omega. $


问:
Prove that if $f \circ g$ is one-to-one, then $g$ if one-to-one
Let $f$ and $g$ be functions and have domain $\mathbb{R}$. I have to prove or disprove two claims:

(1) if $f \circ g$ is one-to-one, then $f$ is one-to-one

(2) if $f \circ g$ is one-to-one, then $g$ is one-to-one

For (1), could you give me an intuition as to why the statement is wrong? What counter-example can be used?

For (2), how should I prove or approach the question? I have attempted direct proof, but I couldn't link the "IF" and "THEN" part.

答:
The first statement is false: take $g(x)=e^x$ and $f(x)=x^2$. The $f\circ g$ is injective, but $f$ isn't.

But the second statement is true. If $f\circ g$ is injective and $x,y\in\Bbb R$ are such that $g(x)=g(y)$, then $f\bigl(g(x)\bigr)=f\bigl(g(y)\bigr)$, and therefore $x=y$. So, $g$ is indeed injective.


答:
$1/(n+1)$ decreases with $n$, so that
$$
 x_n = \frac{n}{n+1} = 1 - \frac{1}{n+1}
$$
increases with increasing $n$.

Another option is to use real analysis: $f(x) = \frac{x}{x+1}$ is increasing on $[0, \infty)$ because
$$
 f'(x) = \frac{1}{(x+1)^2} > 0
$$

Title: Monotone Sequence $\frac{n}{n+1}$


问:
Non-connected covering spaces
Assume that $X$ is a path-connected, locally path-connected, and semi locally simply-connected space. Then there is a way to know all path-connected covering spaces of $X$ by looking at the subgroups (and their conjugates) of $\pi_1(X)$. And we can combine different covering spaces to generate new (non-connected) covering spaces.

Can we generate all covering spaces of $X$ in this way?

答:
Yes. Doing it this way allows for a cleaner statement of the Galois correspondence, which naturally generalizes even to the case that $X$ is not assumed to be path-connected:

The category of not-necessarily-connected covering spaces of $X$ is equivalent to the category of sets equipped with an action of $\pi_1(X)$.

The connected covering spaces correspond to the transitive $\pi_1(X)$-sets, which correspond to conjugacy classes of subgroups. But this way of stating the Galois correspondence tells you not only what non-connected covering spaces look like but also what maps between covering spaces look like, in a nice clean way. The resulting category is also much more nicely behaved than if we restrict to the connected case, e.g. it now has products and coproducts.

To generalize to the case that $X$ is not path-connected we replace the fundamental group $\pi_1(X)$ with the fundamental groupoid $\Pi_1(X)$.

Title: Non-connected covering spaces


问:
Definite integration of a trigonometric function

How to integrate $$\int_0^{\pi/2}\!\dfrac{2a \sin^2 x}{a^2
\sin^2 x +b^2 \cos^2 x}\,dx $$

my first step is $$\frac{2}{a} \int_0^{\pi/2}\!\dfrac{a^2 \sin^2 x}{a^2 +(b^2 - a^2) \cos^2 x}\, dx $$

I would kind of want to do some sort of $u=\cos x$ substitution, to get at $\arctan u $ but no idea what to do with the sine in the numerator.

答:
$$\displaystyle \int_0^{\dfrac{\pi}{2}} \dfrac{2a \sin^2 x}{a^2\sin^2 x +b^2 cos^2 x} dx$$

$$\displaystyle 2a\int_0^{\dfrac{\pi}{2}} \dfrac{\csc^2x }{a^2\csc^2 x +b^2 \cot^2 x \csc^2 x} dx$$

$$\displaystyle 2a\int_0^{\dfrac{\pi}{2}} \dfrac{\csc^2x }{(\cot^2 x+1)(a^2+b^2\cot^2 x)} dx$$

Let $u=\cot x$

$$\displaystyle 2a\int_0^{\infty} \dfrac{du }{(u^2 +1)(a^2+b^2u^2)} $$

$$\displaystyle 2a\int_0^{\infty} \left(\dfrac{1}{(u^2+1)(a^2-b^2)}-\dfrac{b^2 }{(a^2 -b^2)(a^2+b^2u^2)}\right) du $$

$$\displaystyle \frac{2a}{a^2-b^2}\int_0^{\infty} \left(\dfrac{1}{(u^2+1)}-\dfrac{b^2 }{(a^2+b^2u^2)}\right) du $$

And I hope you'll take it from here...


问:
A question on self-adjoint operator on Hilbert space
On page 348 of Infinite Dimensional Dynamical Systems in Mechanics and Physics by Roger Temam, there is something I don't understand. I abstracted the question as follow:
Let $T:H \rightarrow H$ be a bounded self-adjoint operator on Hilbert space $H$. If $T$ is nonnegative and have bounded inverse $T^{-1}$ then why
$$
\inf_{\| \phi \|=1} \langle T\phi , \phi \rangle >0.
$$
is valid?

I found a similar question here Bounded Self-adjoint Operator on Hilbert Space, however it only provide answer when $T$ is positive, is it true for nonnegative operator as well?

答:
Let $m:=\inf_{\| \phi \|=1} \langle T\phi , \phi \rangle $ and $M:=\sup_{\| \phi \|=1} \langle T\phi , \phi \rangle $.

It is well known that for the spectrum of $T$ we have

$$ \sigma(T) \subseteq [m,M].$$

Suppose that that $m=0$, then $0 \in \sigma(T),$ a contradiction, since $T$ has a bounded inverse.


问:
Can I infer that $\sum_{i=1}^n p(A_i|A_j)=1$ from $\sum_{i=1}^n p(A_i)=1$ and $\sum_{j=1}^n p(A_i|A_j)=1$?
I have $n$ events $\{A_i\}_{i=1}^n$ such that
$$\sum_{i=1}^n p(A_i)=1,$$
i.e. the sum over $i$ of the probability that event $A_i$ occurs is $1$, and
$$\sum_{j=1}^n p(A_i|A_j)=1.$$
Can I infer that also $\sum_{i=1}^n p(A_i|A_j)=1$?

答: $1=\sum_j P(A_i|A_j)=1+\sum_{j\neq i} P(A_i|A_j)$ since $P(A_i|A_i)=1$. This gives $P(A_i \cap A_j)=0$ whenever $i \neq j$. Hence $1=\sum_i P(A_i|A_j)=1+\sum_{i\neq j} P(A_i|A_j)=1+0=1$.


问:
Convergence Jacobi and Gauss-Seidel iterations and PDE
How would you do to show that the Jacobi and Gauss-Seidel iterations will always converge towards the exact solution of the discrete version of the PDE? I'm given this information:

The solving of following partial derivative equation, in two dimensions:

$v_{xx}(x, y) + v_{yy}(x, y) = v(x, y) + f(x, y)$ ; where $f(x, y) = \sin(\pi x) \sin(\pi y)$

at any point $(x, y)$ of the square domain $[0,1] \times [0,1]$, with the boundary conditions that $v (x, y) = 0$ along all four sides of the square.

The exact answer is $v(x,y) = − \cfrac{sin(\pi x) sin(\pi y)}{1+2\pi^2}$

We also obtain a system of equations for the $v_{i,j}$ by approximating the second derivatives by finite differences :

$(4 + h^2)v_{i,j} − (v_{i-1,j} + v_{i+1,j} + v_{i,j-1} + v_{i,j+1}) = −h^2f_{i,j}$

Thank you for your help!

答:
The matrix is strictly diagonally dominant, since $4+h^2 > 1+1+1+1$. This ensures that the iteration matrices, in both Jacobi and Gauss-Seidel methods, have a norm which is less then 1 (you can take for instance the $\infty$-norm), making the sequence convergent. In the case of Jacobi's method, we have that
$$
x^{(n+1)} = \underbrace{D^{-1} b}_{g} - \underbrace{D^{-1}(L+U)}_{C}x^{(n)}=g-C x^{(n)}
$$

and, denoting as $z$ the exact solution,

$$
\|x^{(n+1)}-z\| = \|g-C x^{(n)} - (g-Cz))\|\leq \|C\| \|x^{(n)}-z\|\leq \cdots \leq \|C\|^{n+1}\|x^{(0)}-z\|
$$

Since $\|C\| < 1$, $\lim\|x^{(n)}-z\| \leq \lim \|C\|^n \|x^{(0)}-z\| = 0$.


问:
double integrals and iterated integrals
Give an example (if any) for a non-integrable function $f:\mathbb{R\times R}$ $\to$
$\mathbb{R}$ with domain in $[0,1]^2$ such that both iterated integrals exists(i.e. in both order of integration).
Here is what I have got:
$$
f(x,y) =
\begin{cases}
e^{-xy}\sin x \sin y, & \text{if }x,y \geq 0 \\
0, & \text{otherwise }
\end{cases}
$$
 Does this function work for my case? I found this in :http://www.mathnet.or.kr/mathnet/kms_tex/80630.pdf. It says that iterated integrals exist but not double integrals. I am not sure if this implies Riemann integral does not exist.</p>

答:
Consider the function

$$ f(x,y)= \frac{x^2-y^2}{(x^2+y^2)^2}. $$

Now, if you evaluate the integral

$$ \int_{0}^{1}\int_{0}^{1}f(x,y)dydx = \frac{\pi}{4},$$

and if you consider the other order, you get

$$ \int_{0}^{1}\int_{0}^{1}f(x,y)dxdy = -\frac{\pi}{4}. $$

So, the iterated integrals exist, but the double integral does not.

Title: double integrals and iterated integrals


问:
Area between two given equations
Let $x^2 + y^2$ = 6 and $x = y^2$. What is the area between them?

I solved the two equations in terms of y and I graph it in Geogebra: https://www.geogebra.org/classic/ahnk4hfw

Solving the two equations in terms of x, the graph is:
https://www.geogebra.org/classic/nwznqk3c

When I integrate in terms of y (the graph is the first one), the bounds are -$\sqrt 2$ and $\sqrt 2$. The integrand is $ \sqrt {6-y^2} $ - $y^2$. I got 4.63 as the area.
$$\int_{-\sqrt 2}^{\sqrt 2} \sqrt {6-y^2} - y^2 \, dy = 4.63 $$
Although, I am not sure if I expressed the integral correctly because it seems that I am missing one function: the -$\sqrt {6-y^2} $.

When I follow the second graph, isn't it that the whole region is bounded by the two functions, and hence the area is the circle's area? Or should I only choose the positive square roots and not the negative ones?

My last question is, what is the best approach of finding the integral of this case? Is it by integrating with respect to y or x? Do I only need to take the smaller region?

答:
The framing of the question makes it clear that the proposer has not realized there are two regions bounded by the given curves.  The safest choice (showing knowledge and mastery of the material) is to compute both areas in detail.

We plot the equations to understand the region in question.

The two curves partition the plane into four regions.  Two are unbounded.  Two are bounded, labelled $A$ and $B$.  $A$ has both curves on its boundary and $B$ has both curves on its boundary.  Therefore, both $A$ and $B$ meet the specification "Let $x^2+y^2 = 6$ and $x=y^2$. What is the area between them?"

Region $A$ is horizontally simple, with $y \in [-\sqrt{2}, \sqrt{2}]$ bounded to the right by $x = \sqrt{6 - y^2}$ and to the left by $x = y^2$.  The area of $A$ is \begin{align*}
A &= \int_{-\sqrt{2}}^{\sqrt{2}} \; \left(\sqrt{6 - y^2}\right) - \left(y^2\right) \,\mathrm{d}y  \\
    &= \int_{-\sqrt{2}}^{\sqrt{2}} \; \sqrt{6 - y^2} \,\mathrm{d}y - \left. \frac{y^3}{3} \right|_{y=-\sqrt{2}}^{\sqrt{2}}  \\
    &= \int_{-\sqrt{2}}^{\sqrt{2}} \; \sqrt{6 - y^2} \,\mathrm{d}y  \\
    &\quad {}- \left( \left( \frac{(\sqrt{2})^3}{3} \right) - \left( \frac{(-\sqrt{2})^3}{3} \right)  \right)  \\
    &= \int_{-\sqrt{2}}^{\sqrt{2}} \; \sqrt{6 - y^2} \,\mathrm{d}y - \frac{2 \sqrt{2}}{3}  \\
    & \qquad \begin{bmatrix}
        y = \sqrt{6} \sin \theta  &  \theta(\sqrt{2}) = \sin^{-1}(\sqrt{2}/\sqrt{6}) = \sin^{-1}(1/\sqrt{3})  \\
        \mathrm{d}y = \sqrt{6} \cos \theta \,\mathrm{d}\theta  &  \theta(-\sqrt{2}) = \sin^{-1}(-\sqrt{2}/\sqrt{6}) = \sin^{-1}(-1/\sqrt{3})
\end{bmatrix}  \\
    &= \int_{\sin^{-1}(-1/\sqrt{3})}^{\sin^{-1}(1/\sqrt{3})} \; \sqrt{6 - 6 \sin^2 \theta} \sqrt{6} \cos \theta \,\mathrm{d}\theta - \frac{2 \sqrt{2}}{3}  \\
    &= 6 \int_{\sin^{-1}(-1/\sqrt{3})}^{\sin^{-1}(1/\sqrt{3})} \; \sqrt{1 - \sin^2 \theta} \cos \theta \,\mathrm{d}\theta - \frac{2 \sqrt{2}}{3}  \\
    &= 6 \int_{\sin^{-1}(-1/\sqrt{3})}^{\sin^{-1}(1/\sqrt{3})} \; |\cos \theta| \cos \theta \,\mathrm{d}\theta - \frac{2 \sqrt{2}}{3}  \\
\end{align*}
Since the range of arcsine is $[-\pi/2,\pi/2]$ and cosine is nonnegative on that interval, so $|\cos \theta| = \cos \theta$ in this integral.
\begin{align*}
A &= 6 \int_{\sin^{-1}(-1/\sqrt{3})}^{\sin^{-1}(1/\sqrt{3})} \; \cos^2 \theta \,\mathrm{d}\theta - \frac{2 \sqrt{2}}{3}  \\
    &= 6 \left( \left( \frac{2(\sin^{-1}(1/\sqrt{3})) + \sin(2(\sin^{-1}(1/\sqrt{3})))}{4} \right) - \left( \frac{2(\sin^{-1}(-1/\sqrt{3})) + \sin(2(\sin^{-1}(-1/\sqrt{3})))}{4} \right) \right) - \frac{2 \sqrt{2}}{3}
\end{align*}
Now for $-\pi/2 \leq r \leq \pi/2$, \begin{align*}
\sin(2 \sin^{-1}(r)) &= 2 \sin\sin^{-1}(r) \cos \sin^{-1}(r)  \\
    &= 2 r \sqrt{1-r^2}  \text{,}
\end{align*}
so
$$  \sin(2 \sin^{-1}(1/\sqrt{3})) = 2 \frac{1}{\sqrt{3}} \sqrt{ \frac{2}{3} } = \frac{2\sqrt{2}}{3} $$
and
$$  \sin(2 \sin^{-1}(-1/\sqrt{3})) = -2 \frac{1}{\sqrt{3}} \sqrt{ \frac{2}{3} } = \frac{-2\sqrt{2}}{3}  \text{.}  $$
Continuing,
\begin{align*}
A &= 6 \left( \left( \frac{2(\sin^{-1}(1/\sqrt{3})) + \frac{2\sqrt{2}}{3}}{4} \right) - \left( \frac{2(\sin^{-1}(-1/\sqrt{3})) + \frac{-2\sqrt{2}}{3}}{4} \right) \right) - \frac{2 \sqrt{2}}{3}  \\
    &= 6 \left( \left( \frac{2(\sin^{-1}(1/\sqrt{3})) + \frac{2\sqrt{2}}{3}}{4} \right) - \left( \frac{-2(\sin^{-1}(1/\sqrt{3})) + \frac{-2\sqrt{2}}{3}}{4} \right) \right) - \frac{2 \sqrt{2}}{3}  \\
    &= 6 \left(\frac{4(\sin^{-1}(1/\sqrt{3})) + 4\frac{\sqrt{2}}{3}}{4} \right) - \frac{2 \sqrt{2}}{3}  \\
    &= 6 \left(\sin^{-1}\left( \frac{1}{\sqrt{3}} \right) + \frac{\sqrt{2}}{3} \right) - \frac{2 \sqrt{2}}{3}  \\
    &= 6 \sin^{-1}\left( \frac{1}{\sqrt{3}} \right) + \frac{4\sqrt{2}}{3}  \\
    &= 4.635\,687\,293\,604\,387\,412{\dots} \text{.}
\end{align*}

The region $B$ is neither horizontally simple nor vertically simple.  However, it is the disk of radius $\sqrt{6}$ with region $A$ removed, so its area is
\begin{align*}
B &= \pi (\sqrt{6})^2 - A  \\
    &= 6 \pi - \left( 6 \sin^{-1}\left( \frac{1}{\sqrt{3}} \right) + \frac{4\sqrt{2}}{3} \right)    \\
    &= 14.213\,868\,627\,934\,372\,018{\dots}  \text{.}
\end{align*}

The sum of the areas of regions $A$ and $B$ is that of the disk of radius $\sqrt{6}$, which is $6 \pi$.

[ Alternatively, we can decompose region $B$ into three vertically simple regions:

the left half-disk on $[-\sqrt{6},0]$ bounded above by $\sqrt{6 - x^2}$ and below by $-\sqrt{6 - x^2}$,

the portion in quadrant I, on $[0,2]$ bounded above by $\sqrt{6-x^2}$ and below by $\sqrt{x}$, and

the portion in quadrant II, on $[0,2]$ bounded above by $-\sqrt{x}$ and below by $-\sqrt{6-x^2}$.

These last two regions are equal by symmetry, so maybe we avoid evaluating one of them. ]

[ Alternatively, we can decompose region $B$ into
three horizontally simple regions

on $[-\sqrt{6}, -\sqrt{2}]$ right bounded by $\sqrt{6-y^2}$ and left bounded by $-\sqrt{6-y^2}$,

on $[-\sqrt{2},\sqrt{2}]$, right bounded by $y^2$ and left bounded by $-\sqrt{6-y^2}$, and

on $[\sqrt{2}, \sqrt{6}]$, right bounded by $\sqrt{6-y^2}$ and left bounded by $-\sqrt{6-y^2}$.

The first and third of these regions have the same area by symmetry so maybe we avoid evaluating one of them. ]

Title: Area between two given equations


答:
Opening the summation you get P(Ai|A1)+...+P(Ai|Ai)+...=1
But P(Ai|Ai)= 1.
Sum of other probabilities is 0 means they are all individually 0. And this holds for all i. So all these events are mutually exclusive and exhaustive events (as sum of individual probabilities is 1).
Hence, in the final inference sum, open the summation to get 0+0+..+1+...0= 1.
So yes, your inference is correct.</p>

Title: Can I infer that $\sum_{i=1}^n p(A_i|A_j)=1$ from $\sum_{i=1}^n p(A_i)=1$ and $\sum_{j=1}^n p(A_i|A_j)=1$?

Body:
I have $n$ events $\{A_i\}_{i=1}^n$ such that
$$\sum_{i=1}^n p(A_i)=1,$$
i.e. the sum over $i$ of the probability that event $A_i$ occurs is $1$, and
$$\sum_{j=1}^n p(A_i|A_j)=1.$$
Can I infer that also $\sum_{i=1}^n p(A_i|A_j)=1$?


问:
Solve :- $|x^2 - 1| < 3x + 3|x-2|$
I was trying to solve this inequality with two absolute values (see below), I don't know why I got the wrong answer, because this is the standard method for this type of exercises.
I've got four different systems of inequalities, but two (those in the middle) of them don't exist in real set of solutions.
The result must be $-\sqrt7<x<5$, but I got another result.
Before doing this one, I've done other similar exercises, therefore I think I've got an error in this exercise because of frivolous errors I didn't take into account, but let me know.

My attempt to solve it:
$$|x^2 - 1| < 3x + 3|x-2|$$
$$\rightarrow|x^2 - 1| \rightarrow (x \leq -1) /(x \geq 1)$$
$$\rightarrow |x-2| \rightarrow x \geq 2$$

If ($x < -1$), then I've got:-=

$$x^2 - 1 < 3x -3x + 6$$
$$\rightarrow -\sqrt7 < x < \sqrt7$$

If ($-1 < x < 2$), then both of them are negative and in this interval I've got no set of solutions.

If ($x>2$), then I got:-
$x^2 - 1 < 3x + 3x -6$
$\rightarrow 1<x<5$.

Now, I've put on the number line in these interval in order to check all the possible solutions. I got:- $$\rightarrow 2<x<\sqrt7$$
But it's not the correct result. In fact, it must be :- $$\rightarrow-\sqrt7<x<5$$

Can anyone help me?

答:
If I wasn't wrong, I think what you did was let $x^2 - 1 > 0$ and $x - 2 \geq 0$, then take the solution of those two inequalities as intervals to check the sign in the absolute and solve the original inequality, and that would be wrong.

Let look at the inequality again. At first we have:

$$|x^2-1| =
\begin{cases}
x^2 - 1  & \text{if $x^2-1 \geq 0 \implies x \in (-\infty, -1] \cup [1, +\infty)$} \\
1 - x^2 & \text{if $x^2-1 < 0 \implies x \in (-1, 1)$}
\end{cases}
$$

$$|x - 2| =
\begin{cases}
x - 2  & \text{if $x \geq 2$} \\
2 - x & \text{if $x < 2$}
\end{cases}
$$

Within that, we should consider $4$ different cases:

Case $1 \,(x \leq -1) : $

$$\begin{align}
& x^2 - 1 < 3x + 3(2-x) \\
\implies & x^2 < 7 \\
\implies & -\sqrt{7} < x < \sqrt{7}
\end{align}$$
With $x \leq -1 \implies -\sqrt{7} < x \leq -1 \, (1)$

Case $2 \,(-1 < x < 1) : $

$$\begin{align}
& 1 - x^2 < 3x + 3(2-x) \\
\implies & x^2 + 5 > 0 \,\, \text{(always true)}
\end{align}$$

Therefore $-1 < x < 1$ is a solution for case $2$ $(2)$

Case $3 \,(1 \leq x < 2) : $

$$\begin{align}
& x^2 - 1 < 3x + 3(2-x) \\
\implies & x^2 < 7 \\
\implies & -\sqrt{7} < x < \sqrt{7}
\end{align}$$
With $1 \leq x < 2 \implies 1 \leq x < 2 \, (3)$

Case $4 \,(x \geq 2) : $

$$\begin{align}
& x^2 - 1 < 3x + 3(x - 2) \\
\implies & x^2 - 6x + 5 < 0 \\
\implies & 1 < x < 5
\end{align}$$
With $x \geq 2 \implies 2 \leq x < 5 \, (4)$

From $(1), (2), (3), (4)$, you get the sotultion to the inequality:

$$-\sqrt{7} < x < 5$$

Title: Solve :- $|x^2 - 1| < 3x + 3|x-2|$


问:
Proving that at least one of G and its dual graph G* has a vertex v of degree at most 3.
Recently, I read one literature and saw a property about the planar graph. I thought about it for a long time without thinking out the reason:

Let $G$ be a planar graph. Then  at least one of $G$ and its
dual graph $G^*$ has a vertex $v$ of degree at most 3.

I know some  conclusions that may be useful.

Every planar graph has a vertex of degree at most $5$.

The number of edges in a maximal planar graph is $3n-6$.

Definition of dual graph:

Given a plane graph $G$, one can define a second graph $G^∗$ as follows. Corresponding to each face $f$ of $G$ there is a vertex $f^∗$ of G∗ , and corresponding to each edge $e$ of $G$ there is an edge $e^∗$ of $G^∗$ . Two vertices $f^∗$ and $g^∗$ are joined by the edge $e^∗$ in $G^∗$ if and only if their corresponding faces $f$ and $g$ are separated by the edge $e$ in $G$.
Observe that if e is a cut edge of G, then $f = g$, so $e^∗$ is a loop of G∗; conversely,if e is a loop of $G$, the edge $e^∗$ is a cut edge of $G^∗$. The graph $G^*$ is called dual of G.

答:
Let $n$ be the number of vertices of $G$, $m$ the number of edges of $G$, and $r$ the number of faces of $G$. We use the notation $\mathcal{E}(f)$ for the number of edges on the face $f$.

We will assume that $G^*$ has no vertex of degree $3$ or less, and prove that $G$ must then have a vertex of degree at most $3$.

If $G^*$ has no vertex of degree $3$ or less, than every face of the original graph $G$ has at least $4$ edges on its boundary.
Since every edge appears twice on a face boundary (either once on two different faces, or twice on one face if it is a cut edge), we get that:

$$\sum_{f\text{ a face of $G$}} \mathcal{E}(f) = 2m$$

We have assumed that $G^*$ has minimum degree at least $4$, so this gives:

$$2m = \sum_{f\text{ a face of $G$}} \mathcal{E}(f) \geq 4r$$

By substituting $r\leq \frac{m}{2}$ into Euler's Formula ($n-m+r = 2$), we get that:

$$2 = n-m+r \leq n-m + \frac{m}{2}$$

Rearrange this to bound the number of edges as $m\leq 2n-4$.

Since $2m = \sum_{v\in V(G)}d(v)$, we get that the sum of the degrees is bounded by:

$$\sum_{v\in V(G)}d(v) = 2m \leq 4n-8 < 4n$$

The sum of the degrees of $G$ is less than $4n$, so the average degree of $G$ is less than $4$ (the average degree is the sum of the degrees divided by $n$). This means there is at least one vertex with degree less than $4$, completing the proof.


问:
Challenging Integration, cannot get anywhere with by parts or by sub.
Doing some math for a project of mine where I have used this equation for a line of best fit:

$$ (1-(1-x)^a)^{\frac{1}{a}} $$

I need to find the area underneath this curve, between $0$ and $1$, for all real values of a between $0$ and $1$.

Does anyone know how to do this efficiently or elegantly?

Body:
Starting from @Greg Martin's solution
$$I(a)=\left(1+\frac2a\right)B(1+\frac1a,1+\frac1a)$$ you can  also write it as
$$I(a)=\sqrt{\pi }\left(1+\frac2a\right) 2^{-\frac{a+2}{a}}\frac{ \Gamma \left(1+\frac{1}{a}\right)}{\Gamma
   \left(\frac{3}{2}+\frac{1}{a}\right)}$$ which, may be, is more familiar to you.

Now, we can make approximations.

Close to $a=0$, we have
$$I(a)=\sqrt{\pi a}\left(1+\frac2a\right) 2^{-\frac{a+2}{a}}\Big[1+\sum_{k=1}^n c_k\, a^k \Big]$$ where the $c_k$'s make the sequence
$$\left\{-\frac{3}{8},\frac{25}{128},-\frac{105}{1024},\frac{1659}{32768},-\frac{6237
   }{262144},\frac{50765}{4194304},-\frac{242385}{33554432}\right\}$$

Trying for a few values of $a$
$$\left(
\begin{array}{ccc}
a & \text{approximation} & \text{exact} \\
 0.15 & 0.000452 & 0.000452 \\
 0.20 & 0.003968 & 0.003968 \\
 0.25 & 0.014286 & 0.014286 \\
 0.30 & 0.033065 & 0.033065 \\
 0.35 & 0.059604 & 0.059604 \\
 0.40 & 0.092039 & 0.092039 \\
 0.45 & 0.128318 & 0.128319 \\
 0.50 & 0.166664 & 0.166667 \\
 0.55 & 0.205700 & 0.205707 \\
 0.60 & 0.244436 & 0.244451 \\
 0.65 & 0.282193 & 0.282227 \\
 0.70 & 0.318535 & 0.318605 \\
 0.75 & 0.353194 & 0.353328 \\
 0.80 & 0.386022 & 0.386266 \\
 0.85 & 0.416948 & 0.417373 \\
 0.90 & 0.445946 & 0.446661 \\
 0.95 & 0.473017 & 0.474179 \\
 1.00 & 0.498163 & 0.500000
\end{array}
\right)$$

Edit

We can improve the approximation for "large" values of $a$, writing
$$\frac{ \Gamma \left(1+\frac{1}{a}\right)}{\Gamma
   \left(\frac{3}{2}+\frac{1}{a}\right)}=\frac{4}{3 \sqrt{\pi }}+\sum_{k=1}^n d_k\, (a-1)^k $$
$$d_1=\frac{4 (5-6 \log (2))}{9 \sqrt{\pi }} \qquad d_2=\frac{52-6 \pi ^2+12 \log (2) (6 \log (2)-4)}{27 \sqrt{\pi }}$$

Update

We can have a significant improvement writing
$$I(a)=\sqrt{\pi a}\left(1+\frac2a\right) 2^{-\frac{a+2}{a}}\color{red}{\frac{ \Gamma \left(1+\frac{1}{a}\right)}{\sqrt a \,\Gamma
   \left(\frac{3}{2}+\frac{1}{a}\right)}}$$ and build for the "red" term its $[4,4]$ Padé approximant.

This will write
$$\frac{ \Gamma \left(1+\frac{1}{a}\right)}{\sqrt a \,\Gamma
   \left(\frac{3}{2}+\frac{1}{a}\right)}=
\frac {1+\sum_{k=1}^4 \frac {b_k } {c_k }a^k } {1+\sum_{k=1}^4 \frac {d_k } {c_k }a^k }$$ The coefficients $(b_k,c_k,d_k)$ are given below

$$\left(
\begin{array}{cccc}
 k & b_k & c_k & d_k \\
 1 & -45050201133 & 24342404336 & -35921799507 \\
 2 & -157157838285 & 389478469376 & -448758648877 \\
 3 & -3283793920545 & 3115827755008 & -3412530822591 \\
 4 & -13005546423289 & 99706488160256 & -51653145742649
\end{array}
\right)$$

Comparison
$$\left(
\begin{array}{ccc}
a & \text{approximation} & \text{exact} \\
 0.15 & 0.000452 & 0.000452 \\
 0.20 & 0.003968 & 0.003968 \\
 0.25 & 0.014286 & 0.014286 \\
 0.30 & 0.033065 & 0.033065 \\
 0.35 & 0.059604 & 0.059604 \\
 0.40 & 0.092039 & 0.092039 \\
 0.45 & 0.128326 & 0.128319 \\
 0.50 & 0.166669 & 0.166667 \\
 0.55 & 0.205709 & 0.205707 \\
 0.60 & 0.244455 & 0.244451 \\
 0.65 & 0.282234 & 0.282227 \\
 0.70 & 0.318614 & 0.318605 \\
 0.75 & 0.353342 & 0.353328 \\
 0.80 & 0.386286 & 0.386266 \\
 0.85 & 0.417402 & 0.417373 \\
 0.90 & 0.446702 & 0.446661 \\
 0.95 & 0.474235 & 0.474179 \\
 1.00 & 0.500075 & 0.500000
\end{array}
\right)$$ which is much better that before.

Notice, that the difference between this Padé approximant and the Taylor series of the red term is about $\frac{a^9}{447}$.

Encouraged by this improvement, a nonlinear regression was made based on $1001$ equally spaced data points (there is a constraint on coefficient $c_4$ to match the end point). Making the coefficients rational, the result is
$$\frac{ \Gamma \left(1+\frac{1}{a}\right)}{\sqrt a \,\Gamma
   \left(\frac{3}{2}+\frac{1}{a}\right)}=
\frac{1+\frac{6616 }{4209}a+\frac{2610 }{2143}a^2+\frac{343 }{1313}a^3+\frac{5}{793}a^4 } {1+\frac{4727 }{2428}a+\frac{3104 }{1771}a^2+\frac{619 }{966}a^3+\frac{173}{3250}a^4 }$$ and, as shown below the fit is extremely good with very highly significant coefficients.

$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\
 b_1 & 1.5718699 & 0.00091 & \{1.57008,1.57366\} \\
 b_2 & 1.2179188 & 0.00088 & \{1.21620,1.21964\} \\
 b_3 & 0.2612339 & 0.00026 & \{0.26073,0.26174\} \\
 b_4 & 0.0063051 & 0.00001 & \{0.00628,0.00633\} \\
 c_1 & 1.9468699 & 0.00091 & \{1.94508,1.94866\} \\
 c_2 & 1.7526822 & 0.00122 & \{1.75029,1.75507\} \\
 c_3 & 0.6407868 & 0.00054 & \{0.63973,0.64184\} \\
\end{array}$$

Using the above, the maximum absolute error on $I(a)$ is smaller than $\color{red}{4\times 10^{-9}}$.


问:
Are these graph coloring algorithms equivalent?
Suppose you want to color the vertices of a graph in a greedy fashion, given a predetermined order of these vertices.

I am wondering if these two algorithms are equivalent:

Algorithm 1: Consider each vertex (in the given order) and assign the smallest color available.

Algorithm 2: While all vertices are not colored, sequentially build color classes by trying to include vertices (in the given order) in the current class.

I am almost sure that these two algorithms are equivalent, but a confirmation would be great! Thanks.

答:
Assuming that your second algorithm does the following: For each colour $i\in [1,c]$ in order, go through the vertices $v_1,\ldots, v_n$ in order and assign to a vertex the colour $i$ if it is not already coloured, and no neighbour of it has colour $i$.

Let us suppose for contradiction that algorithm II doesn't produce the same colouring as algorithm I, let us call these colourings $c_1$ and $c_2$. There is some smallest colour $i$ which appears in the wrong place, and some smallest vertex $v_j$ given the colour $i$ by algorithm II where algorithm I give $v_j$ a different colour. That is $c_2(v_j) = i$ and $c_1(v_j) \neq i$. That is, we assume that for all $i'<i$ and $v_k$ if $c_1(v_k)=i'$, then $c_2(v_k)=i'$ and also if $v_k < v_j$ and $c_1(c_k) = i$, then $c_2(v_k)=i$.

However, since $c_1(v_j) \neq i$ either there is a neighbour $v_k$ of $v_j$ with $v_k < v_j$ and $c_1(v_k)=i$ or $c_1(v_j)=i' < i$ and so $v_j$ has no neighbour coloured $i'$ in $c_1$. In the first case, $c_2(v_k)=i$ and so we can't also colour $c_2(v_j)=i$, contradicting our assumption. In the second case we can conclude that $v_j$ has no neighbour coloured $i'$ in $c_1$ and hence no neighbour coloured $i'$ in $c_2$, and so when we tried to colour $v_j$ with colour $i'$ in algorithm II, which happens before we try to colour it with colour $i$, we were succesful, again contradiction our assumption.


问:
How are continued fractions useful?
On Wolfram Alpha, I see continued fractions being listed in the results. Although I understand continued fractions, and how they can be used for approximations, what is a better approximation than a decimal representation of the number?

For example,

31.999999999664

tells me so much more than

So, I do not get why continued fractions are important. Are they supposed to tell me something which I am missing all these days?

答:
How are continued fractions cool? Let me count the ways:

If you know how to get the partial terms of the continued fraction expansion of an irrational number, you essentially have a way to find approximate values of the irrational number. This gives you one possible way to get a floating-point estimate for $\sqrt{2}$, $\pi$, etc.

It is known that a number is irrational if and only if it has an infinite continued fraction expansion. This makes it a sort of irrationality test.

Continued fraction expansions of irrational numbers exhibit sometimes surprising regularity. $\pi$ and $e$ have continued fraction representations that are simple, which is strange seeing as they are transcendental numbers. This can sometimes tell you things about these numbers, and other times the existence of such a regular expansion is just plain cool.

As mentioned in the comments, continued fractions can be used to solve certain equations like the Pell equation.

And others. Continued fractions solve some other problems - a famous story relates that Mahalanobis and Ramanujan shared a room, and one day Mahalanobis posed Ramanujan a problem which he instantly solved for all cases via continued fractions.

Title: How are continued fractions useful?


问:
Monotone Sequence $\frac{n}{n+1}$
I have to proof this sequence :
$(x_n) := \frac{n}{n+1}$ for $n\in \Bbb N^+$ is a monotone sequence.
I saw some examples of people using the induction method, so I tried it, but I got stuck when I was trying to prove the $A(k+1)$ is true.

Here's what I have done so far

Let $A(n) := x_n \le x_{n+1}$

$A(n)$ is true for all $n\ge1$. Assume that $A(k):=x_{k}\ge x_{k+1}$ is true, then show that $A(k+1)$ is also true.

I don't know where I should go next. I've been trying to add both side, but can't find the desired end. Should I change the proving method? For some backgrounds, before this question I had to prove that $(x_n)$ is convergent and bounded.

答:
As others said, induction is not the preferred way. Let us try anyway.

Base case:

$$\frac{0}{0+1}\le\frac{1}{1+1}$$ is true.

Inductive step:

We have to show that

$$\frac n{n+1}\le\frac{n+1}{n+2}\implies\frac{n+1}{n+2}\le\frac{n+2}{n+3}$$ or equivalently, subtracting $1$,

$$-\frac 1{n+1}\le-\frac1{n+2}\implies-\frac1{n+2}\le-\frac1{n+3}.$$

Then taking the opposite of the inverse,

$$n+1\le n+2\implies n+2\le n+3$$ which amounts to the trivial truth

$$1\le2\implies 2\le 3.$$

[By the way, this proof also contains the non-inductive proof.]

Title: Monotone Sequence $\frac{n}{n+1}$


问:
Show that the triangle $ADC$ is isosceles
Let $K$ be a circle with center $M$ and $L$ be a circle that passes through $M$ and intersects $K$ in two different points $A$ and $B$ and let $g$ be a line that goes through $B$ but not through $A$. Let $C$ be the second point of $g$ on $L$ and let $D$ be the second point of $g$ on $K$ with $C=B$ or $D=B$ if $g$ is the tangent through $B$ on the circle $K$ or $L$. Show that the triangle $ADC$ is isosceles.

For that do we have to show that that the length of each side of triangle is equal? Or how can we show that?

$$$$

EDIT:

We get the following, or not?

答:
Let's look at the two tangent cases also : when $g$ is tangent to $K$ (call it $g_1$) and when $g$ is tangent to $L$ (then call it $g_2$). Both can be shown on same diagram.

Let $\angle ADB = \alpha$. So $\angle AMB = 2\alpha$ and $\angle ACB=180-2\alpha$.

Now as $g_2$ is tangent to red circle, $\angle ABD = \angle ACB=180-2\alpha$ by alternate segment theorem. Similarly as $g_1$ is tangent to black circle, $\angle ABC= \angle ADB=\alpha$.

The third angles can be calculated. $\angle DAB = \alpha = \angle BAC$.

We have found both $\triangle ABD$ and $\triangle ACB$ are isosceles and similar.

Note :

According to question statement, in case of $g_1$, $B=C$ , $\triangle ADB$ in diagram is $\triangle ADC$. While in case of $g_2$, $B=D$, $\triangle ABC$ in diagram is $\triangle ADC$. So we have shown in both cases $\triangle ADC$ (according to question) is isosceles.


答:
Here is another possible construct -

Please note the proof is very similar. Quadrilateral $AMBC$ is cyclic so

$\angle ACD = 180^0 - \angle AMB$

$\angle ADB = 180^0 - \frac{1}{2} \angle AMB \implies \angle ADC = \frac{1}{2} \angle AMB$

Given $\angle ACD$ and $\angle ADC \,$ in $\triangle ADC, \angle CAD = \frac{1}{2} \angle AMB$.

So $\triangle ADC$ is isosceles.

Body:
Let $K$ be a circle with center $M$ and $L$ be a circle that passes through $M$ and intersects $K$ in two different points $A$ and $B$ and let $g$ be a line that goes through $B$ but not through $A$. Let $C$ be the second point of $g$ on $L$ and let $D$ be the second point of $g$ on $K$ with $C=B$ or $D=B$ if $g$ is the tangent through $B$ on the circle $K$ or $L$. Show that the triangle $ADC$ is isosceles.

For that do we have to show that that the length of each side of triangle is equal? Or how can we show that?

$$$$

EDIT:

We get the following, or not?


问:
Proof that the sequence of functions $f_n(x) = \sin(nx)$ doesn't have a convergent subsequence in $C[0,2 \pi]$.
Proof that the sequence of functions $f_n(x) = \sin(nx)$ doesn't have a convergent subsequence in $C[0,2 \pi]$.

I have came across this question in Rudin's PRINCIPLES OF
MATHEMATICAL ANALYSIS (Chapter 7) but the proof is using dominated convergence theorem. See below

Is there any proof which doesn't use Measure Theory?

答:
I'd recommend to put the problem into the body of your question, it's not good that it's only in the title.

Not knowing that book, I can't imagine why anybody would need the dominated convergence theorem for a proof: convergence in $C[0,2 \pi]$ is uniform convergence, so integrals of such functions converge as well, without any measure theory.

Whatever: let $g_k(x)=\sin n_k x$. If that's a convergent sequence, you must have $\|g_k-g_l\|<\epsilon$ for $k,l>N_\epsilon$, and that implies $\int^{2\pi}_0(g_k(x)-g_l(x))^2\,dx<2\pi\epsilon^2$, while in reality $\int^{2\pi}_0(g_k(x)-g_l(x))^2\,dx=\int^{2\pi}_0g_k(x)^2\,dx+\int^{2\pi}_0g_l(x)^2\,dx=2\pi$, since $2\int^{2\pi}_0g_k(x)g_l(x)\,dx=\int^{2\pi}_0(\cos(n_k-n_l)x-\cos(n_k+n_l)x)\,dx=0$.


问:
Converting base 10 fractions into other bases
How do you convert

$14\frac{8}{13}$

into base 3?

I was able to convert $\frac{3}{7}$ into base 3 by constantly multiplying by 3 and dividing the numerator by denominator until I finally got a repetition, but this method doesn't seem to work for $14\frac{8}{13}$.

The correct answer is supposed to be $112.\overline{121}_{3}$

Body:
You convert 8 and 13 into base 3, and do a long division in that base,

          0.202           8 = 101
      ------------           -----
  111 ) 101              13 = 111
         22.2
         -----
          1.100
          0.222
            101  (repeats on three places)

So the answer is 112.202 202 202


问:
Hartshorne Prop II.6.6
I'm trying to understand an argument in Hartshorne's proof that if the local rings of dimension 1 in $X$ are regular, then so are the local ring of dimension 1 in $X \times \mathbb{A}^1$. My problem is in the "first type" points:

We can reduce to the affine case $X = \operatorname{Spec}\left(R\right)$, and in that case first type points are of the form $x := \mathfrak{p}\left[t\right] \subset R \otimes \mathbb{Z}\left[t\right] \simeq R\left[t\right]$ where $y := \mathfrak{p} \subset R$ is a point of dimension 1 in $X$. In that case, the claim is that the local ring $\mathscr{O}_x$ is isomorphic to $\mathscr{O}_y[t]_{\mathfrak{m}_y}$, and then since $\mathscr{O}_y$ is a DVR it is "obvious" that so is $\mathscr{O}_x$.

First, I guess $\mathscr{O}_y[t]_{\mathfrak{m}_y}$ means $\mathscr{O}_y[t]_{\mathfrak{m}_y\left[t\right]}$, am I right? Then why is it obvious that $\mathscr{O}_x$ is a DVR?

答: Yes, you're right that the $\mathfrak{m}_y$ in  $\mathcal{O}_y[t]_{\mathfrak{m}_y}$ means $\mathfrak{m}_y[t]$ (or equivalently, $(\mathfrak{m}_y)$, the ideal generated by $\mathfrak{m}_y$ inside $\mathcal{O}_y[t]$) - we invert everything not in the prime ideal generated by $\mathfrak{m}_y\subset \mathcal{O}_y[t]$ (this ideal is prime because the quotient is $\kappa(y)[t]$, a domain). The discrete valuation on $\mathcal{O}_y[t]_{\mathfrak{m}_y[t]}$ is given by defining the valuation to be the valuation on $\mathcal{O}_y$ for any element of $\mathcal{O}_y[t]$ which in that subring and setting $v(t)=0$, then extending by the properties of valuations.

Title: Hartshorne Prop II.6.6


问:
Number of $n \to p \bmod n$ before getting to 0
There isn't much background context, but is there any estimations on how many iterations of $n \to p\% n$ are needed before $n$ becomes 0? Percentage sign is modulo. ($p$ is fixed, prime in my context but not sure if it matters).

For example, when $p=10^9+7$, the maximum number of iterations needed is $50$. When $p=998244353$ (a common prime in competitive programming) the number of iterations is $45$. I don't see any proof that it requires approximately $\log p$, but it seems like so.

Thank you for your time and help.

答:
Let us remove the constraint that $p$ is a prime number. Given a strictly decreasing sequence $1=a_0 < a_1 < \ldots < a_k$ and consider the set

$$ S(\textbf{a}) := \{p: p \equiv a_j \pmod{a_{j+1}} \ \ \forall j = 0 \ldots k-1\} $$

This is exactly the set of numbers that will produce the sequence $\textbf{a}$ during this process. Note that this set is stable for shifting by $\text{lcm }\{ a_1, \ldots, a_k\}$, and that the set of solutions is determined by applying the chinese remainder theorem.

We will consider two particular sequences to get the following result:
(1) there exist arbitrarily large (prime) numbers that ends in $2$ iterations;
(2) there exist arbitrarily large numbers that ends in at least (about) $(\log p) /2 $ iterations.

For the first just consider the sequence $3>1$ and take a (prime) number $p\equiv 1 \pmod{3}$.

The second is more interesting. Consider the sequence $k > \ldots > 1$. This means that we have $p$ such that $p\equiv  m-1 \equiv  -1 \pmod{m}$ for all $m\le k$. Solutions of this system are exactly numbers that are $\equiv -1$ modulo $L_k:= \text{lcm}\{1, \ldots, k\}$. The smallest such positive number is exactly $L_k -1$. We can give an estimate of $\log L_k$ as starting from the equation

$$ L_k = \prod_{p \le k} p^{[\log k/\log p]}$$

Indeed, any prime will appear with the maximum exponent it can appear with. Taking the logarithm we get

$$ \log L_k = \sum_{p\le k} (\log p) [\log_p k] \le \sum_{p \le k} (\log p) (\log_p k +1) $$
$$ = \sum_{p\le k} \log k + \log p \le \sum_{p\le k} 2 \log k \approx (2 \log k) \frac{k}{\log k} = 2k $$

Using the prime counting theorem $|\{p\le k\}| \approx k/\log k$. Recall that $k$ is the number of iterations relative to the input number $p=L_k-1$. Thus we get
$$(
 \log p )/2 \approx ( \log L_k)/2 < k$$

Giving the desired solution.


问:
Sum of $\frac{\Lambda(n)}{\log(n)}$ over divisors
I am wondering if $\sum_{d\mid n}\frac{\Lambda(d)}{\log(d)}$ (which evaluates to $1/k$ for $d=p^k$ and $0$ otherwise) has any interesting significances or bounds.

For $n = p_1^{k_1}\dots p_s^{k_s}$, we would have

$$\sum_{d\mid n}\frac{\Lambda(d)}{\log(d)} = \sum_i H_{k_i}$$
where $H_k$ is the $k$-th harmonic number, but is there a good way to approximate this?

Here is the plot for $(n, \sum_{d\mid n}\frac{\Lambda(d)}{\log(d)})$ for $1\le n \le 50000$:

答:
You can for example derive the inequalities
$$
\sum\limits_{d|n} {\frac{{\Lambda (d)}}{{\log d}}} = \sum\limits_i {H_{k_i } }  \le \sum\limits_i {(\log (k_i  + 1) + \gamma )}  = \log d(n) + \omega (n)\gamma
$$
and
$$
\sum\limits_{d|n} {\frac{{\Lambda (d)}}{{\log d}}} = \sum\limits_i {H_{k_i } }  \ge \sum\limits_i {\log (k_i  + 1)}  = \log d(n).
$$
Here $\gamma$ is the Euler$-$Mascheroni constant, $d(n)$ is the divisor function and $\omega(n)$ counts each distinct prime factor of $n$. In particular,
$$
\log 2 \le \mathop {\lim \sup }\limits_{n \to  + \infty } \frac{{\log \log n}}{{\log n}} \sum\limits_{d|n} {\frac{{\Lambda (d)}}{{\log d}}}  \le \log 2 + \gamma .
$$


问:
What is the coefficient of $y$ in the equation of the tangent line? $y = 2x^2 + 4x$; at $(-2,0)$
Find the general equation of the line tangent to the equation below at any given point.

What is the coefficient of $y$ in the equation of the tangent line? $y = 2x^2 + 4x$; at $(-2,0)$

Please elaborate every exact part such as: What is the constant? What is the slope? And what is the coefficient of $x$ if its ever even there?

The quizzes I took are under the stupid assumption that I know all of these terms.

Edit:  The derivative is $4x + 4$.

答:
Strategy:  The tangent line to a curve at the point is a line passing through the point whose slope is equal to the derivative at that point.

Find the derivative of the function.

Evaluate the derivative at the point.  That is the slope of the tangent line.

If the point is $(a, b)$, then the equation of the tangent line is $$y - b = m(x - a)$$ where $m = f'(a)$.

In this case, $(a, b) = (-2, 0)$.  You correctly found that $f'(x) = 4x + 4$, so $m = f'(-2)$.  I'll leave the remaining details to you.


问:
Trying to solve a non-linear differential equation
How would I go about solving for x(t) in the equation below?

$\dfrac{\dot{x}}{\sqrt{1+\dot{x}^2}}=-\dfrac{t}{\lambda}+ C$

It seems as though it is a nonlinear differential equation and I cant seem to find a method that may suitably solve this equation.

I am told that:

The solution consists of circles and is obtained by the substitution of $\frac{dx}{dt}= \tan{\psi}$ or through direct integration

I tried doing the $\tan$ substitution but dont know how to follow through on it. Any guidance would be appreciated.

答:
Just use the direct integration
$$\dfrac{x'}{\sqrt{1+(x')^2}}=f(t)\implies x'=\pm\frac{f(t)}{\sqrt{1-\big[f(t)\big]^2}}$$
$$x=\pm \int \frac{f(t)}{\sqrt{1-\big[f(t)\big]^2}}\,dt+K$$
If $f(t)$ is a linear function of $t$, this not difficult. Say $f(t)=at+b$, make $at+b=y$ to face
$$x=\pm \frac 1a \int \frac{y}{ \sqrt{1-y^2}}\,dy+K=\pm\frac 1a \sqrt{1-y^2}+K$$


问:
Convergence in probability is induced by a distance
I'd like to prove the following : $X_n \overset{P}{\longmapsto} X \iff d(X_n,X) = 0$ where the distance is defined as $d(X,Y) = \mathbb{E}[\lvert X-Y\rvert \wedge 1]$

For $[\Leftarrow]$ I think we can use markov inequality to state that :

$$P(\lvert X_n -X \rvert > \varepsilon) \leq P(\lvert X_n -X \rvert > \varepsilon \wedge 1) \leq P(\lvert X_n -X \rvert\wedge 1 > \varepsilon \wedge 1) \overset{\text{Markov}}{\leq} \frac{d(X_n,X)}{\varepsilon^2} \longmapsto 0$$

But I'm stuck with $[\Rightarrow]$, in particular i'm struggling to understand the following proof found :

$$\mathbb{E}[\lvert X_n -X \rvert \wedge 1] = \int_{0}^1 P(\lvert X_n -X \rvert > t)\text{dt}  + P(\lvert X_n -X \rvert > 1)$$

And to pass the limits inside the integral on the first term we can use dominated convergence theorem and for the second term we note that goes to $0$ with $n$ by hyphotesis.

I don't why Cavalieri's principle would leave the term $P(\lvert X_n -X \rvert > 1)$ since we are using Lebesgue measure, i.e $\int_{1}^{+\infty} P(\lvert X_n -X \rvert \wedge 1 > t)\text{dt}  = P(\lvert X_n -X \rvert > 1)$.

答: $E(|X_n-X| \wedge 1)= E[(|X_n-X| \wedge 1) I_{|X_n-X | >1})]+E[(|X_n-X| \wedge 1) I_{|X_n-X | \leq 1})]$. The first term is $P(|X_n-X| >1$. The second term is $\int_0^{\infty} P((|X_n-X|\wedge 1) I_{|X_n-X | \leq 1} >t )dt $ But this is equal to the  integral form $0$ to $1$.