问:
Impose PDE itself as Boundary Condition?
Consider, for example, the elliptic PDE $u_{x}+u_y+u_{xx}+u_{yy}=0$ for $(x,y)\in[0,\infty)^2$.

Solution methods often require me to impose boundary conditions. Often, these arise naturally from applications (physics, biology, economics etc.). But sometimes, one may not know ex ante how the function should behave on a particular boundary.

Suppose my modeling setting imposes a behavior on $u$ as $x\to\infty$, $y=0$ and $y\to\infty$. However, my application in mind does not give me any priors on what $u$ does as $x=0$. Nonetheless, I have to impose something on this boundary. Can I simply impose that, as $x\to0$, the following inhomogeneous ODE holds
$$u_x(0,y)+u_y(0,y)+u_{xx}(0,y)+u_{yy}(0,y)=0.$$
Because we are on the lower boundary of the $x$-domain, we can impose this conditions via finite (forward) differences,
\begin{align*}
\frac{u_{1,j}-u_{0,j}}{\Delta x}+\frac{u_{0,j+1}-u_{0,j-1}}{2\Delta y}+\frac{u_{2,j}-2u_{1,j}+u_{0,j}}{(\Delta x)^2}+\frac{u_{0,j+1}-2u_{0,j}+u_{0,j-1}}{(\Delta y)^2}=0,
\end{align*}
where $u_{i,j}=u(x_i,y_j)$.

So, instead of coming up with some Dirichlet or Neumann conditions, I essentially just impose the PDE itself on one (or several) of the boundaries. Is this something one has heard of? Do you know of problems with similar boundary conditions?

答:
The underlying problem is that the boundary condition you have provided does not assist in evaluating the solution.

Consider that, for your PDE, if $u=f(x,y)$ is a solution, then so is $u=f(x,y)+C$ for some constant, $C$. This constant cannot be determined using your boundary condition.

Indeed, consider that the following expression is a solution for any choice of constants $C_i$:
$$
u=C_1+C_2(x-y)+C_3e^{-x}+C_4e^{-y}+C_5e^{-x-y}
$$
None of these constants can be determined by the boundary condition as provided.

That being said, there are often implied conditions that can be used to obtain boundary conditions. For example, if you are working in a real-world context, you may need to apply an energy-conservation requirement. Alternatively, boundedness and related concepts often imply some form of boundary condition. For instance, consider
$$
xy'+y=0
$$
In the limit as $x\to0$, we get $y=0$ unless $y'$ tends to infinity. Indeed, the general solution is $y=C/x$, and so if $y$ must remain bounded, then $y(0)=0$, and thus $y=0$.

However, this is actually a separate condition on the problem that generates a boundary condition. You cannot simply apply the DE to the boundary, as it provides no additional information, and boundary conditions are applied to provide information that is otherwise missing.

The mistake you make is the claim that you must apply something at the boundary. This is untrue. You must apply a condition that allows you to identify the correct solution (unless you want a family of solutions, that is).

The nature of the condition will depend on the goal of the differential equation. As mentioned above, sometimes you need to apply some energy-conservation condition. Other conditions are possible - minimisation conditions are common. For example, you might want the solution to your problem that has the least amount of variation (that is, minimise $\iint u_{x}^2+u_{y}^2\ dA$).

There is one other category - cases where your choice won't matter. This is much like how, when performing integration by parts, you don't need to include the constant of integration for the "$\int dv$" - it will cancel out anyway. In this situation, pick the simplest boundary condition for the problem - for instance, if $u(0,0)=1$ and $u(0,\infty)=0$, then you might pick $u(0,y)=e^{-y}$.

EDIT: Here, I'll show what your forward-difference application of the PDE at the boundary is actually assuming for the boundary. Assuming you use central difference for first derivatives when not at the boundary, you have...
$$
\begin{align*}
\frac{u_{1,j}-u_{0,j}}{\Delta x}+\frac{u_{0,j+1}-u_{0,j-1}}{2\Delta y}+\frac{u_{2,j}-2u_{1,j}+u_{0,j}}{(\Delta x)^2}+\frac{u_{0,j+1}-2u_{0,j}+u_{0,j-1}}{(\Delta y)^2}&=0\\
\frac{u_{2,j}-u_{0,j}}{2\Delta x}+\frac{u_{1,j+1}-u_{1,j-1}}{2\Delta y}+\frac{u_{2,j}-2u_{1,j}+u_{0,j}}{(\Delta x)^2}+\frac{u_{1,j+1}-2u_{1,j}+u_{1,j-1}}{(\Delta y)^2}&=0
\end{align*}
$$
With a bit of algebra, we can cancel the $u_{2,j}$ terms to get
$$\begin{align*}
\left(\frac12+\frac{2+\Delta x}{(\Delta y)^2}\right)u_{0,j} - \left(\frac12+\frac2{(\Delta y)^2}\right)u_{1,j} & \\
+ \left(\frac{\Delta x}{4\Delta y} - \frac{\Delta x}{2(\Delta y)^2} + \frac{1}{2\Delta y} - \frac{1}{(\Delta y)^2}\right)u_{0,j-1} - \left(\frac{1}{2\Delta y} - \frac{1}{(\Delta y)^2}\right)u_{1,j-1}  & \\
 - \left(\frac{\Delta x}{4\Delta y} + \frac{\Delta x}{2(\Delta y)^2} + \frac{1}{2\Delta y} + \frac{1}{(\Delta y)^2}\right)u_{0,j+1} + \left(\frac{1}{2\Delta y} + \frac{1}{(\Delta y)^2}\right)u_{1,j+1} & =0
\end{align*}$$
By Taylor expansion around the $(0,j)$ point, letting $u$ be the function at that point and stopping at $O(\Delta)$ (treating both $\Delta x$ and $\Delta y$ as proportional to $\Delta$), we get
$$\begin{align*}
\frac{\Delta x}{(\Delta y)^2} u - \frac{\Delta x}2 (u_x+u_y+2u_{xy}) - \frac{\Delta x^2}4 u_{xx} + O(\Delta) & =0
\end{align*}$$
where I have left in the terms that come out explicitly... but note that this is equivalent to
$$
\frac{\Delta x}{(\Delta y)^2} u + O(\Delta) = 0
$$
And so, if $\Delta x$ and $\Delta y$ are both scaled down in proportion with each other, your boundary condition is effectively $u(0,y)=0$.

However, be careful, as this analysis applies for keeping $\Delta x$ and $\Delta y$ in proportion. I suspect that some issues may arise if you reduce $\Delta x$ without reducing $\Delta y$.


问:
Showing that an absolute integrable monotone decreasing function $f: [1,\infty[ \rightarrow \mathbb{R}$ is in $L^p([1,\infty[)$
For an exercise in my analysis course, I have to show that: if $f: [1,\infty[ \rightarrow \mathbb{R}$ is monotone decreasing and $f \in L^1([1,\infty[)$, then $f \in L^p([1,\infty[)$ for every $p > 1$.

Define $f_+ = \max(f,0)$ and $f_{-} = \max(-f,0)$

I tried several things: My guess is that I have to use the integral of $|f|$ to provide an upper bound for the integral of $|f|^p$. I tried the following:

$|f| = f_+ + f_-$. Then $f_+$ is monotone decreasing and $f_-$ is monotone increasing. Moreover $|f|^p = f_+^p + f_{-}^p$ since $f^+$ and $f_{-}$ are never nonzero together. Hence
$$\int_{[1,\infty[} |f|^p d\lambda = \int_{[1,\infty[} f_+^p d\lambda + \int_{[1,\infty[} f_+^p d\lambda$$

Now I want to try to get the monotonicity into play. Let $g_k = f_+ \cdot \chi_{[k,k+1[}$, then $f_+ = \sum^\infty_{k=0} g_k$. We can find a similar sequence for $f_{-}$, let's call that one $h_k$.

\begin{align*}\int_{[1,\infty[} f_+^p d\lambda + \int_{[1,\infty[} f_+^p d\lambda &= \int_{[1,\infty[}(\sum^\infty_{k=0} g_k)^p d\lambda+ \int_{[1,\infty[}(\sum^\infty_{k=0} h_k)^p d\lambda \\&= \int_{[1,\infty[} \sum^\infty_{k=0} g_k^p d\lambda+ \int_{[1,\infty[}\sum^\infty_{k=0}h_k^p d\lambda
\\& = \sum^\infty_{k=0} \int_{[1,\infty[} g_k^p d\lambda+ \sum^\infty_{k=0}\int_{[1,\infty[}h_k^p d\lambda\end{align*}
We are allowed to bring in the exponent because for every $x$ only one of them is nonzero.

\begin{align*}
\sum^\infty_{k=0} \int_{[1,\infty[} g_k^p d\lambda+ \sum^\infty_{k=0}\int_{[1,\infty[}h_k^p d\lambda &\leq \sum^\infty_{k=0} \int_{[1,\infty[} f_{+}(k)^p\chi_{[k,k+1[} d\lambda+ \sum^\infty_{k=0}\int_{[1,\infty[}f_{-}(k+1)^p\chi_{[k,k+1[} d\lambda\\
&=  \sum^\infty_{k=0}(f_{+}(k)^p + f_{-}(k+1)^p)\end{align*}

Now I am not sure how to proceed. Am I looking into the wrong direction? Please only provide hints!

答:
If your function is monotone and decreasing on $[1,\infty[$ then your function must be positive (if it's negative at any moment, it will not be integrable).

So I don't know (yet) the way to proceed, but I'm not sure you're on the right direction.

I hope this could help.

ContentLicense: CC BY-SA 3.0


问:
Is The Inverse Laplace Transform of $e^{st}\operatorname{Log}\left(\frac{s+1}{s}\right)$ doable using inversion formula?
I'm trying to solve inverse laplace transform using inversion formula and given by this integral:
$$\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i \infty} e^{st}\operatorname{Log}\left(\frac{s+1}{s}\right)\,\Bbb ds.$$

Here is my contour, since the branch points of $\operatorname{Log} \left(\frac{s+1}{s}\right)$ are $0$ and $-1$

First, i want to show integral on $L_u\cup L_d$ is $0$ by bounding the integral with ML and then take the limit when $R$ goes to $\infty$.

By letting $L_u,\, L_d: s= \xi\pm iR,0\leq \xi\leq \gamma$, where $\gamma$ is the real number that the vertical line of the given contour passed by.

Since the $L$ is $\lvert e^{t(\xi\pm iR)} \rvert$, then i have ML inequality as below:

$$\lvert F(s)e^{st} \rvert \leq M_R \lvert e^{t(\xi\pm iR)} \rvert = M_R e^{\xi t} \leq M_R e^{at}$$

Next, i need to find $M_R$ and take the limit.
$$\begin{align}
\left|F(s)\right| &= \left|\operatorname{Log}\left(\frac{s+1}{s}\right)\right|\\
&= \left|\operatorname{Log}\left(\frac{\xi\pm iR+1}{\xi\pm iR}\right)\right| = M_R
\end{align}$$

And by taking the limit of the last expression when $R$ goes to infinity yields $0$. Meaning the integrals along those lines are $0$.

So, from here, am i doing this right? I'm not sure my work is correct. Maybe there are some mistakes there. Help me please!

Edit:
Working with my $L_u$ with $ML$ inequality, i have $L=\gamma$. Assuming $-\pi<\operatorname{arg}{s}\leq \pi$ and parametrizing $s=-\xi+iR$, $\xi\in [-\gamma,0]$:

$\begin{align}
\left|\int_{L_u}\right|
&\leq \left|e^{st} \log\left(1 + \frac 1s\right)\right|\\
&\leq \left|e^{-\xi t}\right| \left|e^{iRt}\right|\left|\ln\left|1+ \frac{1}{-\xi+iR}\right| + i\pi\right|\\
&\leq 1\cdot 1 \cdot \ln\left(1+\frac 1R\right) + \pi\\
&\approx \frac 1R + \pi
\end{align}$

Combining the $ML$ i have
$$\frac{\gamma}{R} +\gamma\pi$$
Which does NOT approach to $0$. Why? Please spot my mistake. I can't think about how to make it goes to $0$ since yesterday. Hope you kind to help me.

New Edit: Now my main question is not about ML. I managed to set the big and small arc goes to $0$. My main question now is "is it possible to evaluate this by using inversion formula (without differentiate both sides) and use log form instead?"

What i mean by differentiate both sides is:
$$\begin{align}
\mathcal{L}^{-1} &= \operatorname{Log}\left(\frac{s+1}{s}\right)\\
(\mathcal{L}^{-1})' &= \frac{1}{1+s} - \frac{1}{s}
\end{align}$$

I don't want this kind of solution. What i really want is evaluating $\operatorname{Log}\left(\frac{s+1}{s}\right)$ by inversion formula. That's it.
because I keep getting integrals on the contour lines above and below the branch cut cancel each other and make everything 0, which of course I don't want it.

Attempt:

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\LARGE\left. a\right)}$

One idea, in this particular case, is to get rid of the $\ds{\ln}$-function by introducing an integral representation of it. Namely,
\begin{align}
&\bbox[5px,#ffd]{\left.{1 \over 2\pi\ic}
\int_{\gamma\ -\ \ic\infty}^{\gamma\ +\ \ic\infty} \expo{st}\ln\pars{s + 1 \over s}
\,\dd s\,\right\vert_{\,\gamma\ >\ 0}}
\\[5mm] = &\
\int_{\gamma\ -\ \ic\infty}^{\gamma\ +\ \ic\infty} \expo{st}\
\overbrace{\int_{0}^{1}{\dd x \over x + s}}
^{\ds{\ln\pars{s + 1 \over s}}}\
\,{\dd s \over 2\pi\ic}
\\[5mm] = &\
\int_{0}^{1}\int_{\gamma\ -\ \ic\infty}^{\gamma\ +\ \ic\infty}
{\expo{st} \over s + x}\,{\dd s \over 2\pi\ic}\,\dd x
\\[5mm] = &\
\int_{0}^{1}\bracks{t > 0}\expo{-xt}\,\dd x =
\bbx{\bracks{t > 0}\,{1 - \expo{-t} \over t}} \\ &
\end{align}

$\ds{\LARGE\left. b\right): {\large Contour Integration}}$'

Note that $\ds{{s + 1 \over s} < 0}$ whenever
$\ds{s \in \pars{-1,0}}$: It indicates that the $\mbox{$\ds{\ln}$-branch-cut}$ lies along $\ds{\bracks{-1,0}}$. Also,
$\ds{{s \pm \ic\epsilon + 1 \over s \pm \ic\epsilon}
\,\,\,\stackrel{{\rm as}\ \epsilon\ \to\ 0^{+}}{\sim}
\,\,\, \pars{1 + s \over s} \mp {\epsilon \over s^{2}}\,\ic}$ such that the $\ds{\ln}$-$\ds{arg}$ is $\pars{-\pi}$ when $\ds{s}$ is above the real axis and $\ds{\pi}$ when it's below the real axis.
\begin{align}
&\bbox[5px,#ffd]{\left.{1 \over 2\pi\ic}
\int_{\gamma\ -\ \ic\infty}^{\gamma\ +\ \ic\infty} \expo{st}\ln\pars{s + 1 \over s}
\,\dd s\,\right\vert_{\,\gamma\ >\ 0}}
\\[5mm] = &\
-\int_{-\infty}^{-1}\ln\pars{s + 1 \over s}\expo{ts}
{\dd s \over 2\pi\ic}\label{1}\tag{1}
\\[2mm] - &\
\int_{-1}^{0}\bracks{\ln\pars{-s - 1 \over s} - \ic\pi}\expo{ts}
{\dd s \over 2\pi\ic}
\\[2mm] - &\
\int_{0}^{-1}\bracks{\ln\pars{-s - 1 \over s} + \ic\pi}\expo{ts}
{\dd s \over 2\pi\ic}
\\[2mm] - &
\int_{-1}^{-\infty}\ln\pars{s + 1 \over s}\expo{ts}
{\dd s \over 2\pi\ic}\label{2}\tag{2}
\end{align}
I included (\ref{1}) and (\ref{2}) "by completeness" but, indeed, it's not necessary because the $\ds{\ln}$-branch-cut lies along $\ds{\bracks{-1,0}}$
albeit it guarantees the vanishing out of the integration along the"big-arc".

Then,
\begin{align}
&\bbox[5px,#ffd]{\left.{1 \over 2\pi\ic}
\int_{\gamma\ -\ \ic\infty}^{\gamma\ +\ \ic\infty} \expo{st}\ln\pars{s + 1 \over s}
\,\dd s\,\right\vert_{\,\gamma\ >\ 0}}
\\[5mm] = &\
-\int_{0}^{1}\bracks{\ln\pars{-s + 1 \over s} - \ic\pi}\expo{-ts}
{\dd s \over 2\pi\ic}
\\[2mm] &\
\,\, + \int_{0}^{1}\bracks{\ln\pars{-s + 1 \over s} + \ic\pi}\expo{-ts}
{\dd s \over 2\pi\ic}
\\[5mm] = &\
\int_{0}^{1}\expo{-ts}\,\dd s =
\bbx{1 - \expo{-t} \over t} \\ &
\end{align}

ContentLicense: CC BY-SA 4.0


问:
Interesting Weighted Sum of Binomial Coefficients
In my research, I've come across the following weighted sum of binomial coefficients:
$$\sum_{i=1}^k i\cdot{2k \choose k+i}.$$
According to Wolfram Alpha, this equals exactly the quantity
$$\frac{1}{2}(k+1){2k \choose k+1}.$$
After trying to prove this, I've come up shorthanded. I have two questions.

Is there a direct procedure (or a combinatorial argument) that yields this expression?

Is there a general class of weighted sums that could be solved in a similar way?

答:
I'll give a different solution from the two in the linked question.

Now to view it as computing $E|X-k|$ where $X \sim Bin\left(2k,\frac 12\right)$, it's quite natural to do the following:

$$\sum_{i=1}^k i\cdot{2k \choose k+i} = \sum_{i=1}^k (k+i-k) \binom{2k}{k+i} \\
= \sum_{i=1}^k (k+i) \binom{2k}{k+i} - k \sum_{i=1}^k \binom{2k}{k+i}\\
= 2k \sum_{i=1}^k \binom{2k-1}{k+i-1} - k \cdot \frac{2^{2k} - \binom{2k}{k}}{2}\\
= 2k \cdot \frac{2^{2k-1}}{2} - k\cdot \frac{2^{2k} - \binom{2k}{k}}{2}\\ = \frac{k}{2} \binom{2k}{k} = \frac{1}{2}(k+1){2k \choose k+1}. \blacksquare
$$

Note the above quantity  is actually equal to $2^{2k-1}E|X-k|$, not $E|X-k|$.


问:
How to solve $\int\sqrt{1+x\sqrt{x^2+2}}dx$
I need to solve

$$\int\sqrt{1+x\sqrt{x^2+2}}dx$$

I've chosen the substitution variables

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

However, I am completly stuck at

$$\int\sqrt{1+xu} dx$$

Which let me believe I've chosen wrong substitution variables.

I've then tried letting $u=x^2+2$ or simply $u=x$, but it does not help me at all solving it.

Would someone please give me an hint on this ?

Thanks.

Body:
Substitute $x=\frac{2-t}{2\sqrt t}$. Then, $dx=-\frac {2+t}{4t\sqrt t}\>dt$ and

\begin{align}
&\int\sqrt{1+x\sqrt{x^2+2}}\>dx \\
= & -\frac18 \int \frac{t+2}{t^2} \sqrt{ 4+4t-t^2}\>dt
= -\frac18 \int \frac{\sqrt{ 4+4t-t^2}}{t-2}d\left( \frac{(t-2)^2}t\right)\\
=& -\frac{t-2}{8t }\sqrt{ 4+4t-t^2}- \int \frac1{t\sqrt{ 4+4t-t^2}}dt\\
=& -\frac{t-2}{8t }\sqrt{ 4+4t-t^2}+\frac12 \tanh^{-1} \frac{t+2}{\sqrt{ 4+4t-t^2}}+C
\end{align}


问:
How to integrate sinc function numerically
How to compute (numerically)

$$
F(x) = \int_{-\infty}^x \dfrac{\sin(t)}{t} dt
$$

答:
$$F(x) = \int_{-\infty}^x \dfrac{\sin(t)}{t} dt=\text{Si}(x)+\frac{\pi }{2}$$

For the computation  of the sine integral function, you will find subroutines in Numerical Recipes (have a look here).


问:
How to find the basis of intersection of subspaces $V$ and $W$?
The text of the problem:

Find the basis for the sum and the intersection of subspaces $V$ and $W$:

$V=\left\{v_1=\begin{pmatrix} 2 & 1 & 0 \end{pmatrix},v_2=\begin{pmatrix} 1 & 2 & 3 \end{pmatrix},v_3=\begin{pmatrix} -5 & -2 & 1 \end{pmatrix}\right\}$

$W=\left\{w_1=\begin{pmatrix} 1 & 1 & 2 \end{pmatrix},w_2=\begin{pmatrix} -1 & 3 & 0 \end{pmatrix},w_3=\begin{pmatrix} 2 & 0 & 3 \end{pmatrix}\right\}$.

What I did do:

First I'd formed the matrix $M_V$ and found the row echelon form of the matrix where it shows that there are two independent vectors in $V$, i.e. dim $V$=2 and formed the set $V'=\{v_1,v_2\}$. Similiarly, I found that dim $W$=2, formed $W'=\{w_1,w_2\}$ and then formed the basis $B_{V+W}=\{v_1,v_2,w_1\}$. Again, I checked if the vectors in it are linearly independant and found that indeed $\textrm{dim } (V+W)$=3.

After this, I get kinda stuck. I know that

dim $V\cap W$=dim $V$+dim $W$-dim $(V+W)=1$

i.e. the basis $B_{V\cap W}$ only has one element $x$ such that $x\in V$ and $x\in W$. Therefore, we have $$x=\sum_{i=1}^3\alpha_iv_i=\sum_{j=1}^3\beta_jw_j,$$
so I formed a set of equations
$$2\alpha_1+\alpha_2-5\alpha_3=\beta_1-\beta_2+2\beta_3$$
$$\alpha_1+2\alpha_2-2\alpha_3=\beta_1+3\beta_2$$
$$2\alpha_2+\alpha_3=2\beta_1+3\beta_3,$$
and I have no clue where to go from here. Any help is appreciated.

答: Observe that $V$ is the plane $x-2y+z=0$ while $W$ is the plane $x-2y+z=0$. Finding $V \cap W$ means identifying the intersection of these two planes; this amounts to solving the system of equations $$x-2y+z=0$$ $$-3x-y+2z=0$$ By setting up and solving an augmented system, we see it has infinitely many solutions, and they are $$(x,y,z)=(3,5,7)t: t\in \mathbb{R}$$ In other words, $$V \cap W = \text{span} \{(3,5,7)\}$$


问:
Solve equations $\sqrt{t +9} - \sqrt{t} = 1$
Solve equation:  $\sqrt{t +9} - \sqrt{t} = 1$

I moved - √t to the left side of the equation $\sqrt{t +9} = 1 -\sqrt{t}$

I squared both sides $(\sqrt{t+9})^2  = (1)^2 (\sqrt{t})^2$

Then I got $t + 9 = 1+ t$

Can't figure it out after that point.

The answer is $16$

答:
$$\sqrt{t +9} - \sqrt{t} = 1$$

Multiplying by $\sqrt{t +9} + \sqrt{t}$ you get

$$9=\sqrt{t +9} +\sqrt{t} $$

Now adding

$$\sqrt{t +9} + \sqrt{t} =9$$
$$\sqrt{t +9} - \sqrt{t} = 1$$

you get

$$\sqrt{t+9}=5 \Rightarrow t=25-9 $$


问:
Proving $f$ is continuous at $1,-1$
Question: given

$$
f(x)=\begin{cases}
         x^4 & \text{if $x\in\mathbb{Q}$;}\\
        2x^2-1 & \text{if $x\in\mathbb{Q}^c$.}
     \end{cases}
$$

I need to prove that given function continuous only at points $1,-1$

My attempt Using sequential criterion of continuity I had proved that, $f$ is 'not' continuous on $\mathbb{R}-\{1,-1\}$.

But, I am facing difficulty in proving $f$ is continuous at points $1,-1$.

I considered, $|f(x)-f(1)|=|f(x)-1|≤sup\{|x^4-1|, |2x^2-1|\}$

How to show this is less than $\epsilon$. Please help me, in proving $f$ is continuous at points $1,-1$

答: Since $x^4$ is continuous at $1$ we know that for an $\epsilon$ there is a $\delta'$ such that $|x - 1| < \delta' \implies |x^4 - 1| < \epsilon$. Likewise there is a $\delta''$ for $2x^2 - 1$ such that $|x - 1| < \delta'' \implies |2x^2 - 2| < \epsilon$. Now take $\delta = inf(\delta', \delta'')$, then $|x - 1| < \delta \implies |x^4 - 1| < \epsilon$ and $|2x^2 -1| < \epsilon \implies sup\{|x^4 - 1|, |2x^2 -2|\} < \epsilon$. This proves continuity at $1$.

Title: Proving $f$ is continuous at $1,-1$


问:
Is this an acceptable proof that $x^2<y^2$ when $x$ and $y$ are positive reals with $x<y$?
The question is

Prove the following statement:

Suppose $x, y \in \mathbb{R}$ and that $x$ and $y$ are positive. If $x < y$ then $x^{2} < y^{2}$.

Proof:

Suppose there is a value $r$ between $x$ and $y$ so that
$$x < r < y \tag1$$ where $r \in \mathbb{R}$ and $ r > 0$. Therefore,
$$x<r \Rightarrow x^{2}<xr \tag2$$ and
$$r<y \Rightarrow yr < y^{2} \tag3$$
Moreover, if we multiply equation $(1)$ by $r$, we get
$$xr < r^{2} < yr \tag4$$
Combining equations $(2)$, $(3)$, and $(4)$, we get
$$x^{2} < xr < r^{2} < xy < y^{2} \tag5$$
Therefore, $x^{2} < y^{2}$. $\blacksquare$

If necessary, we can even define $r=$ $x+y\over{2}$

Is there anything wrong with this approach? I'm aware of the normal method of direct proof.

答:
Note that this is like proving $1 < 3$ by showing $1 < 2$ and $2 < 3$. You basically end up proving the result along the way except you wrote down a bunch of extra steps.

As soon as you have $x^{2} < xr$, you have the result immediately from the hypothesis:
$x^{2} < xr < yr < y^{2}$ as none of them are zero.

Just in case you didn't prove this concisely using a direct proof, you need to prove that $(y+x)(y-x)=y^{2}-x^{2} > 0$ where $y,x > 0$ and $y-x > 0$ which is clear since $y+x > 0$.


问:
Given isosceles triangles $\triangle ABC$ and $\triangle DBF$ (all spherical chords), identify the chord $\overline {DF}$ so that $|AD| = |DF| = |FC|$
tl;dr: As shown in the image below, find the chord $\overline {DF}$ so that $|\overline {AD}| = |\overline {DF}|$, and have the answer be in the form of the ratio between $|\overline {AC}|$ and $|\overline {DF}|$. Ideally, I'm actually looking for a slightly different ratio (which is explained below), but this is the basic problem I believe that needs to solved. Only the isosceles triangle $\triangle ABC$ may be used as input  (the image is slightly misleading, as they all should be straight lines and $\overline DF$ should be slightly higher than $\triangle ABC$ and not intersecting it).

The actual problem I'm trying to solve is to subdivide a dodecahedron into smaller faces (known as a chamfered dodecahedron), and approximate a sphere using that polyhedron. This is done in 3 steps:

Scale down each face from its center by some scale factor (what I'm trying to solve for) so that the hexagons constructed in the next step are planar (they will not be equiangular, however)

Construct hexagons in between the newly subdivided faces (goldberg polyhedrons will always have 12 pentagons, no matter how subdivided they are)

Scale the new face from the origin so that each vertex has the same magnitude of the original face (this is done by normalizing each vertex, then scaling by a specified radius - this will make the hexagons non-planar)

My problem is that I'm stuck on how to identify how to scale factor in the first step, as there doesn't seem to be a linear relationship between the scale factor needed and the number of subdivisions that are performed. This is shown in the two images below, where I'm adding the new faces over the original dodecahedron (radius of 1 in my case). I manually calculated approximately $0.40706$ (or $40.706\%$) for the first subdivision, which allowed the construction of the [green] hexagon to have satisfyingly equivalent sides. However, when I tried scaling the new pentagons again with the same scale factor (shown with the yellow pentagons), the hexagons in-between were not equilateral...

I suspect that the scale factor will change the more I subdivide these faces - indicating that there must be a non-linear relationship between side length and the length of the connecting chord. To attempt to solve my problem, I reconstructed the issue to what I hope is a solvable math problem, shown in the images below (although lines $\overline {AB}$ and $\overline {BC}$ should be chords, not arcs - thanks to @Blue in the comments). However, I frankly have no idea where to start... I feel like solving this problem is beyond my knowledge. I then tried condensing this to a simpler case with an isosceles triangle in 2D space, but I encountered a similar block as well. I know that I need "scale" as the output (specifically when scale was used in the first step), and my known values are the pentagon/hexagon side lengths, its position in 3D space, and the magnitude of each of the points. I would be very grateful for your guidance on this issue!

Edit:

This is the result I'm trying to achieve, using the following manually approximated scale factors (note that the hexagons are not planar):

```<code>Subdivision | Pentagons | Hexagons
------------|-----------|---------
     0      |     -     |    -
     1      |  0.40706  |    -
     2      |  0.44950  |  0.4874
     3      |    ...    |   ...
</code>```

答:
This isn't a complete solution but so far I have the essential conventions in place.
To make things more specific, I'll consider the sphere to have radius 1 with $B$ at the north pole, $O$ as the origin, and the points $A,D$ lying in the positive $xz$-half plane. Additionally, I'll denote
\begin{align}
\theta&=\angle ABC=\angle DBF,\\
\phi_1&=\angle BOD =\angle BOD,\\
\phi_2&=\angle AOB=\angle BOF. \\
\end{align}
(An aside: Note that these denote the arc-lengths of the various segments of the spherical figure, which is always larger than the Euclidean distance between the two vertices. This isn't crucial to the problem posed: If two pairs of vertices have the same arc-length then they have the same Euclidean distance, so we might as well consider arc-length. But it will matter if we consider ratios, since the relationship between these distance measures isn't linear.)

As such, the Cartesian coordinates for points $A,B,C,D,F$ are given as

\begin{align}
A&=(\sin\phi_2,0,\cos\phi_2),\\
B&=(0,0,1),\\
C&=(\sin\phi_2\cos\theta,\sin\phi_2\sin\theta,\cos\phi_2),\\
D&=(\sin\phi_1,0,\cos\phi_1)\\
F&=(\sin\phi_1\cos\theta,\sin\phi_1\sin\theta,\cos\phi_1),\\
\end{align}

The condition of interest is then $\angle DOF=\phi_2-\phi_1$, which is imposed by via the dot product of the position vectors for $D,F$:

$$\cos(\phi_2-\phi_1)=\cos \angle DOF=\vec{D}\cdot \vec{F}=\sin^2\phi_1\cos\theta+\cos^2\phi_1.$$
In principle, we can use $\phi_2,\theta$ to determine the angle $\phi_1\in [0,\pi]$.
Similarly, the angle $\angle AOC$ is determined via the vectors for $A,C$:

$$\cos\angle AOC = \vec{A}\cdot \vec{C}=\sin^2\phi_2\cos\theta+\cos^2\phi_2$$
which is determined by $\phi_2,\theta$ directly. So in that sense there is a functional relationship between $\phi_1$, $\phi_2-\phi_1$ and $\angle AOC$, but at present they seem...unpleasant. I'll see if I can find any clever way to reduce this. Alternatively things may simplify if we restrict to a certain case of interest. For instance, when $\theta=2\pi/5$ (as for the icosahedron) one has $\cos\theta=(2\phi)^{-1}$ where $\phi$ is the golden ratio. So it's plausible that a specific case may be more tractable than the generic.


问:
Arbitrary subset of quotient group or ring
Let $R$ be a group/ring and $I$ a normal subgroup/ideal, and form the quotient group/ring $R/I$.
Is is legitimate to write either of the following?

$S/I$ is an arbitrary subset of $R/I$, where $S \subseteq R$.

$S/I$ is an arbitrary subset of $R/I$, where $I \subseteq S \subseteq R$.

If not, is there a way to represent an arbitrary subset (or subgroup or subring) of $R/I$ without just calling it $A$ or some other single variable? My question stems from wanting to find a way to represent a subset of the quotient ring before I actually check that it's an ideal.

答: Some people like to use that notation, and some people hate it. I personally think the best way to go is to take your arbitrary subset $A \subseteq R/I$ and define $S = \pi^{-1}(A)$ where $\pi : R \to R/I$ is the canonical projection. Then $S/I$ is guaranteed to be $A$ and you don't have to worry about people disliking your notation.


问:
How many integer points are contained within a cube's intersection with plane?
Consider the integral points (x, y, z)(x,y,z) on the plane $35 x + 55 y + 77 z = 1$. How many are contained within a cube with side length 30 centered at (0, 0, 0)?

https://brilliant.org/problems/points-in-a-box-2/#!/solution-comments/

The solution by the author is not very comprehensive. I understand it vaguely until the end where the author ends with "part two: solving inequalities". How would you go about turning this into an inequality?

Thanks to everyone answering (-:

(Extension: Would it be possible to do this with a 3D space and 4D hypercube? Just wondering.)

答:
Hint

You have the system

$$35x + 55y + 77z = 1$$

$$-15 \leq x,y,z \leq 15$$

Let $x+15 = X, y+15=Y, z+15 = Z$

The system becomes

$$35X + 55Y + 77Z = 2506$$

$$0 \leq X,Y,Z \leq 30$$

Can you solve it now? It's a basic combinatorial problem you can solve using stars and bars method


问:
Points to keep in mind while solving a complex equation
I was solving this complex equation, $z^2 = -i$, where z is a complex number. I solved it to get $z = \sqrt{-i}$. But my teacher solved it by assuming $z = x + iy$ and he got the answer $z = \frac{1}{\sqrt2} - \frac{i}{\sqrt2} \space$ and $z = -\frac{1}{\sqrt2} + \frac{i}{\sqrt2}$ which is very different from my answer. Where am I wrong? Why cannot I solve it like that? What else should I keep in mind while solving a complex equation?

答:
By writing that the solution is $\sqrt{-i}$ you have not really solved the problem, which requires you to express it without a square root.

I would solve the problem by first converting to polar coordinates ($-i=e^{-i\pi/2}$), taking the square root by halving the exponent ($\sqrt{-i}=\pm e^{-i\pi/4}$) and then converting back to Cartesian form. Note that the complex equation $z^n=a$ always has $n$ solutions if $a\ne0$.


问:
What is the probability of winning in a dice game
John and Lucy take turns rolling two dice:

If John rolls a pair larger than or equal to $10$ he wins and the game stops.

If Lucy rolls a pair strictly smaller than $10$, she wins and the game stops.

They keep rolling in turn until one of them wins.

Suppose John rolls first. What is the probability that Lucy wins $?$.

My attempt:

First, we have to figure out the probability of John winning on any particular round. That is if he rolls: (5,5), (5,6), (6,5), (4,6), (6,4), (6,6). Thus, his chances of winning are $\frac{6}{36}=\frac{1}{6}$. Then since Lucy wins if she rolls everything else, that means she has a $5/6$ chance of winning on any particular round.

Now to figure out the probability of Lucy winning if John goes first. For this to happen, John has to lose his first roll and Lucy has to win after that. We know that P(John Losing)=$\frac{5}{6}$ and P(Lucy winning)=$\frac{5}{6}$. Thus, we have $(\frac{5}{6})(\frac{5}{6})=\frac{25}{36}$ chance of Lucy winning the game.

However, I don't think that this is right because it does not account for playing additional rounds (i.e. she doesn't win on the second round but on some nth round afterward). Could I get some help? Thanks.

答:
Please see the below. It is a geometric distribution.

John can win by rolling $(6,4), (4,6), (5, 5), (5,6), (6,5)$ or $(6,6)$.

So as you calculated his probability of winning in a given roll $ p = \frac{6}{36} = \frac{1}{6}$

Lucy's probability of winning in a given roll $ q = \frac{5}{6}$

If Lucy's chances of winning the game is $P(W)$, we have

$P(W) = (1-p) \times q + (1-p) \times (1-q) \times P(W)$

$P(W) = \frac{5}{6} \times \frac{5}{6} + \frac{5}{6} \times \frac{1}{6} \times P(W)$

$P(W) = \frac{25}{31}$


问:
logs Challenge between two students >>be smart
two student were given the equation  $2^{4x+6} = 3^{6x-3}$

1.steve rearranged to get  $2^{4x+6} - 3^{6x-3} =0$

then wrote $\log (2^{4x+6} - 3^{6x-3}) = \log0$

are these legal steps ? if not explain what is wrong with them

2 Ali wrote $\log( 2^{4x+6}) =\log(3^{6x-3})$  then $\log( 2^{4x+6}) - \log(3^{6x-3})=0  $

finally $2^{4x+6} / 3^{6x-3}$

are these legal steps ? if not explain what is wrong with them

答: The first is not legal as you cannot take the log of $0$.  The second is legal, but I do not see it as progress toward finding $x$.  Ali should use the laws of logarithms to pull down the powers.  His finally can be reached in one step from the initial equation and does not give a value for $x$.</p>


问:
Let $X,Y,Z$ be i.i.d $U(0,1)$ random variables. Then $P(X>Y+Z)$

Let $X,Y,Z$ be i.i.d $U(0,1)$ random variables. Then $P(X>Y+Z)$

Saw this question but I am not sure how to utilize it here.

If it was $2$D random variable question I would've easily calculated. $P(X>Y)$ which is inside red outlined shape.

Which is $\dfrac{1}{2}$

but its $3$D now and I am not sure how to setup limits of integration. Since there joint density is unity.

$P(X>Y+Z)=\int_{0}^{1}\int_{?}^{?}\int_{y+z}^{1}1 dxdydz$

Also what other way I can use to solve this ?

答:
If we rename the variables so that the problem is $P(Z>X+Y)$, the region of the unit cube $[0,1]^3$ satisfying this condition is a tetrahedron with vertices $(0,0,0),(0,0,1),(0,1,1),(1,0,1)$. The volume of this tetrahedron, hence the final probability since the cube has volume $1$, is $\frac16$.

As an integral this would be
$$\int_{x=0}^1\int_{y=0}^x\int_{z=0}^{x-y}1\,dz\,dy\,dx$$
since $X>Y+Z$ implies $X>Y$ and $X-Y>Z$.


问:
How to find Integral $\int_{0}^{1/2} (e^x - 1)/x \,dx$?
can you give me any advice how to caluclate $\int_{0}^{1/2} \frac{e^x - 1}{x} \,dx$ ? I need to use series. I tried to split it as
$$
\int_{0}^{1/2} \frac{e^x}{x} \,dx - \int_{0}^{1/2} \frac{1}{x} \,dx$$ and then I can use a series for the first equation but I get another $\int_{0}^{1/2} \frac{1}{x} \,dx$. Can I somehow evaluate these integrals or I need to use a completely different approach. Thanks!

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{1/2}{\expo{x} - 1 \over x}\,\dd x} \,\,\,\stackrel{x\ \mapsto\ -x}{=}\,\,\,
-\int_{-1/2}^{0}{\expo{-x} - 1 \over x}\,\dd x
\\[5mm] = &
\lim_{\Lambda \to \infty}
\bracks{-\int_{-1/2}^{\Lambda}{\expo{-x} - 1 \over x}\,\dd x
+ \int_{0}^{\Lambda}{\expo{-x} - 1 \over x}\,\dd x}
\\[5mm] = &\
\lim_{\Lambda \to \infty}
\left[-{\rm P.V.}\int_{-1/2}^{\Lambda}
{\expo{-x} \over x}\,\dd x\right.
\\[2mm] &\ \phantom{\lim_{\Lambda \to \infty}\,\,\,}
+ \int_{-1/2}^{\Lambda}{\dd x \over x} +
\ln\pars{\Lambda}\pars{\expo{-\Lambda} - 1}
\\[2mm] &\
\left.\phantom{\lim_{\Lambda \to \infty}\,\,\,}+ \int_{0}^{\Lambda}\ln\pars{x}\expo{-x}\dd x\right]
\\[5mm] = &\
\bbx{\on{E_{i}}\pars{1 \over 2} + \ln\pars{2} - \gamma}
\approx 0.5702 \\ &
\end{align}

$\ds{\on{E_{i}}}$: Exponential Integral Function.
$\ds{\on{E_{i}}\pars{z} \equiv -{\rm P.V.}\int_{-z}^{\infty}{\expo{-x} \over x}\dd x}$.

$\ds{\gamma}$: Euler-Mascheroni Constant which is equal to $\ds{\phantom{\gamma:}-\int_{0}^{\infty}\ln\pars{x}\expo{-x}\dd x}$.


问:
Monty Hall Problem with Multiple Players?
I understand the common Monty Hall Problem and why switching provides a 2/3 chance of winning, but I'm having trouble wrapping my head around how the probabilities work when multiple players are involved, as their probabilities seem to be contradictory.

Let's say we have two players and four doors.
Each door has a 1/4 probability of hiding a prize.
Let's say contestant 1 chooses door A and contestant 2 chooses door B.
Door D is then revealed to be empty, leaving A, B, and C.
From contestant 1's perspective the odds are 1/4 for door A, 3/8 for door B, and 3/8 for door C.
But from contestant 2's perspective the odds are 3/8 for door A, 1/4 for door B, and 3/8 for door C.

What are the actual odds for each door hiding the prize, does it change depending on if they know each other's choices, and does it differ from the normal example of the problem? If so why, and if not, why not?

答:
If the players know what the other chose then the options are  (and the actual probabilities that we know):

Prize Behind Door A: C is shown.  $1$ in $8$.  (Didn't happen)

Prize Behind Door A: D is shown.  $1$ in $8$.

Prize Behind Door B: C is shown.  $1$ in $8$. (Didn't happen)

Prize Behind Door B: D is shown. $1$ in $8$.

Prize Behind Door C: D is shown. $1$ in $4$.

Prize Behind Door D: C is shown. $1$ in $4$. (Didn't happen).

Of the ones that happened. Prize behind door A, or B are each $1/4$ and behind $C$ is $1/2$.

But if the players don't know what the other picked then from Player 1s perspective the probability as he knows is:

Prize behind Door A; other player choose A: D (one of 3 options) is shown:  1 in 48

Prize behind Door A; other player choose B: D (one of 2 options) is shown: 1 in 32

Prize behind Door A; other player choose C: D (one of 2) is shown: 1 in 32

Prize behind Door B; other player choose A; D (one of 2 options) is shown: 1 in 32

Prize behind Door B; other player choose B; D (one of 2 options) is shown:1 in 32

Prize behind Door B; other player choose D; D (only option) is shown: 1 in 16

Prize behind Door C; other player choose A; D (one of 2 options) is shown: 1 inn 32
Prize behind Door C; other player choose B; D (only option) is show: 1 in 16

Proze behind Door C; other chose C; D (one of two) is shown: 1 in 32

So the weighted options are   A:  1 in 4

B: 3 in 8

C: 3 in 8

Why that differs from the actual probability is simply the player is lacking information we have.

Knowing information always changes probability.


问:
Stationary points of $\ln(1+x²y)$
The stationary points of the function lie along with $(0,y)$ for all $y$. The second derivative test is inconclusive.

I understand that the origin is the saddle point. But how do we categorize the other points without looking at the graph? In general, what is the approach to take in case of inconclusive second derivative tests?

答:
Generally speaking, if the second derivative test is inconclusive one needs to assume an epsilon ball around the point of interest, and if the function changes sign within the ball, no matter how small, then the function is a saddle point

From the first derivative test, we know that all points of the form $(0,y)$ are stationary points. Let us look at the function behaviour centered at a point $(0,y_0)$

For $(x,y)$ such that $$||(x,y)-(0,y_0)|| < \epsilon$$

$$f(x,y) = \ln(1 + x^2y)$$

$$f(0,y_0) = 0$$

Now, if I move in any direction other than along $y$ axis - $x^2y >0$ for $y>0$ and $x^2y <0$ for $y<0$

Hence, If $y>0$ - then $$f(x,y) \geq 0 = f(0,y_0)$$

If $y<0$ then

$$f(x,y) \leq0 = f(0,y_0)$$

Can you tell which point is a minima and which is a maxima now?

Title: Stationary points of $\ln(1+x²y)$


问:
Show that $\{x\}$ is a closed set
Let $y \in \mathbb{R} \backslash\{x\},$ the number $\varepsilon=|y-x|$ is positive. If $z \in(y-\varepsilon, y+\varepsilon),$ then $|z-y|<\varepsilon$ hence
$$
\varepsilon=|x-y|=|x-z+z-y| \leq|x-z|+|z-y|
$$
so that $|z-x| \geq \varepsilon-|z-y|>0$ and then  $z \neq x,$ so $z \in \mathbb{R} \backslash\{x\} .$ This proves that  $(y-\varepsilon, y+\varepsilon) \subset \mathbb{R} \backslash\{x\}.$

This is my question: Why if $z \in\mathbb{R}-\{x\}$ implies $(y-\varepsilon, y+\varepsilon) \subset \mathbb{R} \backslash\{x\}$ i dont understand that step

答:
A set $Z$ is open if for each $y\in Z$, there exists an $\epsilon>0$ such that the open ball $B(y;\epsilon)\subset Z$

In $\Bbb R, B(y;\epsilon)=(y-\epsilon,y+\epsilon)$

Here, $Z=\Bbb R\setminus \{x\}$, and we take $\epsilon=|y-x|>0$, giving us that $y+\epsilon=x$ or $y-\epsilon=x$

Since the open ball does not contain it's end points, we have that $x\not\in B(y;\epsilon)\subset \Bbb R\implies B(y;\epsilon)\subset \Bbb R\setminus\{x\}$

And so the set is open, and it's complement, $\{x\}$, is closed

Title: Show that $\{x\}$ is a closed set


问:
Finding the smallest $n$ that satisfies $\left(\frac{1-i}{\sqrt{2}}\right)^n=1$
Maybe I'm just stupid, but right now I'm in a dilemma so please help. I had the question

$$\left(\frac{1-i}{\sqrt{2}}\right)^n=1$$ where I need to find the smallest number for $n$ ($n> 0$) that would satisfy the problem.

My first step was to $$\left[\left(\frac{1-i}{\sqrt{2}}
\right)^2\right]^\frac{n}{2}=1$$ and than $$(-i)^\frac{n}{2}=1$$ after that just ignore the minus because its powered by a even number and so on so forth $$i^\frac{n}{2}=1$$ $$\frac{n}{2}=4$$the  answer is $$n=8$$. But here is where the dilemma came if $$-i=i^3$$ than it would be $$\left(i^3\right)^\frac{n}{2}=1$$ which turns to $$i^\frac{3n}{2}=1$$ and the $$\frac{3n}{2}=4$$. This seems alright in the head but than the answer turns out to be $n=\frac{8}{3}$.

I used latex. Sorry if bad latex i'ts my first time.

If you know the answer please please tell me.

By the way I'm 17 and dumb. So please don't hate

Body: $Z=((1-i)/\sqrt{2})^n=1 \implies e^{-in\pi/4}=1 \implies (e^{-i\pi})^{n/4}=1 \implies (-1)^{n/4}=1 \implies n=8.$


问:
Complex Exponents
What does it mean to raise a number to a complex exponent, and why? A lot of the explanations that I've seen involve e, why is this?

I'm looking for an intuitive answer describing to how exponentiation works on a number plane (i.e. the complex number plane) rather than just a number line (the real numbers), where I understand exponentiation quite well. I'm not looking for a breakdown of Taylor series - I believe the proofs, I just don't grok them.

For example, if I was asking how sine worked, I'd want this, not this.

答: There are various possible approaches, but I prefer to start with the exponential function $e^z$ as solution to the differential equation $f'(z) = f(z)$ with $f(0)=1$.  The other exponentials are derived from this: $a^z = e^{z \log(a)}$ is a solution to $f'(z) = \log(a) f(z)$ with $f(0)=1$.</p>

Title: Complex Exponents


答:
Rewrite you equation as
$$
\bigg(\frac{2^2}{3^3}\bigg)^{2x}=3^{-3}2^{-6}
$$

Can you handle from here?

Title: logs Challenge between two students >>be smart

Body:
two student were given the equation  $2^{4x+6} = 3^{6x-3}$

1.steve rearranged to get  $2^{4x+6} - 3^{6x-3} =0$

then wrote $\log (2^{4x+6} - 3^{6x-3}) = \log0$

are these legal steps ? if not explain what is wrong with them

2 Ali wrote $\log( 2^{4x+6}) =\log(3^{6x-3})$  then $\log( 2^{4x+6}) - \log(3^{6x-3})=0  $

finally $2^{4x+6} / 3^{6x-3}$

are these legal steps ? if not explain what is wrong with them


答:
If $Y,Z\sim U(0,1)$ then $S=Y+Z$ is distributed as $$f_S(s)=\left\{\begin{array}{cc} s & 0<s<1\\ 2-s,& 1<s<2,\\ 0&\text{else}\end{array}\right.$$ (See for instance Density of sum of two independent uniform random variables on $[0,1]$. ) For intuition, note that the diagonal range of your blue/red square starts at zero in one corner, rises linearly towards the middle of the square, and then returns to zero. As such, it should be reasonable that $S$ has a triangular distribution.

As such, our problem is now to compute $\text{Pr}(X>S)$ where $X\sim U(0,1)$ and $S$ is distributed as above. The corresponding blue/red picture is similar, but the vertical width of the blue stripe (corresponding to the smaller variable) is $2$ rather than $1$. Moreover, we have to account for the variable density of $S$. Life is made simpler, however, by the fact that we only care about $s<x<1$ and so can take $f_S(s)=s$. Keeping this in mind, we have

$$\text{Pr}(X>S)=\int_{x=0}^1\int_{s=0}^x f_S(s)f_X(x)\,ds\,dx=\int_0^1\int_0^x  s^2\,ds\,dx=\int_0^1 \frac13 x^2\,dx=\frac{1}{6}$$
in agreement with the geometric answer.

Body:
Let $X,Y,Z$ be i.i.d $U(0,1)$ random variables. Then $P(X>Y+Z)$

Saw this question but I am not sure how to utilize it here.

If it was $2$D random variable question I would've easily calculated. $P(X>Y)$ which is inside red outlined shape.

Which is $\dfrac{1}{2}$

but its $3$D now and I am not sure how to setup limits of integration. Since there joint density is unity.

$P(X>Y+Z)=\int_{0}^{1}\int_{?}^{?}\int_{y+z}^{1}1 dxdydz$

Also what other way I can use to solve this ?


问:
Is it possible to write $\prod_{x=1}^5 \sin(x) $ as a factorial?
$$\prod_{x=1}^5 \sin(x) $$ is actually equal as writing $$\sin(5)\cdot\sin(4)\cdot\sin(3)\cdot\sin(2)\cdot\sin(1)$$
I think there is no way of writing it as factorial? Or is there any?

Please help, Thanks

答:
Using Pochhammer symbols, you have

$$P_n=\prod_{x=1}^n \sin(x)=2^{-(n+1)} e^{-\frac{1}{2} i n (n-\pi +1)} \left(-1;e^i\right){}_{n+1}
   \left(e^i;e^i\right){}_n$$ and so, you look for $k$ such that
$$k_n!=P_n \qquad \text{or better} \qquad \Gamma(k_n+1)=P_n$$

This is a difficult numerical problem since, as @K.defaoite already commented, the solution is not unique. But this is doable (do not wait for a closed form solution).


答:
Well, to begin with we $\sqrt{-i}$ isn't a valid answer.  Which complex number, expressed as $a + bi$, where $a,b \in \mathbb R$, is that?

This is a bit like answering Solve $x^2- 17 = 64$ and answering $x = \sqrt{64+17}$.  You just aren't finished.

Second of all.  All complex numbers (except $0$) could have multiple square roots and $\sqrt{-i}$ doesn't indicate which one you mean.  $\sqrt{complex number}$ is not well-defined and you shouldn't use it.

The question is... which complex number(s) $a + bi$ is it that $(a+bi)^2 = -i$?

.........

Your instructor did that $(x + yi)^2 = x^2 + 2xyi + y^2i^2 = (x^2 -y^2) + 2xy i = -i=0 + (-1)i$ so $x^2 -y^2 = 0$ and $2xy = -1$.

With that we have $x^2 = y^2$ so $x= \pm y$.

And so $2xy = \pm 2x^2 = -1$.  As $x^2 \ge 1$ (because $x$ is real) we must have $2xy =-2x^2$ and $x = -y$ and $-2x^2 = -1$ so $x^2 = \frac 12$ and $x = \pm \frac 1{\sqrt 2}$ and $y = \mp \frac 1{\sqrt 2}$.

So $x+yi = \pm \frac 1{\sqrt 2} \mp \frac 1{\sqrt 2} i = \pm (\frac 1{\sqrt 2} -\frac 1{\sqrt 2} i)$.

Title: Points to keep in mind while solving a complex equation

Body: I was solving this complex equation, $z^2 = -i$, where z is a complex number. I solved it to get $z = \sqrt{-i}$. But my teacher solved it by assuming $z = x + iy$ and he got the answer $z = \frac{1}{\sqrt2} - \frac{i}{\sqrt2} \space$ and $z = -\frac{1}{\sqrt2} + \frac{i}{\sqrt2}$ which is very different from my answer. Where am I wrong? Why cannot I solve it like that? What else should I keep in mind while solving a complex equation?


问:
Is it true that $M$ has $0$ as eigenvalue with multiplicity exactly $1$.
I am stuck on the following question: Its a true/false statement.

It says : Let $M$ be an $n\times n$ matrix. If none of the entries of $M$ are 0, but the row sum is zero for each row and no two rows are scalar multiples of each other, is it true that $M$ has $0$ as eigenvalue with multiplicity $1$.

My try:  Since $M$ is an $n\times n$ matrix with row sum  $0$ for each row, so $0$ is an eigenvalue of $M$ with multiplicity atleast $1$.
But is it true that it is an eigenvalue with multiplicity exactly $1$. I am stuck. I am getting it false. Is it correct?

What shall i do ? Can someone help.

答:
The multiplicity of $0$ as an eigenvalue is equal to the size of the matrix minus its rank (since it's the dimension of its null space). So you can "create" $0$ eigenvalues by, say, putting the same elements in different rows of the matrix (this will give you linearly dependent rows, which increase the null space).

So basically, if you wish to construct a counter example, you can take a matrix who's sum of elements in the first row is equal to $0$ (although the row itself is non-zero), and just make all other rows identical to the first row (you need at least $3$ rows for this trick to work). Can you now try to construct a counter example?


答:
Denote the column vectors of $A$ as $v_1, v_2,....v_n$ so that
$c_1v_1+c_2v_2+.....+c_nv_n=0$ has a unique non-zero solution given by the vector $(1,1,......,1)$. This tells you that the geometric multiplicity ($GM$) of eigenvalue $0$ is one. Now what does the relationship $AM\ge GM$ say?
You can easily construct a counterexample by taking all the row vectors of the form $(a,a,-2a), a\ne0$ in a $3\times 3 $ matrix $A$.

Title: Is it true that $M$ has $0$ as eigenvalue with multiplicity exactly $1$.

Body:
I am stuck on the following question: Its a true/false statement.

It says : Let $M$ be an $n\times n$ matrix. If none of the entries of $M$ are 0, but the row sum is zero for each row and no two rows are scalar multiples of each other, is it true that $M$ has $0$ as eigenvalue with multiplicity $1$.

My try:  Since $M$ is an $n\times n$ matrix with row sum  $0$ for each row, so $0$ is an eigenvalue of $M$ with multiplicity atleast $1$.
But is it true that it is an eigenvalue with multiplicity exactly $1$. I am stuck. I am getting it false. Is it correct?

What shall i do ? Can someone help.


问:
Generalized Schanuel Lemma
This is on page 128, ex 3.15, of Rotman's AIHA,

(Schanuel) Let $B$ be a left $R$-module over some ring $R$ consider two exact sequences,
      $$ 0 \rightarrow K \rightarrow P_n \rightarrow \cdots \rightarrow B \rightarrow 0 $$
      $$ 0 \rightarrow K' \rightarrow P'_n \rightarrow \cdots \rightarrow B \rightarrow 0 $$
      \noindent where $P_*, P'_*$ are projectives, prove that
      $$ K \oplus P'_n \oplus P_{n-1} \oplus \cdots \cong K' \oplus P_n \oplus P'_{n-1} \oplus \cdots $$

I could not really apply the usuall Schanuel's lemma, any hint?

答:
I found the following proof in Lectures on Modules and Rings by T. Y. Lam.

We do an induction on $n$. Assume the claim is true for $n-1$. Write $f$ and $g$ for the arrows $P_{0}\to B$ and $Q_{0}\to B$.
Applying the usual version of Schanuel's lemma to the sequences
\begin{gather*}
0\to\ker f\to P_{0}\to B\to0,\\
0\to\ker g\to Q_{0}\to B\to0,
\end{gather*}
we deduce that $\ker g\oplus P_{0}\cong\ker f\oplus Q_{0}$. Now the induction hypothesis applies to the sequences
\begin{gather*}
0\to K\to P_{n}\to\dots\to P_{2}\to P_{1}\oplus Q_{0}\to\ker f\oplus Q_{0}\to0,\\
0\to K'\to Q_{n}\to\dots\to Q_{2}\to Q_{1}\oplus P_{0}\to\ker g\oplus P_{0}\to0.
\end{gather*}

Title: Generalized Schanuel Lemma


问:
Absolute error from root in false position method
Suppose I know that $f(x_1)$ and $f(x_2)$ have opposite signs, so $f(x)=0$ has a root $x\in[x_1,x_2]$. Then using the false position method, I have a guess for the root
$$x_3=\frac{f(x_2)x_1-f(x_1)x_2}{f(x_2)-f(x_1)},$$
and I can iterate on either $[x_1,x_3]$ or $[x_3,x_2]$ depending on the sign of $f(x_3)$.

Usually we terminate the process when $|f(x_n)|<\epsilon$ for some specified $\epsilon$.

How can I pick $\epsilon$ so that I am certain that my guess for the root $x_n$ is within $\delta$ of the true value of the root, i.e. $|x_n-x|<\delta$?

I am not sure how to pick such an $\epsilon$ when we don't even know the true value $x$ of the root. When $\delta$ is sufficiently small, something like $\epsilon=\delta f'(x)$ could work, but obviously this requires that you (a) know the true value of the root and (b) know the derivative of the function, two assumptions that are definitely not true in general.

答:
You wrote :

Usually we terminate the process when $|f(x_n)|<\epsilon$ for some specified $\epsilon$.

This is not a convergence test. If it was, multiply any function by $10^{-999}$ and any point would be a solution according tho this test.

What you must use to end the process (and you almost wrote it) is
$$|x_{n+1}-x_n| \leq \epsilon$$


问:
Unique Solution of Initial Value Problem
State where in the $xy$-plane the initial-value problem
$$y' = \frac {x − y}{2x + 5y}$$
is guaranteed to have a unique solution.

My intuition for this problem was that if the derivative ($y'$) is $0$, the equation will have a unique solution. But I can't find any theorem supporting my intuition.

Any guidance is greatly appreciated!

答:
The given ODE is:
$$y' = \frac {x − y}{2x + 5y}~~~~(1)$$
Let $y=Vx \implies y'=V+xV'$
the ODE becomes
$$V+xV'=\frac{1-V}{2+5V} \implies xV'=\frac{1-3V-5V^2}{2+5V} \implies \frac{2+5V}{1-3v-5V^2} dV=\frac{dx}{x}~~~~(2)$$ $$= -\frac{1}{2} \left(\int \frac{dV}{5V^2+3V-1}+\frac{3+10V}{5V^2+3V-1} dV \right)=\ln Cx~~~~(3)$$
You may proceed from here to get a general solution as a family of curves with one constant $C$

The uinique solution without a constant:

Treat $V$ as constant (free of $x$), then $y'=V$, then from (1) we get
$$V=\frac{1-V}{2+5V}\implies V=\frac{-3\pm \sqrt{29}}{10}$$
So the unique(singular, essential) solutions of (1) are two fixed lines passing through origin, these are $$y=\frac{-3\pm \sqrt{29}}{10} ~x$$


答:
I hope you'll look past the many responses you've been given that presume calculus.  It is simply much easier dealing with complex numbers after Euler's identity, once you get there.

Your question, however, isn't how to work this problem, but that working it two different ways is giving you different results.  When that happens, it is a good idea to narrow down where the error is.

First of all, $n=8/3$ isn't a valid solution for the original problem.  You can confirm that on any calculator sufficiently advanced to deal with complex numbers.  Using $n=8/3$ doesn't give $1$, but rather is close to $-0.5 - 0.866i$.  So you made a mistake somewhere.

Up through the $(i^3)^{\frac{n}{2}} = 1$ the value $n=8/3$ does not work, and the value of $n=8$ does.  But that changes at $i^{\frac{3n}{2}} = 1$.  Now $n=8/3$ works.  What about this operation is broken?  Something about the move from here:

$$
(i^3)^{\frac{n}{2}} = 1
$$

to here:

$$i^{\frac{3n}{2}} = 1$$

is not as equivalent as you thought.

HINT: The rule $\sqrt{xz} = \sqrt{x}\cdot\sqrt{z}$ only holds for real numbers, not complex numbers.  We have to be careful when taking roots of complex numbers.

HINT #2: Normally, to isolate n, we would end up with $n=log(something)$.  You couldn't have known this yet, but logarithms have to be treated with care in complex numbers.  You'll get to this later, but multiplying a complex number by a real number scales it, whereas multiplying by an imaginary number rotates it (and multiplying by a complex number does both).  From that much information, can you infer why a logarithm might be problematic?

Title: Finding the smallest $n$ that satisfies $\left(\frac{1-i}{\sqrt{2}}\right)^n=1$

Body:
Maybe I'm just stupid, but right now I'm in a dilemma so please help. I had the question

$$\left(\frac{1-i}{\sqrt{2}}\right)^n=1$$ where I need to find the smallest number for $n$ ($n> 0$) that would satisfy the problem.

My first step was to $$\left[\left(\frac{1-i}{\sqrt{2}}
\right)^2\right]^\frac{n}{2}=1$$ and than $$(-i)^\frac{n}{2}=1$$ after that just ignore the minus because its powered by a even number and so on so forth $$i^\frac{n}{2}=1$$ $$\frac{n}{2}=4$$the  answer is $$n=8$$. But here is where the dilemma came if $$-i=i^3$$ than it would be $$\left(i^3\right)^\frac{n}{2}=1$$ which turns to $$i^\frac{3n}{2}=1$$ and the $$\frac{3n}{2}=4$$. This seems alright in the head but than the answer turns out to be $n=\frac{8}{3}$.

I used latex. Sorry if bad latex i'ts my first time.

If you know the answer please please tell me.

By the way I'm 17 and dumb. So please don't hate


问:
Is this an acceptable proof that $x^2<y^2$ when $x$ and $y$ are positive reals with $x<y$?
The question is

Prove the following statement:

Suppose $x, y \in \mathbb{R}$ and that $x$ and $y$ are positive. If $x < y$ then $x^{2} < y^{2}$.

Proof:

Suppose there is a value $r$ between $x$ and $y$ so that
$$x < r < y \tag1$$ where $r \in \mathbb{R}$ and $ r > 0$. Therefore,
$$x<r \Rightarrow x^{2}<xr \tag2$$ and
$$r<y \Rightarrow yr < y^{2} \tag3$$
Moreover, if we multiply equation $(1)$ by $r$, we get
$$xr < r^{2} < yr \tag4$$
Combining equations $(2)$, $(3)$, and $(4)$, we get
$$x^{2} < xr < r^{2} < xy < y^{2} \tag5$$
Therefore, $x^{2} < y^{2}$. $\blacksquare$

If necessary, we can even define $r=$ $x+y\over{2}$

Is there anything wrong with this approach? I'm aware of the normal method of direct proof.

答:
No need to introduce $r$; to wit:

$x = x; \tag 1$

$x < y; \tag 2$

thus

$x^2 < xy; \tag 3$

also,

$y = y; \tag 4$

from (2) and (4),

$xy < y^2; \tag 5$

combining (3) and (5):

$x^2 < xy < y^2. \tag 5$

$OE\Delta$.


问:
Improper integrals Question
I was asked to define the next intergrals and I want to know if I did it right:

$$1) \int^\infty_a f(x)dx = \lim_{b \to \infty}\int^b_af(x)dx$$
$$2) \int^b_{-\infty} f(x)dx = \lim_{a \to -\infty}\int^b_af(x)dx$$
$$3) \int^\infty_{-\infty} f(x)dx = \lim_{b \to \infty}\int^b_0f(x)dx + \lim_{a \to -\infty}\int^0_af(x)dx$$
Thanks.

答:
In some cases, such as Fourier transformation, it is critical that we have

$$\int_{-\infty}^{\infty} dx \, f(x) = \lim_{a \to \infty} \int_{-a}^{a} dx \, f(x)$$

Title: Improper integrals Question


问:
9 points on a quadric hypersurface
I found on some notes the statement

9 points in $\mathbb{P}^3$ always lie on a quadric hypersurface $Q$.

Actually I can't understand if this problem is easy to solve using the -elementary- techniques I got from a course in projective geometry I attended at the first year of my bachelor degree, or it can be solved only by using more advanced tools.

It's just curiosity, I tried some attacks without any success. I'd like to see why this should works, and in general how to attack statements like this. Thanks in advance.

答:
It's also important that for ''generic'' $9$ points there exists a unique quadric containing them.

Fact: if in $k^N$  we have some vectors such that every $N$ of these vectors are linearly dependent, then $\it{all}$ of these vectors lie in a subspace of $N-1$.
Indeed, the span of these vectors is of dimension $\le N-1$.
From here we conclude ( contrapositive):

If $N$ function $\phi_{i} \colon X \to k$,  are linearly independent, then, there there exist $N$ points $(x_j)_{j}$ such that the vectors $(\phi_{i}(x_{j}))_{i,j}$ are linearly independent,  there exist $N$ points $(x_{j})$
such that
$$ \det (\phi_{\alpha}(x_{\beta}) )\ne 0$$

Consider now $(\phi_{\alpha})$ $N$ polynomials that are linearly independent as polynomials. They they are also independent as polynomial functions ( assume $k$ infinite, or take an infinite extension $\bar k$  of $k$).  Therefore, there exist $N$ points $(x_{\beta})$ such that
the determinant $\det (\phi_{\alpha} ( x_{\beta}) \ne 0$.  This determinant as a function of $N$ points $(x_{\beta})$, is a polynomial function, not $0$.
We also see that  for $k$ generic points $(x_{\beta})$, $1\le k \le N$, the matrix $(\phi_{\alpha} ( x_{\beta})$ has rank $k$

Let's consider now $\phi_{i}$ $N$ homogenous polynomials in variable $x$, of same degree $d$, that are linearly independent. For $N-1$ generic points $(x_{\beta})$ the matrix $(p_{\alpha}(x_{\beta}))$ has rank $N-1$. Since the polynomials are homogenous of same degree, a rescaling of the points will just multiply the rows of the matrix by constants. Therefore, we can say: for $N-1$ generic points in the projective space, there exists a unique ( up to a multiplication by a constant) linear combination of $\sum c_{\alpha} \phi_{\alpha}$ that is zero at these points.  The equation equation can be obtained by considering an $N\times N$ determinant with entries $\phi_i(x_j)$, $\phi_i(x)$.

Title: 9 points on a quadric hypersurface


问:
Freidlin-Wentzell theory: clarification of a theorem
I am studying Freidlin and Wentzell large deviation theory from their 2012 book "Random perturbations of dynamical systems" and I have a problem in justifying two steps of the proof of theorem 1.3. The theorem states

Theorem

We assume that there exists a continuous function $\bar{b}(t, x), t > 0,x \in \mathbb{R}^{r}$  such that for any $δ > 0, T > 0, x ∈ R^r$ we have
\begin{equation}
\label{eqn: hp1}
\lim_{\epsilon \to 0} \mathbb{P} \Big( \Big| \int_{t_0}^{t_0+T}b(\epsilon,t,x,\omega)dt - \int_{t_0}^{t_0+T} \bar{b}(t,x)dt \Big| > \delta \Big) = 0
\end{equation}
uniformly in $t_0 \ge 0$. Then the equation
\begin{equation}
\label{eqn: hp2}
\dot{\bar{x}}_t=\bar{b}(t,\bar{x}_t), \qquad \bar{x}_0=x
\end{equation}
has a unique solution and
\begin{equation}
\lim_{\epsilon \to 0} \mathbb{P} \big( \max_{0 \le t \le T}|X_{t}^{\epsilon}-\bar{x}_t| > \delta \big) =0
\end{equation}
for every $T>0$ and $\delta>0$.

Proof
Since the function $\bar{b}(t,x)$ is continuous, by the mean value theorem we have
$
\int_{t}^{t+\Delta}\bar{b}(s,x) ds =\bar{b}(t,x)\Delta+o(\Delta), \qquad \Delta \to 0
$
Taking into account the first hypothesis, we get
\begin{equation}
\begin{split}
|\bar{b}(t,x)-\bar{b}(t,y)| &= \frac{1}{\Delta} \Big| \int_{t}^{t+\Delta} \bar{b}(s,x)ds - \int_{t}^{t+\Delta} \bar{b}(s,y)ds \Big| + \frac{o(\Delta)}{\Delta}\\
& \le  \frac{1}{\Delta} \Big| \int_{t}^{t+\Delta} b(\epsilon,s,x,\omega)ds - \int_{t}^{t+\Delta} b(\epsilon,s,y,\omega)ds \Big| \\
& + \frac{o(\Delta)}{\Delta} +\delta_{\epsilon} \\
& \le K|x-y|+\frac{o(\Delta)}{\Delta} +\delta_{\epsilon}
\end{split}
\end{equation}
where $\delta_{\epsilon}=\delta_{\epsilon}(t,\omega) \to 0$ in probability as $\epsilon \to 0$. And from this follows the Lipschitz condition on $\bar{b}$ and so the existence and uniqueness.

Here I don't understand how to get $\delta_{\epsilon}$, which I don't even know how it is defined.

I tried in this way: from the hypotesis I remark that $-\delta \le  \int_{t_0}^{t_0+T}b(\epsilon,t,x,\omega)dt - \int_{t_0}^{t_0+T} \bar{b}(t,x)dt \le \delta$
and so, after adding and subtracting both $\int_{t}^{t+\Delta}b(\epsilon,t,x,\omega)dt$ and $\int_{t}^{t+\Delta}b(\epsilon,t,y,\omega)dt$ and use the remark above and I obtain
\begin{equation}
\begin{split}
|\bar{b}(t,x)-\bar{b}(t,y)| &= \frac{1}{\Delta} \Big| \int_{t}^{t+\Delta} \bar{b}(s,x)ds - \int_{t}^{t+\Delta} \bar{b}(s,y)ds \Big| + \frac{o(\Delta)}{\Delta}\\
& \le  \frac{1}{\Delta} \Big| \int_{t}^{t+\Delta} b(\epsilon,s,x,\omega)ds - \int_{t}^{t+\Delta} b(\epsilon,s,y,\omega)ds \Big| \\
& + \frac{o(\Delta)}{\Delta} +\frac{2}{\Delta}\delta \\
& \le K|x-y|+\frac{o(\Delta)}{\Delta} +\delta_{\epsilon}
\end{split}
\end{equation}
where $\delta_{\epsilon}=\frac{2}{\Delta}\delta$. But it doesn't seem right since it is not random and so how can I say that it converges in probability as the quantity of the authors?

The other doubt is some steps after this. I have the estimate that follows:
\begin{equation}
\begin{split}
& \int_{0}^{t} [b(\epsilon,s,\bar{x}_s,\omega)-\bar{b}(s,\bar{x}_s)]ds \\
&= \sum_{k=0}^{n-1} \int_{kt/n}^{(k+1)t/n} [b(\epsilon,s,\bar{x}_s,\omega)-\bar{b}(s,\bar{x}_s)]ds \\
&= \sum_{k=0}^{n-1} \int_{kt/n}^{(k+1)t/n} [b(\epsilon,s,\bar{x}_{kt/n},\omega)-\bar{b}(s,\bar{x}_{kt/n})]ds \\
&+ \sum_{k=0}^{n-1} \int_{kt/n}^{(k+1)t/n} [b(\epsilon,s,\bar{x}_s,\omega)-b(\epsilon,s,\bar{x}_{kt/n},\omega)]ds \\
&+ \sum_{k=0}^{n-1} \int_{kt/n}^{(k+1)t/n} [\bar{b}(s,\bar{x}_{kt/n},)-\bar{b}(s,\bar{x}_s)]ds \\
&= \sum_{k=0}^{n-1} \int_{kt/n}^{(k+1)t/n} [b(\epsilon,s,\bar{x}_{kt/n},\omega)-\bar{b}(s,\bar{x}_{kt/n})]ds + \rho_{n,t}^{\epsilon}
\end{split}
\end{equation}
where they say that $|\rho_{n,t}^{\epsilon}| \le C/n$ with $C=C(T,K)$. I tried in this way but I'm stuck and don't know how to go on
\begin{equation}
\begin{split}
|\rho_{n,t}^{\epsilon}| &\le \sum_{k=0}^{n-1} \int_{kt/n}^{(k+1)t/n} |b(\epsilon,s,\bar{x}_s,\omega)-b(\epsilon,s,\bar{x}_{kt/n},\omega)|ds \\
&+ \sum_{k=0}^{n-1} \int_{kt/n}^{(k+1)t/n} |\bar{b}(s,\bar{x}_{kt/n},)-\bar{b}(s,\bar{x}_s|ds \\
& \le 2K \sum_{k=0}^{n-1}\int_{kt/n}^{(k+1)t/n}|\bar{x}_s - \bar{x}_{kt/n}|ds
\end{split}
\end{equation}
Now I don't know how to procede: how can I get the same result as in the book?

Questions

is what I did at the beginning correct? If not, how do I change it?

how do I go on in the second part?
Thanks to anyone who will have the patience to read and answer.

答: For the first question, I think you are almost there. When you wrote something like $$-\delta \le  \int_{t_0}^{t_0+T}b(\epsilon,t,x,\omega)dt - \int_{t_0}^{t_0+T} \bar{b}(t,x)dt \le \delta,$$ you should be aware of the randomness contained in $\omega$, so this statement can only hold with certain probability (here with high probability), so the bound on $|\bar{b}(t,x) - \bar{b}(t,y)|$ you wrote only hold with high probability. For the second question I am not completely sure as well. I guess the authors used something like $|\bar{x}_s - \bar{x}_{kt/n}| \leq C/n$ for $s \in [kt/n,(k+1)t/n]$, this can probably be justified by the Lipschitz continuity enjoyed by $\bar{b}$ (maybe we also need to use that $\bar{b}$ is uniformly bounded on a compact time-space region). Hope this helps!


问:
1D Wave PDE with Nonzero Initial and Boundary Conditions
I'm not sure how to start this PDE since the initial and boundary conditions are nonzero. May someone point me in the right direction?

This is the problem:

$$u_{tt} = u_{xx}$$
$$u(x,0) = \frac{1}{2+ \sin(x)}$$
$$u_t(x,0) = -\frac{\cos(x)}{(2+ \sin(x))^2}$$
$$u(0,t) = u(2\pi,t)= \frac{1}{2+ \sin(t)}$$

答: Hint: The general solution of the 1D wave equation is $u(x,t)=F(x+t)+G(x-t)$.


答:
The value of the function at any point $(0,c)$ on Y-axis is $f(0,c)=0$.

Now observe the value of function at the points falling in the $\epsilon$-neighborhood of $(0,c)$ by sliding left or right keeping $c$ fixed, so that $f(0-\epsilon,c)=f(0+\epsilon,c)=In(1+\epsilon^2. c)>f(0,c)$ for $c>0$$\implies $ maxima for all points on positive Y-axis.

and $f(0-\epsilon,c)=f(0+\epsilon,c)=In(1+\epsilon^2. c)<f(0,c)$ for $c<0$$\implies $ minima for all points on negative Y-axis.

Title: Stationary points of $\ln(1+x²y)$

Body:
The stationary points of the function lie along with $(0,y)$ for all $y$. The second derivative test is inconclusive.

I understand that the origin is the saddle point. But how do we categorize the other points without looking at the graph? In general, what is the approach to take in case of inconclusive second derivative tests?


问:
Showing that $ \int_0^1 \frac{t^{x}-1}{\ln(t)}dt$ exists for all $x$
How do I show that $\displaystyle  \int_0^1 \frac{t^{x}-1}{\ln(t)}\,dt $ exists for every $ x \in \mathbb{R} $?

I have tried bounding it but without success...

答:
Define $$f(x)=\int_0^1\frac{t^x-1}{\log t}\,dt.$$
Then, we have
\begin{align}
f'(x)&=\int_0^1\frac{\partial}{\partial x}\frac{t^x-1}{\log t}\,dt\\
f'(x)&=\int_0^1t^x\,dt\\
f'(x)&=\frac{1}{x+1}, \quad x>-1.\\
\end{align}

Therefore, the statement is wrong.


问:
Newton Raphson for nonlinear overdetermined equations on noisy data
I would like to know, whether any improved Newton Raphson method is available for non-linear overdetermined equations (So we use Jacobian matrix and pseudo inverse). Data used as measurements are noisy. The final outputs will be very close by, for eg: if x1,x2,x3,x4 are outputs it will be 200.5,200.51, 200.61,200.53. When I initialize Newton raphson with this values it diverges. I have to give far values to initialize. How to give a good guess for Newton Raphson.

Number of equations are 6 (non-linear, quadratic), number of outputs are 4.

Please guide

答:
So, you have $n > 4$ equations
$$f_i(x_1,x_2,x_3,x_4)=0 \qquad \text{for} \qquad i=1,2,\cdots,n$$
I should not try to solve the problem using Newton-Raphson but, instead, I should define
$$\Phi(x_1,x_2,x_3,x_4)=\sum_{i=1}^n \big[f_i(x_1,x_2,x_3,x_4)\big]^2$$and I shall try to minimize $\Phi(x_1,x_2,x_3,x_4)$ providing the analytical Jacobian and, if possible, the analytical Hessian.


问:
Existence of a measure space with a given condition
Motivated by this question and this question, among others, I wondered when can we say that there exists a measure space $(X, \mathcal{S}, \mu)$ such that
$$\operatorname{Im}(\mu) = A \cup\left\{0,+\infty\right\},$$
where $A \subset \mathbb{R}$ is a subset of the real numbers. This question may be too general, so in particular I am interested in the cases where $A = \mathbb{Q}$ and $A = \mathbb{R}\setminus \mathbb{Q}$. Thanks in advance!

答:
There is no measure space $(X, \mathcal{S}, \mu)$ such that
$$\operatorname{Im}(\mu) = \Bbb Q^+ \cup\left\{0,+\infty\right\}$$
(where $\Bbb Q^+$ denotes the positive rational numbers).

Proof:

Let $(X, \mathcal{S}, \mu)$ be a measure space. Suppose that $\operatorname{Im}(\mu) = \Bbb Q^+ \cup\left\{0,+\infty\right\}$.

Using a result from  Sierpinski (see for instance, this article in Wikipedia), we have that, if the image of a measure $\mu$ does not contains any interval, then for every $E \in \mathcal{S}$ such that $\mu(E)>0$, $E$ contains (at least) one atom.

Since $\operatorname{Im}(\mu) = \Bbb Q^+ \cup\left\{0,+\infty\right\}$, we can take a $E_0 \in \mathcal{S}$ such that $0<\mu(E_0)<+\infty$. Let $A_0 \subseteq E_0$ be an atom. We have that $0<\mu(A_0)  <+\infty$.

Since $\frac{1}{3}\mu(A_0) \in \Bbb Q^+$, there is $E_1 \in \mathcal{S}$ such that $\mu(E_1)= \frac{1}{3}\mu(A_0)$. Let $A_1  \subseteq E_1$ be an atom, we have that
$\mu(A_1)\leqslant \frac{1}{3}\mu(A_0)$. Since $A_0$ is an atom, we have that $\mu(A_1 \cap A_0)=0$. So we can assume without loss of generality that $A_1 \cap A_0 = \emptyset$ (if necessary, we can replace $A_1$ by $A_1\setminus (A_1 \cap A_0)$).

So by finite induction we can produce a family $\{A_n\}_{n \in \Bbb N}$ of disjoint atoms such that, for $n \in \Bbb N$,
$$\mu(A_{n+1}) \leqslant  \frac{1}{3}\mu(A_n) \tag{1}$$

Now consider the function $f: 2^{\Bbb N} \rightarrow \Bbb R$ defined by: if $N \subseteq \Bbb N$,
$$f(N)=\mu \left(\bigcup_{i \in N}A_i  \right)=\sum_{i \in N}\mu(A_i) $$
Using $(1)$, it is easy to prove that $f$ is injective (one-to-one) and since $ 2^{\Bbb N}$ is uncountable, we have that $\operatorname{Im}(f)$ is uncountable. However, it is clear that  $\operatorname{Im}(f) \subseteq \operatorname{Im}(\mu)$ and  $\operatorname{Im}(\mu)$ is countable. Contradiction.

There is no measure space $(X, \mathcal{S}, \mu)$ such that $$\operatorname{Im}(\mu) = (\Bbb R^+ \setminus \Bbb Q) \cup\left\{0,+\infty\right\}$$
(where $\Bbb R^+$ denotes the positive real numbers).

Proof:

Let $(X, \mathcal{S}, \mu)$ be a measure space. Suppose that $\operatorname{Im}(\mu) = (\Bbb R^+ \setminus \Bbb Q) \cup\left\{0,+\infty\right\}$.

By a result from  Sierpinski, we know that the image of an non-atomic measure is an interval (see for instance, this article in Wikipedia). In fact, from Sierpinski's result, we can easy prove that: if the image of a measure $\mu$ does not contains any interval, then for every $E \in \mathcal{S}$ such that $\mu(E)>0$, $E$ contains (at least) one atom.

Since $\operatorname{Im}(\mu) = (\Bbb R^+ \setminus \Bbb Q) \cup\left\{0,+\infty\right\}$, we can take a $E \in \mathcal{S}$ such that $0<\mu(E)<+\infty$. Let $A \subseteq E$ be an atom. We have that $0<\mu(A)  <+\infty$.

Now take $r\in \Bbb Q$ such that $\mu(A)< r < 2 \mu(A)$. So, we have that
$$0< r-\mu(A) <  \mu(A)$$
Since $\mu(A) \notin \Bbb Q$ and $r\in \Bbb Q$, we have that $ r-\mu(A) \in (0, +\infty) \setminus \Bbb Q^+$. So there is $B \in \mathcal{S}$, such that $\mu(B) = r-\mu(A)$.

Since $A$ is an atom and $\mu(B) < \mu(A)$, we have that $\mu(A\cap B)=0$. So
$$\mu(A \cup B) = \mu(A) + \mu(B)=  \mu(A) + r-\mu(A) = r \in \Bbb Q$$
and $r>0$.
Contradiction to the assumption that $\operatorname{Im}(\mu) = (\Bbb R^+ \setminus \Bbb Q) \cup\left\{0,+\infty\right\}$.

The cases $A = \mathbb{Q}$ and $A = \mathbb{R}\setminus \mathbb{Q}$, that is, $\operatorname{Im}(\mu) = \Bbb Q \cup\left\{0,+\infty\right\}$ and $\operatorname{Im}(\mu) = (\Bbb R \setminus \Bbb Q) \cup\left\{0,+\infty\right\}$.

We have that $\mu$ should be a signed measure. However, a signed measure must have either its negative part ($\mu^-$) or its positive ($\mu^+$) part finite.

So, there is no measure space $(X, \mathcal{S}, \mu)$ such that
$\operatorname{Im}(\mu) = \Bbb Q \cup\left\{0,+\infty\right\}$
and there is no measure space $(X, \mathcal{S}, \mu)$ such that $\operatorname{Im}(\mu) = (\Bbb R \setminus \Bbb Q) \cup\left\{0,+\infty\right\}$.


问:
What is the expectation of $B_s^2B_t^2$ where $B_t$ is a standard Brownian motion?
I am tring to find the variance of the integral $I=\int_0^T B_t^2 dt$.

I have found that $\mathbb{E}[I] = \int_0^T \mathbb{E}[B_t^2] dt = \frac{T^2}{2}$.

Since $\mathbb{E}[I^2] = \mathbb{E}[\int_0^T \int_0^T B_t^2 B_s^2 \ dt\ ds] = \int_0^T \int_0^T \mathbb{E}[B_t^2 B_s^2] \ dt \ ds$, I want to know the value of $\mathbb{E}[B_t^2 B_s^2]$.

Body: For $s<t$ we have $EB_t^{2}B_s^{2}=E((B_t-B_s)+B_s)^{2}B_s^{2}=E((B_t-B_s))^{2}B_s^{2}+EB_s^{4}$ since $E((B_t-B_s)B_s)B_s^{2}E((B_t-B_s)) EB_s^{3}=0$. So $EB_t^{2}B_s^{2}=E((B_t-B_s))^{2}EB_s^{2}+EB_s^{4}=(t-s) s+EB_s^{4}$. $EB_s^{4}=s^{2}m_4$ where $m_4$ is the 4th moment of the standard normal distribution.


问:
Three equilateral triangles
Can you provide a proof for the following claim:

Claim. Given an arbitrary equilateral triangles $\triangle ABC$ and $\triangle BDE$ with common vertex $B$ . The points $H$ and $I$  divide line segments $CE$ and $AD$ respectively in the ratio $2 : 1$ . Let $G_1$ be the centroid of triangle $\triangle ABC$ . Then triangle $\triangle G_1IH$ is an equilateral triangle as well.

GeoGebra applet that demonstrates this claim can be found here.

So far I have managed to prove that $|AE|=|CD|$ and that line segments $AE$ and $CD$ intersect each other at angle of $60^{\circ}$ , but I dont know if this can be of any use. Also, can we apply the fundamental theorem of similarity in some way?

Title: Three equilateral triangles

Body:
Use complex numbers. Put $G_1=0$, $B=1$ and $BD=z$ (so $BE=ze^{i\pi/3}$). Then, letting $w=e^{i\pi/3}$,
$$H=\frac13(w^2+2(1+zw))=\frac13(w^2+2zw+2)$$
$$I=\frac13(w^4+2(1+z))=\frac13(w^4+2z+2)$$
But $w^2+2=w(w^4+2)$, so
$$H=w\cdot\frac13(w^4+2z+2)=wI$$
Since $|w|=1$ and $\arg w=60^\circ$, $G_1H=G_1I$ and $\angle HG_1I=60^\circ$, so $\triangle HG_1I$ is equilateral.


问:
Can anyone help me to write down the derivative of $j$?
Suppose that $f(z)$ is a modular form of weight $k$, therefore for any $z\in \mathbb{H}$, and any matrix in $SL_2(\mathbb{Z})$ we have:
$f(\frac{az+b}{cz+d}) = (cz+d)^k f(z)$. I do not know how to use this relation to compute the derivative of $f$. Now let $G_{2k}$ be an Eisenstein series of weight $2k$ (https://en.wikipedia.org/wiki/Eisenstein_series). How can I compute derivates of $E_4$ and $E_6$ in terms of Eisenstein series? Starting from $G_{2k}(\frac{az+b}{cz+d}) = (cz+d)^{2k} G_{2k}(z)$, I can just find this immediate relation $(\frac{1}{cz+d})^2G'_{2k}(\frac{az+b}{cz+d}) = (cz+d)^{2k} G'_{2k}(z)+2kc(cz+d)^{2k-1} G_{2k}(z)$. How can I calculate the derivative of $j$-invariant?

Edit: In the third comment of this question The roots of $J' = \frac{dJ}{d\tau}$ it is written that $J'/J = -2\pi i E_6(\tau)/E_4(\tau)$. I don't know why this is true. But if we have this relation then from this point it is easy to finish. But I can not realize how to derive that relation, the link does not lead to "Ramanujan's system of ODEs". The crucial problem is that to me $j$ and $E_k$'s are defined as functions of $\tau$, $j(\tau)$ and $E_k(\tau)$. I don't know how to differentiate with respect to $\tau$.

A situation happened to me which is very similar to my old question Prove or give a counter-example about a statement on faithful simple left modules over a finite-dimensional algebra. . I was reading about modular forms. But I faced too many questions, which I can not handle them or I have some doubts about.

By $\mathbb{H}$ I mean the upper half plane. It deals with the $j$ invariant, but I do not know it well enough. It says that $j$ is a holomorphic function that gives a homeomorphism between $\mathbb{H}/PSL_2(\mathbb{Z})$ and $\mathbb{C}$.

Does the function $\tau \mapsto e^{2\pi i \tau}$ gives a holomorphic bijection between the fundamental domain and punctured disk $\mathbb{D}$? If yes, then I am curious if Is it wise to define the inverse map as $q \mapsto \frac{1}{2 \pi i}\log(q)$? Does it make sense?

Let $q=e^{2\pi i \tau}$. If we are able to define this logarithmic map then we can define $J(e^{2\pi i \tau})=J(q)=j(\frac{1}{2 \pi i}\log(q))$, which is a map from the punctured disk $\mathbb{D}$ to $\mathbb{C}$. If we are not able to define that logarithmic map, then how can we define this last bijection between punctured disk $\mathbb{D}$ and $\mathbb{C}$? Can we consider the point at infinity as the corresponding point to the center of the punctured disc? how?

These were my doubts, now this is my question. Why the set of zeros of $j'$, $j'^{-1}(0)$ is finite in the fundamental domain? The crucial problem is that to me $j$ and $E_k$'s are defined as functions of $\tau$, $j(\tau)$ and $E_k(\tau)$. I don't know how to differentiate with respect to $\tau$.

答:
There is no alternative here apart from Ramanujan's system of differential equations for Eisenstein series.

Let $\tau\in\mathbb{H} $ and $q=\exp (2\pi i\tau) $ so that $dq=2\pi iq\, d\tau$ or $$\frac{d} {d\tau} =2\pi iq\frac{d} {dq} \tag{1}$$ Ramanujan defines his functions $P, Q, R$ which are equivalent to Eisenstein series as
\begin{align}
P(q) &=E_2(\tau)=1-24\sum_{n=1}^{\infty} \frac{nq^n} {1-q^n}\tag{2a}\\
Q(q) &=E_4(\tau)=1+240\sum_{n=1}^{\infty} \frac{n^3q^n} {1-q^n}\tag{2b}\\
R(q) &=E_6(\tau)=1-504\sum_{n=1}^{\infty} \frac{n^5q^n} {1-q^n}\tag{2c}\\
\end{align}
Ramanujan used elementary methods to establish the following system of differential equations satisfied by $P, Q, R$
\begin{align}
q\frac{dP(q)} {dq} &=\frac{P^2(q)-Q(q)}{12}\tag{3a}\\
q\frac{dQ(q)} {dq} &=\frac{P(q)Q(q)-R(q)}{3}\tag{3b}\\
q\frac{dR(q)} {dq} &=\frac{P(q)R(q) -Q^2(q)}{2}\tag{3c}\\
\end{align}
Usually the above equations are written without the parameter $q$ so that $P(q) $ is written just as $P$. Ramanujan's proof of the above equations is given in my blog posts (see this post and the next one). Alternatively they can proved using technique of modular forms by knowing that both sides of each relation are in same space of modular forms of a specific dimension. I am not an expert in this area of modular forms so can't really give more details.

The function $J(\tau) $ is defined as $$J(\tau) =\frac{E_4^3(\tau)} {E_4^3(\tau)-E_6^2(\tau)}=\frac{Q^3}{Q^3-R^2}\tag{4}$$ Using Ramanujan's equations above one can easily prove that $$q\frac{d} {dq}\{\log(Q^3-R^2)\}=P\tag{5}$$ and hence
\begin{align}
q\frac{d} {dq} (\log J) &=3q\frac{d\log Q}{dq}-q\frac{d} {dq} \{\log(Q^3-R^2)\}\notag\\
&=3\cdot\frac{PQ-R} {3Q} -P\notag\\
&=-\frac{R}{Q}\notag
\end{align}
Now we have $$\frac{d} {d\tau} (\log J) =2\pi i q\frac{d} {dq} (\log J) =-2\pi i\frac{R} {Q} $$ ie $$\frac{J'(\tau)} {J(\tau)} =-2\pi i\cdot\frac{E_6(\tau)}{E_4(\tau)}$$ as desired.


Title: Find $^{505}C_5-5\cdot{^{404}C_5}+10\cdot{^{303}C_5}-10\cdot{^{202}C_5}+5\cdot{^{101}C_5}-1$

Body:
Find the sum

$$^{505}C_5-5\cdot{^{404}C_5}+10\cdot{^{303}C_5}-10\cdot{^{202}C_5}+5\cdot{^{101}C_5}-1$$

Critiques and other answers are always welcome!

Body:
We have to find the coefficient of $x^5$ in
$$(1+x)^{505}-5.(1+x)^{404}+10.(1+x)^{303}-10.(1+x)^{202}+5.(1+x)^{101}-1=((1+x)^{101}-1)^5=({101}x+O(x^2))^5$$
Multinomially expand the final term and convince yourself that the coefficient of $x^5$ is $101^5$


问:
A problem on square matrices

If $B,C$ are $n$ rowed square matrices and if $A=B+C, BC=CB, C^2=O$, then show that for every $n \in \mathbb N$,
 $$A^{n+1}=B^n(B+(n+1)C)$$

I tried to prove it using mathematical induction. But I could not get $P(1)$ to be true.
$$P(1): A^2 = B(B+2C)$$
I couldn't equate them.

Please offer some assistance.
Thank you. :)

答: Try P(0) it wll be easier. But P(1) should work too.</p>

Title: A problem on square matrices


问:
How to calculate the intersection of two planes?
How to calculate the intersection of two planes ?

These are the planes and the result is gonna be a line in $\Bbb R^3$:

$x + 2y + z - 1 = 0$

$2x + 3y - 2z + 2 = 0$

答:
You have to solve the following system of  equations:
$$ x + 2y + z - 1 = 0 \\
2x + 3y - 2z + 2 = 0. $$

Since these are two equations in three variables, so you don't have solution such as $(a,b,c)$ but you will have a parametric relation between  variables. Let $x=t$, then we have

$$ t + 2y = 1-z \\
2t + 3y  = 2z-2. $$

Solving this system gives

$$ x=t,y=-\frac{4}{7}t,z= 1-\frac{1}{7}t$$

which is   the parametrized equation of the required line.


答:
Notice that the function $f:t\mapsto \frac{t^x-1}{\ln t}$ can be extended by continuity on 1 since by the Hospital rule

$$
\lim_{t\to1}f(t)=\lim_{t\to1}\frac{xt^{x-1}}{1/t}=x,
$$
hence the integral $\int_{1/2}^1f(t)dt$ exists for all $x$. Now for the problem in 0 and by change of variable $u=-\ln t$ the problem becomes at $\infty$
$$\int_0^1 f(t)dt=\int_{0}^\infty\frac{e^{-u}-e^{-(x+1)u}}{u}du$$
so because the integral $\int_1^\infty \frac{e^{-u}}{u}du$ is convergent we see that the given integral converges iff $x+1>0$ i.e. $x>-1$.

Title: Showing that $ \int_0^1 \frac{t^{x}-1}{\ln(t)}dt$ exists for all $x$

Body:
How do I show that $\displaystyle  \int_0^1 \frac{t^{x}-1}{\ln(t)}\,dt $ exists for every $ x \in \mathbb{R} $?

I have tried bounding it but without success...


问:
Find a joint distribution of two random variables $X$ and $Y$ that are not independent and that $Y$ has a marginal distribution $ Y\thicksim Ber(p)$
Find a joint distribution of two random variables $X$ and $Y$ that are not independent and that $Y$ has a marginal distribution $ Y\thicksim Ber(p)$

The problem I have is that for every X with a certain distribution I propose for this question, it ends up fulfilling that X and Y are independent

答:
How about $Y\sim\text{Ber}(p)$ and $X=Y$?

$X=1-Y$ does it too.


问:
For a 2n-bit binary number, how many possible values are there such that half the bits are set to 1 (including leading zeros)??
So for n = 1 you'd have 01 and 10. (2 possibilities)

For n = 2 you'd have 0011, 0101, 0110, 1001, 1010, 1100. (6 possibilities)

How do I find the general formula for any positive integer value of n?

答: You choose $n$ out of $2n$ positions to write a $1$, so $2n\choose n$.


问:
expected value of random variables
Take two random variables $X=a+bX_0$ and $Y=c+dY_0$, and define $T=X-Y=\mu+\sigma Z$ where $\mu$ is the mean of $T$, $\sigma$ its standard deviation and $Z$ is a standardized random variable with mean $zero$, unit variance and continuous PDF $\phi$ and continuous and increasing CDF $\Phi$. NB: $a,b,c,d \in \mathbb{R}$

What does $\Phi(\frac{-\mu}{\sigma})$ mean, either in terms of $T$?  or $X$ and $Y$?

答:
It represents the probability that $T$ is less than $0$ or equivalently the probability that $X$ is less than $Y$.

$$\text{Prob}(T \le 0) = \text{Prob}(\mu+\sigma Z \le 0)=\text{Prob}(Z \le \frac{-\mu}{\sigma})=\Phi(\frac{-\mu}{\sigma})$$

Title: expected value of random variables


答:
Here's a synthetic solution. You have already proven the required facts.

Let $X, Y$ be the midpoints of $\overline{BC}, \overline{BA}$, respectively. Then $\triangle XBY$ is equilateral. Apply your results to $\triangle XBY$ and $\triangle EBD$ to obtain

$\angle(\overline{XD}, \overline{YE}) = 60^{\circ}$.

$XD = YE$.

Now here's the critical point of the argument: $\triangle CYE \sim \triangle CG_1H$ and $\triangle AXD \sim \triangle AG_1I$. Thus $\overline{YE} \parallel \overline{G_1H}$ and $\overline{XD} \parallel \overline{G_1I}$. Furthermore, $G_1H = \lambda YE$ and $G_1I = \lambda XD$ for a scaling factor $\lambda = 2/3$. This, all together, shows

$\angle(\overline{G_1I}, \overline{G_1H}) = 60^{\circ}$.

$G_1H = G_1I$.

We're done.

Title: Three equilateral triangles

Body:
Can you provide a proof for the following claim:

Claim. Given an arbitrary equilateral triangles $\triangle ABC$ and $\triangle BDE$ with common vertex $B$ . The points $H$ and $I$  divide line segments $CE$ and $AD$ respectively in the ratio $2 : 1$ . Let $G_1$ be the centroid of triangle $\triangle ABC$ . Then triangle $\triangle G_1IH$ is an equilateral triangle as well.

GeoGebra applet that demonstrates this claim can be found here.

So far I have managed to prove that $|AE|=|CD|$ and that line segments $AE$ and $CD$ intersect each other at angle of $60^{\circ}$ , but I dont know if this can be of any use. Also, can we apply the fundamental theorem of similarity in some way?


问:
How we can find an admissible change of variables that goes from (0,2) to $(-\infty,0)$?
I find a text on the web about curves and it ask for an admissible change of variables that goes from (0,2) to (-$\infty$, 0), I think about log(-x) or some function like that, but I really don't get it, any hints on this one?

Body: $$\tan\left[\frac{\pi}{2}\left(\frac{x}{2} - 1\right)\right].$$


问:
Prove that if a differential equation $f$ is separable, then $\frac{\partial ^2}{\partial x \partial y} \ln |f|=0$
Prove that if the differential equation
$$\frac {dy}{dx} = f(x, y)$$

is separable, then for all $(x,y)$ such that $f(x,y) \neq 0$,

$$ \frac{\partial ^2}{\partial x \partial y} \ln |f|=0$$

I tried to approach this by starting with
$$ \frac{\partial ^2}{\partial x \partial y} \ln |f|= \frac{\partial}{\partial x} \frac{\partial}{\partial y} \ln |f| = \frac{\partial}{\partial x} \frac{\partial \ln |f|}{\partial f} \frac {dy}{dx} = \frac{\partial}{\partial x} \frac{\partial \ln |f|}{\partial f} f(x, y) = \frac{\partial}{\partial x} \frac{f(x, y)}{ f} = \frac{\partial}{\partial x} 1 = 0$$

However, I am not sure if this proof is valid. Any guidance is greatly appreciated!

Edit: $$ \frac{\partial ^2}{\partial x \partial y} \ln |f|= \frac{\partial}{\partial x} \frac{\partial}{\partial y} (\ln |M(x)| - \ln |N(y)|) = \frac{\partial}{\partial x} (0- \frac{\partial}{\partial y}(\ln |M(x)| - \ln |N(y)|)) =0$$

答:
If a differential equation is separable then $\frac{dy}{dx}=f(x,y)=\frac{M(x)}{N(y)}$.

$\ln|f(x,y)|=\ln|M(x)|-\ln|N(y)|$

$\frac{\partial^2}{\partial x \partial y} \big[\ln|M(x)|-\ln|N(y)| \big] = \frac{\partial}{\partial x}\bigg[0-\frac{\partial}{\partial y}\big[\ln|N(y)|\big]\bigg]=0 $ because $M(x)$ and $N(y)$ are functions of $x$ and $y$ strictly.


问:
If $x^2\equiv 1 \pmod{n}$ and $x \not\equiv \pm 1 \pmod{n}$, then either $\gcd(x-1,n)$ or $\gcd(x+1,n)$ is a nontrivial factor of n
I'm reading elementary number theory and trying to understand the following problem: If $x^2\equiv 1 \pmod{n}$, $n=pq$, $p$ and $q$ are odd primes and $x \not\equiv \pm 1 \pmod{n}$, then either $\gcd(x-1,n)$ or $\gcd(x+1,n)$ is a nontrivial factor of $n$.

EDIT: Andreas Caranti wrote an updated, corrected version of my problem, so I wrote the definition again.

答:
If not, by Euclid's Lemma, $\ (n,x\pm1)=1\,\Rightarrow\, (n,(x\!-\!1)(x\!+\!1)) = 1,\,$ contra $\, n\mid x^2-1$.

Or, more directly,  $\ (x\!-\!1)(x\!+\!1) = pqk,\, $ so, by unique factorization, the prime $\,p\,$ is a factor of one of $\, x\!-\!1,\ x\!+\!1,\,$ so $\,q\,$ is a factor of the other one (else both primes divide the same factor, so $\, n = pq\,$ divides $\,x\!-\!1\ \ {\rm or}\ \ x\!+\!1,\,$ contra hypothesis).


答:
Let $L$ and $M$ be the midpoints of side $AB$ and $BC$ respectively. Draw medians $AM$ and $CL$.The point at which they intersect will be the centroid of $\triangle ABC$, point $G$. Draw $GH$, $LE$, $GI$ and $MD$. Let the points at which $GH$ and $LE$ intersect $MD$ be $P$ and $Q$ respectively.Let the point at which $GI$ intersect $LE$ be $R$. Draw $BQ$.

Observe that, $\triangle MDB\cong \triangle LBE$. $\Rightarrow LE=MD$

Also, observe that, since $\frac {CG}{GL}=\frac {CH}{HE}=\frac {2}{1}$, $GH\parallel LE$ and $GH=\frac 23LE$. Similarly, $GI\parallel MD$ and $GI=\frac 23MD$.

$LE=MD$         $\Rightarrow GH=GI$

Quadrilateral $GPQR$ is a parallelogram since its opposite sides are parallel.

$\Rightarrow \angle HGI=\angle PGR=\angle PQR=\angle EQD$

Notice that, in quadrilateral $QEDB$, $\angle QEB=\angle LEB=\angle MDB=\angle QDB$; Hence, $\angle EQD=\angle EBD=60^{\circ}$.

$\Rightarrow \angle HGI=\angle EQD=60^{\circ}$

$HG=GI$ and $\angle HGI=60^{\circ}$ ; Therefore, $\triangle HGI$ is equilateral.

Title: Three equilateral triangles


问:
How can a sentence (not a formula) describe a set?
What does this sentence mean?

Let $M$ be a model and  $S$ be a set described by a sentence in the language of $M$.
In
Paper

Page 128, part 2.4.

答:
As Eric Wofsey comments, "sentence" should be "formula." I wouldn't call this a typo however; I think it's a deliberate choice to increase readability for a non-logic audience. E.g. consider the line on page $125$:

The first of these principles expresses the fact that ${}^*M$ is a model of $M$.

To a logic audience this would be a terrible misuse of the term "model" (since $M$ is a structure, not a theory). However, to a non-logic audience the notion of one structure modelling another might feel pretty natural.

So really I think the situation is this: some liberties are being taken with respect to technical logic language here in the interest of readability.


问:
$A^2-B$ is positive-definite if and only if $A^{-1}BA^{-1}$ has all eigenvalues $<1$

Let $A,B$ be $n\times n$ real matrices such that $A^2$ is positive-definite and $B$ is symmetric. Show that $A^2-B$ is positive-definite if and only if $A^{-1}BA^{-1}$ has all eigenvalues $<1$.

If we choose invertible matrix $P$ such that $P'(A^2)P=I_n$, then $A^2-B$ is positive-definite implies $P'(A^2-B)P=I-P'BP$ is positive-definite, and eigenvalues of $P'BP$ is $<1$. But it is inconsistent with $|P' A(x I-A^{-1}BA^{-1})AP|=|x P'P-P'BP|$. If $P$ is orthogonal, OK then. However, ...

答:
If $P$ is invertible and $H$ is Hermitian, the matrix product $P^\ast HP$ is said to be congruent to $H$. Positive or negative (semi)definiteness are preserved under congruence. In fact, since $P$ is invertible, $v=Pu$ is a one-to-one correspondence between two vectors $u$ and $v$. Therefore $v^\ast Hv>0$ if and only if $u^\ast P^\ast HPu>0$, and likewise when $>$ is replaced by $\ge,<$ or $\le$.

In your case, let $P=(A^2)^{1/2}$. Then $A^2-B=P^2-B=P(I-P^{-1}BP^{-1})P$ is positive definite if and only if $I-P^{-1}BP^{-1}$ is positive definite, and the latter occurs if and only if all eigenvalues of $P^{-1}BP^{-1}$ are less than $1$.

Now, in general, if $X$ and $Y$ are two square matrices of the same sizes, then $XY$ and $YX$ have the same characteristic polynomials and hence also the same spectra. Therefore $P^{-1}BP^{-1},\,P^{-2}B=A^{-2}B$ and $A^{-1}BA^{-1}$ have the same eigenvalues. Hence the result follows.


问:
Dummit & Foote 1.7.12 - finding a nicer expression for $r^xs^y\cdot a$ for $a\in A$

Assume $n$ is an even positive integer and show that $D_{2n}$ acts on the set consisting of pairs of opposite vertices of a regular $n$-gon. Find the kernel of this action.

Here's my solution to the first part, showing that this is a group action:

Let $$A = \left\{\left\{i, i+\frac{n}{2}\right\}: 1\le i\le \frac{n}{2}\right\}$$
be the set of pairs of opposite vertices, labeled from $1$ to $n$. In words, we are required to first show that two opposite vertices remain opposite after rotations and reflections of the $n$-gon, as dictated by elements of $D_{2n}$. Given this, we can define a map $D_{2n}\times A \to A$, and check that it is a group action. First of all, to ensure that $g\cdot a\in A, \forall g\in D_{2n},\forall a\in A$, we need to check that $r\cdot a, s\cdot a \in A$ for all $a\in A$ (because any $g\in G$ is just $g = r^is^j$).

Consider arbitrary $a\in A$ given by $\{i,i+\frac{n}{2}\}$ for $1\le i\le \frac{n}{2}$. $r\cdot a = \{i+1 \bmod n, i+\frac{n}{2}+1\bmod n\} \in A$. $s\cdot a = \{n+2-i, \frac{n}{2}+2-i\} \in A$.

Also, $1_G\cdot a = a, \forall a\in A$.

Let $g_1 = r^cs^d, g_2 = r^es^f \in D_{2n}$. Consider arbitrary $a\in A$ given by $\{i,i+\frac{n}{2}\}$ for $1\le i\le \frac{n}{2}$. Note that $r^c\cdot a = \{i+c \bmod n, i+\frac{n}{2}+c\bmod n\}$. Also, $s^d\cdot a = a$ if $d$ is even, and $s^d\cdot a = s\cdot a$ if $d$ is odd. Verify $g_1\cdot (g_2\cdot a) = (g_1g_2)\cdot a$. Hence, this is a group action.

The last part in bold is where I need help. I know $r^c\cdot a$ and $s\cdot a$ but to verify $g_1\cdot (g_2\cdot a) = (g_1g_2)\cdot a$ I would need a nice expression for $r^cs^d\cdot a$. Otherwise, it's too much computation that's involved, i.e. assuming $d=0,1$ and going ahead making cases. Is there an easier way to do this?

Also, I can see intuitively see that the $r^{n/2}$ and the identity element of $D_{2n}$ are in the kernel of this action. I will need an expression for $r^xs^y\cdot a$ for $a\in A$ to find other elements in the kernel, or to show that there are none.

Thank you!

答:
The trick is to note that $r^{n/2}$ is in the center, which spares you all the symbolic work. For the Kernel, we use orbit stabilizer to find an upper bound of two possible the elements, which we already know.

I will be overly verbose here, but you can condense the argument a lot.

Recall that the dihedral group $D_{2n}$ has a natural action on the set $V := \{1,\cdots,n\}$ by defining it to be generated by the permutation action $r := (1,\cdots,n)$ and $s(i) := ((i+n/2 -1) \operatorname{mod}\: n) +1$.  Also, $r^n=s^2=(rs)^2=e$.  A simple inductive argument shows that $r^j s = sr^{-j}$ for all integers $j$, which in particular implies that $r^{n/2}$ commutes with both $r$ and $s$, and hence is in the center.

We will denote the aforementioned action with a dot, i.e. by $r.i := r(i)$.

Now, we will define $\overline V$ to be $V$ modulo the antipodal equivalence relation you established, and writing $[i]$ for the equivalence class of $i\in V$, we define a new map $$
  D_{2n}\times \overline V\to \overline V,\qquad g.[i]:= [g.i],
$$ And show that this is well-defined.
Indeed, let $i\neq i'\in [i]$. Since $i'$ is the antipode, we can represent it as $r^{n/2}.i$, and since $r^{n/2}$ is in the center, we have $$
g.[i'] = [g.i'] = [g.(r^{n/2}.i)] = [(gr^{n/2}).i] = [(r^{n/2}g).i] =[r^{n/2}.(g.i)] = [g.i] = g.[i],
$$ where the second to last equality follows from the fact that application of $r^{n/2}$ does not change the equivalence class.
The fact that this does constitute an action follows then directly from the fact that $-.-\colon D_{2n}\times V\to V$ is an action.

To find the Kernel of that action, recall that the orbit stabilizer theorem applied to $x := [1]\in \overline V$ gives us $$
|D_{2n}| = |D_{2n}.x| \cdot |(D_{2n})_x|,
$$ and since the action is transitive (powers of $r$ can map $x=[1]$ to an arbitrary element), the cardinality $|D_{2n}.x|$ of the stabilizer is $|\overline V|=n$, implying the stabilizer has exactly two elements.

Furthermore, we know that the kernel $K$ of the action contains group elements that stabilize all of $\overline V$, implying that those in particular stabilize $x$.  So we have $K\subseteq (D_{2n})_x$.
On the other hand, at least the identity and $r^{n/2}$ do stabilize everything in $\overline V$, which means we have $$
  \{1, r^{n/2}\} \subseteq K = (D_{2n})_x,
$$ and since the right hand side has cardinality two, we have equality.


问:
The restricted product topology of locally compact spaces is locally compact
I want to prove that the restricted product topology of lcoally compact spaces is locally compact. Let $X_i$ locally compact topological spaces, and $K_i$ corresponding compact open subsets. I consider the restricted product $X$ of the $X_i$ with respect to the $K_i$.

The basis of neighborhood for the restricted product is given by open sets of the form
$$\prod_{i \in S} U_i \times \prod_{i \notin S} K_i$$
where $S$ is a finite set, and $U_i$ is open in $X_i$. We can as usual consider the $S$-restricted product
$$X_S = \prod_{i\in S} X_i \prod_{i\notin S} K_i.$$
It is clear that the $X_S$ cover $X$, and I would like to this to prove the local compactness of $X$. Let $x \in X$, this $x \in X_S$ for a certain $S$. Since $X_S$ is locally compact (essentially a finite product), $x$ has a compact open neighborhood $U$ in $X_S$.

I would like to prove (or disprove?) that $U$ is a compact open in $X$. I can see that an open cover of $U$ in $X$ restricts to an open cover of $U$ in $X_S$, thus there is a finite subcover. But what ensures that this subcover also works in $X$, i.e. that $U$ is covered in $X$ by it?

答:
I'm not sure what you're trying to do here, but the result to go for is this:

If $X_i$, $i \in S$ is a collection of non-empty spaces then $X=\prod_{i \in S} X_i$ is locally compact iff all $X_i, i \in S$ are locally compact and there is a finite $F \subseteq S$ such that $X_i$ is compact for all $i \in S\setminus F$.

The left to right implication seems your concern here. If for some $p=(x_i)_{i \in S}$ has a compact neighbourhood $K$ in $X$ it contains a basic open neighbourhood that only depends on finite many coordinates $F$ and then it follows that, as $X_i = \pi_i[K]$ for $i \notin F$, that all those $X_i$ are compact. The rest is routine.


问:
Under what assumption is the following statement true: $f(x) < 0$, then there exists a constant $k > 0$ such that $f(x) \leq -k$
I wish to justify the following statement using a set of weak assumptions, e.g. continuity, closedness. We assume a standard metric space $\mathbb{R}^n$,

Given $f: \mathbb{R}^n \to \mathbb{R}$, suppose  $f(x) < 0, \forall x \in D \subset \mathbb{R}^n$ then there exists some $k > 0$ such that $f(x) \leq -k, \forall x \in \mathbb{R}^n \backslash D$. What property makes this hold true?

I suspect this is either an assumption on $D$, e.g., $D$ is closed and/or bounded, or an assumption on $f$, e.g., $f$ is continuous, or both (continuous over closed).

答:
Even taking $D$ to be closed and $f$ to be continuous is insufficient. If $\Bbb R^n\setminus D$ has non-empty interior, any continuous real-valued function on $\operatorname{cl}D$ can be extended to one that takes a positive value somewhere in $\Bbb R^n\setminus D$. If, on the other hand, $D$ is dense in $\Bbb R^n$, and $\sup_{x\in D}f(x)=0$, there is no $k>0$ such that $f(x)<-k$ for all $x\in\Bbb R^n\setminus D$.

I doubt that any reasonable conditions will give you what you want.


Title: A problem on square matrices

Body:
If $B,C$ are $n$ rowed square matrices and if $A=B+C, BC=CB, C^2=O$, then show that for every $n \in \mathbb N$,
 $$A^{n+1}=B^n(B+(n+1)C)$$

I tried to prove it using mathematical induction. But I could not get $P(1)$ to be true.
$$P(1): A^2 = B(B+2C)$$
I couldn't equate them.

Please offer some assistance.
Thank you. :)

Body:
$$A^2=(B+C)^2=B^2+BC+CB+C^2$$

Now use $CB=BC$ and $C^2=0$.