问:
How can i predict how many participants i will need in hypothesis test trial?
I try to understand if having more serious side effects after vaccine mean higher level of antibodies.
There are 2 groups, divided by the severity of there side effects after the vaccine and there antibodies levels are tested.

Now the $H_0$ is that there is no difference in the amount of antibodies and $H_1$ says there is.

I don't know ahead what the average or variance will be.

How can i know how many participants i will probably need to get significant results with a p-value 0.05?

NOTE: I am not a statistician, i just try to understand how to do a good trial.

Thanks.

答:
Power and sample size computations require some educated guesses in advance. You expect more antibodies among
subjects who have more side effects. That means the alternative hypothesis
is one-sided. So $H_0$ is that there is no difference
and $H_a$ is more antibodies in the high side-effects group.

You say you want to find significance at the 5% level. Possibly you would use a two-sample t test (which requires antibody counts to be approximately normal).
Let's say you want probability 80% or 90% of finding
a meaningful difference.

Maybe the standard deviation $\sigma$ of antibodies will be
about 30 units in each group and you would consider
$\Delta = 10$ units difference to be clinically meaningful. (It is the ratio $\Delta/\sigma$ that matters. So you're
saying it would be important to detect a difference that's $1/3$ the standard deviation.)

Someone connected with
the study must have some information or intuition about
the variability of antibody counts and how big a difference
is of important to detect, or you shouldn't be trying
to plan such a study.

Testing is most efficient when the number in each group is
roughly the same. For planning purposes we assume it will
be reasonable to define 'serious side effects' so that
about half of the subjects have serious side effects. Maybe not true, but it gives a way to start.

With this information one can find the sample size in each group. Relevant output from Minitab software is shown
below. This seems like it might be a phase one clinical
trial where you're trying to get an idea whether the
vaccine is safe and may be effective at a particular dosage. Typically, such
trials use a few hundred volunteer subjects. Altogether,
you would need something like 225 or 300 subjects,
depending on the required power of the test (probability
of detecting a useful difference).

```<code>Power and Sample Size

2-Sample t Test

Testing mean 1 = mean 2 (versus >)
Calculating power for mean 1 = mean 2 + difference
α = 0.05  Assumed standard deviation = 30

            Sample  Target
Difference    Size   Power  Actual Power
        10     112     0.8      0.800098
        10     155     0.9      0.900282

The sample size is for each group.
</code>```

Notes: (1) A clinical
trial for final governmental approval of a vaccine typically
requires several thousand subjects half getting the
vaccine and half getting a placebo. Government requirements
for such trials, including sample sizes, are based on a many considerations,
only some of which are statistical.]

(2) To determine power for a given sample size in a
2-sample pooled t test,
there is a formula involving the non-central t distribution (used in the Minitab procedure above).
It can be used in case you must have different sample sizes in the two groups.

Various statistical programs have similar 'power and sample size' procedures, and there are some
on-line calculators (of varying clarity and accuracy).

If you are
using some other kind of test, you may need to use
simulation to get the power for the assumed specifications.

Here is a simulation in R for sample sizes 80 and 200 in a Welch 2-sample t test (equal variances not required), showing
power about 80%.

```<code>set.seed(2021)
n1 = 80;  n2 = 200;  mu1 = 110;  mu2 =100; sg = 30
pv = replicate(10^5, t.test(rnorm(n1,mu1,sg),
      rnorm(n2,mu2,sg), alt="gr", var.eq=T)$p.val)
mean(pv <= .05)
[1] 0.8058
</code>```

Body:
Power and sample size computations require some educated guesses in advance. You expect more antibodies among
subjects who have more side effects. That means the alternative hypothesis
is one-sided. So $H_0$ is that there is no difference
and $H_a$ is more antibodies in the high side-effects group.

You say you want to find significance at the 5% level. Possibly you would use a two-sample t test (which requires antibody counts to be approximately normal).
Let's say you want probability 80% or 90% of finding
a meaningful difference.

Maybe the standard deviation $\sigma$ of antibodies will be
about 30 units in each group and you would consider
$\Delta = 10$ units difference to be clinically meaningful. (It is the ratio $\Delta/\sigma$ that matters. So you're
saying it would be important to detect a difference that's $1/3$ the standard deviation.)

Someone connected with
the study must have some information or intuition about
the variability of antibody counts and how big a difference
is of important to detect, or you shouldn't be trying
to plan such a study.

Testing is most efficient when the number in each group is
roughly the same. For planning purposes we assume it will
be reasonable to define 'serious side effects' so that
about half of the subjects have serious side effects. Maybe not true, but it gives a way to start.

With this information one can find the sample size in each group. Relevant output from Minitab software is shown
below. This seems like it might be a phase one clinical
trial where you're trying to get an idea whether the
vaccine is safe and may be effective at a particular dosage. Typically, such
trials use a few hundred volunteer subjects. Altogether,
you would need something like 225 or 300 subjects,
depending on the required power of the test (probability
of detecting a useful difference).

Power and Sample Size

2-Sample t Test

Testing mean 1 = mean 2 (versus >)
Calculating power for mean 1 = mean 2 + difference
α = 0.05  Assumed standard deviation = 30

            Sample  Target
Difference    Size   Power  Actual Power
        10     112     0.8      0.800098
        10     155     0.9      0.900282

The sample size is for each group.

Notes: (1) A clinical
trial for final governmental approval of a vaccine typically
requires several thousand subjects half getting the
vaccine and half getting a placebo. Government requirements
for such trials, including sample sizes, are based on a many considerations,
only some of which are statistical.]

(2) To determine power for a given sample size in a
2-sample pooled t test,
there is a formula involving the non-central t distribution (used in the Minitab procedure above).
It can be used in case you must have different sample sizes in the two groups.

Various statistical programs have similar 'power and sample size' procedures, and there are some
on-line calculators (of varying clarity and accuracy).

If you are
using some other kind of test, you may need to use
simulation to get the power for the assumed specifications.

Here is a simulation in R for sample sizes 80 and 200 in a Welch 2-sample t test (equal variances not required), showing
power about 80%.

set.seed(2021)
n1 = 80;  n2 = 200;  mu1 = 110;  mu2 =100; sg = 30
pv = replicate(10^5, t.test(rnorm(n1,mu1,sg),
      rnorm(n2,mu2,sg), alt="gr", var.eq=T)$p.val)
mean(pv <= .05)
[1] 0.8058


问:
Step by step method for $\int\frac{x^2+1}{x^2-1}\,dx$
$$\int\frac{x^2+1}{x^2-1}\,dx$$

What is the step-by-step way of solving this integral problem?
I tried using substitution which was $x^2-1=t^2$, but end up with an even more complicated equation.
Substituting trigonometric functions for example :$x^2=\sec^2(a)$ so that $x^2-1$ part becomes $\tan^2(a)$ did not help either.

答:
This can be evaluated using partial fraction decomposition. Observe how

$$\begin{align*}\frac {x^2+1}{x^2-1} & =\frac {x^2-1+2}{x^2-1}\\ & =1+\frac 2{x^2-1}\end{align*}$$

The final fraction can be broken down by assuming the partial fractions follow the form

$$\frac 2{(x+1)(x-1)}=\frac A{x+1}+\frac B{x-1}$$

Multiplying both sides by $(x+1)(x-1)$ and setting $x=\pm1$ gives $A=-B=-1$. Or, in other words,

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

Integrate piece by piece to get

$$\int\mathrm dx\,\frac {x^2+1}{x^2-1}\color{blue}{=x+\log(x-1)-\log(x+1)+C}$$

ContentLicense: CC BY-SA 4.0


问:
given $a+b+c=3$ prove that $abc(a^2+b^2+c^2) \leq 3$
I am preparing for inmo and I came accross this problem while solving a worksheet, but couldn't solve it, pl help me...

Problem-

Prove that if a,b,c are non negative real numbers such that a+b+c=3, then $abc(a^2+b^2+c^2)\leq3$ ... (0)

Developments-

Firstly applying am-gm on $a+b+c$ to get $abc\leq1$

Then applying quadratic mean arthematic mean to get

$a^2+b^2+c^2 \ge 3$

Then writing the expansion of $(a+b+c)^2$ we get

$a^2+b^2+c^2 = 9-2(ab+bc+ca)$ ...(I)

Thn applying am-hm on $ab+bc+ca$ to get

$ab+bc+ca \ge 3abc$

Then substituting in (I) we get

$a^2+b^2+c^2 \leq 9-6(abc)$

Then we substituting in (0)

$LHS \leq abc(9-6abc)$

Whose max value is $27/8$ but is not less than three, so I am stuck for this point onwards

Body:
WLOG $a\le b\le c$ so $\frac{b+c}{2}\ge 1$

let $$f(a,b,c)= abc(a^2+b^2+c^2)$$ now $$f(a,b,c)-f(a,\frac{b+c}{2},\frac{b+c}{2})=-\frac{1}{8} a (b - c)^2 (2 a^2 + b^2 - 2 b c + c^2)\le 0$$  let $t=\frac{b+c}{2}\ge 1$ So it suffices to show  $$f(a,t,t)\le 3 \iff f(3-2t,t,t)\le 3$$ but $$f(3-2t,t,t)-3=-3 (t - 1)^2 (4 t^3 - 6 t^2 + 2 t + 1)\le 0$$ which is true  because $4t^3-6t^2+2t=2t(t-1)(2t-1)\ge 0$

Done!


问:
Finding $\mathbb{E}(XY|XY>\mathbb{E}(X)\mathbb{E}(Y))$

Assuming case where both $X$ and $Y$ are independent and normally distributed with different mean and variance how do I calculate $\mathbb{E}(XY|XY>\mathbb{E}(X)\mathbb{E}(Y))$.

Assuming $f(xy) = f(x)f(y)$,

I can't seem to integrate xy on the new $f(xy)$ as the $x$ and $y$ are separated most of the time.

And I am uncertain of the lower limit of integrating $f(x)f(y)xy$ as I am unsure of what lower limits for both $x$ and $y$ would guarantee $XY>\mathbb{E}(X)\mathbb{E}(Y)$.

答:
As you say the x and y are separated most of the time, but you can get a very good approximation. We want to find $E\left(XY|XY>E(X)E(Y)\right)$. Because E(X)E(Y) is just a constant, let's call it C. Let's assume C>0, then the bounds are $x>0,y>\frac 3 x$ and $x<0, y<\frac 3 x$. So there are two regions to integrate and they are indeed separate, making things more difficult.

We first find $E(XY|XY>C,X<0)$. This is $$H_1=\int_{-\infty}^0\int_{-\infty}^{\frac 3 x} xy\frac{f(x,y)}{A}dydx$$ with $A=\int_{-\infty}^0\int_{-\infty}^{\frac 3 w}  f(z,w)dzdw$. That is, we reweighted the density - the denominator is just a constant and is the probability that x<0 and y>3/x.

The $E(XY|XY>C,X>0)$ can be calculated the same way as follows

$$H_2=\int_{0}^\infty\int_{\frac 3 x}^{\infty} xy\frac{f(x,y)}{B}dydx$$

with $B=\int_{0}^\infty\int_{\frac 3 x}^{\infty}  f(z,w)dzdw$ being another reweighting of the density.

Once you get these two expected values, the expected value of the overall $E(XY|XY>E(X)E(Y))$ must be between these two. But we can do better than just find bounds. Take the weighted average:

$$E(XY|XY>E(X)E(Y))\approx \frac{H_1A+H_2B}{A+B}$$

I am actually not sure if it is approximate or in fact equal.


问:
Parametrization of curve in $\Bbb R^3$
I’m trying to find the parametrization of a curve in $\Bbb R^3$ satisfying $p_1=(0,1,1)$ and $p_2=(1,0,0)$ that parallel projects to the x-y plane onto $\log(x)\log(y)=1.$ I think that maybe I can do a rotation on the x-y plane curve and then scale its length to satisfy the endpoint conditions? Also, I must note that $p_1,p_2$ are undefined for the function $\log(x)\log(y)=1.$ So I don’t require that the function has to be defined exactly at the two points $p_1,p_2.$ Any help is appreciated! Thanks.

答:
HINT:

Let

$$ \log x\log y=\text{ a constant =} \log p\log q \to \log_p x \log_ q y = 1$$

from which again let

$$ \log_p x =u,\; \log_ q y = 1/u $$

$z$ can be arbitrarily chosen. So a parameterization can be

$$(x,y,z)= (p^u,q^{1/u}, v \frac13 \cos u ) $$

```<code>> p = 0.8; q = 1.25; ParametricPlot3D[{p^u,    q ^(1/u), (v Cos[u]/3)},
> {u, -4, 2}, {v, -2, 4},   PlotPoints -> {40, 60}, PlotStyle -> Yellow]
</code>```

We have yet to choose constants to make the surface  contain the given points.

Title: Parametrization of curve in $\Bbb R^3$

Body:
HINT:

Let

$$ \log x\log y=\text{ a constant =} \log p\log q \to \log_p x \log_ q y = 1$$

from which again let

$$ \log_p x =u,\; \log_ q y = 1/u $$

$z$ can be arbitrarily chosen. So a parameterization can be

$$(x,y,z)= (p^u,q^{1/u}, v \frac13 \cos u ) $$

> p = 0.8; q = 1.25; ParametricPlot3D[{p^u,    q ^(1/u), (v Cos[u]/3)},
> {u, -4, 2}, {v, -2, 4},   PlotPoints -> {40, 60}, PlotStyle -> Yellow]

We have yet to choose constants to make the surface  contain the given points.


问:
rank function on Spec (help with definition)
one definition of the line bundle over a ring is: a finitely generated projective A-module such that the rank function Spec A → N (positive integers) is constant with value 1. We call A itself the trivial line bundle.

so here i think that spec is equipped with zariski topology and N with the discrete one. Does this mean that in general the rank function is not continuous? Does one know about a basic example of non constant rank function to illustrate the peculiarities implied by the definition above?

Many thanks

答:
I'm not quite sure I understand the question:

Does this mean that in general the rank function is not continuous?

How general is "in general"?  Anyway, here are some facts which taken together may answer your question.

A function $f$ from a topological space $X$ to a discrete space is continuous iff it is locally constant, i.e., for all $x \in X$ there exists a neighborhood $U$ of $x$ such that $f|_U$ is constant.  For whatever reason, it is more common to speak of "locally constant" rank functions than continuous ones.

In order to have a rank function, an $R$-module $M$ must have the property that for all prime ideals $\mathfrak{p}$, the localization $M_{\mathfrak{p}}$ is finitely generated and free.

Certainly all the localizations will be finitely generated if $M$ itself is finitely generated (I'm blanking a little on conditions for the converse at the moment), so let's restrict to the case of finitely generated modules $M$.

For a finitely generated module $M$ with a well-defined rank function -- i.e., such that $M_{\mathfrak{p}}$ is free for all prime ideals $\mathfrak{p}$, the following are equivalent:

(i) The rank function is locally constant (equivalently, continuous).

(ii) $M$ is projective.

This takes some work: see e.g. $\S 13.9$ of my commutative algebra notes.

There are conditions on $R$ which force the rank function of any finitely generated locally free module to be constant: e.g. this holds if $R$ has a unique minimal prime ideal ($\S 13.9$), so in particular if $R$ is a domain.  It also holds if $R$ is Noetherian ($\S 7.8$).  However there are examples of finitely generated locally free modules with nonconstant rank function: see this MO question and its answer.

ContentLicense: CC BY-SA 3.0


问:
when you see the math expression 4^(1/2) , how do you know if it means to give just the positive answer, or both positive and negative answers
So, is $4^{1/2} =$ just $+2$ , or is it $+2$ and $-2$ ?
In high school I learnt that it was just $+2$.

i see already in the 1st answers to my questions that this is a startlingly difficult question to answer!  normally in maths, there is no such ambiguity, and i would like to understand what sort of symbology should be used to remove the ambiguity of this.

Note that i came to this question originally from having found issues in this matter in complex numbers.

there i came across the expression "Principle Root", perhaps for strictly clear math symbology, a little plus suscript for that might be

what is needed to indicate that root?

And if you expect 2 answers (is ALL the roots, rather than the one with the largest real component), what symbology would you use then to make that clear?

i am looking for a general rule here, the power of a half is a simple case.  what if it was power of some other fraction?  how should the question be expressed so as to indicate if the principle root is requested, or all roots?

答: By definition (when $a> 0$), $a^{1/2}$ and $\sqrt a$ both mean the positive square root of $x$. If you wanted to include both numbers which square to give $a$, you would need to write $\pm a^{1/2}$ or $\pm\sqrt{a}$.


答:
$4^{1/2}$ is evaluated to be $2$.

Fundamentally, the function $f(x)=x^{1/2}$ has a range of $[0,\infty)$.

In comparison, if we were looking for the solutions to the equation $$x^2=4$$ instead, then in solving the quadratic equation we would get $x=\pm2$.

Title: when you see the math expression 4^(1/2) , how do you know if it means to give just the positive answer, or both positive and negative answers

Body:
So, is $4^{1/2} =$ just $+2$ , or is it $+2$ and $-2$ ?
In high school I learnt that it was just $+2$.

i see already in the 1st answers to my questions that this is a startlingly difficult question to answer!  normally in maths, there is no such ambiguity, and i would like to understand what sort of symbology should be used to remove the ambiguity of this.

Note that i came to this question originally from having found issues in this matter in complex numbers.

there i came across the expression "Principle Root", perhaps for strictly clear math symbology, a little plus suscript for that might be

what is needed to indicate that root?

And if you expect 2 answers (is ALL the roots, rather than the one with the largest real component), what symbology would you use then to make that clear?

i am looking for a general rule here, the power of a half is a simple case.  what if it was power of some other fraction?  how should the question be expressed so as to indicate if the principle root is requested, or all roots?


问:
Total variation distance of two random vectors whose components are independent
Let $X^n=\left(X_1,\ldots,X_n \right)$ and $Y^n=\left(Y_1,\ldots,Y_n \right)$ be such that all $X_i$'s are independent and all $Y_i$'s are independent. I am trying to prove the following:
$$
d_{TV} \left(\mathbb{P}_{X^n},\mathbb{P}_{Y^n} \right) \leq \sum_{i=1}^n d_{TV} \left( \mathbb{P}_{X_i},\mathbb{P}_{Y_i} \right).
$$

Can anyone help me how to start ? Or provide a useful hint to proceed in the right direction ?

答:
This is a variation on @zhanxiong's answer. Let $Z^i= (Y_1,\dots,Y_i,X_{i+1},\dots,X_n)$. In this way $Z^0=X$ and $Z^n=Y$;
the neighboring random vectors $Z^{i-1}$ and $Z^{i}$ differ only on $i$-th coordinate.

By the triangle inequality
$$ d_{TV}(X,Y) \leq \sum_{1\leq i \leq n} d_{TV}(Z^{i-1},Z^{i}). $$
In the following we will show that the total variation distance between the neighboring random vectors
$$ d_{TV}(Z^{i-1},Z^{i}) = d_{TV}(X_{i},Y_{i}) $$
is equal to the total variation distance of the coordinate on which they differ.

By the symmetry of the problem we may assume that $i=1$.
By merging the coordinates $2,\dots,n$ into a single vector
$\mathbf{X_2}=(X_2,\dots,X_n)$ we see that it is enough to consider the case $n=2$. In other words, we need to show that
$$ d_{TV}\big( (X_1,X_2), (Y_1,X_2) \big) = d_{TV} (X_1,Y_1). $$

We proceed as in @zhanxiong's answer:
\begin{align}
& |(\mu_1 \times \mu_2)(A) - (\nu_1 \times \mu_2)(A) | \\
\leq &  \left|\int_\mathbb{R} \mu_1(A_y) \mu_2(dy) -
\int_\mathbb{R} \nu_1(A_y) \mu_2(dy)\right| \\
\leq &  d_{TV}(\mu_1, \nu_1).
\end{align}
Above  $A_y = \{x: (x, y) \in A\}$ denotes the section set of $A$ for $y$.
This shows that
$$ d_{TV}\big( (X_1,X_2), (Y_1,X_2) \big) \leq d_{TV} (X_1,Y_1). $$

On the other hand, for the product set $A= A' \times \mathbb{R}$ we have
$$ \mathbb{P}\big\{ (X_1,X_2) \in A' \big\} - \mathbb{P}\big\{ (Y_1,X_2) \in A' \big\}= \mathbb{P}\big\{ X_1 \in A \big\} - \mathbb{P}\big\{ Y_1 \in A \big\};  $$
it follows therefore that
$$ d_{TV}\big( (X_1,X_2), (Y_1,X_2) \big) \geq d_{TV} (X_1,Y_1). $$


问:
Absolute maximum and minimum values of a function over a region
The prompt is to find the absolute maximum and minimum values of the function $f(x, y) = x^2 + 2x - y^2$ on the upper half of a unit disc, $D =\{(x, y): x^2 + y^2 \leq 1, y \geq 0 \}$

I began by finding the critical points of the region
$$f_x = 2x + 2$$
$$f_y = -2y$$
$$f_{xx} = 2$$
$$f_{yy} = -2$$
$$f_{xy} = 0$$

Finding critical points from first and second equations,
$$2x + 2 = 0$$
$$-2y = 0$$
$$x = -1, y = 0$$

I'm not sure how to proceed further, any help would be appreciated.

答:
From your calculation, you have shown that there is no critical point in the interior of the half-disk, but the left endpoint of the semicircular boundary is one.  The index of that point is

$$ D \ \ = \ \ f_{xx}·f_{yy} \ - \ (f_{xy})^2 \ \ = \ \ 2·(-2) \ - \ 0^2 \  =  \ -4 \ \ < \ \ 0 \ \ , $$

which identifies $ \ (-1 \ , \ 0 ) \ $ as a saddle-point.  This makes sense since the function is "concave upward" in the $ \ x-$ direction about $ \ x \ = \ -1 \ $ , due to the $ \ x^2 \ + \ 2x \ = \ (x + 1)^2 \ - \ 1 \ $ terms, but is "concave downward" in the $ \ y-$ direction because of the $ \ -y^2 \ $ term.  As we will need it for comparisons later, we compute that   $ f(-1,0) \ = \ (-1)^2 \ + \ 2(-1) \ - \ 0^2 \ = \ -1 \ \ . $

The lower boundary of the region is the $ \ x-$ axis over the interval $ \ [ \ -1 \  , \ 1 \ ] \ $ .  Inserting $ \ y \ = \ 0 \ $ into the function expression gives   $ \ f(x,0) \ = \ (x + 1)^2 \ - \ 1 \ \  . $  We have already found $ \ f(-1,0) \ $  to be the minimum on this line segment; the right endpoint has the value   $ f(1,0) \ = \ (1+1)^2 \  - \ 1 \ = \ 3 \ \ . $

It remains to examine the upper semicircular boundary.  There are a number of ways we might set this up, but to avoid having to work with a square-root or trigonometric identities, we will use the circle equation to write $ \ y^2 \ = \ 1 \ - \ x^2 \ $ and express our function on this section of the boundary as

$$ \ \phi(x) \ =  \ (x + 1)^2 \ - \ 1 \ - \ (1 - x^2) \ \ =  \ \ (x + 1)^2 \ + \ x^2 \ - \ 2 \ \ .  $$

[This gives us the correct values for the function at the endpoints of the region:

$$  \phi(-1) \ \ =   \ \ ([-1] + 1)^2 \ + \ (-1)^2 \ - \ 2 \ \ = \  -1 \ \ \ , \ \ \ \phi(1) \ \ =   \ \ (1 + 1)^2 \ + \ 1^2 \ - \ 2 \ \ = \ \ 3 \ \ . \ ] $$

We find $ \ \phi'(x) \ =  \ 2·(x + 1)   \ + \ 2x  \ \ = \ \ 0 \ \ \ \Rightarrow \ \ \ 4x \ \ = \  -2 \ \ \ \Rightarrow \ \ \ x \  = \  -\frac{1}{2} \ \ , $ for which $  \phi \left( -\frac{1}{2} \right) \ \ =   \ \ \left( \ \left[-\frac{1}{2} \right] + 1 \ \right)^2 \ + \ \left(-\frac{1}{2}  \right)^2 \ - \ 2 \ \ = \ \ \frac{1}{4} \ + \ \frac{1}{4} \ - \ 2 \ \ = \  -\frac{3}{2} \ \ . $

Hence, we observe that the absolute maximum of the function on this closed half-disc is  $ f(1,0) \  = \ 3 \ $ and the absolute minimum is $  f \left( -\frac{1}{2} \ , \ \frac{\sqrt{3}}{2} \right) \  =   \   -\frac{3}{2} \ \ . $  The saddle point at $ \ (-1 \ , \ 0 ) \ $ has no special role in this region.


问:
Equivalences of bounded sets
$\newcommand{\diam}{\operatorname{Diam}}$
$\newcommand{\R}{\mathbb{R}}$

Definition: Let $X$ be a metric space, $M \subseteq X$ is bounded if $\diam(M) < \infty$

Let $(X,d)$, $X \not=  \emptyset$, be a metric space and $M \subseteq X$. Show that the following are equivalent:

(i) $M$ is bounded.

(ii) $\exists_{x \in X}\left\{\exists_{r > 0}\left[ M \subseteq B_{r}(x)\right]
\right\}$

(iii) $\forall_{x \in X}\left\{\exists_{r > 0}\left[ M \subseteq B_{r}(x)\right]\right\}$

Attempt (following suggestions from you):

Suppose $M$ is bounded. We want to see that $\exists_{x \in X}\left\{\exists_{r > 0}\left[ M \subseteq B_{r}(x)\right]\right\}$.

(iii) $\rightarrow$ (ii)

Suppose that $\forall_{x \in X}\left\{\exists_{r > 0}\left[ M \subseteq B_{r}(x)\right]\right\}$. We want to see that $\exists_{x \in X}\left\{\exists_{r > 0}\left[ M \subseteq B_{r}(x)\right]\right\}$.

By hypothesis it is true for any $x \in X$. Therefore, in particular, it is true for some $x \in X$. Hence, $\exists_{x \in X}\left\{\exists_{r > 0}\left[ M \subseteq B_{r}(x)\right]\right\}$.

(ii) $\rightarrow$ (i)

Suppose that $\exists_{x \in X}\left\{\exists_{r > 0}\left[ M \subseteq B_{r}(x)\right]\right\}$. We want to see that $M$ is bounded. Suppose by contradiction that $M$ is not bounded.

By hypothesis above, we have that $\diam(M) = \infty$. So there exists $m_{1},m_{2} \in M$ such that they are arbitrary far. It is a contradiction! Because the ball $B_{r}(x)$, which has a finite radius $r$, can not contain its points $m_{1}$ and $m_{2}$ but they are inside $M$.

How can I deduce (i) $\rightarrow$ (iii)?

答:
The case $M=\emptyset$ is boring and irrelevant. Leave it out.

BTW, $\text{diam}(\emptyset) = -\infty$ or is undefined so your original definition of "bounded = finite diameter" also has an exception clause for the empty set.
So bounded= "empty" or finite diameter", presumably. If $M=\emptyset$ then (ii) and (iii) are both automatically true, for any $x \in X$, $r>0$, because $\emptyset \subseteq A$ is true for any $A$. So the equivalence holds and we move on to assume that $M \neq \emptyset$ throughout.

The circular steps then can become:

(i) $\implies $ (ii):
Let $R=\text{diam}(M)$, a finite real in $[0,+\infty)$. Pick any $x \in M$. Then (ii) holds for this $x$ and $r=R+1>0$: if $y \in M$ then $d(x,y) \le \text{diam}(M) = R < R+1$, so $M \subseteq B(x, r)$.

(ii) $ \implies $ (iii). Fix $x_0, r_0>0$ as in (ii). Let $x \in X$ be arbitrary. Define $r=d(x,x_0) + r_0 >0$. Then (iii) holds with this choice of $r$ for our $x$: let $m \in M$ then by (ii) we have $d(m,x_0) < r_0$ so  $$d(m, x) \le d(m, x_0) + d(x_0,x) < r_0 + d(x_0,x)=r$$ and so $M \subseteq B(x,r)$.

(iii) $\implies $ (i): let $m \in M$ and apply (iii) to get $r>0$ with $ M \subseteq B(m,r)$. It follows that for all $x,y \in M$ we have $d(x,y) \le d(x,m) + d(m,y) < 2r$ and so $\text{diam}(M) \le 2r < +\infty$ so $M$ is bounded.

Title: Equivalences of bounded sets


问:
Finding all Laurent expansions of $f=\frac{1}{z}$
I have the following homework problem I need help with:

Let $G=\{z\in\mathbb{C}:\, z\neq0\}$ and define $f:\, G\to G$ by
  $f(z)=\frac{1}{z}$.

Find all possible Laurent expansions of $f$ which are not Taylor
  expansions and for each such expansion specify in what set the Laurent
  series converges.

What I tried:

For the point $z=z_{0}$ there are Laurent expansions in $$E_{1}:=0<|z|<R$$
and in $$E_{2}:=|z|>R$$

Where $R>0$ is real.

I am unsure about in which set the Laurent series converges to $f$
, but it seems that in both cases the Laurent series of $f$ is simply
$\frac{1}{z}$ and it converges in all points of $E_{1}$or $E_{2}$
(according to the case).

Am I correct ?

For $z_{0}\ne0$ the Laurent expansions which is not Taylor is in
$$E_{1}:=0\leq|z|<R$$ where $R>|z_{0}|$ and $$E_{2}:=|z|>R$$ where
$R<|z_{0}|$.

I am not sure about where I should of used $<$ or $\leq$ in the
description of $E_{1},E_{2}$.

I am also having difficulty finding the Laurent series at $E_{i}$
and telling where it is convergent.

Can someone please help me out with the case $z\neq z_0$ and tell me if I did correctly the first case ($z=z_0$) ?

答:
What you have is a bit confused. When $z_0=0$, then there is only the one expansion with $0<|z_0|$, and as you said, the Laurent expansion is just $f(z) = \dfrac 1z$.

When $z_0\ne 0$, you have two possible regions: (1) $0<|z-z_0|<|z_0|$ and (2) $|z_0|<|z-z_0|<\infty$. You need an expansion in powers of $z-z_0$. So you start like this:
$$\frac1z = \frac1{z_0+(z-z_0)}=\frac1{z_0}\cdot\frac1{1+\frac{z-z_0}{z_0}}\,.$$
When $|z-z_0|<|z_0|$, substitute $u=\frac{z-z_0}{z_0}$ and note that $|u|<1$. Now use geometric series. When $|z-z_0|>|z_0|$, this doesn't work, so you factor out the larger term:
$$\frac1z = \frac1{z_0+(z-z_0)}=\frac1{z-z_0}\cdot\frac1{1+\frac{z_0}{z-z_0}}\,.$$
As before, substitute $u=\frac{z_0}{z-z_0}$, note that $|u|<1$, and proceed.


问:
Show $x$-coordinate of stationary point satisfy the equation $x=\frac32+ \frac{x}{e^{2x}}$
I solved the equation but when it satisfies it spouse to have opposite signs i think  this is the question

Question 1 The curve with equation $y=\dfrac{e^{2x}+x}{x^3}$ has a stationary point with $x$-coordinate lying between $1$ and $2$.

a. Show that the $x$-coordinate of this stationary point satisfies the equation $x=\dfrac32+\dfrac{x}{e^{2x}}$.

b. Using a suitable iterative process based on the qeuation given in part a, find the value of the $x$-coordinate of this stationary point correct to 4 decimal places.  Give each iteration to 6 decimal places.

the question is 1a this is my working out. I have no teacher cso any help would be much appreciated. Thank you.

答:
$y = \frac{e^{2x} + x}{x^3} = \frac{e^{2x}}{x^3} + \frac{x}{x^3} = e^{2x} x^{-3} + x^{-2}$. Thus you can just use the product rule:

$$\frac{dy}{dx} = 0 \Rightarrow e^{2x} \cdot -3x^{-4} + 2e^{2x} \cdot x^{-3} - 2x^{-3} = 0$$

$$\Rightarrow e^{2x} \cdot -3 + 2e^{2x} \cdot x - 2x = 0$$

then divide both sides by $e^{2x}$.


答:
By definition, taking square root (aka arithmetic root) results in only non-negitive answer. You simply ignore the fact that $(-2)^2 = 4$.

But if you want to get all possible "answers", you should solve an equation like this one:

$$ x^2 = 4 $$

Both left and right parts of this equation are positive so you can take square root on both of them. Since $\sqrt{x^2} = |x|$ we get the following equation:

$$ |x| = 2 $$

The absolute value $|x|$ tells us that both $-2$ and $2$ are roots, not only $2$.


问:
discrete random variable with cumulative distribution function
Consider a discrete random variable X with CDF (Cumulative
Distribution Function) () specified below:

I am just new to the course of statistics and wonder if only Pr(X=1), Pr(X=2), Pr(X=3) and Pr(X=5)are all the probabilities that can be drawn out from this cdf and other remaining probabilities like Pr(X=4)etc are equal to 0. Am I correct?

Based on my understanding, I have tried to compute the following, are they correct?
(a)For the value of Pr(2 <  ≤ 3) and Pr( = 3), I think that both Pr(2 <  ≤ 3) and Pr( = 3)are equal to find the value of Pr( = 3) and the value is 0.2.
(b)For the value of Pr( = 4)and Pr( = 6), both of the answers are 0.
(c)Pr(3 ≤  ≤ 5)= 1-0.6=0.4
(d)Pr(3 <  ≤ 6)=0.2
(d)Is the population mean  = E() equal to the sum of product of all possible values (which are X= 1, 2, 3, 5)with their corresponding probabilities?

Thanks for your precious time solving my problem!
Thank you!!!!

答:
Yes, absolutely right.

You can also derive the pmf, if you want being

$$P(X=x)=F_X(x)-F_X(x^-)$$

Thus you immediately realize that your rv takes the values

$$\{1;2;3;5\}$$

with probabilities

$$\{0.3;0.3;0.2;0.2\}$$

respectively


问:
Insignificant value
Let us say that a line is defined as
$$ y=\frac{\sqrt{c^2-a^2}}{a}x $$
and $a<c$. Can we tell when the value of $a$ becomes insignificant with respect to $c$?

答:
You probably mean that if we keep increasing $c$ and keep $a$ constant, when can we ignore $a$. This is usually done in science  where we need to approximate things and it us upto us what degree of accuracy we want. There is no such standard where we can say that $c >>a$.

However, suppose if $c=100$ and $a=1$, for calculation purposes, we may simply write your expression as $$y=\dfrac{\sqrt{10000-1}}{1}x\approx \sqrt{10000}x=100x$$

Which differs from actual answer by roughly $0.005x.$ Now it is upto you to accept it or not.

Title: Insignificant value


答:
Assume $a=\min\{a,b,c\} \rightarrow 0\le a \le 1.$
By AM-GM, we have $$abc\left(a^2+b^2+c^2\right) =\dfrac{1}{3} a\cdot
3bc\cdot\left(a^2+b^2+c^2\right)$$
$$\le \dfrac{1}{12} a \left[a^2+bc+\left(b+c\right)^2\right]^2\le \dfrac{1}{12} a \left[a^2+\dfrac{\left(3-a\right)^2}{4}+\left(3-a\right)^2\right]^2\le 3,$$
so it suffices to prove
$$\left( 9\,{a}^{2}-33\,a+64 \right)  \left(1-a \right) ^{3}\ge 0,$$
which is obvious since $0\le a\le 1.$

Done!

Title: given $a+b+c=3$ prove that $abc(a^2+b^2+c^2) \leq 3$

Body:
I am preparing for inmo and I came accross this problem while solving a worksheet, but couldn't solve it, pl help me...

Problem-

Prove that if a,b,c are non negative real numbers such that a+b+c=3, then $abc(a^2+b^2+c^2)\leq3$ ... (0)

Developments-

Firstly applying am-gm on $a+b+c$ to get $abc\leq1$

Then applying quadratic mean arthematic mean to get

$a^2+b^2+c^2 \ge 3$

Then writing the expansion of $(a+b+c)^2$ we get

$a^2+b^2+c^2 = 9-2(ab+bc+ca)$ ...(I)

Thn applying am-hm on $ab+bc+ca$ to get

$ab+bc+ca \ge 3abc$

Then substituting in (I) we get

$a^2+b^2+c^2 \leq 9-6(abc)$

Then we substituting in (0)

$LHS \leq abc(9-6abc)$

Whose max value is $27/8$ but is not less than three, so I am stuck for this point onwards


答:
I will use a well known inequality for my solution -

$a^2b^2+b^2c^2+c^2a^2 \geq abc(a+b+c)$

$\implies (ab+bc+ca)^2 \geq 3abc(a+b+c) = 9abc$

Now using AM-GM,

$a^2+b^2+c^2+2 (ab + bc + ca) \geq 3 [(a^2+b^2+c^2)(ab+bc+ca)^2]^{1/3}$

$9 \geq 3[(a^2+b^2+c^2)(ab+bc+ca)^2]^{1/3}$

$3^3 \geq 9 abc(a^2+b^2+c^2)$

$3 \geq abc(a^2+b^2+c^2)$

On the inequality I used to answer comes from $(ab-bc)^2+(bc-ca)^2+(ca-ab)^2 \geq 0$.


问:
Question involving the definition of the derivative
We know the the derivative of a function at $x_0$, $f’(x_0)$, can be defined as $\lim_{x\to x_0} \frac{f(x_0)-f(x)}{x_0-x}$.

Can $f’(-x_0)$ be defined as $\lim_{x\to x_0} \frac{f(-x)-f(-x_0)}{x-x_0}$, or does this contradict the definition of the derivative in some way?

Body: $$f'(-x_0)=\lim_{y\to-x_0}\frac{f(y)-f(-x_0)}{y-(-x_0)}=\lim_{-y=t\to x_0}\frac{f(-t)-f(-x_0)}{\color{red}{x_0-t}}$$


问:
Combinatorial proof of an identity of Stirling number of first kind

Let $s\left(n,k\right)$ be the Stirling numbers of the first kind. Prove that
  \begin{align}
\left(x\right)_n = \sum_{k=0}^n s\left(n,k\right) x^k ,
\end{align}
  where
  \begin{align}
\left(x\right)_n = x\left(x-1\right)\left(x-2\right)\cdots\left(x-n+1\right) .
\end{align}

I can prove this identity using induction but i was looking for a combinatorial proof for this identity regarding stirling numbers of first kind. How should i proceed?

答:
Here is a combinatorial proof that does not require any knowledge of groups. Let ${n \brack k}=(-1)^{n-k}s(n,k)$. The Stirling number ${n \brack k}$ may be interpreted as the number of ways to seat $n$ people at $k$ identical round tables where two arrangements are considered the same if each person has the same left neighbor in both arrangements. The coefficient of $x^k$ in the product $x(x+1)\cdots (x+n-1)$ is a sum of products. Each of these products counts the number of seatings of $n$ people at $k$ tables under certain constraints.

An example will serve to illustrate the general case. Suppose $n=8$ and $k=3$. Consider the number of seatings of 8 `people', say $0,1,\ldots,7$, at 3 tables such that the minimum numbers at each of the tables are $0,3$ and $6$ respectively. First seat $0,3,6$ at different tables. Now succesively seat the other numbers $1,2,4,5,7$.

$1$ can only be seated at the table with $0$: 1 way.

$2$ can only be seated at the table with $0$: 2 ways.

$4$ can be seated at the table with $0$ in 3 ways or the table with $3$ in one way: 4 ways.

$5$ can be seated at the table with $0$ or at the table with $3$: 5 ways.

$7$ can be seated at any of the three tables: 7 ways.

Note that, at each stage, the number of ways to seat the number $i$ is exactly $i$ for each $i\in \{1,2,4,5,7\}$. Thus the total number of seatings corresponding to the minimum values $\{0,3,6\}$ is precisely the product of elements in $\{1,2, 4, 5, 7\}$. This idea can be generalized to show that, given $n,k$ and a subset $S\subseteq \{0,1,\ldots,n-1\}$ of size $k$, the number of seatings of $\{0,\ldots,n-1\}$ in which the minimum elements at the tables are given by the subset $S$ is the product of those $n-k$ elements in $\{0,1,\ldots,n-1\}$ that are not in $S$. Summing up all such products as $S$ varies over $k$-element subsets, we find that the Stirling number ${n \brack k}$ is the coefficient of $x^k$ in the product $$x(x+1)\cdots (x+n-1).$$


问:
Prove that $P(A|B) = P(A|B,C)P(C) + P(A|B,\overline C)P(\overline C)$
Let $(Ω,S,P)$ be a probability space. Let $A, B, C ∈ S$ with $P(B)$ and $P(C) > 0$. If $B$ and
$C$ are independent show that
$$P(A|B) = P(A|B,C)P(C) + P(A|B,\overline C)P(\overline C)$$

$\textbf{My Solution:}$

[Note: All the arrows used in the solution represent the direction of flow of inference in a graphical model.]

[Note: I have used standard notation which are generally in the literature of probabilistic graphical models. Take a look. https://ermongroup.github.io/cs228-notes/representation/directed/]

It is given that A depends on B. So we can start with the fact that $B \rightarrow A$.

Take RHS
$$P(A|B,C)P(C) + P(A|B,\overline C)P(\overline C)$$
$$ = \frac{P(A,B,C)}{P(B,C)}P(C) + \frac{P(A,B,\overline C)}{P(B,\overline C)}P(\overline C)$$
$$ = \frac{P(A,B,C)}{P(B)P(C)}P(C) + \frac{P(A,B,\overline C)}{P(B)P(\overline C)}P(\overline C)$$
$$ = \frac{P(A,B,C)}{P(B)} + \frac{P(A,B,\overline C)}{P(B)}\ \ \ \text{(eq. 1)}$$

$\textbf{Case I}: A \perp C$

from $(eq. 1)$
$$= \frac{P(A,B,C)}{P(B)} + \frac{P(A,B,\overline C)}{P(B)}$$
$$= \frac{P(A,B)P(C)}{P(B)} + \frac{P(A,B)P(\overline C)}{P(B)}$$
$$= \frac{P(A,B)}{P(B)}$$
$$= P(A|B) = LHS$$

$\textbf{Case II}: C\leftarrow A$

from $(eq. 1)$
$$= \frac{P(A,B,C)}{P(B)} + \frac{P(A,B,\overline C)}{P(B)}$$
$$= \frac{P(B)P(A|B)P(C|A)}{P(B)} + \frac{P(B)P(A|B)P(\overline C|A)}{P(B)}$$
$$= P(A|B)P(C|A) + P(A|B)P(\overline C |A)$$
$$= P(A|B) = LHS$$

$\textbf{Case III}: C\rightarrow A$

from $(eq. 1)$
$$= \frac{P(A,B,C)}{P(B)} + \frac{P(A,B,\overline C)}{P(B)}$$
$$= \frac{P(B)P(C)P(A|B,C)}{P(B)} + \frac{P(B)P(\overline C)P(A|B,\overline C)}{P(B)}$$
$$= P(A|B,C)P(C) + P(A|B,\overline C)P(\overline C)$$
How to proceed from here?

答:
I am kind of confused of your notations: what do you mean by $A\rightarrow C$ and $A\perp C$?

Actually from your (eq.1) there is only one line further to go:
$$RHS=P(A,C\mid B)+P(A,\overline{C}\mid B)=P((A\cap C)\cup(A\cap\overline{C})\mid B)=P(A\mid B),$$

where the second equality follows from the fact that $(A\cap C)\cap(A\cap \overline{C})=\emptyset$.


问:
Prove that there exists no bijective function $f: \Bbb{N} \to \Bbb{N}$ such that $f(mn)=f(m)+f(n)+3f(m)f(n)$ for $m,n \geqslant1.$

Prove that there exists no bijective function $f: \Bbb{N} \to \Bbb{N}$ such that $$f(mn)=f(m)+f(n)+3f(m)f(n)$$ for $m,n \geqslant1.$

This was a problem from a Putnam practice book and I couldn't seem to figure out how to show this. Initially, I started to wonder that if the function were bijective then it has to be injective as well as surjective. So trying to determine if its injective let $a,b \in \mathbb{N}$ then if it were injective $$f(mn)=f(ab) \implies f(m)+f(n)+3f(m)f(n) =f(a)+f(b)+3f(a)f(b)$$ but this doesn't seem to help since I don't have explicitly $f$. How should I look at this problem?

答:
Let's assume $0\in\mathbb{N}$, since otherwise the condition $m,n\ge1$ would be redundant. Then, $m=n=1$ implies $f(1)=0$. If we define $g(n)=3\,f(n)+1$, we obtain $$g(mn)=g(m)g(n).\tag{mult}$$ So we have a completely multiplicative bijection between $\mathbb{N}$ and $\mathbb{M}=\{m\in\mathbb{N}: m=1\bmod3\}$. Now the numbers $10,34,55,187$ all belong to $\mathbb{M}$, so there must be natural numbers $n_1, n_2, n_3, n_4$ with $$g(n_1)=10,\quad g(n_2)=34,\quad g(n_3)=55,\quad g(n_4)=187.$$ The $n_i$ must be different, and they must be primes, because $d|n_i$ implies $g(d)|g(n_i)$ through (mult), and $g(n_i)$ by definition don't have non-trivial divisors in $\mathbb{M}$. But then, we have
$$g(n_1n_4)=g(n_1)\,g(n_4)=g(n_2)\,g(n_3)=g(n_2n_3),$$ and since $g$ is injective, $$n_1n_4=n_2n_3,$$ but this is impossible for four different primes.


问:
Prove existence of the saddle point for a Lagrangian implies the strong duality holds
In "Convex Optimization" of Boyd, a saddle-point $(v_0, x_0)$ for a Lagrangian L is the one that satisfies the below inequalities, with $v$ and $x$ as variables of the dual and primal problems respectively (assuming the primal problem does not have equality constraint),

$$L(v_0, x) <= L(v_0, x_0) <= L(v, x_0) \space | \space \forall (x,v) $$

The optimal values of the dual and primal problem can be represented as the two expressions repspectively,
$$
d^* = \sup_v \inf_x L(v,x) \\
p^* = \inf_x \sup_v L(v, x)
$$

Strong duality holds, if
$$ p^* = d^* $$

How to prove the existence of the saddle point implies the below equations?
$$ p^* = d^* = L(v_0, x_0) $$

答:
Suppose $(v_0, x_0)$ is a saddle point. Note that
\begin{align*}
\forall x, L(v_0, x) \le L(v_0, x_0) &\implies \sup_x L(v_0, x) = L(v_0, x_0) \\
&\implies \inf_v \sup_x L(v, x) \le L(v_0, x_0).
\end{align*}
Suppose the inequality is strict. Then there exists some $v_1$ such that
$$\sup_x L(v_1, x) < L(v_0, x_0).$$
But then,
$$L(v_0, x_0) \le L(v_1, x_0) \le \sup_x L(v_1, x) < L(v_0, x_0).$$
This is a contradiction, hence $p^* = L(v_0, x_0)$.

We can show the other equality similarly, or simply by replacing $L$ with $-L$, and swapping the roles of $v$ and $x$.


问:
Why the codomain of the norm function for quadratic integer rings is $\mathbb{Z}_{\ge 0}$, but not $\mathbb{Z}_{>0}$?
Let $R=\mathbb{Z}[\sqrt{d}] = \{ s+t\sqrt{d} : s,t \in \mathbb{Z} \}$ with $d$ square-free and $d \not \equiv 1 \pmod{4}$ be a quadratic integer ring. The most usual definition of norm is $N: R \setminus \{ 0 \} \rightarrow \mathbb{Z}_{\ge 0}$ by setting $s+t\sqrt{d} \mapsto |s^2 - dt^2 |$.

Apparently, if (1) most (all?) definitions require $d$ be squared-free and (2) $N(0)$ is undefined, then $N(s+t\sqrt{d})$  can't be zero. Why the codomain of $N$ is $\mathbb{Z}_{\ge 0}$, but not $\mathbb{Z}_{>0}$?

答:
It's honestly not important and you shouldn't waste too much thought on it.

If you're really interested in details though, here they are:

Despite your title, your $\mathbb{Z}[\sqrt{D}]$ are "quadratic integer rings" only when $D \not\equiv 1 \pmod{4}$. That phrase refers to the ring of integers of the quadratic field $\mathbb{Q}(\sqrt{D})$, which is a bit bigger than you'd naively expect when $D \equiv 1 \pmod{4}$.

Norms are defined quite generally in field theory as the product of the Galois conjugates; e.g. see here.

That standard general definition of the norm $N \colon \mathbb{Q}(\sqrt{D}) \to \mathbb{Q}$ is simply $N(a+b\sqrt{D}) = a^2-Db^2$. Obviously when $D<0$ this restricts to $N \colon \mathbb{Z}(\sqrt{D}) \to \mathbb{Z}_{\geq 0}$, with $N(0) = 0$ and $N(a+b\sqrt{D}) > 0$ for $a, b \in \mathbb{Z}$ not both zero.

Most people typically use the general norm definition in this context without tossing out $0$ or using absolute values e.g. Wikipedia, William Stein. However, when you're talking about whether the norm makes the ring of integers a Euclidean domain and $D>0$, the absolute value signs become important.

Euclidean functions require the codomain to be $\mathbb{Z}_{\geq 0}$ or $\mathbb{Z}_{>0}$ depending on your axiomatization. Restricting to a half-open interval is done to make sure the Euclidean algorithm terminates in finitely many steps, so that explains the absolute value signs.

But why are Euclidean functions often left undefined at $0$? See this MO thread on the topic. The short answer is for, say, $\mathbb{Q}[x]$, we want to use $\deg$ as the Euclidean function, but $\deg(0) = -\infty$ is the best definition, which conflicts with requiring the codomain to be $\mathbb{Z}_{\geq 0}$.


答:
An expression or function evaluation has only one value (or "answer") - or perhaps none in case it is not defined. Even though the equation $x^2=9$ has two real solutions $x=3$ and $x=-3$, the expression $\sqrt 9$ has only one value $3$.

What "is" $a^b$, anyway? In order to know what is meant by $a^b$, we need to define the expression $a^b$ for suitable $a$ and $b$.
Unfortunately, we don't do this all at once, but first only for a few cases and then extend the definition. And even more unfortunately, sometimes we encounter different routes to extend along and have to keep an eye on consistency. Here's an overview.

For

non-negative integers $a,b$,

i.e., when these can be interpreted as the cardinalities of finite sets, we can readily define $a^b$ as the number of maps from a set $B$ of cardinality $b$ to a set $A$ of cardinality $a$. For example, simple counting then gives us $3^2=9$, $2^3=8$, $0^0=1$, and so on. One can even show that this implies
$$\tag1 a^0=1,\quad a^{n+1}=a\cdot a^n$$
and more general relations such as
$$\tag2 a^{b+c}=a^ba^b,\quad a^{bc}={(a^b)}^c,\quad (ab)^c=a^cb^c.$$
We can use $(1)$ to extend our definition (namely, by using recursion as definition) to the case

$a$ an arbitrary number, $b$ a non-negative integer.

When $a\ne0$, we can even use the recursion in $(1)$ to go backwards and thereby extend to

$a\ne0$, $b\in\Bbb Z$.

Note that we cannot nicely define $0^b$ when $b$ is a negative integer. Well, theoretically we can define anything - but a definition should be useful. For example, we want to be able to make use of properties $(2)$ without having to check for applicability. Thus any extension of our definition should make sure that $(2)$ keeps holding. This is called the Principle of Permanence. Therefore any useful definition of $0^{-n}$ should make $0=0^{-n}\cdot 0=0^{-n}\cdot 0^n=0^0=1$, which is absurd.

But can we extend to rational exponents $b=\frac nm$, $n\in \Bbb Z$, $m\in \Bbb Z_{>0}$ (respecting the Principle of Permanence)?
That is, $x=a^{\frac nm}$ should be a number with the property $x^m=a^n$. Sometimes there is no (real) number with this property, (namely when $a<0$, $m$ even, $n$ odd; or when $a=0$ and $\frac nm<0$). Sometimes there is exactly one real solution (namely when $m$ is odd; also when $a=0$ and $\frac nm>0$). And sometimes there are several solutions (namely when $a>0$, $m$ even, $n$ odd). In the latter case we have to remember that we want to define the value of $a^b$ as a function of $a$ and $b$. The convention is to take the positive value, which has a huge advantage when we extend even further. But first let's summarize that we managed to deal with

$b\in\Bbb Q$, provided $a>0$, or $a=0$ and $b\ge0$, or $a<0$ and denominator of $b$ is odd.

Apparently, things are already becoming somewhat convoluted.

Thanks to our choice odf positive solution above, we have a new property: If $a>0$, then the map $\Bbb Q\to\Bbb R$, $b\mapsto a^b$ is continuous! This allows us to extend to

$a>0$, $b\in\Bbb R$

simply by continuity! This cannot work for negative $a$ because of the gaps at rationals with even denominator. And it almost works for $a=0$ by defining $0^b=0$ for all $b>0$, but neither can we define this for negative exponents, nor can we avoid the discontinuity at $b=0$, i.e., $0^0=1\ne 0=\lim_{b\to0^+}0^b$.

Once we know enough calculus, we can use a different way to extend our definition by setting $a^b=\exp(b\ln a)$ using the exponential function and natural logarithm. This agrees with the above as long as $a>0$ (in particular, we still have $4^{\frac12}=2$, period). It cannot deal with the case of $a=0$ where the logarithm is not defined and also has serious trouble when $a<0$. On the other hand, it allows us to use complex exponents $b$.
The goal now seems to be to extend the definition of the $\ln$ function. There are nice was to do so for any open subset $U$ of $\Bbb C$ that is a simply connected neighbourhood of $1$. While there is a somewhat natural choice $U=\Bbb C\setminus (-\infty,0]$ for a maximal such $U$, different choices may sometimes be more useful (and do ultimately lead to different values for some $a^b$!). Note that this standard choice (the "principal branch") ignores the cases with $a<0$, and just combining the two methods is also no good idea regarding Permanence.


问:
Same order subgroup of $\mathbb{Z}_{n_1} \oplus \mathbb{Z}_{n_2}$ where $n_1 \mid n_2$
Let $G$ be a group isomorphic to $\mathbb{Z}_{n_1} \oplus \mathbb{Z}_{n_2}$ for some integers $n_1$ and $n_2$ with $n_1 \mid n_2$.

Can this group $G$ have two subgroups $H_1$ and $H_2$ such that $|H_1| = |H_2|$ and $H_1 \neq H_2$

Any hint to show existence or non-existence?

答: Sure.  You can find a subgroup $H$ isomorphic to $\Bbb Z_{n_1}$ inside $\Bbb Z_{n_2}$.  So you have $\Bbb Z_{n_1} \oplus \{0 \}$ as one subgroup and $\{0\} \oplus H$ as the other.


问:
Conditional Probability Example using permutations

You are dealt three cards.  The events of interest concern the number of face cards that you are dealt (0, 1, 2, or 3).  What is the conditional probability that you are dealt at least 2 face cards given that the last card dealt to you was a face card?

There 12 face face cards (1 jack, queen, king * 4 suits) and 40 non face cards.

Let F= face card, and R = regular or nonface card.

So the question is asking us to find the probability of getting {RFF, FRF, FFF}.  I thought I would approach this question using permutations:

$\Pr(\text{@ least 2 face cards | last card has face}) = \Pr(\{RFF, FRF, FFF\})=\cfrac{(40)(12)(11) + (12)(40)(11) + (12)(11)(10)}{(52)(51)(50)} = 0.089593$

but the answer is 0.3882.  Can anyone please tell me why this is not working?  My understanding is that when you're solving probability you can consider ordering or not consider it.  In other words, you could use permutations or combinations.

Thank you in advance.

Here is a tree diagram.  Let A, B, and C be events where a face card is obtained.

答:
Last or first or second makes no difference. It will feel clearer with first.

We want the probability of at least one face card out of two, from a deck in which one face card is gone.

It is easier to find the probability of no face card in two picks. This is $\frac{40}{51}\cdot\frac{39}{50}$.

So our answer is $1-\frac{40}{51}\cdot\frac{39}{50}$.

Another way: There are fancier ways to proceed. Let $F$ be the event "the third card is a face card," and let $A$ be the event "at least two face cards." We want the conditional probability $\Pr(A|F)$. By the definition of conditional probability, we have
$$\Pr(A|F)=\frac{\Pr(A\cap F)}{\Pr(F)}.\tag{$1$}$$
Note that $\Pr(F)=\frac{12}{52}$.

You calculated the top of $(1)$ correctly, and did not divide by the bottom.

Remark: Remember that we are given that the third card is a face card. So the sample space is restricted to those draws in which the third was a face card. What you calculated was the probability that there are at least two face cards and the third is a face card. That is not what the problem asked for, buut once we have found it, the answer is one step away.


问:
how come square root of 2 times itself equals 2
since square root of $2$ is a irrational number, which we know or assume its decimal part is not a finite number, or doesn't terminate, how come we say that this infinite number (not in terms of being a large number) times itself is equals $2$.  It is like saying $0.928041\dots$ times itself is some finite number (like $1.64$ not going $3$ dots in the end).

答:
It's by definition. $\sqrt 2$ is the unique positive number that, when squared, results in $2$. Are you also surprised that $\frac13+\frac13+\frac13=1$ even though you are adding a number repeatedly to itself that has a decimal representation that is non-terminating? (Similar to above, $\frac13$ is defined as the unique number that, when multiplied by $3$, results in $1$)

Do not confuse a number with the representation of a number by a decimal expansion.


答:
There is no such thing as an infinite number. If you mean that the decimal expansion of $\sqrt2$ never ends, then you are right. But that also applies to the decimal expansion of $\frac13$; however, I suspect that you see no problem with the assertion that $3\times\frac13=1$.

The fact that the decimal expansion of $\sqrt2$ never ends is something that concerns the specific way that we use to represent numbers, which was developed centuries after square roots started to be studied. It is not something about the number $\sqrt2$ itself, which is defined as the only real number greater than or equal to $0$ whose square is $2$.

After having defined it this way, we can try to compute its decimal expansion. For instance, $1^2<2<2^2$, and therefore $1<\sqrt2<2$. Furthermore, $\left(\frac75\right)^2<2<\left(\frac85\right)^2$ and therefore $\frac75<\sqrt2<\frac85(1.4<\sqrt2<1.6)$.

Title: how come square root of 2 times itself equals 2

Body: since square root of $2$ is a irrational number, which we know or assume its decimal part is not a finite number, or doesn't terminate, how come we say that this infinite number (not in terms of being a large number) times itself is equals $2$.  It is like saying $0.928041\dots$ times itself is some finite number (like $1.64$ not going $3$ dots in the end).


问:
$a\cdot x \mod n = 0$, if I know $a$ and $n$, what can I say about $x$?
I have the equation

$ax \mod n = 0$

$a,x,n$ are integers and only $x$ is unknown.

Are there any properties of $x$ that follow from this equation?

I tried the following myself, but I am stuck:

Because of the $\mod n $ it follows that $ax = pkn$  with $k,p$ being other unknown integers, thus:

$x=p\frac{kn}{a}$

And this means that I know that $x$ has the property that it can be divided by $\frac{kn}{a}$.

But is that all that I can derive? And how can I calculate $\frac{kn}{a}$?

答:
$x = k\dfrac{n}{a}$, where $k \in \mathbb{Z}$

you said you know $n$ and $a$... just put $k$ as any integer to get a value for $x$.

and since you wanna restrict $x$ to be an integer you must find $k \in \mathbb{Z}: k\dfrac{n}{a} \in \mathbb{Z}$

also if you are looking for the minimum value of $x: x>0$ then set $k = \dfrac{a}{\gcd(n,a)}$


问:
law of large numbers for $P\{X_k = \pm 2^k\} = \frac{1}{2}$;
I want to show that the law of large numbers holds/doesn't hold for the sequence of independent random variables $P\{X_k = \pm 2^k\} = \frac{1}{2}$

A sufficient condition is $\frac{s_n}{n} \rightarrow 0$ where $s_n$ is the sum of the variances of the variables. However, this is clearly not true so I'm trying to show that the law of large numbers fails to hold.

答:
Assume they are independent, and $S_n=\sum\limits_{k=1}^n X_k$.

Actually for this simple example we can list all possibilities of the outcome: using induction or binary system, we know $S_n$ is symmetric with $2^{n-1}$ positive outcomes with equal probability $1/2^{n}$: $2,6,10,\cdots,2^{n+1}-2$. So there are at most $\lfloor (n+2)/4\rfloor<n$ number of the outcomes can be larger than $n$, which means
$$P(|\frac{S_n}{n}|>1)\geq 1-\frac{n}{2^{n}}\rightarrow 1$$

if $n\rightarrow\infty$. So it cannot satisfy the weak law of large numbers.

If they are dependent, I wonder whether there is an example that WLLN holds--will edit later if I find one or disprove it.


问:
Proving that sequence $a_n = \sqrt{x \sqrt{x \sqrt{x \sqrt{\cdots}}}} = x^{1-2^{-n}}$
Let $x>0$.

For sequence $a_n$, such that $n$ denotes the $n$th term:

$$\begin{align} a_1&=  \sqrt{x}\\
a_2&= \sqrt{x \sqrt{x}}\\
a_3&= \sqrt{x \sqrt{x \sqrt{x}}}\\
a_4&= \sqrt{x \sqrt{x \sqrt{x \sqrt{x}}}}\\
&\vdots\\
a_{n-1}&= \sqrt{x \sqrt{x\sqrt{... \sqrt{x}}}}\\
a_n&= \sqrt{x \sqrt{x\sqrt{... \sqrt{x \sqrt{x}}}}}\end{align}$$

How could one prove that:

$${a_n =  x^{1-2^{-n}}}?$$

答:
Hint:
\begin{align}
a_4&=\sqrt{x \sqrt{x\sqrt{{x \sqrt{x}}}}}\\&=\sqrt{x \sqrt{x\sqrt{{x^{\frac{3}{2}} }}}}\\&=\sqrt{x \sqrt{x^{\frac{7}{4}}}}\\&=\sqrt{x^{\frac{15}{8}}}\\&=x^{\frac{15}{16}}\\&=x^{1-2^{-4}}
\end{align}

Can you use induction in these footsteps?


问:
Is this a valid proof of the fact that a subspace of an $n$-dimensional vector space has a basis?
Is the following proof of the statement "Any vector subspace $M$ of a $n$-dimensional vector space $L$ has a finite basis" valid ?
Proof: If $M$ consists of only zero vector then we are done because the empty set will be a basis. Now suppose there exists a nonzero vector $x_{1}\in M$. Now we do the following. Label set $A_{1}=\{x_{1}\}$

Step $1$: If  $\forall y\in M\setminus A_{1}: \{x_{1},y\} \text{is linearly dependent}$, then we conclude that $A_{1}=\{x_{1}\}$ forms a basis, because it is maximal linear independent set.
If it is not the case that $\forall y\in M\setminus A_{1}: \{x_{1},y\} \text{is linearly dependent}$, meaning $\exists y\in M\setminus A_{1}: \{x_{1},y\} \text{is linearly independent}$, then we label $x_{2}=y$ and form a set $A_{2}=A_{1}\cup\{x_{2}\}=\{x_{1},x_{2}\}$.

Step $2$: If  $\forall y\in M\setminus A_{2}: \{x_{1},x_{2},y\} \text{is linearly dependent}$, then we conclude that $A_{2}=\{x_{1},x_{2}\}$ forms a basis, because it is maximal linearly independent set.
If it is not the case that $\forall y\in M\setminus A_{2}: \{x_{1},x_{2},y\} \text{is linearly dependent}$, meaning $\exists y\in M\setminus A_{2}: \{x_{1},x_{2},y\} \text{is linearly independent}$, then we label $x_{3}=y$ and and form a set $A_{3}=A_{1}\cup\{x_{3}\}=\{x_{1},x_{2},x_{3}\}$.

Step $j$: If  $\forall y\in M\setminus A_{j}: \{x_{1},x_{2},..,x_{j},y\} \text{is linearly dependent}$, then we conclude that $A_{j}=\{x_{1},x_{2},..,x_{j}\}$ forms a basis, because it is maximal linearly independent set.
If it is not the case that $\forall y\in M\setminus A_{j}: \{x_{1},x_{2},..,x_{j},y\} \text{is linearly dependent}$, meaning $\exists y\in M\setminus A_{j}: \{x_{1},x_{2},y\} \text{is linearly independent}$, then we label $x_{j+1}=y$ and and form a set $A_{j+1}=A_{j}\cup\{x_{j+1}\}=\{x_{1},x_{2},..,x_{j},x_{j+1}\}$.

Because of the fact that $M$ is a subspace of a $n$-dimensional vector space $L$. If we will get to the step $n$, the set $A_{n}$ will neccesarily be maximal linearly independent set (and so will be a basis for $M$), because any set of $n+1$ vectors in $L$ is linearly dependent. Therefore the process will terminate and we will form a basis for $M$. Q.E.D.

My worries about this proof are that it does something like a proof by cases but the number of cases is not fixed and is dependent on $n$. So is this proof valid ?

Thanks in advance.

答:
It looks okay but there's an easier way of doing this.

Let $V$ be a vector space such that $dim(V) = n$. Then, it has a basis $(x_1,\ldots,x_n)$. This basis generates $V$ and, therefore, it must generate $M$. So, $M$ is finitely generated. Then, there exists a finite list of vectors $(v_1,\ldots,v_m)$ that generate $M$. If this list is linearly independent, then we have a basis for $M$. Otherwise, there is a vector in the list that can be written as a linear combination of the other vectors. So, remove that vector and check if the resulting list is linearly independent.

If it is, then we're done. Otherwise, continue this process of removing vectors until a linearly independent list is found. Indeed, we are guaranteed to find a linearly independent list because if we reduce the list size to $1$, we will certainly have a linearly independent list. Moreover, this list is guaranteed to span $M$ at each stage of the process. Hence, $M$ has a basis. $\Box$

Actually, your argument seems to be a "backwards" version of the argument I've put forth above, where you've constructed a basis from ground up. What I've done is to take a list that generates $M$ (we are guaranteed that at least one such list exists) and then I've reduced it to a basis of $M$. You've also been rather formal and that's a good thing too.


问:
Continuous Images of $GL_n(\mathbb{R})$
Which of the following is a continuous image of $GL_n(\mathbb{R})$?

The real line $\mathbb{R}$

The subspace $\{(x,\frac1x):x\in \mathbb{R},x \ne 0\}$ of $\mathbb{R}^2$

$\mathbb{R}^2 \setminus \{(x,\frac1x):x\in \mathbb{R},x \ne 0\}$

$\{(x_1,x_2,...,x_n):x_1^2+x_2^2+...+x_n^2 \le 1\}$ of $\mathbb{R}^n$

How do I proceed?

$GL_n(\mathbb{R})$ is not compact, not connected, so I cannot rule out some options using topological properties.

答:
The determinant $\det : GL_n(\mathbb R) \to \mathbb R$ is continuous and has image $\mathbb R \setminus \{0\} = (-\infty,0) \cup (0,\infty)$. But $(0,\infty)$ is homeomorphic to $\mathbb R$; for example $\ln : (0,\infty) \to \mathbb R$ is a homeomorphism.

Yes. Let $f = \phi \circ \det$, where $\phi \mid_{(0,\infty)}$ is a homeomorphism and $\phi \mid_{(-\infty,0)}$ is any map. Lubin's comment takes $f = \ln \circ \lvert det \rvert$ which is a special case of this construction.

Yes. $\phi : \mathbb R \setminus \{0\} \to \mathbb R^2, \phi(x) = (x,1/x)$, is continuous and has the desired image. Take $f = \phi \circ \det$. Note that $\phi$ is a homeomorphism onto its image.

No. $X  = \mathbb{R}^2 \setminus \{(x,\frac1x):x\in \mathbb{R},x \ne 0\}$ has three components, but $GL_n(\mathbb R)$ has only two. See How many connected components does $\mathrm{GL}_n(\mathbb R)$ have?

Yes. Let $g : GL_n(\mathbb R) \to \mathbb  R^n$ be the map assigning to each matrix $M$ its first column vector. This is a continuous map with image $\mathbb  R^n \setminus \{0\}$. Now define
$$r : \mathbb R^n \setminus \{0\} \to \mathbb R^n, r(x) = \begin{cases} x & \lVert x \rVert \le 1 \\ (2- \lVert x \rVert)\frac{x}{\lVert x \rVert} & 1 \le \lVert x \rVert \le 2 \\ 0 & \lVert x \rVert \ge 2 \end{cases}$$
This a continuous map and $f = r \circ g$ has the desired image.

Title: Continuous Images of $GL_n(\mathbb{R})$


问:
Dummit's Abstract Algebra book, Exercise 9, Sec.3.2.
$ G $ is a finite group and $p$ is a prime dividing $|G|$,
$
S=\left \{ \left ( {x}_{1},{x}_{2},...,{x}_{p} \right ) | x_{i}\in G\space and\space x_{1}x_{2}...x_{p} = 1 \right \}
$
.
Show that $S$ has $|G|^{p-1}$ elements.

This problem is the first step of proving Cauchy's Theorem in Abstract Algebra.
The complete proof is due to James McKay which was published on Amer.Math.Monthly, 66(1959),p119.

I have tried to construct an epimorphism from $G^{(p)}$ onto $G$ sending $(x_{1},x_{2},...,x_{p})$ to $x_{1}x_{2}...x_{p}$, but I don't know how to define a multiplication fitting the condition $\varphi(xy)=\varphi(x)\varphi(y)$.
If we let $A_{g_{i}} = \left \{ \left ( {x}_{1},{x}_{2},...,{x}_{p} \right ) | x_{i}\in G\space and\space x_{1}x_{2}...x_{p} = g_{i} \right \}$, then it's clear that the set of $A_{g_{i}}$ is a group by defining $A_{g_{i}} \cdot A_{g_{j}} = A_{g_{i}\cdot g_{j}}$ . But this is still helpless.

答:
The set $\;S\subset G^p:=\overbrace{G\times\ldots\times G}^{p\;\text{times}}\;$ is not a subgroup of the cartesian product (=direct product, in this case) $\;G^p\;$, and I can't understand what do you want some homomorphism from $\;G^{(p)}:=\langle x^p\;|\;x\in G\rangle\;$ to $\;G\;$ for at all... If your problem is to prove that $\;|S|=|G|^{p-1}\;$ then you can argue as follows:

An element $\;(x_1,...,x_p)\in G^p\;$ belongs to $\;S\;$ iff

$$x_1\cdot\ldots\cdot x_p=1\iff x_p=(x_1\cdot\ldots\cdot x_{p-1})^{-1}$$

and thus we can see that an element in $\;S\;$ is uniquely and completely determined by its first $\;p-1\;$ coordinates, and thus $\;|S|=|G|^{p-1}\;$ ...as easy as that.


问:
Ideals of quotient ring of a principal ideal domain are also principal ideals.
Let $R$ be a PID. I have to show that every ideal of a quotient ring of $R$ is a principal ideal. I am able to visualize this problem by taking $R=Z$. I took ideal $(20)$ of $Z$ for example. Then I got quotient ring as:
$R/(20)=\{(20),1+(20),2+(20),.......,19+(20)\}$.
When I consider an ideal of this quotient ring e.g. $\{(20),2+(20),4+(20),.....,18+(20)\}$,

I can see that this ideal is  generated by a single element of the ideal i.e. $2+(20)$. Thus it is a principal ideal. Similar thing can be said about other ideals of the quotient ring. But when it comes to formal proof of this, I do not have a clue as to how to do that. Please suggest.

答:
Hint:

Use that ideals in a quotient ring $R/I$ correspond bijectively to ideals $J$ of $R$ that contain $I$ by the correspondance $J\longmapsto J/I$. In  terms of P.I.D., this means that a generator of the ideal $J$ will divide the generators of $I$.


问:
How do the sets $\emptyset\times B,\ A\ \times \emptyset, \ \emptyset \times \emptyset $ look like?
If we have a function $f:A \rightarrow B$, then one way to give meaning, I think, to this function, in terms of set theory, is to say, that $f$ is actually a binary relation $f=(A,B,G_f)$, where $G_f \subseteq A \times B$ is the graph of the function. Now my question is: what is $f$ if

$\bullet \ A=\emptyset, \ B\neq\emptyset$,?

$ \bullet \ B=\emptyset, \ A\neq\emptyset$ ?

$ \bullet \ B=\emptyset, \ A=\emptyset$ ?

(Another way to formulate this, I think, would be: How do the sets $\emptyset\times B,\ A\ \times \emptyset, \ \emptyset \times \emptyset $ look like? Are they all $\emptyset$ ?)

答:
A slightly more formal rephrase of the other answer is to calculate the elements in $A \times \emptyset$: for any $p$,
\begin{align}
& p \in A \times \emptyset \\
\equiv & \;\;\;\;\;\text{"definition of $\times$ on sets"} \\
& \textrm{isPair}(p) \;\land\; \textrm{first}(p) \in A \;\land\; \textrm{second}(p) \in \emptyset \\
\equiv & \;\;\;\;\;\text{"$x \in \emptyset \equiv \textrm{false}$ for any $x$; simplify"} \\
& \textrm{false} \\
\equiv & \;\;\;\;\;\text{"$x \in \emptyset \equiv \textrm{false}$ for any $x$"} \\
& p \in \emptyset \\
\end{align}
and therefore, by set extensionality, $A \times \emptyset = \emptyset$.  A very similar proof of course goes for $\emptyset \times B = \emptyset$, and obviously either of these implies $\emptyset \times \emptyset = \emptyset$.


问:
Hypergeometric differential equation for $c=1$ and $a+b+1=0$
I would like to find the base of solutions for the following differential equation,
$$z(1-z)f''(z)+f'(z)+ \alpha \cdot f(z)=0$$
where $\alpha$ is a parameter and the prime indicates derivative w.r.t z.  The above equation is the hypergeometric one say, $$z(1-z)f''(z)+(c-z(a+b+1))f'(z) -ab\cdot f(z)=0$$ in the case with $c=1$ and $a+b+1=0$. BUT it is not clear at all what are the two independent solutions. One has to be the hypergeometric but  the other one I do not know beacuse of the degeneracy of the parameters [ $c=1$ and $a+b+1=0$].

Many thanks!!!

答:
If you want just some expression in terms of special functions for the general solution, then you can use as two independent solutions
$$f_\pm(z)= z^{\frac12\pm\beta}{}_2F_1\left(-\tfrac12\mp\beta,-\tfrac12\mp\beta,1\mp2\beta,z^{-1}\right), $$
where $\alpha=\beta^2-\tfrac14$.

Explanation: For hypergeometric equation, the singular points $z=0,1,\infty$ play similar roles. Your parameters are "pathological" for the bases "adapted" to $z=0$ and $z=1$ but for the basis "adapted" to $z=\infty$ they are okay.

On the other hand, if you do want the bases adapted to the description of local behavior of solutions at $z=0$ or $z=1$, you may find them in the book of Abramowitz-Stegun, formulas 15.5.16-15.5.17; or equation 15.10.8 in DLMF.


问:
Possibility of throwing exactly one six
I am reading the following question:

What is the probability of throwing precisely one six when 6 dice are
rolled?

So when we have one dice the probability of having one six is $\frac{1}6$

When we have six dice we can have 6 cases that the six can come i.e. first dice or second or third etc.

The total possible outcomes of throwing 6 dice is: $6\cdot6\cdot6\cdot6\cdot6\cdot6=6^6$

So the probability is: $\frac{6}{6^6}=\frac{1}{6^5}$

But the solution says:

The probability of throwing precicely $1$ six out of $6$ dices is:

$\binom{6}{1}\cdot\frac{1}6\cdot\frac{5}6\cdot\frac{5}6\cdot\frac{5}6\cdot\frac{5}6\cdot\frac{5}6\cdot\frac{5}6$

Could someone help me understand why is $\binom{6}{1}$ being used here? That equals to $6$ as I also reasoned but why would we use choice formula here? What does it mean?

Also I assume the $\frac{5}6$ is the probability of not throwing a $6$ but why does it have to be multiplied $6$ times.

If I apply this formula to the case of throwing $1$ six when we roll $2$ dice I get:

$\binom{2}{1}\cdot\frac{1}6\cdot\frac{5}6\cdot\frac{5}6 = 2\cdot\frac{1}{6}\cdot\frac{5^2}{6^2}=\frac{25}{108}$

But if I actually count the cases we have:

Dice 1
Dice 2

1
6

2
6

3
6

4
6

5
6

6
1

6
2

6
3

6
4

6
5

this is $\frac{10}{36}$.

So what am I doing wrong here?

答:
Expanding what @N.F.Taussig said  in the comments of the question.

Instead of thinking about $6$ dice think about one single dice rolled six times. In six rolls you want to have one exaclty $6$ and all the other results must be different.

You can get a $6$ on the first throw with probability $\frac{1}{6}$ and then the probability that in the next $5$ rolls you do not get a $6$ is $\left(\frac{5}{6}\right)^5$

On the opposite hand, at the first throw you can get a number different from $6$ with probability $5/6$ and get a $6$ at the second throw. Then all the other 4 throws must end up with a number different from a $6$. This gives you again $\frac{1}{6}\left(\frac{5}{6}\right)^5$

Go on and count the probability you can get the $6$ at the $i$-th throw for $i = 3,4,5,6$.

This lead you to the result that the probability of getting exactly one $6$ in $6$ different rolls is
$$\binom{6}{1}\cdot\frac{1}{6}\left(\frac{5}{6}\right)^5$$

If the book gives $\left(\frac{5}{6}\right)^6$ then it is an error.

Title: Possibility of throwing exactly one six


问:
calculating area of triangle APE
I'm calculating the area of the triangle APE.

Given that $\vec{AP} = \vec{PE}$. My assumption is that P would represent an 'identity', for A and E, suggesting that A = E. Although, this is probably not the case.

How would I tackle this question?

The area of a triangle for ABC = $\frac{1}{2} |\vec{AB}\times \vec{AC}|$

I thought that APE would be something like: $\frac{1}{2} |\vec{AP}\times \vec{AE}|$ given that A=E, although I get 0.

What did I miss?

Given that $\vec{AP}=(3, 5, 0) - (1, 5, 2) = (2, 0, -2)$

How would I find more than one possible coordinate for E?

If a point E lies on the line $l_1$, which has the equation:

$\pmatrix{1 \\ 5 \\ 2} + \lambda \pmatrix{-5 \\ 4 \\ 3}$ Given $\lambda$ = 1, does this change the significance above?

答:
If $\vec{AP} = \vec{PE}$, we have a degenerate triangle.

$$\vec{AP} = \vec{PE} $$
$$\Rightarrow \vec{P}-\vec{A}=\vec{E}-\vec{P}$$
$$\Rightarrow \vec{P} = \frac{\vec{A}+\vec{E}}{2}$$

Since the direction of $\vec{AP}$, $\vec{PE}$ is same, $P$ is collinear with $A,E$ and is shown to be the midpoint of $AE$. The area is thus, zero.

Title: calculating area of triangle APE


问:
Proving the logic problem. would you check please???
Prove that $A \cup (B \cap C)= (A \cup B ) \cap (A \cup C)$

$x \in A$ or $x \in B$ and $x \in A$ or $x \in C$.

Since $A \cup B$ is $x \in A$ or $x \in B$, $x \in A$ is in $x \in A$ or $x \in B$

Since $A \cup C$ is $x \in A$ or $x \in C$, $x \in A$ is in $x \in A$ or $x \in C$

I would like to learn how to prove better than what i just wrote.

Can you give me a trick or technique to write better? I just started learning logic and I'm having difficulty time with writing mostly.

答:
Your way is essentially fine: By considereing element relations, we reduce this set equality to a logical equivalence:
$$ a\lor(b\land c)\iff (a\lor b)\land(a\lor c).$$
This can be proved by truth tables if one has no better idea. Or by deductions involving standard techniques such as case distinctions etc.</p>


问:
what are the rules regarding $x,a,b$ for this expression to be true: $(x^a)^b = (x^b)^a$ (i am considering only for when $a,b$ are real)?
in particular i am asking for the case when one of the powers $a$ or $b$ is a fraction.  in such a case, i believe the maths expression then may be ambiguous, as when you do to the power of a fraction, it's not clear if the required answer is the principal root, or if it is all roots.

eg i know $(4^{\frac12})^2 = 4$
but is $(4^2)^{\frac12} = 4$?  in this case, i think it's ambiguous, some will say also $-4$ is a solution , so we should add some notation that we want the "principal root" , and then it is true

答:
If $n\in\Bbb N$, and $x\geqslant0$, $x^{1/n}$ is the only number $y\geqslant0$ such that $y^n=x$ . In particular, $(4^2)^{1/2}=16^{1/2}=4$.

And, if $x>0$, and $a,b\in\Bbb R$, $\left(x^a\right)^b=\left(x^b\right)^a$. Actually, both numbers are equal to $x^{ab}$. To see why, I will assume that you define $x^a$ as $e^{a\log x}$. Then\begin{align}\left( x^a\right)^b&=e^{b\log\left(x^a\right)}\\&=e^{ab\log x}\\&=x^{ab}.\end{align}


问:
Prove equality for every positive integer by induction
The task is to prove this equation by induction:
$$\left(1-\frac {1}{n^2}\right)\left(1-\frac {1}{(n+1)^2}\right)...\left(1-\frac {1}{(2n-1)^2}\right)=1 - \frac {1}{2n-1}$$
I've done the base case, but I got stuck here:
$$\left(1-\frac {1}{(n+1)^2}\right)...\left(1-\frac {1}{(2n+1)^2}\right)=\left(1-\frac {1}{2n-1}\right)
\left(\frac {1}{1-\frac {1}{n^2}}\right)\left(1-\frac {1}{(2n)^2}\right)\left(1-\frac {1}{(2n+1)^2}\right)$$
What should be done now?

Any help appreaciated.

Body:
Assume that
$$A = \left(1-\frac {1}{n^2}\right)\left(1-\frac {1}{(n+1)^2}\right) \ldots \left(1-\frac {1}{(2n-1)^2}\right)=1 - \frac {1}{2n-1}$$ holds for some $n \in \mathbb N$. We shall show it for $n+1$:
$$\left(1-\frac {1}{(n+1)^2}\right)\left(1-\frac {1}{(n+2)^2}\right) \ldots \left(1-\frac {1}{(2n+1)^2}\right) = A \cdot \frac{\left(1-\frac {1}{(2n)^2}\right) \cdot \left(1-\frac {1}{(2n+1)^2}\right)}{\left(1-\frac {1}{n^2}\right)}$$ $$= \left(1 - \frac {1}{2n-1} \right ) \cdot \frac{\left(1-\frac {1}{(2n)^2}\right) \cdot \left(1-\frac {1}{(2n+1)^2}\right)}{\left(1-\frac {1}{n^2}\right)} = \frac{2n-2}{2n-1} \cdot \frac{\frac{(2n-1)(2n+1)}{4n^2} \cdot \frac{(2n)(2n+2)}{(2n+1)^2}}{\frac{(n-1)(n+1)}{n^2}} $$ $$= \frac{2n}{2n+1} = 1-\frac{1}{2n+1}$$ after some cancelation. So, we are done.


答:
In the real world, you only want to be accurate up to a point.  Within an inch, or a millimetre, or much less if you are building a computer chip.

Two people might agree that two numbers are close enough to being equal, but a third might disagree.

If two numbers are so close that everyone agrees they are close enough - so they agree to every decimal point - then we say they are equal.  So $1/3=0.333333333...$


问:
For $x\in [0,1]$, what is the maximal value of the following function?
For $x\in [0,1]$, what is the maximal value of the following function

$$\max_{x\in [0,1]}c_1x^{-1}\exp(-c_2 x^{-1})$$

I just plot the graph of $x^{-1}\exp(-1/x)$ as follows:

So it seems that the maximal value is obtained as $x=1$.

答:
$$\frac{d}{dx}\left(\frac{c_1e^{-\frac{c_2}{x}}}{x}\right) =
-\frac{c_1(x-c_2)e^{-\frac{c_2}{x}}}{x^3} \overset{!}{=}0\qquad \text{only at}\ x=c_2$$
So if $c_2>1$ the maximum value is at $x=1 \to c_1e^{-c_2}$
and for $c_2<1$ it's at $x=c_2 \to \frac{c_1}{c_2e}.$


问:
NP hard/complete
I have never been very clear on this concept. Please help:

At the end of the day, we should want to identify useful problems for which we don't have polynomial solution so far and only have exponential solutions.We want to keep finding if we can find a polynomial solution to these problems and the use of reductions is only to be able to identify new only exponential solvable problems with help of existing known exponential problems.??

Why is concept of determinism and non determinism brought into this whole concept of exponential and polynomial solvable problems?? How is this notion useful?

What do we call the problems for which we have only exponential solution so far, but we haven't been able to find reduction to a known NP - hard/complete problems..??

Thanks,

答: I would say that the concept of 'nondeterminism' is one of the great red herrings in the discussion of NP-complete problems.  To my mind it's much better to think of a problem's being in NP as a criterion for confirming solutions; 'if we're given a hypothetical solution instance for a problem, can we verify it in polynomial time?'.  Seen this way, a lot of problems in NP become a lot clearer: for instance, Satisfiability (of a boolean expression) is verified just by plugging in the purported assignment to all the variables of the expression and confirming that the result is true; the Traveling Salesman Problem can be verified by confirming that a proposed route gets to all cities, doesn't go to any city more than once, and has total length under the requisite bound; etc.  All of these obviously take polynomial time to confirm.  'Exponential solvability' is really a red-herring; it happens that there's an obvious exponential-time algorithm for NP-complete problems (use our polynomial-time checking algorithm on each of the exponentially-many possible solution instances), but the exponentiality isn't really at the heart of these problems; it's just an expression of the size of the search space.  Seen this way, the P vs. NP question becomes 'if we can efficiently confirm a solution, can we efficiently find a solution?' - but this has nothing to do with exponential-time problems in and of itself.

Title: NP hard/complete


问:
What does the notation "$\;(f·g)(x)\;$" mean?
Suppose I have two functions $f(x)$ and $g(x)$. What does it mean to have $(f·g)(x)$?

This notation can be found in a paper on Partial Differential Equations and Image Processing:

What we call a multiscale analysis is a family of transforms $(T_t)_{t>0}$ of an image $u^o(x)$, such that $u(t,x) = (T_t \cdot u^o)(x)$ can be seen as the image $u^o$ at a coarser scale $t$.

答:
In the most general setting, considering an arbitrary set $A$, a magma $(M, \cdot)$ -- which is a structure obtained by equipping the arbitrary set $M$ with an arbitrary binary operation $\cdot \colon M \times M \to M$ -- and two maps $f, g \colon A \to M$ one can define the product of the two maps as the map given by:
$$\begin{align}
fg \colon A &\to M \\
\left(fg \right)(x)&＝f(x)g(x),
\end{align}$$
where the product mentioned on the right-hand side of the second line refers of course to the binary operation $\cdot$ on $M$.

If one considers the diagonal direct product:
$$\begin{align}
f \underline{\times} g \colon A &\to M \times M \\
\left(f \underline{\times} g\right)(x)&＝\left(f(x), g(x)\right),
\end{align}$$
then we have the relation $fg＝\cdot \circ \left(f \underline{\times} g\right)$ expressing the product with respect to the diagonal direct product and the original underlying multiplication.

P.S. Upon closer inspection it seems that the notation used in the article you are referring to has nothing to do with the algebraic construction described above and is most likely an example of vaguely and sloppily used syntax, a custom which to the disappointment of some has become quite widespread nowadays...


问:
Reflexivity and homeomorphisms
Let $ X_{1}, X_{2} $ standard spaces where one of them is reflective. If there is a linear homeomorphism between these spaces, that is, a bijective linear operator $T: X_{1} \to X_{2} $ continuous with continuous inverse, then the other space is also reflective.

答: Scheme of proof: If $T:X_1\to X_2$ is an isomorphism, and $J:X_1\to X_1^{**}$, $x\mapsto x^{**}$, is an isomorphism, then let $S:X_1^{**}\to X_2^{**}$ be the isomorphism induced by $T$. Then $$SJT^{-1}:X_2\to X_1\to X_1^{**}\to X_2^{**}$$ is the required isomorphism.

Title: Reflexivity and homeomorphisms


问:
Transcendence basis as subset of generators
Let $K \subset L = K(a_1,...,a_n)$ be a field extension finitely generated as $K$-algebra with
transcendence degree $\operatorname{Trdeg}_K(L):= m \le n$. It is well
known that the choice of a transcendence basis is hightly not unique.
The question is why it is always possible to find a transcendence basis
which is a subset of the set of generators $\{a_1,...,a_n\}$ of $L$?

答:
Note that if the set $a_1,\dots,a_n$ was algebraically independent that set already forms a transcendence basis for $L/K$. If not then there is some non-trivial polynomial $p$ over $K$ with $$p(a_1,\dots,a_n)=0$$
At least one of the $a_i$ has to appear on the left side. Wlog assume that $a_n$ appears there. Then $a_n$, and therefore also $L$, is algebraic over $K(a_1,\dots,a_{n-1})$. Now continue in the same way until you are left with algebraically independent elements $a_1,\dots,a_{n-i}$, so that the other elements $a_{n-i+1},\dots,a_n$ are algebraic over $K(a_1,\dots,a_{n-i})$. Then $\{a_1,\dots,a_{n-i}\}$ will be your transcendence base.

Note that this argument heavily relied on the fact that we have a finite set (otherwise the process might not stop). Here is another argument (which is actually not that different) which also works in the infinite case:

Let $S$ be the set of all algebraically independent subsets of $A:=\{a_1,\dots,a_n\}$. Note that this set is not empty as $\emptyset\in S$. As $S$ is finite $S$ has a maximal element $B$ with respect to the inclusion (in the infinite case we would need to invoke Zorn's lemma here to get the existence of a maximal element). I claim that this $B$ is a transcendence base for $L/K$. By definition $B$ is algebraically independent. Now for any $a\in A\setminus B$ by maximality of $B$ the set $B\cup \{a\}$ is algebraically independent. It follows that there is a non-trivial polynomial $p$ over $K$ with $$p(B,a)=0$$
As $B$ is algebraically independent $a$ has to appear on the left side, so $a$ is algebraic over $K(B)$. It follows that $L=K(A)$ is algebraic over $K(B)$, so $B$ is indeed a transcendence base.

You might see that the reasoning above is very similar to that in linear algebra when dealing with bases of vector spaces. The principle that is behind these two things is the notion of a Dependence relation.


问:
What is $\lim_{n\to\infty} 1+\frac{i}{n}$

What is $$\lim_{n\to\infty} 1+\frac{i}{n}\;\;\;\;,i=\sqrt{-1}$$

Can we say that this limit is $1$? Can a limit in the complex world be a real number? If not what is the limit(if it exists) ?

答:
Evaluating:
$$\lim_{n \to \infty} (1 + \frac{i}{n}) = 1 + i \lim_{n \to \infty} \frac{1}{n}$$
Where:
$$\lim_{n \to \infty} \frac{1}{n} = 0$$
So:
$$1 + i 0 = 1$$

Have to remember that "$i$" is the number like any other.


问:
Inequality with quadratic and maximum
Let $x,y$ be real numbers in $[0,1]$ such that $$x \ge y \geq \max\left\{x^2, \frac{2x}{3}, \frac{1+x}{4}, \frac{1}{3}\right\}.$$ Is it true that $x^2+2y \ge 2x$?

In order to prove this, the term $x^2$ is currently on the "wrong" side, between the condition and the needed inequality. Using a linear combination of the conditions $y\geq \dots$ is not sufficient.

答:
nope not true...
let $x = 0.6, y=0.4, \max\left\{(0.6)^2, \frac{2*0.6}{3}, \frac{1+0.6}{4}, \frac{1}{3}\right\}=\max\left\{0.36, 0.4, 0.4, 0.33\right\} =0.4\\$

$\therefore x \geq y \geq \max\left\{x^2, \frac{2x}{3}, \frac{1+x}{4}, \frac{1}{3}\right\}$
holds when x = 0.6 and y = 0.4. Now check if $x^2+2y \geq 2x \implies 1.16 \geq 1.2$ which is false! Hence it is not true that $x^2+2y \geq 2x$

Title: Inequality with quadratic and maximum


问:
Can this Diophantine equation have infinitely many solutions?
I came across this as a intermediate step while solving another problem in analysis.

Suppose $m$, and $n$ are integers satisying,

$$\frac{1}{n} - \frac{1}{m} = \alpha$$ where $\alpha$ can be any non zero real.

I essentially need to show that only finitely many $(n,m)$ pair solutions are possible to complete my proof. How can I proceed? Clearly, we only need to check when $\alpha$ is rational.

答:
This is obviously not right if $\alpha=0$, but let's say we are interested in $\alpha\ne 0$. Then, WLOG we can assume $\alpha>0$. Now, even though one of $m,n$ can be negative, by a suitable change in signs we can reduce the equation to $\frac{1}{n}\pm\frac{1}{m}=\alpha$ with $m,n>0$. Thus, we have two cases:

"minus" case: $\frac{1}{n}-\frac{1}{m}=\alpha$ means that $\frac{1}{n}>\alpha$, i.e. $n<\frac{1}{\alpha}$. There are only finitely many such $n$. For each of them, the (integral) solution for $m$ is unique, if it exists (as it is given by $m=\frac{1}{\frac{1}{n}-\alpha}$).

"plus" case: $\frac{1}{n}+\frac{1}{m}=\alpha$. In that case, WLOG let $n\le m$, so $\alpha\le\frac{2}{n}$ i.e. $n\le\frac{2}{\alpha}$. This again has finitely many solutions in $n$, each of which again having at most one integral solution in $m$, as before.


问:
Probability of Yahtzee straight with strict re-roll rules
Assuming you have 5 fair dice and up to 3 re-roll attempts (during which you can re-roll any dice, as per standard Yahtzee rules) what is the probability of succesfully rolling a "Small Straight" (aka 1-2-3-4, 2-3-4-5, or 3-4-5-6)? Also, what is the probability of rolling a "Large Straight" (aka 1-2-3-4-5 or 2-3-4-5-6)?

In either case, assume the following rules for keeping and re-rolling dice:

Keep exactly one of each 2, 3, 4, and 5 result

Only keep a 1 or 6 result if it is already part of a Small Straight or Large Straight

Unfortunately I'm quite unfamiliar with even basic probability problems let alone something with this level of complexity, and online research hasn't helped much. I'm much more comfortable backing into probabilities by simulating roll outcomes like this in Python, though I've hit a wall with those efforts as well for straights in particular, even with these defined rules. Would deeply appreciate some help.

Thanks!

答:
(I have edited the answer)

Large Straight

Let me see if I got the strategy rules right. Does the following simulation code (in Python) model the situation correctly?

```<code>import random

keepers = set([2,3,4,5])

def simuLS(doPrint=False):
    kept = set()
    for roundI in range(3):
        ds = [random.randint(1, 6) for _ in range(5-len(kept))]
        for d in ds:
            if d in keepers: kept.add(d)
        if 1 in ds and all(x in kept for x in [2,3,4]):
            kept.add(1)
        if 6 in ds and all(x in kept for x in [3,4,5]):
            kept.add(6)
        gotLS = len(kept)==5
        if doPrint:
            print ("Rolled", ds)
            print ("Kept are ", kept)
            print ("Got LS: ", gotLS)
        if gotLS: return True
    return False

#Let's see how it goes
#simuLS(doPrint=True)

simuN = 10000
p = sum(1.0 for _ in range(simuN) if simuLS())/simuN
print (float(p))
</code>```

The output is around 0.24.

Are you familiar with Markov chains? We can construct a Markov chain with states all the combinations of the keeper-numbers $\{1,2,3,4,5,6\}$ at most of length 4 and having 1 or 6 only if 2,3,4 and 3,4,5 are present respectively. We also have one absorbing state 'LS' that means we have gotten a large straight. We can calculate the transition probabilities by brute force by considering all possible rolls of the not kept dice (there are 5 - length of the state of them) and seeing what additional keepers we got or did we in fact get a LS. I wrote the following code in SageMath, since it handles matrices and rational numbers without any efforts.

```<code>import itertools

keepers = set([2,3,4,5])

states = [tuple(S) for S in Subsets(range(1,7)) if len(S)<5 and
          (1 not in S or all(x in S for x in [2,3,4]))
          and (6 not in S or all(x in S for x in [3,4,5]))]
startState = tuple() #the empty set
targetState = 'LS' #have gotten large straight
states.append(targetState)
stateToInd = {s: i for i,s in enumerate(states)}
startI = stateToInd[startState]
targetI = stateToInd[targetState]
mat = matrix(QQ, len(states))

for i1,s1 in enumerate(states[:-1]): #don't do for the LS-state
    d = 5-len(s1) #how many dice are (re-)thrown
    s1Set = set(s1)
    p = 1/6**d #probability for each outcome
    for xs in itertools.product(*[range(1, 7) for _ in range(d)]):
        s2Set = s1Set.union(set(xs)&keepers)
        if 1 in xs and all(x in s2Set for x in [2,3,4]): s2Set.add(1)
        if 6 in xs and all(x in s2Set for x in [3,4,5]): s2Set.add(6)

        if len(s2Set)==5: #got LS
            mat.add_to_entry(i1, targetI, p)
        else:
            s2 = tuple(sorted(s2Set))
            mat.add_to_entry(i1, stateToInd[s2], p)
mat.add_to_entry(targetI, targetI, 1) #absorbing state

#look at the transition matrix
print (len(states))
print (states)
show(mat)

#print ([sum(row) for row in mat]) #sanity check

sol = (mat**3)[startI][targetI]
print (sol, "=", float(sol))
</code>```

You can run Sage code online: Sage Cell Server

The output is $\frac{295073737}{1224440064} = 0.24098667274578822$.

I'm not sure the grouping together of states, that I mentioned in my answer before editing, can be done anymore, since now the transitions to the states containing 1 or 6 depend on what we already have. At least not for all same-size-groups. Anyhow, we have a $19\times 19$-matrix, so it is doable.

Small Straight

We can use the previous Markov chain also here:

```<code>startRow = (mat**3)[startI]
endsForSmall = [stateToInd[s] for s in states if len(s)>=4] + [targetI]
smallSol = sum(startRow[j] for j in endsForSmall)
print (smallSol, "=", float(smallSol))
</code>```

Answer: $\frac{2966131613}{4897760256} = 0.605609800799527$

For simulation:

```<code>import random

keepers = set([2,3,4,5])

def simuSS(doPrint=False):
    targets = [(1,2,3,4),
               (2,3,4,5),
               (3,4,5,6)]
    kept = set()
    for roundI in range(3):
        ds = [random.randint(1, 6) for _ in range(5-len(kept))]
        for d in ds:
            if d in keepers: kept.add(d)
        if 1 in ds and all(x in kept for x in [2,3,4]):
            kept.add(1)
        if 6 in ds and all(x in kept for x in [3,4,5]):
            kept.add(6)
        gotSS = any(all(x in kept for x in y) for y in targets)
        if doPrint:
            print ("Rolled", ds)
            print ("Kept are ", kept)
            print ("Got SS: ", gotSS)
        if gotSS: return True
    return False

#Let's see how it goes
#simuSS(doPrint=True)

simuN = 10000
p = sum(1.0 for _ in range(simuN) if simuSS())/simuN
print (float(p))
#--> 0.60
</code>```

Body:
(I have edited the answer)

Large Straight

Let me see if I got the strategy rules right. Does the following simulation code (in Python) model the situation correctly?

import random

keepers = set([2,3,4,5])

def simuLS(doPrint=False):
    kept = set()
    for roundI in range(3):
        ds = [random.randint(1, 6) for _ in range(5-len(kept))]
        for d in ds:
            if d in keepers: kept.add(d)
        if 1 in ds and all(x in kept for x in [2,3,4]):
            kept.add(1)
        if 6 in ds and all(x in kept for x in [3,4,5]):
            kept.add(6)
        gotLS = len(kept)==5
        if doPrint:
            print ("Rolled", ds)
            print ("Kept are ", kept)
            print ("Got LS: ", gotLS)
        if gotLS: return True
    return False

#Let's see how it goes
#simuLS(doPrint=True)

simuN = 10000
p = sum(1.0 for _ in range(simuN) if simuLS())/simuN
print (float(p))

The output is around 0.24.

Are you familiar with Markov chains? We can construct a Markov chain with states all the combinations of the keeper-numbers $\{1,2,3,4,5,6\}$ at most of length 4 and having 1 or 6 only if 2,3,4 and 3,4,5 are present respectively. We also have one absorbing state 'LS' that means we have gotten a large straight. We can calculate the transition probabilities by brute force by considering all possible rolls of the not kept dice (there are 5 - length of the state of them) and seeing what additional keepers we got or did we in fact get a LS. I wrote the following code in SageMath, since it handles matrices and rational numbers without any efforts.

import itertools

keepers = set([2,3,4,5])

states = [tuple(S) for S in Subsets(range(1,7)) if len(S)<5 and
          (1 not in S or all(x in S for x in [2,3,4]))
          and (6 not in S or all(x in S for x in [3,4,5]))]
startState = tuple() #the empty set
targetState = 'LS' #have gotten large straight
states.append(targetState)
stateToInd = {s: i for i,s in enumerate(states)}
startI = stateToInd[startState]
targetI = stateToInd[targetState]
mat = matrix(QQ, len(states))

for i1,s1 in enumerate(states[:-1]): #don't do for the LS-state
    d = 5-len(s1) #how many dice are (re-)thrown
    s1Set = set(s1)
    p = 1/6**d #probability for each outcome
    for xs in itertools.product(*[range(1, 7) for _ in range(d)]):
        s2Set = s1Set.union(set(xs)&keepers)
        if 1 in xs and all(x in s2Set for x in [2,3,4]): s2Set.add(1)
        if 6 in xs and all(x in s2Set for x in [3,4,5]): s2Set.add(6)

        if len(s2Set)==5: #got LS
            mat.add_to_entry(i1, targetI, p)
        else:
            s2 = tuple(sorted(s2Set))
            mat.add_to_entry(i1, stateToInd[s2], p)
mat.add_to_entry(targetI, targetI, 1) #absorbing state

#look at the transition matrix
print (len(states))
print (states)
show(mat)

#print ([sum(row) for row in mat]) #sanity check

sol = (mat**3)[startI][targetI]
print (sol, "=", float(sol))

You can run Sage code online: Sage Cell Server

The output is $\frac{295073737}{1224440064} = 0.24098667274578822$.

I'm not sure the grouping together of states, that I mentioned in my answer before editing, can be done anymore, since now the transitions to the states containing 1 or 6 depend on what we already have. At least not for all same-size-groups. Anyhow, we have a $19\times 19$-matrix, so it is doable.

Small Straight

We can use the previous Markov chain also here:

startRow = (mat**3)[startI]
endsForSmall = [stateToInd[s] for s in states if len(s)>=4] + [targetI]
smallSol = sum(startRow[j] for j in endsForSmall)
print (smallSol, "=", float(smallSol))

Answer: $\frac{2966131613}{4897760256} = 0.605609800799527$

For simulation:

import random

keepers = set([2,3,4,5])

def simuSS(doPrint=False):
    targets = [(1,2,3,4),
               (2,3,4,5),
               (3,4,5,6)]
    kept = set()
    for roundI in range(3):
        ds = [random.randint(1, 6) for _ in range(5-len(kept))]
        for d in ds:
            if d in keepers: kept.add(d)
        if 1 in ds and all(x in kept for x in [2,3,4]):
            kept.add(1)
        if 6 in ds and all(x in kept for x in [3,4,5]):
            kept.add(6)
        gotSS = any(all(x in kept for x in y) for y in targets)
        if doPrint:
            print ("Rolled", ds)
            print ("Kept are ", kept)
            print ("Got SS: ", gotSS)
        if gotSS: return True
    return False

#Let's see how it goes
#simuSS(doPrint=True)

simuN = 10000
p = sum(1.0 for _ in range(simuN) if simuSS())/simuN
print (float(p))
#--> 0.60


问:
A differential inequalites related to Riccati equation
I am studying the paper Kwong, Man Kam, Uniqueness of positive solutions of (\Delta u-u+u^ p=0) in (R^ n), Arch. Ration. Mech. Anal. 105, No. 3, 243-266 (1989). ZBL0676.35032.", and I am stucked in  the proof of Lemma 4. The author stated the following result:

Let $U$ and $V$ are the solutions of the following equalitions
\begin{equation}
U''(x)+f(x)U'(x)+g(x)U(x)=0,~~\forall x\in (a,b)
\end{equation}
\begin{equation}
V''(x)+F(x)V'(x)+G(x)V(x)=0,~~\forall x\in (a,b)
\end{equation}
where $f,F,g,G$ are continuous function. Let $(c,d)$ be a subinterval in which $V(x)\neq 0$ and $U(x)\neq 0$ on $(c,d)$, and in which the comparison conditions $G(x)\geq g(x)$ and $F(x)\geq f(x) $ for all $x\in (c,d)$ holds. If $U'(x)\geq 0$ for all $x\in (c,d)$ and $$\frac{V'(c)}{V(c)}\leq \frac{U'(c)}{U(c)},$$
then we have
$$\frac{V'(x)}{V(x)}\leq \frac{U'(x)}{U(x)}~~~~\forall~ x\in (c,d).$$

In the proof of Lemma 4, the author let $r(x)=\frac{U'(x)}{U(x)}\geq 0$ and $R(x)=\frac{V'(x)}{V(x)}$. Then, $r(x)$ and $R(x)$ satify the following Riccati equations:
\begin{equation}
r'(x)=-(f(x)r(x)+g(x)+r^2(x))\geq -(F(x)r(x)+G(x)+r^2(x)),
\end{equation}
and
\begin{equation}
R'(x)=-(F(x)R(x)+G(x)+R^2(x)),
\end{equation}
where $r(c)\geq R(c)$ according to the condition above, the author then claim that $r(x)\geq R(x)$ for all $x\in (c,d)$ by the theory of differential inequalities. I feel very confuse in this step, can someone give a complete proof of this step? Thank you very much!

答:
This is just applying the following comparison inequality:

Theorem: Suppose $u,v$ are continuous $[a,b]\to\mathbb{R}$ and differentiable on $(a,b]$, $u(a)<v(a)$, $f\colon\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ is continuous and either case (1):
\begin{equation}
u'(x)-f(x,u(x))<v'(x)-f(x,v(x))\quad\forall x\in(a,b]\tag{$\star$}\label{cond:star}
\end{equation}
or case (2): with nonstrict inequality in \eqref{cond:star} and $f$ satisfies the Lipschitz condition on the second variable $\lvert f(x,y_1)-f(x,y_2)\rvert\leq K\lvert y_1-y_2\rvert$ for all $x\in[a,b]$.

Then $u<v$ on $[a,b]$.

Proof: For (1), suppose $u\geq v$ somewhere on $[a,b]$.  Then there exists $c\in(a,b)$ such that $u<v$ on $[a,c)$ and $u(c)=v(c)$.  But then $u'(c)\geq v'(c)$ and $f(c,u(c))=f(c,v(c))$.  This violates \eqref{cond:star}.  So no such $c$ exists.

For (2), let $A=v(a)-u(a)>0$ and let $w(x):=u(x)+\frac{A}2\exp(-2K(x-a))$.  Then $w>u$ on $[a,b]$ and
\begin{align*}
w'(x)-f(x,w(x)) & = u'(x)-AKe^{-2K(x-a)}-f(x,u(x))\\
&\qquad +[f(x,u(x))-f(x,w(x))]\\
&\leq[u'(x)-f(x,u(x))]+K\lvert w(x)-u(x)\rvert\\
&\qquad -AKe^{-2K(x-a)}\\
&\leq v'(x)-f(x,v(x))+\frac{AK}2e^{-2K(x-a)}-AKe^{-2K(x-a)}\\
&<v'(x)-f(x,v(x))
\end{align*}
and we have $w(a)<v(a)$.  So by (1), $w<v$ on $[a,b]$ and hence the result follows. $\square$

Note: we can replace the global Lipschitz condition with a local one since we only need it on $[a,b]\times[\inf_{[a,b]}\min\{u,v,w\},\sup_{[a,b]}\max\{u,v,w\}]$.

So you can apply case (2) with $f(x,y)=-(F(x)y+G(x)+y^2)$ being polynomial in $y$ is locally Lipschitz, and $r'(x)+f(x,r(x))\geq 0=R'(x)+f(x,R(x))$.  If $r(c)=R(c)$, consider $R_\epsilon$ with $R_\epsilon(c)=R(c)-\epsilon$ and satisfying the same DE as $R$.  We have $r>R_\epsilon$ for all $\epsilon>0$ and take limit (since we have local Lipschitz condition we know $R_\epsilon\to R$ pointwise on $[c,d]$).


问:
Inequality Problem in a Circle
so here is a question I came across scanning thru geometry questions:

$A$ and $B$ are points on a circle and $C$ is the midpoint of arc $\overset{\LARGE \frown}{AB}$. P is an arbitrary point IN the circle where $ PA<PB$ prove that $\angle BPC< \angle APC$.

I tried extending $BP$ and $AP$ and calculating the angles using arc relations but I still get stuck and can't find a good way to use the fact that $ PA<PB$ . Any ideas or solutions? Seemed simple at first but I've tried all the ideas I had but can't figure it out.

答:
Extend $CP$ to $Q$.

We have $\angle AQC = \angle BQC$ and $AP < BP$.

We also have $\dfrac {\sin\angle PAQ}{PQ} = \dfrac {\sin \angle AQP}{AP} > \dfrac {\sin \angle BQP}{BP} = \dfrac {\sin\angle PBQ}{PQ}$.

Additionally, note that $\angle CBQ$, and hence $\angle PBQ$, is always acute.

Hence $\angle QAP > \angle QBP$, and thus $\angle APC > \angle BPC$ by considering exterior angles and $\angle AQC = \angle BQC$.

Title: Inequality Problem in a Circle