问:
Set of branch points isn't discrete, but branch points are isolated?
I refer to Chapter II.4 of Rick Miranda - Algebraic curves and Riemann surfaces, which I understand says that the branch points of a nonconstant holomorphic map $F: X \to Y$ between Riemann surfaces $X$ and $Y$, both of which are not necessarily compact, form a discrete subset of range $Y$ of the map $F$. (Not sure if 'connected' is part of the definition of Riemann surfaces in other textbooks, but it is here.)

Question 1: Do I understand correctly, and is this indeed true?

What else I understand:

$A$ is a discrete subspace/subset (I just assume everything here is the usual subspace topology) of $B$ if and only if every singleton subset of $A$ is open in $A$ if and only if every point of $A$ is isolated.

This overflow question (Basic question about branch points on Riemann surfaces) seems to suggest this is not true but is true if $F$ is proper. However, I'm not sure the definitions of Riemann surface are the same (maybe some answers use definitions where Riemann surfaces are not necessarily connected or something).

2.1. I know $F$ is 'discrete' (meaning that its fibres are discrete; here, i refer to a different text): see definition before Lemma 3.1, on p. 7), by Proposition II.3.12, and open, by open mapping theorem, which is Proposition II.3.8, but I'm not sure these help show $F$ is proper or are otherwise helpful. Ostensibly 'discrete' maps don't map discrete subsets to discrete subsets.

This stackexchange question (Are branch points always isolated?) seems to suggest this is true.

I think this is true if $F$ is injective or at least something like locally injective. (Please don't make me type up all the stuff I did for this part.) However based on the question in (3) and based on the '3.2. Branch points' on p. 7 in text by Armin Rainer in (2.1), which I think uses the term 'branch points' to mean the same thing as what Miranda means by 'ramification points' (and thus is different from Miranda's 'branch points'), it seems that no neighbourhood $U$ of a ramification point $p \in U \subseteq X$ of $F$ is such that the restriction $F|_U$ is injective.

Kinda related: If $F$ has a ramification point, then is $F$ necessarily not injective?

This page (rigtriv: Hurwitz’s Theorem) says

Now, the ramification and branch points must form a discrete set

I don't think the page assumes anything like $F$ is proper.

The Armin Rainer text in (2.1) actually seems to assume proper based on the '3.7. Proper holomorphic maps' part, but said part refers to 'Lemma 3.17', so I think proper can be replaced with closed. Maybe $F$ is closed or something.

Kinda related (see Question 2 below): When closed and continuous maps discrete to discrete

S. K. Donaldson - Riemann Surfaces says

If $F$ is proper then the image $\Delta = F(R)$ is discrete in $Y$.

This seems to have all the same assumptions as with Miranda. Ostensibly, either Miranda is wrong or there's some higher level machinery that allows us to omit $F$ proper.

Question 2: To possibly generalise this, what are the minimum requirements on $F, X$ and $Y$ to make $F$ map discrete subsets to discrete subsets? I mean, I'm not sure we really need 'holomorphic' here. I have a feeling this applies perhaps to maps that are just open/closed, continuous and discrete and spaces that are just Hausdorff and locally compact or something.

(again) Kinda related: When closed and continuous maps discrete to discrete

答:
In general for any map $F: X \to Y$ of any topological spaces $X$ and $Y$ with $X$ compact and $Y$ Fréchet/T1 and for any closed discrete subspace $A$ of $X$, we have $F(A)$ discrete.

Proof: Closed discrete subspaces $A$ of compact is finite $\implies$ $A$ is finite $\implies$ $F(A)$ is finite $\implies$ $F(A)$ is discrete because finite subspaces of Fréchet/T1 are discrete. QED

Apply this to the case of $A=Ram(F)$ when $F$ is a non-constant holomorphic map between connected Riemann surfaces with $X$ compact (and thus $F$ is surjective, open, closed and proper and $Y$ is compact) to get $F(A)=Branch(F)$ is discrete.

In particular, this means we do not use that $F$ is proper, closed, open, surjective, non-constant or holomorphic or that $X$ is connected or that $Y$ is connected. We can relax this to $X$ compact (and not necessarily Riemann surface) and $Y$ Fréchet/T1 (and not necessarily Riemann surface, Hausdorff/T2 or compact).

I think finding conditions to make $F$ map discrete to discrete is another story. Actually, what we want is for $F$ to map closed discrete to closed discrete or at least closed discrete to discrete.


问:
About branch points of a holomorphic map
Let $F:X \to Y$ be a holomorphic map between Riemann surfaces. $q \in Y$ is a branch point if it is the image of a ramification point. How to prove that the set of branch points is a discrete subset of $Y$. This is from Rick Miranda's Algebraic curves and Riemann surfaces. Thank you.

答:
You forgot to say $F$ is non-constant. Then again, I guess $Ram(F)$ is not defined for $F$ non-constant.

In general for any map $F: X \to Y$ of any topological spaces $X$ and $Y$ with $X$ compact and $Y$ Fréchet/T1 and for any closed discrete subspace $A$ of $X$, we have $F(A)$ discrete.

Proof: Closed discrete subspaces $A$ of compact is finite $\implies$ $A$ is finite $\implies$ $F(A)$ is finite $\implies$ $F(A)$ is discrete because finite subspaces of Fréchet/T1 are discrete. QED

Apply this to the case of $A=Ram(F)$ when $F$ is a non-constant holomorphic map between connected Riemann surfaces with $X$ compact (and thus $F$ is surjective, open, closed and proper and $Y$ is compact) to get $F(A)=Branch(F)$ is discrete.

In particular, this means we do not use that $F$ is proper, closed, open, surjective, non-constant or holomorphic or that $X$ is connected or that $Y$ is connected. We can relax this to $X$ compact (and not necessarily Riemann surface) and $Y$ Fréchet/T1 (and not necessarily Riemann surface, Hausdorff/T2 or compact).

ContentLicense: CC BY-SA 4.0


问:
Find the x and y coordinates after traveling some distance around a circle
Just a disclosure I am not a huuuuge math person. I am trying to find points on circles if I was to travel some distance around those circles.

For example, if I have this circle:

$$\left(x-0\right)^{2}+\left(y-0\right)^{2}=2^{2}$$

This circle's center is at (0,0) and the "top" of the circle is at (0,2). Let's say I now wanted to travel from the top, in a clockwise direction around the circle. Let's say I wanted to only go 1/5th of the circumference, what would be the x and y coordinates for where I would stop?

In practice, any circle I use would always be centered at (0,0) but could be any size and the distance I would travel would also vary (sometimes 1/5th, sometimes 1/2, etc).

As mentioned previously, I am not a big math person, but I did try googling how to do this and was unsuccessful. The concept of arc lengths looked promising, but I couldn't understand how the central angle works in my case.

答:
The central angle is this case is $\theta=2\pi/5$. Whatever the fraction of the circumference you want to travel, in angle is the same fraction out of $2\pi$.

If your circle has radius $r$, $r=2$ in this case, and you measure the angle in the clockwise direction, from the vertical line, the position on the circle is given by $$x=r\sin\theta\\y=r\cos\theta$$
If the center of the circle is not at $(0,0)$, then $$x=r\sin\theta+x_c\\y=r\cos\theta+y_c$$


问:
Show that $\sum_{i=1}^k|a_ib_i|c_i\leq \left(\sum_{i=1}^k|a_i|^pc_i\right)^{1/p}\left(\sum_{i=1}^k|b_i|^qc_i\right)^{1/q}$, where $c_1,...c_k >0$
I know that this immediately follows from Holder's inequality if all the $c_i=1$, and I take $\mu$ as the lebesgue measure and define $f(x)=\sum a_i I_{[i-1,i)}(x)$, $g(x)=\sum b_i I_{[i-1,i)}(x)$. But I'm not sure how to handle the $c_i$'s.

Is there perhaps some way to show that $$\int|fg|h\leq\left(\int|f|^ph\right)^{1/p}\left(\int|g|^qh\right)^{1/q}$$

if $h>0$, and $f\in L^p,g\in L^q$?

Maybe some well known identity/inequality that would help push the initial problem further, or perhaps some intuition that would help me think of it problem more clearly?

Thanks in advance.

答: This is just the usual Holder's inequality applied to $(a_ic^{1/p})$ and $(b_ic^{1/q})$.


问:
Why do I get a negative value for this integral?
I am trying to compute the integral: $$\int_{4}^{5} \frac{dx}{\sqrt{x^{2}-16}}$$
The question is related to hyperbolic functions, so I let $x = 4\cosh(u)$ therefore the integral becomes: $$-\int_{0}^{\ln(2)}\frac{4\sinh(u)}{\sqrt{16-16\cosh^{2}(u)}}du = -\int_{0}^{\ln(2)}1du = -\ln(2)$$

The answer is $\ln(2)$ so if someone could point out where I went wrong that would be great, thanks

ContentLicense: CC BY-SA 3.0

Body:
use this formula
$$\int \frac{dx}{\sqrt{x^{2}-a^2}}=\log|x+\sqrt{x^2-a^2}|+c$$
$$\int_{4}^{5} \frac{dx}{\sqrt{x^{2}-16}}$$
$$\int_{4}^{5} \frac{dx}{\sqrt{x^{2}-4^2}}$$
$$(\log|x+\sqrt{x^2-16}|)_4 ^5$$
$$\log|8|-\log|4|$$
$$\log2$$


问:
Prove that parallels of sides of a triangle drawn from the centroid, divides each sides in three parts.
I have been trying for hours, but I don't know if I should approach it through areas or Euclid's Fifth Postulate.

答: Consider side a. Centroid is where all three medians of triangle cross. So  median from A  on opposite side  a is also the median from the centroid D, which is the vertex of small triangle DIE its base IE on that side a of triangle, that is we have subtracted a certain measure (IJ=JE)from half of the side a(BJ=JC) which results remain portions JC and BI must be equal.Now draw a line from E  parallel with side c and mark its intersection with side b as K. KD is equal to the base of small triangle IE. it is also equal to segment DH  (because the median of triangle bisect all lines parallel with base a). Also HD=BI. Hence all three segments of  side a , i.e. BI, IE and EC are equal. This is also true for other two sides.


问:
How to manipulate $a^2b + a^2c + ab^2 + b^2c + ac^2 + bc^2$ into a form for substituting polynomial relations of roots
The polynomial is $2x^3 - x^2 + 14x - 30 = 0$. Through relations of roots (eg.$a + b + c = \frac{-b}{a}$ etc.), I coudl substitute in these values to evalute the value of this express. However, I'm gaving troubel manipulating it into a nice form. I tried taking out $a^2, b^2, c^2$, and $a,b,c$ - even $abc$ (then $a^{-1}, b^{-1}, c^{-1}$ accordingly. None of these ways work. What am I missing?

答:
using  $$a+b+c=-1/2$$ we have to find  $$a^2(b+c)+b^2(c+a)+c^2(a+b)=\dfrac{a^2+b^2+c^2}{2}-(a^3+b^3+c^3)$$
Can you end it now?

Hint:use vietas formula along with  $$\displaystyle a^3+b^3+c^3=(a+b+c)({(a+b+c)}^2-3ab-3bc-3ca)+3abc$$,


问:
when $\lim_{\epsilon\to 0}\int_{B_\epsilon / \{0\}} f(x)dx = 0$ holds
Let $f$ be a measurable function from $\mathbb{R}^n/\{0\}\to \mathbb{R}$,under which condition the integral below near the origin goes to zero?

$$\lim_{\epsilon\to 0}\int_{B_\epsilon / \{0\}} f(x)dx  = 0$$

I try to show $\frac{1}{|x|^s}$ is not integrable near to origin (inside $B_1(0)$) when $s\ge n$ , and here uses the fact to split the ball into two parts and the small $\epsilon$ part is negligible.  For general $f(x)$ under which condition the  integral above is negligible?

I try to gives an ansewer using Lebesgue differentiation theorem ,which state for Locally integrable function,with $x$ a.e. we have: $$\lim _{{B\rightarrow x}}{\frac  {1}{|B|}}\int _{{B}}f\,{\mathrm  {d}}\lambda $$

the question then is why 0 it one of the desired point?

答: If $f$ is integrable on $B_r \setminus \{0\}$ for some $r>0$ then this is true. Take any sequence $\epsilon_n$ decreasing to $0$ and apply Monotone Convergence Theorem to the sequence $|f|\chi_{B_{\epsilon_k\setminus B_{\epsilon_{k+n}}}}$. [This sequence increases to $|f|\chi_{B_{\epsilon_k}}$].


问:
if $f(x)\geq0$ for all $x$, then $f+f'+f''+f'''+ \cdots + f^n \geq0$
Suppose $f$ is a polynomial of degree $n$, and $f\geq0$ for all $x$, prove that $f+f'+f''+f'''+\cdots +f^n\geq0$. (so just functions of the form $ax^n+bx^{n-1}+.....c$. We are not going to include irrational functions, exponential, logs, etc.)

So for this problem, I looked at a few functions and graphed them to see why this statement would be true. The first thing I found out is that the polynomial's highest degree must be even, otherwise there would be some $x$ where $f(x)$ is negative. Now this implies the odd numbered derivatives will have an odd number for the term with the highest degree.

Having figured that out, I was wondering whether the sum of every successive would change the function so that $f+f'$ has negative values, and $f+f'+f''$ would have positive values and so on, and this does turn out to be true for the ones I've tried. If this is true in general, that would mean that $f+f'$ has some negative values, $f+f'+f''$ has not negative values, and $f+f'+f''+f'''+....f^{n-1}$ has negative values until we get to $f+f'+f''+f'''+....f^{n-1}+f^n$ which has no negative values.

Ok but having figured this out, I'm not sure as to how to proceed with the proofs because the above assumptions I've made are not lemmas or proofs that I was given.

答:
It is known that a polynomial that is nonnegative on all reals must be the sum of squares of two polynomials. As a result, it suffices to prove the claim when the given polynomial $f$ can be written as $g^2$ for some polynomial $g$. We may show that
$$\frac{d}{d^k}g^2=\sum_{i=0}^k\binom ki g^{(i)}g^{(k-i)}$$
by induction on $k$, and so, if $g$ has degree $m=n/2$, the sum we desire is
$$\sum_{k=0}^{2m}\sum_{i=0}^k\binom kig^{(i)}g^{(k-i)}.$$
As $g^{(i)}=0$ for $i>m$, this sum can be rewritten as
$$\sum_{j_1=0}^m\sum_{j_2=0}^m\binom{j_1+j_2}{j_1}g^{(j_1)}g^{(j_2)}.$$
For a given $x$, write $\frac{1}{j!}g^{(j)}(x)=x_j$, so $x_0=x$. We will show that, for any reals $x_0,\dots,x_m$,
$$\sum_{j_1=0}^m\sum_{j_2=0}^m(j_1+j_2)!x_{j_1}x_{j_2}\geq 0,$$
which will finish the proof. We note that we have
$$k!=\int_0^\infty e^{-t}t^kdt,$$
so our sum is
$$\int_0^\infty \sum_{j_1=0}^m\sum_{j_2=0}^mx_{j_1}x_{j_2}t^{j_1+j_2}e^{-t}dt=\int_0^\infty e^{-t}\left(\sum_{j=0}^m x_jt^j\right)^2dt;$$
for each $t$, the integrand is nonnegative, and so the whole integral is nonnegative, finishing the proof.


问:
Quick Finding Relative Max Question
So, I was studying for a test and came across the following question:

"Let $$g(x)$$ be the function by $g(x)= \int_0^x (\frac{-3}{4}+t+cos(\frac{pi}{4}*t^2+t))$ for $0<x<4$. At what value of $x$ does $g$ attain a relative maximum?"

MathJax doesn't seem to wanna work on my computer, so here's a screenshot of the prompt as well:

I'm a bit confused on how I'd go about tackling something like this, would it involve converting the integral into terms of $x$? Any help would be appreciated!

答:
The fundamental theorem of calculus "says" that
$$g(x)=\int_0^x f(t) \, dt \implies g'(x)=f(x)$$ So, you first need to find the zero('s) of
$$g'(x)=-\frac{3}{4}+x+\cos\left(\frac{\pi}{4}x^2+x\right)$$ This is a transcendental equation which, by definition, will not show analytical solutions. As a consequence, some numerical methods will be required and then a reasonable starting guess.

If you plot the function, there is a root which is very close to $x=1.0$ and another close to $x=1.6$

Now, you are ready for Newton method which will generate the following iterates
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 1.000000 \\
 1 & 1.024501 \\
 2 & 1.024508
\end{array}
\right)$$
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 1.600000 \\
 1 & 1.616232 \\
 2 & 1.615677
\end{array}
\right)$$
One of them will be a maximum and the other will be a minimum.

Title: Quick Finding Relative Max Question


问:
Rotation equivalent to what two reflections
What two reflections are equivalent with the rotation about $(0,-1)$ by $180^\circ$?

Can I say one of the reflections is about the $y$-axis? If so, what would the second reflection be? Would it be $y=-1$?

答:
Consider a point given by $re^{i\theta}$ to be rotated around the origin (first translate the point by rotational origin minus actual origin, then rotate, and then translate the new point back).

Rotation is given by $re^{i(\pi+\theta})$.

Reflection in the $x$-axis is $re^{-i\theta}$.

Reflection in the $y$-axis is $re^{i(\pi-\theta)}$.

Putting the two reflections together gives $re^{i(-\pi+\theta)}$

which is the same as the rotation because $e^{i\pi}=e^{-i\pi}=-1$.

So your two reflections are about the $x$-axis and $y$-axis, translated to the point $(0,-1)$, i.e. the lines of reflection are $x=0, y=-1$.


问:
Subtracting two infinities
I am Curious if the following is mathematically correct:

Let $a$ be the infinite set of all nonnegative integers $0,1,2,3...$.

I take from $a$ some of its elements, say integers $10$, $11$, and $12$ only.

So now we have a new set $a'$ that is the infinite set of all nonnegative integers $0,1,2,3...$ except for $10$, $11$, and $12$.

If I subtract: $a-a'$ the result is a set comprised of $10$, $11$, $12$ only. Is this correct? Can one subtract infinities like this?

If yes, does this mean that $a \gt a'$(despite that both are infinite)?

答:
Subtraction between infinite cardinals cannot be well-defined. This is a good example why.

We know that $|\Bbb N|$ and $|\Bbb Z|$ are both of the same cardinality, but so is $|\Bbb{Z\setminus N}|$. On the other hand, $|\Bbb N\setminus\{k\in\Bbb N\mid k>2\}|=3$ (zero is a natural number here).

So we have $\aleph_0-\aleph_0=\aleph_0$ but at the same time $\aleph_0-\aleph_0=3$ and we can easily engineer this to be any other result between $0$ and $\aleph_0$ as well. This is indeterminate much like $\infty-\infty$ is an indeterminate form in calculus, although $\infty$ and $\aleph_0$ are not directly related (these are two different, and mostly incompatible, notions of infinity).

The reason is that for infinite sets being a proper subset does not imply having strictly smaller cardinality like it does with finite sets. Our intuition from finite mathematics usually does not apply to infinite objects.

On the other hand, subtracting two concrete sets is perfectly doable and legitimate mathematically. But we cannot conclude anything on the cardinality of the result, unless we know what are the specific sets (or at least more than just "two sets", which is the most general notion).

Read more about this:

Why do the rationals, integers and naturals all have the same cardinality?

Is there a way to define the "size" of an infinite set that takes into account "intuitive" differences between sets?

Why the principle of counting does not match with our common sense

Cardinality of subtraction of sets

Cardinal number subtraction (See my answer, as it is related to the comment below by user14111.)

Title: Subtracting two infinities


问:
Show $\lim_{n \to \infty} \sup_{x \in [0,1]} (e^{-nx} - (1-x)^n) = 0$?
I am trying to show that
$$\lim_{n \to \infty} \sup_{x \in [0,1]} (e^{-nx} - (1-x)^n)=0.$$
I have tried expanding $e^{-nx}$ with its Taylor series and $(1-x)^n$ with Newton's binomial theorem. The resulting Taylor series is $\sum_{k=2}^{\infty}  (-1)^k \left(\frac{n^k}{k!} - {n \choose k}\right)x^k.$

I also tried taking the first derivative with respect to $x$ to find the point in $(0,1)$ where $e^{-nx} - (1-x)^n$ is maximized (in $x$ as a function of $n$), but I was only able to find the hard-to-solve $\frac{n}{1-n} = \frac{\ln(1-x)}{x}.$

答:
Use the result:
$$\color{red}{0\le e^{-nx}-(1-x)^n\le nx^2 e^{-nx} \qquad (0\le x\le 1),}$$
which can be refered in key inequality.
When $n>2$, $f_n(x)=nx^2 e^{-nx}$ has maximum at $\frac{2}{n}\in(0,1)$.
So
$$\sup_{x\in[0,1]}\left(e^{-nx}-(1-x)^n\right)\leq\frac{4}{n}e^{-2}\to 0.$$


答:
Identically we have
$$
r^2s+s^2r+s^2t+t^2s+t^2r+r^2t
=
(r+s+t)(rs+st+tr)-3rst
$$
and if $r,s,t$ are the roots of the polyomial $ax^3+bx^2+cx+d$, then by Vieta's formulas we have
$$
\left\lbrace
\begin{align*}
&r+s+t=-\frac{b}{a}\\[4pt]
&rs+st+tr=\frac{c}{a}\\[4pt]
&rst=-\frac{d}{a}\\[4pt]
\end{align*}
\right.
$$
so you have all the ingredients to finish.

Title: How to manipulate $a^2b + a^2c + ab^2 + b^2c + ac^2 + bc^2$ into a form for substituting polynomial relations of roots

Body: The polynomial is $2x^3 - x^2 + 14x - 30 = 0$. Through relations of roots (eg.$a + b + c = \frac{-b}{a}$ etc.), I coudl substitute in these values to evalute the value of this express. However, I'm gaving troubel manipulating it into a nice form. I tried taking out $a^2, b^2, c^2$, and $a,b,c$ - even $abc$ (then $a^{-1}, b^{-1}, c^{-1}$ accordingly. None of these ways work. What am I missing?


问:
Help with Real Analysis
Is the sequence ($x_n$)= $a^n(n^2)$ , $0<a<1$ convergent?

Proof that the sequence is unbounded and hence divergent:

Assume it is bounded, then there exists $M>0$, such that $M\geq|x_n|$ for all $n \in \Bbb N$. We note that since $a>0$ and $n>0$, all the terms of the sequence are positive. So $M\geq x_n$. Since $a^n$ is positive, then its reciprocal must also be positive. Hence by multiplying both sides by $1/a^n$, we obtain $M/a^n\geq n^2$ for all $n \in \Bbb N$.

By the Archimedean property, there exists $n_1 \geq M/a^n$, since $M/a^n$ is positive. Let $n_2 = n_1 +1$. Then $n_2$ is a natural number, and $n_2 > M/a^n$, so
$$
n_2^2 > n_2  M/a^n > n_1 M/ a^n \geq M/ a^n
$$
so $n_2^2 > M/ a^n$. But then $M/a^n \geq n_2^2$ since $n_2 \in \Bbb N$. Since $n_2^2$ cannot be more than $M/a^n$ and less than or equal to $M/a^n$ at the same time, we have derived a contradiction and hence the sequence is unbounded.

Proof that the sequence converges:
We employ the Cauchy ratio test.

Take the ratio $x_{n+1} / x_n$. We obtain
$$
(a^{n+1} (n+1)^2)/a^n n^2 = a(n+1)^2 / n^2 = a(n^2/n^2 +2n/n^2 + 1/n^2) = a(1+2/n+ 1/n^2).$$
Since the sequences $1/n^2$ and $1/n$ both converge to $0$, the sequence converges to $a(1)+ a(2)(0)+a(0)= a$. Since $a<1$, then the sequence converges.

Could someone help me, which of these proofs is false? Am I making any mistakes, any areas I can improve on? Thank you!

答:
The first proof is wrong. When it says "but then $M/a^n\ge n_2^2$ since $n_2\in\mathbb N$" it's taking the true proposition
$$\forall n\in\mathbb N:M/a^n\ge n^2$$
and then substituting $n_2$ for $n$ only in one of the two places, which is not allowed. The sequence $(x_n)$ is indeed convergent to $0$.

Title: Help with Real Analysis


问:
What do we call this property when we go from $a<b$ to $a ≤b$ and from $a=a$ to $a≤a$?
The title is self-explanatory.

Let $a=1$, $b=2$

Is there a property we can state when we go from  $1<2$ to $1 ≤2$?

Is there a property we can state when we go from $2=2$ to $2≤2$?

Can I say something like universal generalization or something?

This question stems from the fact that you can't go from $a≤b$ to $a<b$ but can go from $a<b$ to $a≤b$ so I assumed there is a name

答:
You could say something along the lines of one binary relation being "coarser" (weaker?) than the other. Thinking of $<, \leq, =$ as a subsets of $\mathbb{Z}^2$ (i.e., all those pairs where those relations are true), what you're stating is $=\subseteq \leq, <\subseteq \leq$. But now you can ask the same question about $\subset, \subseteq, =$ relations between subsets.

Also it comes from the definition of $\leq $being $<$ or $=$ (this is the same as the subset notion above).


问:
Are solutions to a second order differential equation unique in this case?
In my math book, a general solution to the differential equation $y'' + y = f(t)$ where $f(t)$ is continuous on $(-\infty, \infty)$ is given by $y = A\cos t + B\sin t + \int_0^t f(s)\sin(t-s)ds$.

Taking as an example $y'' + y = t$, I get the following result when I use undetermined coefficients: $y = A\cos t + B\sin t + t$

When I use the formula given above, I get $y = A\cos t + B\sin t + t - \sin t$

Both solutions also work when you plug them back into $y'' + y = t$, so is this to say that both solutions are valid and the only difference would be in the values of the constants?

答:
This will help if you have studied linear algebra before. Usually in that class you work with finite-dimensional vector spaces in one guise or another: $\mathbf{R}^n$ with standard addition and scalar multiplication. You have seen linear transformations, and these have kernels.  Usually, you think about them in terms of matrices, but forget about those.

Now, let's consider the $\mathrm{C}^\infty$ functions; functions defined on all of $\mathbf{R}$ that are infinitely differentiable, together with addition and multiplication by real constants. These functions form a vector space. Now, consider the operator $\mathrm L$ defined by $\mathrm Lu = u'' + u.$  This is a map from $\mathrm{C}^\infty \rightarrow \mathrm{C}^\infty.$ It has a kernel: the linear combinations of the cosine and the sine functions. So, if $\mathrm Lu = f$ and $\mathrm Lv = 0$, then $\mathrm L(u+v) = f$ too, just as $\mathbf{Au} = \mathbf{x}$ and $\mathbf{Av} = \mathbf0$ imply $\mathbf{A(u+v)} = \mathbf x$.  One solution becomes a whole set of solutions.


问:
$\lim_{x\to6^-}\big[\frac{\sqrt{36-x^2}}{x-6}\big]$ using L'Hopital's rule?
Disclaimer: this is a homework problem, I at least need a bit of direction, I can't seem to get anywhere, I must be overlooking something.

$\lim_{x\to6^-}\biggl[\frac{\sqrt{36-x^2}}{x-6}\biggr]$
I've tried applying L'Hopital's rule since we end up with an indeterminate form if we evaluate at $x=6$

I took the first, second, and third derivative but the denominator will always force division by 0. I'm not sure what to do. I'd love to put in the first and second derivatives into this question but, I'm a bit sloppy with the 'Ol Mathjax!

EDIT: Found the big wet mistake: For L'Hopital's rule, I have been taking the derivative of the ENTIRE function, instead of the derivative of the numerator, divided by the derivative of the denominator! Don't do that folks!

答:
The answer is -infinity. It's not in 0/0form neither in infinity/infinite form.

To get that form divide both numerator and denominator by sqrt(6-x) and u will get the desirable answer


答: I think you can do it without L'Hospital Rule. Try to multiply with $\frac{\sqrt{36-x^2}}{\sqrt{36-x^2}}$. You will then get $\frac{36-x^2}{(x-6)(\sqrt{36-x^2})}$. Then you can decompose $36-x^2$ to $(6-x)(6+x)$ and eliminate the $x-6$. You can then apply the limit to get $-\infty$.

Title: $\lim_{x\to6^-}\big[\frac{\sqrt{36-x^2}}{x-6}\big]$ using L'Hopital's rule?

Body:
Disclaimer: this is a homework problem, I at least need a bit of direction, I can't seem to get anywhere, I must be overlooking something.

$\lim_{x\to6^-}\biggl[\frac{\sqrt{36-x^2}}{x-6}\biggr]$
I've tried applying L'Hopital's rule since we end up with an indeterminate form if we evaluate at $x=6$

I took the first, second, and third derivative but the denominator will always force division by 0. I'm not sure what to do. I'd love to put in the first and second derivatives into this question but, I'm a bit sloppy with the 'Ol Mathjax!

EDIT: Found the big wet mistake: For L'Hopital's rule, I have been taking the derivative of the ENTIRE function, instead of the derivative of the numerator, divided by the derivative of the denominator! Don't do that folks!


问:
How many ways of arranging $7$ examinees are possible?

In an examination hall, there are $20$ chairs in a row. In how many ways the seats of $7$ examinees can be arranged such that no two examinees can occupy two consecutive chairs?

I tried doing it by using the binary string method adn got the answer to be $\dfrac{14!}{7!}$, but the correct answer is the book is $8^7$. i just needed to verify the answer. If the correct answer is $8^7$, then please explain the solution.

My Approach:

Let each arrangement of states of chairs (empty or non empty) correspond to a binary string of length $20$.

For example, If chair no. $1,5,7,9,13,15,19$ are non empty while all other chairs are empty, then this must correspond to string $10001010100010100010$

Now for 7 examinees to sit, there would be $7$ non empty chairs and $13$ empty chairs. This means that the string will consist of $7$ ones and $13$ zeroes. Since no examinees must sit in consecutive chairs, no two $1s$ must appear consecutively in the string. Thus the string can be like this:
$$\_0\_0\_0\_0\_0\_0\_0\_0\_0\_0\_0\_0\_0\_$$
There are $14$ "gaps" where one can appear. We need to choose $7$ gaps out of $14$ for $1$ to appear at whereas the remaining gaps will be removed from the final string. This can happen in $\binom{14}{7}$ ways.

This gives us the arrangement of chairs. Now in any given arrangement, $7$ examinees can be permuted in $7!$ ways.

$\therefore$ The total number of ways of seating $7$ examinees in $20$ chairs such that no two examinees sit on consecutive chairs are $\binom{14}{7}\cdot 7!=\dfrac{14!}{7!}$

THANKS

答: I did a brute force calculation, and it confirms that you are right.  I counted all the numbers from $1$ to $2^{20}$, whose binary representations have $7$ one-bits, no two of which are consecutive.  There are $3432$ such.  This gives the number of ways to choose the seats.  Then I multiplied by $7!$ to count the ways to seat the students.  The product turned out to be $$\frac{14!}{7!}$$ in agreement with your answer.


问:
Prove that $\overline{A} - A \subset A'$
Prove that $\overline{A}-A \subset A'$

I did the following:

Let $x \in \overline{A}$ $\wedge x \not \in A' $

$\forall r_x>0$ $\{B(x,r_x) \cap \overline{A} \not = \emptyset\}$

$\implies \forall r_x>0$ $\{B(x,r_x) - \{x\}\cap \overline{A} \not = \emptyset\}$

$\implies \forall r_x>0$ $\{B(x,r_x) - \{x\}\cap A' \not = \emptyset\}$ Because $A' \subset \overline{A}$

$\implies x \in A'$

$\therefore \overline{A}-A \subset A'$

答:
$\overline A= A\cup A^{'}$.

Since $x\in \overline A$ but $x\notin A$, so $x\in A^{'}$.

Thus $\overline A\setminus A\subset A^{'}$.


问:
Is central X% of distribution of average is $E(X)\pm qnorm(0.X)\cdot \sigma$ in exponential distribution?
Question: It is claimed that the expected of time some computer part may work before requiring a reboot is 26 days. In order to examine this claim 80 identical parts are set to work. Assume that the distribution of the length of time the part can work (in days) is Exponential. The central region that contains 70% of the distribution of the average is of the form $E(X)\pm c$, where $E(X)$ is the expectation of the sample average. The value of c is (You are asked to apply the Normal approximate to the distribution of the average of the 80 parts that are examined. The answer may be rounded up to 3 decimal places of the actual value.):

Its answer: 3.012796

My thought

70% of the distribution of the average is: $\mu \pm$ qnorm(0.75) $\cdot \sigma$

$\mu = E(X) = 26$

$\sigma = \frac {\sqrt{E(X)^2}} {\sqrt{80}} = 2.90688837075$

With given $E(X) \pm c$ and since $E(X)$ is $\mu$,

$c=$ qnorm(0.75) $\cdot \sigma$

$c=0.6744898 \cdot 2.90688837075 =  1.960666$

How come 3.012796 is the correct answer?

答:
Your error is the same you did here.

The central region that contains 70% of the distribution is the one that excludes 15% in the left tail and another 15% in the right one.

Thus the quantile you are interested in is qnorm(0.7+0.15)=qnorm(0.85)1.036433

Concluding

$$1.036433\times 2.906888=3.012796$$

as claimed


问:
What do I need to read Philip Griffths
I had taken course in complex analysis and self-studied differential manifolds and basic topology from Loring Tu Introduction to Manifold. What else do I need (minimally) to prepare myself to read the Principal of Algebraic Geometry by Philip Griffiths? (Do I need Huybrechts Complex Geometry or Rick Miranda's Algebraic curve and Riemann surface?)</p>

答:
Also too long for a comment. So I too will put it as an answer.

I was told in a comment to one of my deleted questions that Griffiths' Introduction to Algebraic Curves is a prerequisite to griffiths harris principles of algebraic geometry.

I used Rick Miranda's Algebraic curve and Riemann surface (chapters I - VIII) as an alternative to Griffiths' Introduction to Algebraic Curves.

It appears that griffiths harris principles of algebraic geometry somewhat overlaps with both Rick Miranda's Algebraic curve and Riemann surface (chapters IX onwards) and Huybrechts Complex Geometry.

P.S. I actually have to study Griffiths' Introduction to Algebraic Curves, Huybrechts Complex Geometry and chapters 0 and 1 of griffiths harris principles of algebraic geometry, and I'm hoping it will suffice to study Rick Miranda's Algebraic curve and Riemann surface (chapters I - VIII), chapters 0 and 1 of griffiths harris principles of algebraic geometry and all the complexification I studied in external sources (see my questions eg this one or this one)

Title: What do I need to read Philip Griffths


问:
Finding constants so that inequality holds
Find $b > 0$ such that $b^x \geqslant x + 1$ holds for all $x$ values of $x$.

This is apparently related to differentiation but since you can’t differentiate inequalities, I don’t know how this works.

答:
Well, on how to use derivatives here, note we want to show $f(x) = b^x-x-1\geqslant 0$.  So we want the minimum of $f$ to be non-negative.

Extrema are when $f'=0$.  As $f(0)=0$, you want an extremum to be at $x=0$, and that too it needs to be a minimum, so we need $f'(0) = 0, \; f''(0)>0$.
$f'(x) = \log b \cdot b^x -1 \implies  f'(0) = 0$ iff $b=e$.  Now $f''(0) = e^0>0$ and so we have a unique solution.


问:
A question with infinity
I am a sophomore in high school and my math teacher did a very short lesson on infinity, here's how it went:  (Try to solve each part yourself the first two are easy)

Part 1

You have an inf. number of boxes each labeled like so

```<code>[1][2][3][4].....
</code>```

and that there is a person in each one of these boxes.  But you have another set of five boxes like so

```<code>[1][2][3][4][5]
</code>```

You want to only have one set of boxes what do you do?  You must be able to tell each person what box they should goto next.

Spoiler

Each person is told to move down five boxes.  Or n+5 is you new box, and the five people from the other boxes are able to move into the first five boxes or just n+0.

Part 2

Well you actually have two of these inf. box sets, but you really only want one.  What now? (same rules apply)

Each person from set one is told to go to box 2n and each person from set two is told to box 2n-1

Part 3

Now you have an inf. number of these inf. box sets arranged like so

```<code>1: [1][2][3][4][5]...
2: [1][2][3][4][5]...
3: [1][2][3][4][5]...
4: [1][2][3][4][5]...
5: [1][2][3][4][5]...
</code>```

and you want to fit them all into one inf. box set, how do you do it.  She refused to tell us the answer and no one in my class figured it out, but I think I just did (correct me if I'm wrong, I'm also looking for alternate solutions).

How I went about solving this problem

Step 1

I assumed that there had to be some sort of pattern so I tried a few

Step 2

I picked the following patter.

1: [1][3][6][10][15]...

2: [2][5][9][14][]...

3: [4][8][13][][]...

4: [7][12][][][]...

5: [11][][][][]...

The numbers in the box represent the new box of each person

Step 3

I started to create a formula for the people to find there new room based of there current room number and box set number.  I found the following sequence in for people in the first row, 1 3 6 10 15.  Or +2 +3 +4 +5.  Which told me that it was quadratic so the equation had to be ax^2 + bx + c = new room number

Step 4

I found three equation by plugging in room numbers and their correct new room numbers to find these three equations:

(1^2)a + 1b + c = 1 or a + b + c = 1

(2^2)a + 2b + c = 3 or 4a + 2b + c = 3

(3^2)a + 3b + c = 6 or 9a + 3b + c = 6

Step 5

After solving this system I got the following equation 1/2x^2 + 1/2x + 0 = new room number where x = current room number.  This of course only tell the people in the boxes on the bottom where to go. Now I needed to find how to place the rest of the people.  The first thing I noticed was that people on the same diagonal as seen bellow were very much related to each other when you look at the rooms they will end up in.

[x-4][][][][]

[][x-3][][][]

[][][x-2][][]

[][][][x-1][]

[][][][][x]

The solution

Working this knowledge into my equation was fairly easy (I was able to do it while going for a run so I won't explain how I got my equation).  The equation is:

 (1/2(r+c-1)^2 + 1/2(r+c-1) + 0)-r+1 = new room number | r = row | c = column

Now my question to all of you is

Is my equation correct?

Are there any alternate solutions to this problem?

How could you fit a 3D array of boxes (inf boxes of people in all direction) into a 1D line of boxes?

Can you find a way to fit a n dimensional array of boxes into a 1D line of boxes?

For any of you into JAVA here's a a short snippet of code I made to prove my answer

```<code>    package main;

public class Main {

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        int f = 2;
        int r = 1;
        int c = 1;
        while(true){
            double x = (.5*(r+c-1)*(r+c-1) + .5*(r+c-1) + 0)-r+1;
            System.out.print(x +", ");
            c++;
            r--;
            if(r < 1){
                r = f;
                f++;
                c = 1;
            }
        }
    }
}
</code>```

答:
You have a great mathematical future!! Do you have a geometric way of coming up with your counting principle, rather than algebraic?

The answers to 3. and 4. are, indeed, YES. Get to work!!

Then: 5. What about an infinite sequence of such arrays (i.e., one for each positive integer)?

Title: A question with infinity


问:
Best way to spend three months before entering masters program: Review fundamentals, or pre-study topics in the program?
Recently I got an offer letter for a master program (with courses like algebraic topology, algebraic geometry (first chapter of Hartshorne), differential geometry, representation theory, etc... ).

Now I have three months before I go to there and start my program. My question is, should I use these three months to review the fundamental subjects in mathematics, i.e Abstract Algebra, Topology, Analysis, and Linear Algebra or should I pre-study those master programs before I go?

To be honest, I am not firm enough in those fundamental subjects as I didn't do exercises at all when I took those courses (but I managed to score well, mainly because my school only gives questions in the lecture notes). Also, I have almost forgotten some results in linear algebra (like the change of basis etc.)

At first, I would prefer to revise (by doing exercise) first before going to some advanced courses. But then a friend of mine suggested me to study further courses, where he says if I am stuck somewhere, then I go back for related topics. (For example, if I need a change of basis in the study of manifolds, then I go back to revise it.)

Which one would be better for my case, given the time constraint and the mathematical background?

答:
Pre-study and then that will give you the idea of what to look out for in review.

Mostly, you never really get the basics from studying the basics over and over. I recall reading that you never really understand topic X until you get to topic X+1. I doubt a calculus student will really understand limits from studying the calculus book again and again. Said student will actually understand limits better from starting to study elementary real analysis, where the student learns the concepts of supremum and infimum.


问:
Mean & variance of normal distribution
Let $Y = \ln(X)$, where $Y$ follows a normal distribution with mean $μ$ and variance $σ^2$. What is the mean and variance of $X$?

I know that if $Z$ follows a normal distribution with mean $μ$ and variance $σ^2$
, then $M_z(t) = e^{μt+ \tfrac{1}{2}σ^2 t^2}$

答:
If you know that a normal distribution $Y \sim \mathcal{N}(\mu,\sigma^2)$ has a moment generating function of $M(t) = \mathbb{E}[e^{tY}] = e^{\mu t + \sigma^2t^2/2}$, then you can use that to calculate the mean and variance of $X = e^Y$ as follows:

$$\mathbb{E}[X] = \mathbb{E}[e^Y] = M(?) \cdots$$

$$\text{Var}[X] = \mathbb{E}[X^2] - \mathbb{E}[X]^2 = \mathbb{E}[e^{2Y}] - \mathbb{E}[e^Y]^2 = M(?) - M(?)^2 = \cdots$$

I'll let you fill in the $?$'s and continue the calculations from there.

Title: Mean & variance of normal distribution


问:
Structural differences between $2 \mathbb{Z}$, $3 \mathbb{Z}$ as rings.
Is usual to find in abstract algebra books this exercise:

Show that $2 \mathbb{Z}$ and $3 \mathbb{Z}$ are isomorphic as groups (with usual sum) but they aren't as rings (with usual sum and addition).

Its solution is based in consider, for example, the images f(4)=f(2+2)  and f(4)=f(2*2) and conclude the "non existence" of such an isomorphism.

But, my question is, what are the structural differences (ring properties) that avoid this equivalence as rings?

Thanks in advance.

答:
For such a homomorphism: $f(2)+f(2) = f(2+2) = f(4) = f(2*2) = f(2)*f(2)$.

Here $f(2) = 3*m$ for some nonzero integer $m$.

Thus $3*m + 3*m = (3*m)*(3*m)$, i.e., $2*3*m = 3^2*m^2$. Shortening gives $2 = 3*m$ which is a contradiction.


问:
Show that $f (x) = ax + 1 \in R[x]$ is invertible in $R[x]$.
Let $R$ be a commutative ring with unit. Consider $a ∈ R$ such that $a^2 = 0$. Show that $f (x) = ax + 1 \in  R[x]$ is invertible in $R[x]$.

I have no idea how to do it, can you give me any tips?

Body: Hint: $(ax+1)(1-ax) = 1-a^2x^2  = 1$.


问:
PDF of sufficient statistic of a normal distribution $N(0, \theta)$
Given:

$x_1, x_2....x_n$ is a random sample and an iid with normal distribution $N(0,\theta)$ where variance is the unknown paramter.

I calculated the sufficient statistic for this case and it comes out to be:
$$Y = \sum_{i=1}^n x_i^2$$

Now, I want to find the pdf of this Sufficient statistic Y and the distribution of Y?

I know $Y/\theta$ is equal to $Y = \sum_{i=1}^n x_i^2/\theta$ is a chi-square of distribution with n degrees of freedom. How can I get the pdf:

Using the change of variables

Using moment generating function (MGF)

Can anyone help me know how to do this? What are the steps? Thanks.

答:
First observe that $\frac{X}{\sqrt{\theta}}\sim N(0;1)$ and thus

$$ \bbox[5px,border:2px solid black]
{
\frac{X^2}{\theta}\sim \chi_{(1)}^2=Gamma\Big(\frac{1}{2};\frac{1}{2}\Big)
\qquad (1)
}
$$

Now it is evident that

$$ \bbox[5px,border:2px solid black]
{
T=\frac{1}{\theta}\sum_{i=1}^{n}X_i^2\sim Gamma\Big(\frac{n}{2};\frac{1}{2}\Big)
\qquad (2)
}
$$

Concluding, $Y=\theta T$,

thus

$$ \bbox[5px,border:2px solid black]
{
Y\sim Gamma\Big(\frac{n}{2};\frac{1}{2\theta}\Big)
\qquad (3)
}
$$

To prove (1) and (3) use the fundamental transformation theorem (change of variable)

To prove (2) use MGF's properties

Some hints for the proofs:

For (1), use the change of variable

$Z=\frac{X^2}{\theta}$ then

$$F_Z(z)=\mathbb{P}[Z\leq z]=\mathbb{P}[X^2\leq z\theta]=\mathbb{P}[-\sqrt{z\theta}\leq X \leq \sqrt{z\theta}]=F_X(\sqrt{z\theta})-F_X(-\sqrt{z\theta})$$

derivate it and get you first PDF:

$$f_Z(z)=\frac{1}{\sqrt{2\pi}}z^{-\frac{1}{2}}e^{-\frac{z}{2}}$$

This can be rewritten in the following way:

$$f_Z(z)=\frac{\Big(\frac{1}{2}\Big)^{\frac{1}{2}}}{\Gamma\Big(\frac{1}{2}\Big)}z^{\frac{1}{2}-1}e^{-\frac{z}{2}}$$

...and the first step is done! $f(z)$ is evidently a $Gamma\Big(\frac{1}{2};\frac{1}{2}\Big)$

For step (2) very easily multiply the n identical MGF's

For step (3) same procedure as (1): change of variable.


问:
Help interpret exercise 4.2 in Probability Essentials, Jacod & Protter
Some years ago there was a question here regarding this exercise. I find it a bit confusing and I'm struggling to understand what is in fact being asked here.

My issue is with the first part of question 4.2.

I want to prove $q_k$ are the probabilities of singletons for a $Binomial(1-p,n)$.

In order to check if what is being asked can be true I computed the singleton probabilities $p_k$ for a $Binomial(0.3, 4)$ in Julia and subtracted each $p_k$ to $1$ to get the corresponding $q_k$. I then computed the singleton probabilities of a $Binomial(1-0.3, 4)$. As you can see results are not the same:

```<code>using Distributions

n = 4
p = 0.3

d1 = Binomial(n, p)
p_ks = pdf(d1)  # returns the probabilities of singletons for a B(p=0.3, n=4)
q_ks = 1 .- p_ks

d2 = Binomial(n, 1-p)
d2_p_ks = pdf(d2) # returns the probabilities of singletons for a B(p=0.7, n=4)
</code>```
```<code>julia> q_ks
5-element Array{Float64,1}:
 0.7599
 0.5884
 0.7353999999999998
 0.9244
 0.9919
</code>```
```<code>julia> d2_p_ks
5-element Array{Float64,1}:
 0.008100000000000003
 0.07560000000000003
 0.2646000000000001
 0.4115999999999999
 0.24009999999999992
</code>```

Is the exercise wrong or am I missing something? Summing every $q_k$ doesn't even add up to 1.

答:
Of course there is something going wrong.

Suppose that $p_k$ is the probability of exactly $k$ successes, which is what $P(\{k\})$ seems to indicate, and what the other answers talk about. Then, in particular we have $\sum_{k=0}^n p_k = 1$, because in a binomial experiment we can have exactly $0,1,2,...,n$ successes with probability $p_0,p_1,...,p_n$ respectively, and these events are mutually disjoint and exhaust the entire sample space.

However, if this is true, then $\sum_{k=1}^n q_k = \sum_{k=1}^n (1-p_k) = \sum_{k=0}^n 1 - \sum_{k=0}^n p_k = n$. So how can $q_k$ be probabilities, if their sum isn't $1$ always? (Note , for example that the entries in the first array $1-p_k$ add up to $4$.)

In that case there is an issue with the exercise. Here's how to rectify it.

You have written two arrays down. The first is $1 - P(\{k\})$ where $P$ is the probability assignment for $Bin(p,n)$. The second array is $P'(\{k\})$ where $P'$ is the probability assignment for $Bin(1-p,n)$.

I want you to observe that for each $k$, we have $P'(\{4-k\}) + (1-P(\{k\})) = 1$ for each $k$. In words, if you add up the $k$th first entry and $k$th last entry of the first and second array respectively, you always get $1$. For example, $0.7599..+0.2400.. = 1$ (first entry and last entry of first and second array resp.) and $0.9244... + 0.756... = 1$ (fourth and fourth last entry of first and second array resp.).

From this we get $P'(\{4-k\}) = P(\{k\})$. In words : if I change $p$ to $1-p$, then the probability of $k$ successes in the first case, equals the probability of $n-k$ successes in the second case, for any $0 \leq k \leq n$.

To correct the exercise :

If $p_k = P(\{k\})$ are the success probabilities for $Bin(p,n)$ and $\boxed{q_k = p_{n-k}}$, then $q_k$ are the success probabilities for the random variable $Bin(1-p,n)$.

To answer this question, the exercise is wrong, and the first array is not to be computed at all, rather a different expression was required.

EDIT : The corrected exercise is almost a formality, and follows from the equivalence of $\binom nk = \binom n{n-k}$ for any $0 \leq k \leq n$. The Poisson approximation follows by noticing similar complement relations between $Bin(p,n)$ and $Bin(1-p,n)$.


问:
Calculating acceleration based on a starting and ending distance
I have been trying to figure this problem out and just can't.

The answer is supposed to be (A): 1.0 m/s/s.

Can anyone tell me how they got that answer?  I thought this problem was based on the formula: Acceleration $\displaystyle = \frac {(v2 - v1)}{t}$, but I can't get that answer.  What am I doing wrong? Or is the correct answer not listed?  Thanks for any help.

Problem:
  A Pinewood Derby car rolls down a track.   The total distance travel is measured each second.   Use the chart to calculate the acceleration of the car.

Time(s) |  Dist(m)

0       |     0

1       |     1

2       |     3

3       |     6

4       |    10

A) 1.0  meters/sec/sec

B) 1.5  meters/sec/sec

C) 2.0 meters/sec/sec

D) 2.5 meters/sec/sec

答: Your equation acceleration = (v2 - v1)/t is correct (if the acceleration is constant), but you don't have measurements of velocity.  You should also have one that says distance=acceleration*t^2/2 (also for constant acceleration and a standing start).  You do have measures of distance, and any of the measurements of distance will work.  You also have to note that the measurements you have are distance in the last second.  If you want to use the data from 4 seconds, you need the total distance traveled.</p>


问:
Largest area of a trapezoid in a semicircle
the question reads:

find the maximum area of a trapezium inscribed in a semicircle(radius= R), if the lower base in on the diameter.

The last statement(in bold) is what confuses me. Does this mean the base HAS to be the diameter? because thats how the solution in the book was given. I was hoping for a more generalized and perhaps intuitive approach to solve this problem. We were supposed to use calculus by the way.

答:
The question only states the lower base has to lie on the diameter. However, you can immediately infer that the top has to be a chord parallel to the diameter. Which means that for any given base less than the diameter, you can just expand the base till it equals the diameter, and the area will be higher than what you started with. This implies that any maximal solution will require the base to be the entire diameter.

You need a convenient parameter to work with. I suggest drawing a radial line from the center of the circle to the top right corner of the trapezoid. If you let the angle between that radial line and the diameter be $\theta$ (measured counterclockwise), then you can easily show the area $A = r^2(1+\cos\theta)\sin\theta$, which you should be able to maximise.


问:
What are some alternatives to Griffiths Harris 'Principles of algebraic geometry" chapters 0 and 1?
I liked using Rick Miranda's Algebraic Curves and Riemann Surfaces chapters I to VIII as an alternative to Griffiths' Introduction to Algebraic Curves. I notice they overlap in most topics.

I notice Rick Miranda's Algebraic Curves and Riemann Surfaces IX onwards overlaps with Griffiths Harris "Principles of algebraic geometry" chapters 0 and 1, but not really for most topics.

What are some alternatives to Griffiths Harris "Principles of algebraic geometry" chapters 0 and 1 that preferably are like Rick Miranda's Algebraic Curves and Riemann Surfaces chapters I to VIII as an alternative to Griffiths' Introduction to Algebraic Curves?

Here's a screenshot of the table of contents for Griffiths Harris "Principles of algebraic geometry" chapters 0 and 1

P.S. Not sure if this relevant, but: I was told in a comment to one of my deleted questions that Griffiths' Introduction to Algebraic Curves is a prerequisite to griffiths harris principles of algebraic geometry.

Edit: Maybe related:

What to study after Miranda's "Algebraic curves and Riemann surfaces"?

What do I need to read Philip Griffths

答:
I think pretty much everything you need is in Huybrechts' book titled Complex Geometry. This has the benefit of having many exercises and being written for someone new to the subject, whereas I think that Griffiths and Harris assumes that you already have had exposure to the materials in chapter $0$.

The book is available through Springer.


答:
Let $x_1,...,x_6$ be the gaps (number of empty seats) between successive students.

Let $x_0$ be the number of empty seats to the left of the left-most student, and let $x_7$ be the number of empty seats to the right of the right-most student.

Then the number of legal choices for the seats to be used by the students is the number $S$ of $8$-tuples $(x_0,...,x_7)$ of integers satisfying the equation
$$
x_0+\cdots+x_7=13
$$
where $x_1,...,x_6$ are positive and $x_0,x_7$ are nonnegative.

Equivalently, letting
$$
\;\;\;\;\,
\left\lbrace
\begin{align*}
y_0&=x_0+1\\[4pt]
y_k&=x_k\text{, for}\;1\le k\le 6\\[4pt]
y_7&=x_7+1\\[4pt]
\end{align*}
\right.
$$
$S$ is the number of $8$-tuples $(y_0,...,y_7)$ of positive integers satisfying the equation
$$
y_0+\cdots+y_7=15
$$
By the Stars-and-Bars method we get
$$
S=\binom{15-1}{8-1}=\binom{14}{7}
$$
hence, since for a given choice of seats for the students, there are $7!$ ways to seat the students, it follows that number of legal seatings is
$$
7!{\,\cdot\,}\binom{14}{7}=\frac{14!}{7!}
$$
matching your answer.

Thus, the answer key's answer of $8^7$ is not correct for the problem as stated.

Title: How many ways of arranging $7$ examinees are possible?

Body:
In an examination hall, there are $20$ chairs in a row. In how many ways the seats of $7$ examinees can be arranged such that no two examinees can occupy two consecutive chairs?

I tried doing it by using the binary string method adn got the answer to be $\dfrac{14!}{7!}$, but the correct answer is the book is $8^7$. i just needed to verify the answer. If the correct answer is $8^7$, then please explain the solution.

My Approach:

Let each arrangement of states of chairs (empty or non empty) correspond to a binary string of length $20$.

For example, If chair no. $1,5,7,9,13,15,19$ are non empty while all other chairs are empty, then this must correspond to string $10001010100010100010$

Now for 7 examinees to sit, there would be $7$ non empty chairs and $13$ empty chairs. This means that the string will consist of $7$ ones and $13$ zeroes. Since no examinees must sit in consecutive chairs, no two $1s$ must appear consecutively in the string. Thus the string can be like this:
$$\_0\_0\_0\_0\_0\_0\_0\_0\_0\_0\_0\_0\_0\_$$
There are $14$ "gaps" where one can appear. We need to choose $7$ gaps out of $14$ for $1$ to appear at whereas the remaining gaps will be removed from the final string. This can happen in $\binom{14}{7}$ ways.

This gives us the arrangement of chairs. Now in any given arrangement, $7$ examinees can be permuted in $7!$ ways.

$\therefore$ The total number of ways of seating $7$ examinees in $20$ chairs such that no two examinees sit on consecutive chairs are $\binom{14}{7}\cdot 7!=\dfrac{14!}{7!}$

THANKS


问:
What is the meaning of the parentheses in $\phi^{-1}\left[\{\phi(g)\}\right]=gH=Hg$?
I am studying homomorphisms is groups and i saw a theorem saying:

For $g$ in a group $G$, the cosets $gH$ and $Hg$ are the same, and collapsed onto the single element $\phi(g)$ by $\phi$. That is, $\phi^{-1}\left[\{\phi(g)\}\right]=gH=Hg$.

I am confused about the parentheses here. I think $\{\phi(g)\}$ denotes the set, formed by the elements of the mapping, but i am not sure what $[\ ]$ represent. Can anyone tell how should i interpret those parentheses? Or explain the theorem in other words?

Thank you

答:
Some use the notation $f^{-1}[X]$ to denote the preimage of a set $X$ under a map $f:A\to B\supseteq X$: apparently the brackets prevent ambiguity in set theory, as one of my instructors explained to me.

Here is an example justifying the need. Suppose we build the natural numbers $\bf N$ as ordinals:

$0:=\{\}$

$1:=\{0\}=\{\{\}\}$

$2:=\{0,1\}=\{\{\},\{\{\}\}\}$

$3:=\{0,1,2\}=\{\{\},\{\{\}\},\{\{\},\{\{\}\}\}\}$

$4:=\{0,1,2,3\}=\cdots$

$\cdots$

$n:=\{{\rm ordinals}<n\}$

$\cdots$

Now let $f:\bf N\to N$ be an injective function. Suppose $f(0)=1$ and $f(1)=0$. Then the inverse function $f^{-1}$ applied to $0$ is $1$, and applied to $1$ is $0$. That is,

$f^{-1}(0)=1$,

$f^{-1}(1)=0$.

However, the preimages under $f$ of these as sets of ordinals are

$f^{-1}[0]=f^{-1}[\varnothing]=\varnothing=0$,

$f^{-1}[1]=f^{-1}[\{0\}]=\{f^{-1}(0)\}=\{1\}$.

But $0,1,\{1\}$ are all distinct sets, so $f^{-1}(0)\ne f^{-1}[0]$ and $f^{-1}(1)\ne f^{-1}[1]$. So these sorts of clashes can occur in situations where you have set-valued functions of sets and thus need to differentiate between applying inverse functions and preimages. It's worth mentioning that the ambiguity may arise in the forward direction with plain-old images just as well, so this is just a general feature of set-valued mappings of sets: $f(x)$ and $\{f(u):u\in x\}$ need not be the same set.

However elsewhere in mathematics it is generally unlikely one happens across such issues, so most don't need to bother with brackets as no real ambiguity will arise.


问:
Let $(X,A,\mu,T)$ be a measure preserving dynamical system. Equality of integrals after composing with $T$?
Let $(X, A, \mu)$ be a measure space, let $T:X \rightarrow X$ be a measure preserving map, and let $f:X \rightarrow \mathbb{R}$ be an $L^1$ function. Is it true that for any measurable subset $B \subset X$, $$\int_B f \ d\mu = \int_{T^{-1}(B)} f \circ T \ d\mu \ ?$$ If so, hints on how to prove it would be preferred over full solutions. If it isn't true, are there any additional reasonable hypotheses that would make it true? So far I know it's true when $f$ is an $L^1$ simple function, but I haven't seen a way to make the leap to all $L^1$ functions yet.

答:
Well there is a natural transition of showing to positive $L^1$ functions and then deducing from linearity of the integral, that it is true for general $L^1$ functions.

To show this for positive $L^1$ functions, you know by the simple function approximation theorem that there exists a sequence of simple functions, $\{ \phi_n \}_{n=1}^\infty$, to $f$. The equality holds for each of them and there is equality of the limit by either the DCT or MCT.


问:
Is the limit $\lim_{t \to 0} \left(t^2\ln t , t, \ln(t^3)\right) $ defined?
Find the limit or prove that it doesn't exist
$$\lim_{t \to 0} \left(t^2\ln t , t,  \ln(t^3)\right) $$
I got
$$(0,\space 0, \space -\infty)$$
does this mean the limit is defined ?

Thanks in advance

答:
It depends upon the definition we are referring to.

A first issue is with the limitation $t>0$ which is usually implicitly assumed in order to deal with values within the domain of the function, that is for $t\to 0^+$.

The second issue is about the definition for not existence and usually we consider three main cases

limit exists finite $L \in \mathbb R$, i.e. the function converges

limit exists infinite $L =\infty$ (or $L =-\infty)$, i.e. the function diverges

limit doesn't exist otherwise


问:
Uniformly convergent for functions problem
The problem is as follows.

Let {$f_n$} is defined on $[0,1]$ and suppose that $lim_{n-> infinity}$ $f_n(x_n)=0$ for convergent sequence in $[0,1]$,  {$x_n$}.

Show that {$f_n$} is converges uniformly to $f=0$ on $[0,1]$.

I think using the bolzano-weierstrass theorem but i get stuck here. I'd like to show that
Every sequence {$x_n$} in $[0,1]$, $lim_{n->infinity} |f_n(x_n)|=0$ .
Please give me full explanation.

答: If $f_n$ does not converges to $0$ uniformly the there exist $\epsilon >0$, integers $n_1<n_2...$ and point $x_k$ such that $|f_{n_k}(x_k)| >\epsilon$ for all $k$. There is a subsequence, say $y_k$, of $(x_k)$ which converges to some point $x$. But then the hypothesis is violated since $f_n(y_n)$ does not tend to $0$.


问:
Quadratic equation with multiple parameter
I have the following equation:

$$x^2 +mx+(a−m)^2 =0\text{ ($a$ be there as a fixed parameter)}$$

and I need to calculate for which values of $m$ the equation has exactly one solution for $x.$

I know I have to look the value of $m$ for which the discriminant of this equation is equal to $0.$ But every time when I went this path is got stuck.

Can someone give me some hint I need to search for or look for when trying to calculate this?

Addition:

One thing I forgot to mention is that first of all the goal is to find a value for $m$ and afterwards to calculate the solution(s) fo $x$.

答:
$$x^2+mx+(a-m)^2=0.$$
Since the roots are equal,
$$m^2-4(a-m)^2=0$$
$$m^2-(2a-2m)^2=0$$
$$(m+2a-2m)(m-2a+2m)=0$$
$$m=\frac{2a}{3},2a.$$
Plug in these values of m in original equation, we get two quadratic equations.
$$x^2+2ax+a^2=0$$
and
$$9x^2+6ax+a^2=0$$
Now you can find the roots of these equations by Vieta's formulas.

The roots of these equations are $-a$ and $\frac{-a}{3}$.


问:
Looking for suggestions on how to proceed with showing that:
for $x \ge 2863:$

$$\ln\left(\left\lfloor\frac{x}{6}\right\rfloor!\right) < \sum_{k=5}^{\infty}-\mu(k)\ln\left(\left\lfloor\frac{x}{k}\right\rfloor!\right)$$

I've written a java application which checked from 2,863 to 24,600 which is how I came up with $2,863$.

I'm looking for tips on how to proceed to prove or disprove this comparison.

Since:

$$\ln\left(\left\lfloor\frac{x}{5}\right\rfloor!\right) > \ln\left(\left\lfloor\frac{x}{6}\right\rfloor!\right)$$

The problems comes down to showing that for $x \ge 2863$:

$$\sum_{k=6}^{\infty}\mu(k)\ln\left(\left\lfloor\frac{x}{k}\right\rfloor!\right) \le 0$$

or showing that:

$$\sum_{k=6}^{\infty}\mu(k)\ln\left(\left\lfloor\frac{x}{k}\right\rfloor!\right) \le \ln\left(\left\lfloor\frac{x}{5}\right\rfloor!\right) - \ln\left(\left\lfloor\frac{x}{6}\right\rfloor!\right)$$

Does anyone have any suggestions?

Thanks very much!

-Larry

答:
Did you mean
$$
\sum_{k=6}^\infty \mu(k) \ln\left(\left\lfloor \frac{x}{k} \right\rfloor !\right) \geq 0
$$

If not, I can't reproduce your verification for this inequality using the following Matlab code:

```<code>function s = larry( x )

    % Compute \sum_{k=6}^\infty \mu(k) \ln\left(\lfloor \frac{x}{k} \rfloor !\right)
    s = 0;
    for k = 6:ceil(x/2)
        s = s + moebiusmu(k)*logfactorial( floor(x/k) );
    end

    % Compute \mu(k)
    function mu = moebiusmu(n)

        if n == 1, mu = 1; return; end

        p = factor(n);
        r = histc(p, unique(p));

        if all(r < 2)
            mu = (-1)^length(unique(p));
        else
            mu = 0;
        end
    end

    % Compute \ln\left(\lfloor \frac{x}{k} \rfloor !\right)
    function p = logfactorial(n)
        if n < 2
            p = 0;
        else
            p = sum(log( 2:n ));
        end
    end

end
</code>```

Body:
Did you mean
$$
\sum_{k=6}^\infty \mu(k) \ln\left(\left\lfloor \frac{x}{k} \right\rfloor !\right) \geq 0
$$

If not, I can't reproduce your verification for this inequality using the following Matlab code:

function s = larry( x )

    % Compute \sum_{k=6}^\infty \mu(k) \ln\left(\lfloor \frac{x}{k} \rfloor !\right)
    s = 0;
    for k = 6:ceil(x/2)
        s = s + moebiusmu(k)*logfactorial( floor(x/k) );
    end

    % Compute \mu(k)
    function mu = moebiusmu(n)

        if n == 1, mu = 1; return; end

        p = factor(n);
        r = histc(p, unique(p));

        if all(r < 2)
            mu = (-1)^length(unique(p));
        else
            mu = 0;
        end
    end

    % Compute \ln\left(\lfloor \frac{x}{k} \rfloor !\right)
    function p = logfactorial(n)
        if n < 2
            p = 0;
        else
            p = sum(log( 2:n ));
        end
    end

end


问:
Find the product of all values of $(1+i\sqrt 3)^{\frac{3}{4}}$.
Find the product of all values of $(1+i\sqrt 3)^{\frac{3}{4}}$.

My try:

$(1+i\sqrt 3)^{\frac{3}{4}}=\exp (\frac{3}{4}(Log(1+i\sqrt 3)))=\exp(\frac{3}{4}(\log2+i\frac{\pi}{3}+2n\pi i))$.

I am kinda stuck on how to find the product of all values of  the above expression.
Can someone please help me out.

答:
$$1+i\sqrt3=2e^{i\pi/3}$$
The fourth roots of this number have common magnitude $2^{1/4}$ and arguments $\pi/12,7\pi/12,-5\pi/12,-11\pi/12$. The $\frac34$-powers thus have common magnitude $2^{3/4}$ and arguments $\pi/4,7\pi/4\equiv-\pi/4,-5\pi/4\equiv3\pi/4,-11\pi/4\equiv-3\pi/4$. The producf of all the four possible values of $(1+i\sqrt3)^{3/4}$ therefore has magnitude equal to the product of the magnitudes of each possible value, which is $8$, and argument equal to the sum of the arguments of each possible value, which is $0$. The final answer is $8$.


问:
On the quantity $I(q^k) + I(n^2)$ where $q^k n^2$ is an odd perfect number with special prime $q$
The topic of odd perfect numbers likely needs no introduction.

In what follows, we let $\sigma(x)$ denote the sum of divisors of the positive integer $x$.  Let
$$D(x) = 2x - \sigma(x)$$
denote the deficiency of $x$, and let
$$s(x) = \sigma(x) - x$$
denote the sum of aliquot/proper divisors of $x$.  Finally, denote the abundancy index of $x$ by
$$I(x) = \frac{\sigma(x)}{x}.$$

Euler proved that a hypothetical odd perfect number must have the form $N = q^k n^2$ where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

Since $q$ is prime and $N$ is perfect, we obtain
$$\frac{q+1}{q} \leq I(q^k) < \frac{q}{q-1}$$
so that we get
$$\frac{2(q-1)}{q} < I(n^2) = \frac{2}{I(q^k)} \leq \frac{2q}{q+1}.$$
Using the identity
$$\frac{D(x)}{x} = 2 - I(x)$$
we obtain the bounds
$${q^k}\bigg(\frac{q-2}{q-1}\bigg) < D(q^k) \leq {q^{k-1}}\bigg(q-1\bigg)$$
and
$$\frac{2n^2}{q+1} \leq D(n^2) < \frac{2n^2}{q}.$$
This implies that
$${2q^k n^2}\cdot\bigg(\frac{q-2}{(q-1)(q+1)}\bigg) < D(q^k)D(n^2) < {2q^k n^2}\cdot\bigg(\frac{q-1}{q^2}\bigg).$$
Dividing both sides of the last inequality by $2q^k n^2$, we get
$$\frac{q-2}{(q-1)(q+1)} < \frac{D(q^k)D(n^2)}{2q^k n^2} < \frac{q-1}{q^2}.$$
But we know that
$$D(q^k)D(n^2)=2s(q^k)s(n^2),$$
which can be verified by a direct, brute-force computation.
Thus, the fraction in the middle of the last inequality simplifies to
$$\frac{D(q^k)D(n^2)}{2q^k n^2}=\bigg(\frac{s(q^k)}{q^k}\bigg)\bigg(\frac{s(n^2)}{n^2}\bigg)=\bigg(I(q^k) - 1\bigg)\bigg(I(n^2) - 1\bigg) = 3 - \bigg(I(q^k) + I(n^2)\bigg).$$

We therefore finally have the bounds
$$3 - \bigg(\frac{q-1}{q^2}\bigg) < I(q^k) + I(n^2) < 3 - \bigg(\frac{q-2}{(q-1)(q+1)}\bigg)$$
which does not improve on the known bounds
$$3 - \bigg(\frac{q-2}{q(q-1)}\bigg) < I(q^k) + I(n^2) \leq 3 - \bigg(\frac{q-1}{q(q+1)}\bigg).$$
(See this paper for a proof.)

Here are my:

QUESTIONS Is it possible to improve on the bounds for $D(q^k)$ and $D(n^2)$ (where $q^k n^2$ is an odd perfect number with special prime $q$) to hopefully produce stronger bounds for $I(q^k) + I(n^2)$?  If so, how could this be done?

答:
Let me try to work backwards from
$$3 - \bigg(\frac{q-2}{q(q-1)}\bigg) < I(q^k) + I(n^2) \leq 3 - \bigg(\frac{q-1}{q(q+1)}\bigg).$$
This can be rewritten as
$$\frac{q-1}{q(q+1)} \leq 3 - \bigg(I(q^k) + I(n^2)\bigg) = \frac{D(q^k)D(n^2)}{2q^k n^2} < \frac{q-2}{q(q-1)}.$$
We also have
$$\frac{2n^2}{q+1} \leq D(n^2) < \frac{2n^2}{q},$$
which we can rewrite as
$$q < \frac{2n^2}{D(n^2)} \leq q + 1.$$
We therefore obtain
$$\frac{q-1}{q+1} = \frac{q(q-1)}{q(q+1)} < \bigg(\frac{q-1}{q(q+1)}\bigg)\cdot\bigg(\frac{2n^2}{D(n^2)}\bigg) \leq \frac{D(q^k)}{q^k} < \bigg(\frac{q-2}{q(q-1)}\bigg)\cdot\bigg(\frac{2n^2}{D(n^2)}\bigg) \leq \frac{(q+1)(q-2)}{q(q-1)},$$
which implies that
$$\frac{q-1}{q+1} < 2 - I(q^k) < \frac{(q+1)(q-2)}{q(q-1)},$$
which simplifies to
$$\frac{q^2 - q + 2}{q^2 - q} = 2 - \bigg(\frac{(q+1)(q-2)}{q(q-1)}\bigg) < I(q^k) < 2 - \bigg(\frac{q-1}{q+1}\bigg) = \frac{q+3}{q+1}.$$
Note that both bounds do not improve on the currently known
$$\frac{q+1}{q} \leq I(q^k) < \frac{q}{q-1}.$$


问:
if $f(x)\geq0$ for all $x$, then $f+f'+f''+f'''+ \cdots + f^n \geq0$
Suppose $f$ is a polynomial of degree $n$, and $f\geq0$ for all $x$, prove that $f+f'+f''+f'''+\cdots +f^n\geq0$. (so just functions of the form $ax^n+bx^{n-1}+.....c$. We are not going to include irrational functions, exponential, logs, etc.)

So for this problem, I looked at a few functions and graphed them to see why this statement would be true. The first thing I found out is that the polynomial's highest degree must be even, otherwise there would be some $x$ where $f(x)$ is negative. Now this implies the odd numbered derivatives will have an odd number for the term with the highest degree.

Having figured that out, I was wondering whether the sum of every successive would change the function so that $f+f'$ has negative values, and $f+f'+f''$ would have positive values and so on, and this does turn out to be true for the ones I've tried. If this is true in general, that would mean that $f+f'$ has some negative values, $f+f'+f''$ has not negative values, and $f+f'+f''+f'''+....f^{n-1}$ has negative values until we get to $f+f'+f''+f'''+....f^{n-1}+f^n$ which has no negative values.

Ok but having figured this out, I'm not sure as to how to proceed with the proofs because the above assumptions I've made are not lemmas or proofs that I was given.

答:
Let $g(x) = \sum\limits_{k=0}^n f^{(k)}(x)$. Notice

$$(e^{-x} g)' = e^{-x}(g'-g)
= e^{-x}\sum_{k=0}^n (f^{(k+1)} - f^{(k)})
= e^{-x}(f^{(n+1)} - f)$$
Since $f$ is a non-negative polynomial of degree $n$, we get
$$f^{(n+1)}\equiv 0 \quad\implies\quad (e^{-x}g)' = -e^{-x} f \le 0\quad\forall x \in \mathbb{R}$$
This means the function $e^{-x}g(x)$ is non-increasing.

Since $f(x)$ is non-negative for all $x$, $n$ has to be an even integer and the leading coefficient of $f(x)$ is positive. It is easy to see $g(x)$ is
a polynomial with same degree $n$ and leading coefficient as $f(x)$. This means for sufficiently positive $y$, we will have $g(y) > 0$.

For any $z \in \mathbb{R}$, take a $y > z$ large enough to make $g(y) > 0$.
Using the fact $e^{-x} g(x)$ is non-increasing, we obtain

$$e^{-z}g(z) \ge e^{-y}g(y) > 0\quad\implies\quad g(z) > 0$$

This is slightly stronger than we want to show. Namely, $g(x)$ is not only non-negative for all $x$, it is positive for all $x$.


问:
How to get the center of this circle
I now realize that I have over-simplified the problem in my last post because it is hard to explain in words.

This is the original problem that I am fighting with for more than a week:

drawing

There are two lines originating from the origin. The angle of each line in relation to the diagonal is the same and is known. A circle is centered on the diagonal and only a 90 degrees arc of the circle is drawn between the lines. On the arc of the circle is a point P of which the coordinates are known. The dimension and location of the circle is not known, the only thing that is known is that point P lies on the arc and the arc is a 90 degrees segment of a circle with its center on the diagonal.

How can I find the coordinates of the circle?

答:
First, let us identify all the elements:

The diagonal $d$ is the line $y=x$.

The point $P$ has coordinates $(p_x,p_y)$.

The angles are defined by lines $r_A : y=ax$ and $r_B:y=bx$.

The circle intersects the lines $r_A$ and $r_B$ in points $T : (t_x, t_y)$ and $U : (u_x,u_y)$ respectively (there are two more points of intersection, but you should be able to discard them by inspection).

You need to find the center $C: (c_x,c_y)$ and the radius $r$ of the circle, which has equation:
$$(x-c_x)^2 + (y-c_y)^2 = r^2$$

Now, here there are several key points that you should try to justify:

Since the angles to the diagonal are equal, $b=1/a$. Use the angle to deduce the value of a.

The center lies on the diagonal, so $c_x=c_y$.

$d$ is the bisector of the angle between lines $r_A$ and $r_B$. Use its properties to conclude that $(t_x,t_y)=(c_x,c_x+r)$ and $(u_x,u_y)=(c_x+r,c_x)$.

Finally, replacing coordinates in the equations of the lines $r_A$, $r_B$ and the circle, you should be able to deduce the values of $c_x$, $c_y$ and $r$ in terms of $a$, $p_x$ and $p_y$.

Detailed steps:

The slope of $r_A$ is given by the tangent of the angel with respect to the horizontal axis:
$$a=\tan(\frac{\pi}{4} + A),$$
where $A$ is the angle between $d$ and $r_A$. Analogously, the slope of $r_B$ is
$$b=\frac{1}{\tan(\frac{\pi}{4} + A)} = \tan(\frac{\pi}{4} - A).$$

Since the center $(c_x,c_y)$ lies on $d$ (of equation $y=x$) then it clearly satifies that $c_x=c_y$.

The points of $d$ are equidistant from $r_A$ and $r_B$ because $d$ is the angle bisector. Moreover, $\triangle TCU$ needs to be a rectangular and isosceles triangle (with legs of length $r$). So the only way for this to be possible is that $\overline{TC}$ is parallel to the vertical axis (so that $t_x=c_x$) and $\overline{UC}$ is parallel to the horzontal axis (that is, $u_y=c_y=c_x$).

Note also that $T$ and $U$ lie on the circle, so the distance to $C$ is $r$. Therefore, $t_y=c_x+r$ and $u_x=c_y+r$.

Taking all previous steps into account, knowing that $T$ lies on both $r_A$ an the circle, its coordinates have to satisfy both equations:
$$\begin{cases}
c_x + r &= c_x \tan(\frac{\pi}{4} + A),\\
(p_x-c_x)^2 + (p_y-c_x)^2 &= r^2.
\end{cases}$$

Finally, you need to solve for $r$ in the first equation and replace it in the second one:
$$(p_x-c_x)^2 + (p_y-c_x)^2 = \left(c_x \tan(\frac{\pi}{4} + A) - c_x\right)^2.$$
Expand that expression to obtain a quadratic equation on $c_x$ which should be easy to solve (see the quadratic formula) in terms of $p_x$ and $p_y$. The rest of the values can be obtained by replacing $c_x$ with that value.

Title: How to get the center of this circle


问:
Conditional Probability Problem with Dice
Three distinct fair dice are thrown. The probability that $4$ appears on two dice given that $5$ occurs on atleast one dice.

I counted the number of total cases when atleast one five occurs. This can be counted as (exactly 1+ exactly 2+exactly 3) = $ 75+15+1$. The favourable cases are $3$.
So probability should be $3/91$ according to me. However, the answer that is given in the source from where the question has been taken is something else. So, I just wanted to confirm whether my method is correct.

答:
$A$ is event of $4$ appearing exactly on two of the three fair dice.

$B$ is the event of $5$ appearing on at least one dice.

$P(A \cap B) = {3 \choose 1} \frac{1}{6^3}$ (two $4$ and one $5$)

$P(B) = 1 - (\frac{5}{6})^3$ (this is probability of at least one $5$ appearing).

Desired conditional probability $P(A|B) = \displaystyle \frac{P(A \cap B)}{P(B)} = \frac{{3 \choose 1} \frac{1}{6^3}}{1 - (\frac{5}{6})^3} = \frac{3}{91}$.


问:
Equation of the line that has $x$ and $y$ intercepts at $a$ and $b$.
Please can anyone help me with proving the following problem:

Show that the line that crosses the $X$-axis at $a \neq 0$ and the $Y$-axis at $b \neq 0$ has the equation $$\dfrac{x}a + \dfrac{y}b -1=0$$

I don't have an idea of where I should start from for proving the problem.

答:
Actually there are set of 4 lines. because intercepts are length and its coordinates have set of 4 coordinates.(a,0)(0,b),(a,0)(0,-b),(-a,0)(0,b),(-a,0)(0,-b).
so lines equation will be..
$$ \frac xa + \frac yb -1 =0$$

$$ \frac xa - \frac yb -1 =0$$

$$ -\frac xa + \frac yb -1 =0$$
$$ -\frac xa - \frac yb -1 =0$$


问:
Finite/Infinite groups
Which of the following statements is/are true?

a) There are infinitely many finite groups in which every non-identity element have order $2$.

b)There exists an infinite group in which every non-identity element have order $2$.

c)There exists an infinite group in which there are elements of order $n$ for all $n\in \Bbb N$.

d)There are infinitely many infinite groups in which each non-identity element have finite order.

My attempt: Option 2 seems correct as I have an example of such a group( Power set of Natural numbers under the binary operation of symmetric difference is one such example) but I am unable to conclude anything about other options.

About option 1, it seems that it is true keeping in mind $\Bbb Z_2\times \Bbb Z_2\times\cdots$ but still I am confused. For the last two options, I don't have any conclusive idea which I can apply here. Please guide.

答:
(a) For $n\geq 1$, $\prod_{i=1}^n\Bbb{Z}_2$ is a finite group whose every nontrivial element has order $2$. So there are infinitely many such groups.

(b) $\prod_{i=1}^{\infty}\Bbb{Z}_2$ is the desired group.

(c) You may consider the example $\Bbb{Q}/\Bbb{Z}$. Then for every $n\in \Bbb{N}$, $\frac{1}{n}+\Bbb{Z}$ is element of order $n$ in $\Bbb{Q}/\Bbb{Z}$.

(d) For $n\geq 2$, $\prod_{i=1}^{\infty}\Bbb{Z}_n$ is an infinite group whose every nontrivial element has finite order. So there are infinitely many such groups.

Title: Finite/Infinite groups


问:
$\lim_{x\to6^-}\big[\frac{\sqrt{36-x^2}}{x-6}\big]$ using L'Hopital's rule?
Disclaimer: this is a homework problem, I at least need a bit of direction, I can't seem to get anywhere, I must be overlooking something.

$\lim_{x\to6^-}\biggl[\frac{\sqrt{36-x^2}}{x-6}\biggr]$
I've tried applying L'Hopital's rule since we end up with an indeterminate form if we evaluate at $x=6$

I took the first, second, and third derivative but the denominator will always force division by 0. I'm not sure what to do. I'd love to put in the first and second derivatives into this question but, I'm a bit sloppy with the 'Ol Mathjax!

EDIT: Found the big wet mistake: For L'Hopital's rule, I have been taking the derivative of the ENTIRE function, instead of the derivative of the numerator, divided by the derivative of the denominator! Don't do that folks!

答:
The domain gives $-6\leq x<6,$ which says that our function is not defined  for $x\rightarrow6$ and we can't say about the limit.

We can try to calculate the limit for $x\rightarrow6^-$.

By the  L'Hopital's rule we obtain:
$$\lim_{x\rightarrow6^-}\frac{\sqrt{36-x^2}}{x-6}=\lim_{x\rightarrow6^-}\frac{\frac{-x}{\sqrt{36-x^2}}}{1}=-\infty.$$

Solution without L'Hopital:

$$\lim_{x\rightarrow6^-}\frac{\sqrt{36-x^2}}{x-6}=-\lim_{x\rightarrow6^-}\sqrt{\frac{6+x}{6-x}}=-\infty.$$


问:
Find $\lim\limits_{n \to \infty} n^{\frac 3 4} \Bbb E \left [\left (Y_n - \Bbb E \left [Y_n \right ] \right )^2 \right ].$

Let $X_i,\ i \geq 1$ be i.i.d. discrete random variables with mean $\mu$ and variance $\sigma^2.$ Let $k \gt 1.$ Define the sequence \begin{align*} Y_n : & = \dfrac {X_1 X_2 \cdots X_k + X_2 X_3 \cdots X_{k+1} + \cdots  + X_{n-k+1} X_{n-k+2} \cdots X_n} {n}. \end{align*} Find $\lim\limits_{n \to \infty} n^{\frac 3 4} \Bbb E \left [\left (Y_n - \Bbb E \left [Y_n \right ] \right )^2 \right ].$

How do I find that? Is there any easy way to proceed?

Thanks in advance.

EDIT $:$ What I get is that $$ \Bbb {E}\ \left [ {Y_n}^2 \right ]  = \dfrac {1} {n^2} \left [    \left (n - k + 1 \right )   \left ({\sigma}^2 + {\mu}^2 \right )^k + 2 \left [\sum\limits_{r = 0}^{k-2} \left (n-k-r \right ) \left (\mu^2 \right )^{r+1} \left ({\sigma}^2 + {\mu}^2 \right )^{k-r-1} + \dfrac {\left (n-2k+1 \right ) \left (n-2k+2 \right )} {2} \left ({\mu}^2 \right )^k \right ] \right ].$$ But then $$\begin{align*} \lim\limits_{n \to \infty} n^{\frac {3} {4}} \Bbb {Var}\ [Y_n] & = \lim\limits_{n \to \infty} \left [ \dfrac {\left (n-k+1 \right ) \left (n-k+2 \right)} {n^{\frac {5} {4}}} - \dfrac {\left (n-k+1\right)^2} {n^{\frac {5} {4}}} \right ]  {\mu}^{2k} \\ & = \lim\limits_{n \to \infty} \dfrac {\left (n - k + 1 \right )} {n^{\frac {5} {4}}} {\mu}^{2k} = 0.\end{align*}$$ Am I right? Can anybody please check my computation whether it is correct or not.

Thanks in advance.

Source $:$ This question appeared in ISI PhD entrance test in Mathematics held in $20$th September this year (TEST CODE : MTB) in the afternoon session (Question No. $9$).

答:
It is nice to start with a small problem and then generalize it, so I start with the variance for $k=1$.

First we have that $var(x_jx_{j+1})=\sigma^4$ and $cov(x_jx_{j+1})=\mu^2\sigma^2$ hence $$var(Y_n)=\frac1{n^2}\left((n-1)\sigma^4+2(n-2)\sigma^2\mu^2\right)$$ implying that for this simple case it holds that $\lim\limits_{n \to \infty} n^{\frac {3} {4}} \text {Var}\ [Y_n] = 0$. For $k=2$ we obtain $$var(Y_n)=\frac1{n^2}\left((n-2)\sigma^6+2(n-2)\sigma^2\mu^2(\sigma^2+2\mu^2)+2(n-3)\mu^4\sigma^2\right)$$ for which we have again that $\lim\limits_{n \to \infty} n^{\frac {3} {4}} \text {Var}\ [Y_n] = 0$. These too cases indicate that as long as $k$ is fixed and $n \to \infty$ the limit you are after will be zero.


问:
What is the smallest filter in $\mathcal{B}$ containing $a \in B$?

$\mathcal{B}:=(B,\leq,\lor,\land,^c,0,1)$ is a Boolean algebra. Show that the smallest filter in $\mathcal{B}$ containing $a\in B$ is given by $$F(a) = \{x\in B:a\leq x\}$$

Here's my work:

First, I think it's important to verify that $F(a)$ is a filter. Clearly, $a \in F(a)$ so $F(a)\neq \phi$. $F(a)$ is also upwards closed, i.e. if $x \in F(a), x\leq y$ then $y \in F(a)$. This follows from the definition of $F(a)$ and transitivity of $\leq$, the partial order on $B$. Lastly, we must show that if $x,y\in F(a)$ then $x\land y \in F(a)$. $x\land y$ is the greatest lower bound of $x$ and $y$. Since $x,y\in F(a)$, we know that $a\leq x, a\leq y$. From here, we see that $a\land x =x$ and $a\land y=y$. So $a \land (x\land y) = (a\land x)\land y = a\land y = a$, implying $a \leq x\land y$, so $x\land y\in F(a)$. So F(a) is indeed a filter!

Why is $F(a)$ the smallest filter containing $a$?

P.S. For reference, $F \subseteq B$ is a filter if:

$F \neq \phi$

If $x,y \in F$ then $x\land y\in F$

If $x\in F$ and $x\leq y$ then $y\in F$

答: Hint: any filter to which $a$ belongs is “upwards closed”, so it has to include $F(a)$


问:
Why is finiteness required here (in Coset result)?
I was given the following result:

Suppose that $H$ is a subgroup of $G$. Then the following are true.

(i) For $x \in G$, the map $f_x : H \rightarrow Hx$ defined by $f_x(h) = hx$ is a bijection

(ii) If $H$ is finite, then $\lvert H \rvert = \lvert Hx \rvert$

($Hx$ here is the right coset of $H$ with element $x \in G$)

My question is why is finiteness required in (ii)? Could you not define the same bijection as in (i) even if $H$ is not finite and still obtain that $\lvert H \rvert = \lvert Hx \rvert$?

答:
You are right. Since there is a bijection between $H$ and $Hx$, $|H|=|Hx|$ even if $H$ is not finite.

I guess that the result added the assumption that $H$ is finite maybe because it is used to prove Lagrange's Theorem later on. To prove Lagrange's Theorem, it needs the condition that $H$ is finite.


问:
Show that tanz doesn't assume i, -i
This question is from my assignment in complex analysis and I was unable to solve it.

Prove that tanz doesn't assume the value i, -i. Does this contradict Picard's Theorem.

Attempt- tanz = $ i \frac{e^{iz} -e^{-iz}} { e^{iz}+ e^{-iz}}$  => $e^{iz}=i$ and z=π/2 and z=-π/2 is not in domain. But how does it tells which values will never be attained?  So, how to deduce it?

I think picard theorem is not violated as tanz is not entire, due to cosz being in denominator.

Am I right?

答: No, you want to say that $$\tan z=i\frac{e^{iz}-e^{-iz}}{e^{iz}+ e^{-iz}}=\pm i \,,$$ which becomes equivalent to $$e^{\pm iz}=0\,,$$ which is impossible (the exponential function has no complex zeroes).

Title: Show that tanz doesn't assume i, -i


问:
Examples of finite groups that are not a semidirect product
I'm looking for examples of (families of) finite groups that are not semidirect products.

When first learning group theory, the first such group that one encounters is $Q_8$. In my search for other groups that are not semidirect products, the only examples I could find were simple groups, which clearly cannot be semidirect products since they don't have normal subgroups.

Does anyone have example of non-simple finite groups that are not semidirect products?

答:
The name for what you are looking for is a "splitting-simple group".

If you are looking for the smallest non-nilpotent example, that would be a group of order 48, the tilde covering of $S_4$.  It's actually pretty easy to see that it's the smallest non-nilpotent example.  Due to the existence of Hall subgroups, a solvable group that is not a $p$-group will be a semidirect product if it has a normal Sylow subgroup.  Using Sylow's theorems and the property of the Fitting subgroup mentioned below, you can find that there are three such groups of order less than 48: $S_4$ and its two covering.  However, $S_4$ is a semidirect product in another way ($2^2$ by $S_3$) as is its other double cover (GL(2,3), which is $Q$ by $S_3$).  However, the tilde cover has a unique element of order 2, making it clear it can't be a semidirect product of two groups of even order, and it can't by odd by even or even by odd, either, since the Sylows are not normal and 3*8 is the only way to write 24 non-trivially as even*odd.

The best description of the tilde cover of $S_4$ is here, showing its faithful 2-dimensional representation over GF(9).

With the possible exceptions of orders 32 and 64, the only splitting simple groups of order less than 81 are $A_5$, the tilde cover of $S_4$, the generalized quaternion (including quaternion) groups, and the cyclic groups of prime power order.

To see how the reasoning works: If a splitting-simple group of order 72 exists, its Fitting subgroup must be non-cyclic abelian of order 12, $O_3$ must be central, and the quotient by $O_3$ must be $S_4$.  $O_3$ cannot then be in the derived group, since $S_4$ doesn't have a triple cover.  But then the derived group has order 12 (it must contain the pre-image of $A_4$ without meeting $O_3$).  Thus, all elements of order 2 must be in $O_2$ lest we have a semidirect product of $A_4$ by 6.  But since the derived group has index 6, it is contained in an index 3 normal subgroup that doesn't meet $O_3$--thus the group is a semidirect product after all.

Also, note that not all non-solvable examples are simple.  A cover of a simple group will be splitting-simple, so, for example, at order 120, although $S_5$ is not splitting-simple, $SL(2,5)$, the universal cover of $A_5$, is splitting-simple.  In fact it is not hard to see that $SL(2,5)$ is the only splitting simple group of order 120, as you can rule out solvable examples using the theorem that the Fitting subgroup of a solvable group contains its own centralizer.


答:
Reusing the nice picture and notations of AugSB, you have

$$\begin{cases}
c_x &= (r+c_x)\tan\left(\frac{\pi}{4} -A\right)\\
(p_x-c_x)^2 + (p_y-c_x)^2 &= r^2
\end{cases}$$

as $c_x=c_y=u_y$ and where the angle $A$ is given in radians.

Which is equivalent to

$$\begin{cases}
r &=c_x \frac{1 - \tan\left(\frac{\pi}{4} -A\right)}{\tan\left(\frac{\pi}{4} -A\right)}\\
(p_x-c_x)^2 + (p_y-c_x)^2 &= r^2
\end{cases}$$

$A$ is given. Replacing $r$ from the first equation into the second one you get an equation of the second degree where $c_x$ is the only unknown as $p_x, p_y$ are known. Solving it you get the desired center $C=(c_x,c_x)$.

Title: How to get the center of this circle

Body:
I now realize that I have over-simplified the problem in my last post because it is hard to explain in words.

This is the original problem that I am fighting with for more than a week:

drawing

There are two lines originating from the origin. The angle of each line in relation to the diagonal is the same and is known. A circle is centered on the diagonal and only a 90 degrees arc of the circle is drawn between the lines. On the arc of the circle is a point P of which the coordinates are known. The dimension and location of the circle is not known, the only thing that is known is that point P lies on the arc and the arc is a 90 degrees segment of a circle with its center on the diagonal.

How can I find the coordinates of the circle?


问:
Why is $\frac{d}{dt}[\frac{1}{y}] = \frac{d}{dt}[\cos{x}]$ result in an application of the chain rule in related functions?
Why does the below step result in a chain rule?
Intuitively, I would just differentiate it against y and x resulting in $-y^{-2} = -\sin{x}$.

Khan Acad mentioned something about relating y = y(t) and x = x(t), but my brain is struggling to comprehend why it can just be related like this... (Though I could just be getting it completely wrong)

答:
It might be somewhat complicated, but the essence of the problem is simple.

We know that the variable $x, y$ is unknown functions of $t$, In other words, we know that the some functions $f(t), g(t)$ exist such that $x = f(t), y = g(t)$ but we don't know $yet$ what the actual expression of $f, g$ look like. It maybe look like $f(t) = 2t$, or $f(t) = \tan(t) + 12$ or anything can be. But thanks to the problem, we know some relation between $f$ and $g$.

We know $\frac{1}{y} = \cos(x)$, can be written as $\frac{1}{g(t)} = \cos(f(t))$. You can see that there is no $x, y$ in the relation. So you can just simply differentiate it with $t$(because LHS and RHS are always equal).

According to your idea, just simply differentiate LHS and RHS with $x$, can be also applied, but we can't get $\frac{dy}{dt}$, which we want. Because:
$$\frac{1}{y} = \cos{x}$$
If we differnetiate both side with $\frac{d}{dx}$:
$$-\frac{1}{y^2}\frac{dy}{dx}=-\sin(x)\frac{dx}{dx}$$
Which is same as:
$$\frac{dy}{dx}=y^2 \sin(x)$$


答:
I don't actually think that it is the chain rule. Since we have the value for $\dfrac 1y$ as $\cos{x}$, you can directly substitute it, if you are looking for $\dfrac{d}{dt}[\dfrac 1y]$. (I say this about the title, to make it clearer)

Here's another solution I tried to make. Hope you find this one helpful.

$\because \dfrac{dx}{dt} = -2, dx = -2dt \implies \int{dx} = \int{-2dt} =\implies x = -2t + C$ (preparing stuff for application in chain rule)

(If the correct answer is $0$, let's take $C = 0$, which makes $x = -2t$. Also, it (taking $C = 0$) might work, since no special graph to find the intercept has been given or because of lack of reference to anything that alters the value of $C$. )

$\implies \cos{x} = \cos{(-2t)}$

$\therefore , \dfrac 1y = \cos{x} = \cos(-2t), y = \dfrac 1{\cos{(-2t)}}$

$\therefore, \dfrac{dy}{dt} = \dfrac{d(\dfrac 1{\cos{(-2t)}})}{dt} = \dfrac{-\sin{(-2t)}}{\cos^2{(-2t)}}\times-2 = \dfrac{2\sin(-2t)}{\cos^2{(-2t)}}$ (Here's where we use the chain rule, alongside the quotient rule.)

Now we are looking for the derivative at $t = -\dfrac \pi2$ , that is , $x = \pi$ (since $x = -2t$ as we get from the integration above), so just plug in the values, you get $\dfrac {dy}{dt} = 0$.

For an understandable solution, since we have $x = -2t$, we can write $\dfrac{dy}{dt} = \dfrac{2\sin(-2t)}{\cos^2{(-2t)}}$ as $-2\dfrac{dy}{dx} = \dfrac{2\sin(x)}{\cos^2(x)}$ (we have to remove the $-2$ factor in $x$). Now plug in $x = \pi$ and you get $\dfrac{dy}{dx} = 0$
$\implies \dfrac{dy}{dt} = -2\dfrac{dy}{dx} = 0$

Title: Why is $\frac{d}{dt}[\frac{1}{y}] = \frac{d}{dt}[\cos{x}]$ result in an application of the chain rule in related functions?

Body:
Why does the below step result in a chain rule?
Intuitively, I would just differentiate it against y and x resulting in $-y^{-2} = -\sin{x}$.

Khan Acad mentioned something about relating y = y(t) and x = x(t), but my brain is struggling to comprehend why it can just be related like this... (Though I could just be getting it completely wrong)


问:
Proving origin to be removable singularity(Proof verification)
This particular question is from Ponnusamy Silvermann complex section 9.22 ( question no. 2) and I am unable to solve it.

Let f(z) be analytic in deleted neighborhood of origin and lim $z\to 0$ | zf(z) |=0 . Then show that origin is removable singularity of f(z).

I need to show that lim $z\to 0 f(z) $=M(finite) . But if it were infinite, then  $z \to z_0$ |zf(z) | would be indeterminate.

So, it's removable.

Is my proof correct?

答: A counterexample for your proof would be $f(z)=1/\sqrt{z}$. I think you need to use the property of analyticity in some way to exclude such cases.


问:
How many ways are there of choosing $k$ numbers from $\{1, . . . , n\}$ if $1$ and $2$ can’t both be chosen?
How many ways are there of choosing $k$ numbers from $\{1, . . . , n\}$ if $1$ and $2$ can’t both be chosen? (Suppose $n, k ≥ 2, n ≥ k$).

My answer is $2\binom{n-1}{k} - \binom{n-2}{k}$. First, I counted sets without $1$, then without $2$, then found their sum and, as the formulas are the same, it's just $2$ multiplied by the formula.  Then I subtracted all sets with both $1$ and $2$, as they was already counted. Is it correct?

答:
Your answer is incorrect.

Method 1:  We count directly.

Exactly one of the numbers from the subset $\{1, 2\}$ is selected:  We select one of the two numbers from the subset $\{1, 2\}$ and $k - 1$ of the numbers from the $(n - 2)$-element subset $\{3, 4, 5, \ldots, n\}$, which can be done in
$$\binom{2}{1}\binom{n - 2}{k - 1}$$
ways.

Neither number from the subset $\{1, 2\}$ is selected:  Then we must select $k$ numbers from the $(n - 2)$-element subset $\{3, 4, 5, \ldots, n\}$, which can be done in
$$\binom{2}{0}\binom{n - 2}{k} = \binom{n - 2}{k}$$
ways.

Total:  Since the two cases are mutually exclusive and exhaustive, the number of admissible selections is
$$\binom{2}{1}\binom{n - 2}{k - 1} + \binom{n - 2}{k}$$

Method 2:  We use complementary counting.

There are
$$\binom{n}{k}$$
ways to select a subset of $k$ elements from the set $\{1, 2, 3, \ldots, n\}$.

From these, we must subtract those selections which include both the elements $1$ and $2$.  If we select both $1$ and $2$, we must select $k - 2$ elements from the subset $(n - 2)$-element subset $\{3, 4, 5, \ldots, n\}$, which can be done in
$$\binom{2}{2}\binom{n - 2}{k - 2} = \binom{n - 2}{k - 2}$$
ways.

Hence, the number of admissible selections is
$$\binom{n}{k} - \binom{n - 2}{k - 2}$$

Method 3:  We correct your explanation of your formula.

You correctly found that the number of selections of $k$ elements from the set $\{1, 2, 3, 4, \ldots, n\}$ which exclude $1$ is
$$\binom{n - 1}{k}$$
and that the number of selections of $k$ elements from the set $\{1, 2, 3, 4, \ldots, n\}$ which exclude $2$ is also
$$\binom{n - 1}{k}$$
Notice that the $k$-element subsets which have been counted twice are those which exclude both $1$ and $2$.  Since we only want to count them once, we must subtract them from the total.  Since the subset $\{3, 4, 5,\ldots, n\}$ contains $n - 2$ elements, there are indeed
$$\binom{n - 2}{k}$$
such subsets.  Thus, by the Inclusion-Exclusion Principle, there are
$$\binom{2}{1}\binom{n - 1}{k} - \binom{n - 2}{k}$$
admissible selections, but your explanation for this formula did not make sense.


问:
Losing original values while solving inverse trig equations
While proving, say, that $3tan^{-1}\frac{1}{4} + tan^{-1}\frac{1}{20} = \frac{\pi}{4}- tan^{-1}\frac{1}{1985}.$

There is the possibility that one side would just be n$\pi$ added to the other. So just proving that the tan values of both sides are equal won't be enough, would it? Apparently this method is used a lot

答:
$$\tan\left(\frac{\pi}{4}-\arctan\frac{1}{1985}\right)=\frac{1-\frac{1}{1985}}{1+\frac{1}{1985}}=\frac{992}{993}.$$
Now, we can use the same things for the left side:
$$\tan\left(3\arctan\frac{1}{4}+\arctan\frac{1}{20}\right)=\frac{\frac{3\cdot\frac{1}{4}-\left(\frac{1}{4}\right)^3}{1-3\left(\frac{1}{4}\right)^2}+\frac{1}{20}}{1-\frac{3\cdot\frac{1}{4}-\left(\frac{1}{4}\right)^3}{1-3\left(\frac{1}{4}\right)^2}\cdot\frac{1}{20}}=\frac{992}{993}.$$
Now,  since $$0<\frac{\pi}{4}-\arctan\frac{1}{1985}<\frac{\pi}{4}<\frac{\pi}{2}$$ and $$0<3\arctan\tfrac{1}{4}+\arctan\tfrac{1}{20}<3\arctan(2-\sqrt3)+\arctan(2-\sqrt3)=\tfrac{\pi}{3}<\tfrac{\pi}{2},$$ we obtain our identity.

We see that the case with adding $n\pi$ is impossible here.


问:
dimension of column space in linear algebra
I am confused why the dimension of the column space of a matrix is the number of pivotal columns? Could anyone elaborate it by some concrete examples? BTW, could you also elaborate a bit about why number of pivotal columns = number of pivotal rows? Is it related to A and A^T? I don't quite understand what does pivotal mean here. Thanks a lot!</p>

答: That the dimension of the column space equals that of the row space is a nontrivial theorem.  The pivots are the leading $1$'s ($1$'s with only zeros below it), after row-reduction.  Consider $\begin{pmatrix}1&2&3\\4&5&6\end{pmatrix}$.  After row-reduction/gaussian elimination, we get $\begin{pmatrix}1&2&3\\0&1&-2\end{pmatrix}$.  There are two leading $1$'s, so two pivot columns, and the dimension of the column/row space is two.  Thus we can say the rank is two.


问:
Curly braces in the language of set theory
We usually write sets using curly braces. However the language of set theory(i.e. the first-order language with the only binary relation-symbol $\in$) doesn't have curly braces. How is it justify in mathematical logic?

答:
It is a term-forming operator: when applied to a formula it outputs a term (a "name").

$\{ x \mid \varphi(x) \}$

is the name of the set of all and only those elements that satisfy formula $\varphi(x)$.

We can add it to the basic language through the definition:

$\{ x \mid \varphi(x) \}=y \leftrightarrow [\forall x (x \in y \leftrightarrow \varphi(x))]$,

provided that we have previously proved that :

$\exists y [\forall x (x \in y \leftrightarrow \varphi(x))]$.

The key property of the symbol is that:

$z \in \{ x \mid \varphi(x) \} \leftrightarrow \varphi(z)$.


问:
Evaluate $\lim_{x\to 1^{+}}\left(\frac{x}{1+x}\right)^x$
I tried:
$$\left(\frac{x}{1+x}\right)^x=A$$
$$\ln(A)=x\ln\left(\frac{x}{1+x}\right)$$
$$\lim_{x\to 1^{+}}\frac{\ln x-\ln (x+1)}{\frac1x}=\lim_{x\to 1^{+}}\frac{\frac1x-\frac1{x+1}}{\frac{-1}{x^2}}=\frac{-1}{2}$$
I applied L'Hopital rule in last step.
I get $\lim(A)_{x\to 1^{+}}=e^{\frac{-1}{2}}$
But the final answer provided in the book is $e^{-1}$. where I made a mistake?

答:
Hint

You cannot apply l'Hopital here since it's not indeterminated. Nevertheless, $$\lim_{x\to 1^+}\frac{\ln(x)-\ln(1+x)}{1/x}=-\ln(2).$$
I let you conclude (and the limit won't be $e^{-1}$ but $\frac{1}{2}$).


问:
Total expectation under measure theory
I want to show that $\mathbb {E}(\mathbb {E}(X\mid \mathcal {F}))=\mathbb {E}X$. My thought is that

\begin{align*}\mathbb {E}(\mathbb {E}(X\mid  \mathcal {F}))=\sum_{\Omega_{i}} \frac {\mathbb {E}(X;\Omega_{i})}{\mathbb {P}(\Omega_{i})}\mathbb {P}(\Omega_{i})=\mathbb{E}X,
\end{align*}
where $\Omega_{i}'s$ are either finite or infinitely countable partition of $\Omega$ into disjoint sets, and each one has positive probability. I wonder if it's always true that there exists such $\Omega_{i}'s$ such that $\mathcal {F}=\sigma(\Omega_{1},\Omega_{2},\ldots)$ with the desired properties? Thanks.

答:
Not every sigma field is countably generated and not many sigma fields are generated by countable partitions.

$E(E(X|\mathcal F) I_A)=EXI_A$  for all $A \in \mathcal F$ by the definition of $E(X|\mathcal F)$. Just taking $A =\Omega$ gives $E(E(X|\mathcal F)) =EX$

Title: Total expectation under measure theory


问:
10 to the power of 3.5: $10^{3.5}$
So $10^3 = 10\times 10\times 10 = 1000$, this is really easy to understand.

But what about: $\,10^{3.5}\,?\,$ My logic would suggest this was

$10\times 10 \times 10\times 5 = 5000,\;$ but the calculator says it's 3162.27...

Can someone illustrate how the calculator calculates the power of when the number is with decimals?

Please keep in mind that I'm a newbie to mathematics!

答:
$$10^{3.5}=10^3\cdot 10^{.5}=10^3 \cdot 10^{\frac{1}{2}}=10^3 \sqrt{10}$$

Half power doesn't mean half of the number, it means square root.

Title: 10 to the power of 3.5: $10^{3.5}$


问:
Why does Gaussian elimination not preserve similarity of a matrix?
I am trying to understand reduction of an unsymmetric real square matrix to Hessenberg form from Numerical Recipes Vol. 3.

In it, the author states that one does not use Gaussian elimination for reducing to Hessenberg form because the Gaussian elimination does not preserve similarity and hence ends up changing the eigenvalues of the matrix, which is undesirable.

Why does Gaussian elimination not preserve similarity?

Also, what conditions decide whether a particular matrix transformation preserves similarity?

Heres a Google Books link to where I found it. In case you cant view that, it's on page 594 of Numerical Recipes in C++ (Volume 3).

答:
Gauss Elimination preserves similarity if you right multiply EA by E^-1 as you can see from this example

it's not unitary similarity but similarity. This algorithm was in old EISPACK library but it has been left out of LAPACK. See the beautiful book of Watkins D.S. Fundamentals of matrix computations on page 358.