问:
looking for reference or nice proof of trig lemma
Math people:

I am looking for a reference or a nice proof of the following fact.  I have proven it myself, but my proof is messy: let $\theta \in (0,1]$ and $\alpha \in (0, \frac{1}{2}\theta^2]$.  Let $P$ and $S$ be two points on the unit circle $x^2 + y^2 = 1$ such that the arc length of ${\stackrel{\frown}{PS}}$ is $\theta$.  Let $Q$ and $R$ be on ${\stackrel{\frown}{PS}}$ with $Q$ between $P$ and $R$, such that ${\stackrel{\frown}{PQ}}$ and ${\stackrel{\frown}{RS}}$ have equal arc length, and the arc length of ${\stackrel{\frown}{QR}}$ is $\alpha$.  Let $T$ and $U$ be the points on the chord ${\overline {PS}}$ such that $\Delta PQT$ and $\Delta SRU$ are right triangles. See the picture below.

Obviously I would not include a hand-drawn picture in a paper I would submit to a journal.

I have proven that the the triangles have disjoint closures.

I proved it the following way: I assumed $P = (\cos(\frac{1}{2}\theta),\sin(\frac{1}{2}\theta))$, $S = (\cos(\frac{1}{2}\theta),-\sin(\frac{1}{2}\theta))$.  Clearly it is good enough to take $\alpha = \frac{1}{2}\theta^2$.  Then $Q = (\cos(\frac{1}{4}\theta^2),\sin(\frac{1}{4}\theta^2))$.  I defined $V = (\cos(\frac{1}{2}\theta, 0)$.  It suffices to show $\angle VQP$ is obtuse.  I showed ${\vec{QP}} \cdot {\vec{QV}} < 0$.  This required a long calculation that was not pretty.  I am looking for a reference of this fact or a more elegant proof, perhaps a geometric proof.

答:
No one has answered in a long time, so I'll answer myself.  I'll start as I indicated in my question. It actually is not that bad.  Using standard trig identities,

$${\vec{QP}}\cdot{\vec{QV}} = (\cos(\frac{1}{4}\theta^2)-\cos(\frac{1}{2}\theta))^2-
      (\sin(\frac{1}{2}\theta)-\sin(\frac{1}{4}\theta^2))\sin(\frac{1}{4}\theta^2)=\ldots=
     \frac{3}{2}+\frac{1}{2}\cos(\theta)-\frac{1}{2}\cos(\frac{\theta}{2}+\frac{\theta^2}{4})-\frac{3}{2}\cos(\frac{\theta}{2}-\frac{\theta^2}{4}). $$

Using the Maclaurin series of $\cos$ and properties of alternating series,
$1-\frac{x^2}{2}<\cos x < 1-\frac{x^2}{2}+\frac{x^4}{24}$ for all $x \in (0,1)$.  Both
$\frac{\theta}{2}+\frac{\theta^2}{4}$ and $\frac{\theta}{2}-\frac{\theta^2}{4}$ are between $0$ and $1$.  Therefore

$${\vec{QP}}\cdot{\vec{QV}} <
   \frac{3}{2}+\frac{1}{2}(1-\frac{\theta^2}{2}+\frac{\theta^4}{24}) -
   \frac{1}{2}(1-(\frac{1}{2}(\frac{\theta}{2}+\frac{\theta^2}{4})^2)-
  \frac{3}{2}(1-(\frac{1}{2}(\frac{\theta}{2}-\frac{\theta^2}{4})^2) =
    -\frac{1}{8}\theta^3+\frac{1}{24}\theta^4<0.$$


问:
Is the collection of infinite sequence of 0 and 1 countable if we fix finite slots to be 1 and the rest of the slots are 0
I know the set of all infinite sequence of 0 and 1 is uncountable because they are just binary representation of the reals which are uncountable. Here I am asking whether the collection of all sequence where we have finite number of tuples are 1s and the rest 0s are countable? So for example the sequences (1,0,0,0,0...) , (0,1,0,0,0,0...) and we do this for all tuples. Then we move on to (1,1,0,0,0,0..)..(1,1,0,0,0,0,0). We go through every possible combinations of slots with 2 1s and then move on to 3 1s and 4s and so on and take the union of the entire collection. I believe collection is not equal to the set of all binaries because everything so far is countable and we are only taking countable unions so in the end, this collection is still countable right?</p>

答:
Good question. I think it is countable due to the fact that a countable union of countable sets is countable. The number of sequences with exactly one $1$ is countable. The number of sequences with exactly two $1$s is countable. etc
and there are countably many of these sets.

ContentLicense: CC BY-SA 4.0


问:
find all $a,b \in \mathbb N$ such that $9925-26a^2=b^2$
My Attempt:

$$9925-26a^2=b^2 \iff 9925=b^2 + 26a^2$$
Since $9925$ is odd and $26a^2$ is even $\implies b^2$ is odd $\implies b^2 = 2n+1$.
$$9925=2n+1+26a^2\iff 9925=2(n+13a^2)+1$$
$$ \iff 4962 = n+13a^2$$

Now We can reword this problem to Find all $(n, a)\in \mathbb N$ such that $13\mid 4962+\frac{1-b^2}{2}$ and $$\frac{4962+\frac{1-b^2}{2}}{13} = a^2$$

But the problem is that I don't know how to solve this.

答:
Note that $9925$ equals $6$ modulo $13$. Therefore $b^2$ should also have this remainder modulo $13$, but you can check by hand that this is never possible.

When trying to solve these types of contest math problems with diophantine equations, it's always a good idea to look at it modulo some small (prime) number.


问:
Prove that $\Bigl( 1 - \frac{1}{x} \Bigr)^{\frac{x}{\ln x}} = 1 + O \Bigl( \frac{1}{\ln x} \Bigr)$
It was my first attempt to use
$\lim\limits_{x \rightarrow \infty} \Bigl( 1 - \frac{1}{x} \Bigr)^{x} = e^{-x}$.
But then I didn't know what to do with $\frac{1}{\ln x}$.
On my second try, I wanted to use
\begin{align*}
\Bigl( 1 - \frac{1}{x} \Bigr)^{\frac{x}{\ln x}} - 1 = e^{-x} \Bigl( e^{x \frac{\ln(x-1)}{\ln(x)}} - e^x \Biggr).
\end{align*}
But the attempt was unsuccessful, because I didn't know how to estimate the term to prove the statement.

答:
It suffices to prove that
$$1 - \left(1 - \frac{1}{x}\right)^{\frac{x}{\ln x}} \le \frac{1}{\ln x},\ \forall x \ge \mathrm{e}^2.$$
It suffices to prove that
$$\left(1 - \frac{1}{\ln x}\right)^{\ln x} \le \left(1 - \frac{1}{x}\right)^{x}, \ \forall x \ge \mathrm{e}^2.$$

Let $f(u) = (1 - \frac{1}{u})^u$. We have
$$f'(u) = (1 - \frac{1}{u})^u \left(\ln\left(1 - \frac{1}{u}\right) + \frac{1}{u-1}\right) > 0,
\ \forall u\ge 2.$$

Since $2 \le \ln x < x$ for $x\ge \mathrm{e}^2$, we have $f(\ln x) < f(x)$.

We are done.


问:
Is this function defined in this domain continuous at $x = 0$?
$f$, a function defined on $(-1,1)$, satisfying $|f(x)|\leq \sqrt \mid{x}$

How can I tell if this is continuous at $x = 0$?

答: As $\lim\limits_{x \to 0} \sqrt{\vert x \vert}=0$, you have $\lim\limits_{x \to 0} \vert f(x) \vert =0$ according to the squeeze theorem. Therefore $f$ is continuous at zero.


问:
Dense Countable Subsets of the Unit Disk
Consider a countable dense subset A of the unit disk D in R. Is it true that the sum over elements of A vanishes? Intuitively if we choose a neighborhood U of a point x in D we may find points in A arbitrarily close to x. Similarly we may do this with -x hence we may find points in A whose sum is arbitrarily close to zero. Doing so for each x in D should lead us to the conclusion that the sum over A is zero (or as close as we'd like).</p>

答: If we're thinking of the unit disk $D$ in $\mathbb{R}$, then let's take $S = (0,1) \cap \mathbb{Q}$ as our dense countable subset (everything I'm saying works for any $S$, but this is nice for examples). Let's assume we're summing $S$ in some order and that the sum converges. It's pretty clear that the sum is not absolutely convergent, so by Riemann's rearrangement theorem we can change the order of summation to sum to anything at all. In specific we can order it in such a way that the sum is zero (for instance, $0 + \frac{1}{2} - \frac{1}{2} + \ldots)$. But we can also have it sum to $1$ (try to come up with this one, it's not hard), or $2$, or $\infty$. Basically, there isn't really a notion of just summing a set, unless it's a very well-behaved set (for instance, is composed only of finitely many number of one sign, and the rest of the other).</p>

ContentLicense: CC BY-SA 3.0


问:
What a graph should look like with a diameter at least 5 times the average shortest path length?
I need to find what a graph should look like with a diameter at least 5 times the average shortest path length. I understand that it must be a compete graph with a long tail, cause with longer tail diameter will grow faster than the average shortest path length. But I don't know how to find exact number of nodes of that graph. How many nodes there must be in complete graph and how many in tail? Could I define those numbers using formulas of graph diameter and average shortest path length?

By experiments and iterations, I found that complete graph with 100 nodes and tail with 11 nodes has diameter 12 and  average shortest path length 2.097, so it is 5 times larger, but how to find it using math?

答:
I looked at the minimum complete graph $K_i$ that with two $j$-node tails would produce a ratio of diameter to average shortest path, $r>5$.

I found the average shortest path in five parts:

\begin{array}{|c|c|} \hline
\text{Part of graph} & \text{#paths} & \text{av. length}\\ \hline
\text{Within }K_i & i(i-1)/2 & 1   \\
\text{Tails to base} & 2j  &  (j+1)/2  \\
\text{Tails to other }K_i & 2j(i-1) &  (j+1)/2 + 1  \\
\text{Tails to self} & j(j-1)  & (j+1)/3   \\
\text{Tail to tail} & j^2 &  j+2  \\ \hline
\end{array}

Of course there was a minimum smallest $K_i$ for any given $j>2$. Slightly unexpectedly this was itself minimum at $j=5,6$ (I thought it would be higher before the length of the tails became significant):

\begin{array}{|c|c|} \hline
j & \min i \\ \hline
3 & 52  \\
4 & 39  \\
5 & 36  \\
6 & 36  \\
7 & 37  \\
8 & 38  \\
9 & 40  \\ \hline
\end{array}


问:
Is there an easy way to calculate the probability of rolling a certain number or higher with multiple die.
For example, assuming I have a 6 sided die, a 4 sided die, and 10 sided die. Is there a way to calculate the probability of rolling a total sum of 14 or higher?

I know I could manually add each combination to see if they are equal to or greater than 14, and then divide the number of outcomes which do by the total number of possible outcomes. But that is a bit tedious.

答:
For large number of die you could use the CLT (approximate the sum to a gaussian).

Here, you must compute that numerically.

With Octave or Matlab:

```<code>pa = ones(1,6)/6; % pmf of die A
pb = ones(1,4)/4; % pmf of die B
pc = ones(1,10)/10; % pmf of die C
pab = [0,conv(pa,pb)] % pmf of sum A,B
pabc = [0,conv(pab,pc)]; % pmf of sum A,B,C
sum(pabc(14:end)) % prob of sum being 14 or higher
</code>```

This gives $0.30417$

Body:
For large number of die you could use the CLT (approximate the sum to a gaussian).

Here, you must compute that numerically.

With Octave or Matlab:

pa = ones(1,6)/6; % pmf of die A
pb = ones(1,4)/4; % pmf of die B
pc = ones(1,10)/10; % pmf of die C
pab = [0,conv(pa,pb)] % pmf of sum A,B
pabc = [0,conv(pab,pc)]; % pmf of sum A,B,C
sum(pabc(14:end)) % prob of sum being 14 or higher

This gives $0.30417$


问:
Inverse of the Third Axiom in the Definition of a Sigma Algebra
$\Sigma$ is a $\sigma$-algebra if it satisfies the following three properties:

$X$ is in $\Sigma$, and $X$ is considered to be the universal set in
the following context.

$\Sigma$ is closed under complementation: If $A$ is in $\Sigma$, then
so is its complement, $A^{c}=X-A$.

$\Sigma$ is closed under countable unions: If $A_{1}$, $A_{2}$,
$A_{3}$, ... are in $\Sigma$, then so is $A=A_{1}\cup A_{2}\cup
   A_{3}\cup\ldots$.

If the above three are our definitions.

Can we show that if  $A=A_{1}\cup A_{2}\cup A_{3}\cup\ldots$ and $A$ is in $\Sigma$ then each of the $A_{i}$ are in $\Sigma$?

Do we need any additional conditions to show this or would the three basic axioms suffice?

Notes:

I found some related discussions, but am unable to see if my doubts can be cleared using the pointers from these threads. Listing them below for completeness:

Can every member of a $\sigma$-algebra be represented by a countable union of disjoint members?

Show that any element of a sigma algebra is the union of disjoint sets

I was trying to understand the concept in the question asked since the corresponding property seemed to be required to show that If $E$ has $\sigma$-finite measure, then $E$ is inner regular (Rudin RCA, Definition 2.16, some related threads below):

If $E$ has $\sigma$-finite measure, then $E$ is inner regular

Rudin Real And Complex Definition 2.16

This point is clarified: If $E$ is in a sigma algebra $\Sigma$ and $E$ has $\sigma$-finite measure then it is a countable union of sets, $E_{i}$ with finite measure ($\mu\left(E_{i}\right)<\infty$). Which means the sets that make up the union of $E$ are already in $\Sigma$ since the measure $\mu$ is defined only on the members of the sigma algebra $\Sigma$. Hence the $E_{i}$ belong to the sigma algebra, $\Sigma$

Though my original doubt is clarified, I am still curios about the conditions under which the reverse of the third axiom hold. It is quite possible that I am missing something very basic. Happy to delete this question, if that is the case once my doubt is clarified.

答:
@texmex wrote "I am still curios about the conditions under which the reverse of the third axiom hold".

Let us prove the following result.

Let $X$ be a set and $\Sigma$ be a $\sigma$-algebra of subsets of $X$. Then the condition

if  $A=A_{1}\cup A_{2}\cup A_{3}\cup\ldots$ and $A$ is in $\Sigma$ then each of the $A_{i}$ are in $\Sigma$

is true if and only  $\Sigma = 2^X$.

where $2^X$ means the collection of all subsets of $X$

Proof: $(\Rightarrow)$ We already have tha $\Sigma \subseteq 2^X$. So all we need to prove is that  $2^X  \subseteq \Sigma  $.

Suppose that,

if  $A=A_{1}\cup A_{2}\cup A_{3}\cup\ldots$ and $A$ is in $\Sigma$ then each of the $A_{i}$ are in $\Sigma$.

Let $E$ be any element of $2^X$, that means any subset of $X$. We can take $A_1=E$, $A_2 = X$ and  $A_3=A_4=\ldots = \emptyset$. We have
$$A_{1}\cup A_{2}\cup A_{3}\cup\ldots = E\cup X = X \in \Sigma$$ so, by the condition, we have that each of the $A_{i}$ are in $\Sigma$. In particular $A_1 \in \Sigma$, which means $E \in \Sigma$. Since $E$ is an arbitrary element of $2^X$, we have that $2^X \subseteq \Sigma$. So we have $\Sigma = 2^X$.

$(\Leftarrow)$  If $\Sigma = 2^X$, then all subsets of $X$ are in $\Sigma$ and the condition

if  $A=A_{1}\cup A_{2}\cup A_{3}\cup\ldots$ and $A$ is in $\Sigma$ then each of the $A_{i}$ are in $\Sigma$

becomes trivially true.

Remark: Another way to prove $(\Rightarrow)$ is by the counter-positive, which consists in showing that if $\Sigma \neq 2^X$, then there is always a counter-example to

if  $A=A_{1}\cup A_{2}\cup A_{3}\cup\ldots$ and $A$ is in $\Sigma$ then each of the $A_{i}$ are in $\Sigma$

In fact, suppose $\Sigma \neq 2^X$. then there is $E \in 2^X$ such that $E \notin \Sigma$. Take $A_1=E$, $A_2 = X$ and  $A_3=A_4=\ldots = \emptyset$. We have
$$A_{1}\cup A_{2}\cup A_{3}\cup\ldots = E\cup X = X \in \Sigma$$
but $A_1=E \notin \Sigma$.


答:
You're right; these form a countable set. In the decimal (or rather binary) expansion analogue the ones ending in all $0$ are like the rational numbers (that have eventually repeating expansions).

If $S_n$ are all sequences with exactly $n$ $1$'s (and all others $0$); there are countably many of those: countably many choices for the positions of the $1$'s. And your set is $\bigcup_n S_n$ so countable.

Or see all sequences as all subsets of $\Bbb N$ (via indicator functions) and all your sequences are the finite subsets of $\Bbb N$, a countable set.

Title: Is the collection of infinite sequence of 0 and 1 countable if we fix finite slots to be 1 and the rest of the slots are 0

Body: I know the set of all infinite sequence of 0 and 1 is uncountable because they are just binary representation of the reals which are uncountable. Here I am asking whether the collection of all sequence where we have finite number of tuples are 1s and the rest 0s are countable? So for example the sequences (1,0,0,0,0...) , (0,1,0,0,0,0...) and we do this for all tuples. Then we move on to (1,1,0,0,0,0..)..(1,1,0,0,0,0,0). We go through every possible combinations of slots with 2 1s and then move on to 3 1s and 4s and so on and take the union of the entire collection. I believe collection is not equal to the set of all binaries because everything so far is countable and we are only taking countable unions so in the end, this collection is still countable right?</p>


问:
Expressing Integral as Power Series
I am asked to express the given function that involves an improper integral in a form $f_n(z)=\frac{1}{(n-1)!}\int_0^\infty \frac{x^{n-1}}{\frac{1}{z}e^x+z}dx$ in terms of power series centered at x = 0.
My attempt was to express $e^x$ in terms of its power series, such that:
$e^x =\sum_{n=0}^\infty \frac{x^n}{n!} $ but in that case the sum terms is in the denominator and I am incapable of combining it with the independent value of the numerator to combine it into a single sum. Am I doing it right? I am sort of confused on this one. I've got to a final step of $f_n(z)=\frac{1}{(n-1)!}\int_{0}^\infty \frac{x^{n-1}}{\frac{1}{z}\sum_{n=0}^\infty \frac{x^n}{n!}+z}$. I am not sure how to reduce this whole expression into a series from here, as I've only learnt how to get rid of the integral if the sum term is in the numerator.
Any help would be greatly appreciated.

Title: Expressing Integral as Power Series

Body:
Let's assume that $|z|\leq 1$. If not, then things become more complicated.

$$f_n(z)=\frac{z}{(n-1)!}\int_0^\infty \frac{x^{n-1}}{e^x+z^2}dx$$

$$f_n(z)=\frac{z}{(n-1)!}\int_0^\infty \frac{e^{-x}x^{n-1}}{1+e^{-x}z^2}dx$$

$$f_n(z)=\frac{z}{(n-1)!}\int_0^\infty e^{-x}x^{n-1} \sum_{k=0}^\infty (-1)^k e^{-kx}z^{2k}dx$$

$$f_n(z)=\frac{z}{(n-1)!} \sum_{k=0}^\infty (-1)^k z^{2k} \int_0^\infty x^{n-1}  e^{-(k+1)x}dx$$

$$f_n(z)=\frac{z}{(n-1)!} \sum_{k=0}^\infty \frac{(-1)^k z^{2k}}{(k+1)^n} \int_0^\infty t^{n-1}  e^{-t}dt$$

$$\color{blue}{f_n(z)=z \sum_{k=0}^\infty \frac{(-1)^k z^{2k}}{(k+1)^n}}$$


答:
You say "unit disk in $\mathbb{R}$", which is just the interval $[-1,1]$, but the same argument works in $\mathbb{R}^n$ too. A countable dense subset $S\subseteq[-1,1]$ will necessarily have infinitely many terms of absolute value greater than $\frac{1}{2}$, so the sum
$$\sum_{x\in S}|x|$$
diverges. Riemann's rearrangement theorem implies that the sum
$$\qquad\qquad\sum_{n=1}^\infty x_i,\qquad S=\{x_1,x_2,\ldots\}$$
can therefore be reordered to give any real number whatsoever, and so in particular, any non-zero number. But just to give an example where the "intuitive" sum is not zero, let $S$ be
$$S=(\mathbb{Q}\cap[-1,1])\cup\left\{\tfrac{1}{\sqrt{2}}\right\}.$$
All the rationals "cancel out with their negatives" (metaphorically) and $\tfrac{1}{\sqrt{2}}$ is left over.</p>

Title: Dense Countable Subsets of the Unit Disk

Body: Consider a countable dense subset A of the unit disk D in R. Is it true that the sum over elements of A vanishes? Intuitively if we choose a neighborhood U of a point x in D we may find points in A arbitrarily close to x. Similarly we may do this with -x hence we may find points in A whose sum is arbitrarily close to zero. Doing so for each x in D should lead us to the conclusion that the sum over A is zero (or as close as we'd like).</p>


问:
Doubt in connected and path connected s
My teacher taught me that a set of points is connected if and only if it is path connected. However,the Topologist's sine curve is connected but not path connected. Am i missing something here?</p>

答: No, your teacher is wrong. Even for planar subsets connected does not imply path-connected. In $\Bbb R$ this equivalence does hold, maybe you misheard him and did he mean that instead. Otherwise just plain wrong. Within the class of locally path-connected spaces the equivalence does hold, though. So e.g. in $\Bbb C$, the regions/domains (open connected subsets) are also path-connected (even with piecewise linear paths) which becomes important for path-integration theory.

Title: Doubt in connected and path connected s


问:
Can the quotient of a division of a polynomial act as the residual in the remainder theorem?
The problem is as follows:

The polynomial $p(x)$ is divisible separately by $(x-2)$, $(x-3)$,
$(x-4)$, and by $(x-5)$. If $p(x)$ is divided by $(x-9)$ leaves a
remainder equal to $5040$, what is the independent term of such
polynomial?

The alternatives given in my book are as follows:

$$\begin{array}{cc}
1.&\textrm{600}\\
2.&\textrm{360}\\
3.&\textrm{740}\\
4.&\textrm{720}\\
\end{array}$$

The official solution for this problem according to my workbook in precalculus states the following argument:

$p(x)$ is divisible separately by $(x-2)$, $(x-3)$, $(x-4)$ and $(x-5)$ $\Leftrightarrow$ $p(x)$ is divisible by $(x-2)$, $(x-3)$, $(x-4)$ and $(x-5)$.

Using the division algorithm we have:

$$p(x)= (x-2)(x-3)(x-4)(x-5)\cdot q(x)$$

Therefore $\operatorname{grad}[q(x)]=0$ (comment: here the notation grad indicates the degree of the polynomial)

Hence $q(x)=M$

Using the remainder theorem it is obtained:

$$\frac{p(x)}{x-9}$$

$$p(9)=(9-2)(9-3)(9-4)(9-5)q(x)=5040$$

Then:

$$(7)(6)(5)(5)(4)M=5040$$

$$M=6$$

Then to get the independent term we do evaluate the polynomial in zero as: $p(0)$

$$p(0)=(0-2)(0-3)(0-4)(0-5)(6)=6\cdot 20 \cdot 6 =720$$

which corresponds with the option number $5$.

That's where it ends the official solution.

Now my problem is that I'm confused with the way how it was found the degree of the quotient and why it was earlier declared as such and later used as a remainder.

First and foremost, how do we know the degree of the quotient is zero? I'm only assuming that since this polynomial is divisible by those factors it implies an exact division, hence the quotient can only be a constant term, otherwise it will increase an additional factor for which the polynomial is not divisible with. But is this assumption correct?. Can someone explain this?

The second part for which I'm confused is a weird usage of the polynomial remainder theorem.

If my memory serves me right, the remainder theorem goes as this:

When a polynomial is divided by $(x-a)$ then

$$p(x)=d(x)\cdot q(x)+r$$

$$p(x)=(x-a)q(x)+r$$

When we do evaluate the polynomial in a thus:

$$p(a)=r$$

Hence evaluating the polynomial with that $a$ will result in the remainder of the division which it can be another polynomial, right? But when it is applied to a linear quotient then the remainder is a constant term.

But in this case, the way how it was made such interpretation is some confusing because it uses $M$ and $M$ was earlier declared as the quotient.

Shouldn't it be as:

$$p(x)=(x-9)((x-2)(x-3)(x-4)(x-5)\cdot q(x))+r=5040$$

Because of these inconsistencies. I'd appreciate someone could help me to understand what might the author had intended to do, because I'm lost. Perhaps does it exist a better method of argumentation here?. I appreciate if someone could explain with words the right interpretation of the polynomial remainder theorem because this is the part which I'm struggling the most. How is it being used here?.

The part which it confuses me a lot is why is it treating $M$ as if it were $r(x)$ if it was earlier declared as quotient?. Isn't this a contradiction when evaluates the polynomial on $p(9)$?. So all and all, can someone help me to address these doubts please?

答:
The only use of the remainder theorem is to recognise that $p(9)=5040.$

From the point where $p(x)= M(x-2)(x-3)(x-4)(x-5)$ simply substitute $x=9$ to obtain $5040=7.6.5.4.M$ i.e. $M=6$.

Multiplying out gives the constant term as $2.3.4.5M=720$.

P.S.

Your five lines saying what you remember of the remainder theorem are correct. Don't let the answer in the book confuse you!


答:
In general, path-connected implies connected. The converse is not true as you point out with the topologist's sine curve. As Henno is correctly pointing out, the notion is equivalent in $\mathbb{R}$ since the topology in $\mathbb{R}$ is "so poor" that it is impossible to find a non-path-connected set that is connected. To be able to find one, you need to go on "bigger" topological spaces, in some sense.

I suggest that you go here to see how far you can generalize the equivalence, where it is noticed that in a locally path-connected space, connected open sets are path-connected. Every topological vector space over $\mathbb{R}$ or $\mathbb{C}$ is locally path-connected since the balanced neighbourhoods form a local base at 0.

Title: Doubt in connected and path connected s

Body: My teacher taught me that a set of points is connected if and only if it is path connected. However,the Topologist's sine curve is connected but not path connected. Am i missing something here?</p>


答: Even ignoring any convergence/divergence questions, while it is true that for each $x$, you can find another point arbitrary close to $-x$, there's no guarantee that the points are uniformly distributed in the disc. If there are more points near $x$ than near $-x$, then you can't count the ones near $-x$ twice, so it's possible the sum is not zero.</p>


问:
Partial Fraction Decomposition methods
I have this problem where I have to solve the following integral.

${\displaystyle\int}\dfrac{1}{\left(x+a^2\right)\left(x+b\right)^2}\,\mathrm{d}x$

I checked the solution on https://www.integral-calculator.com/ and the first step of the solution is a trick.

$={\displaystyle\int}\left(-\dfrac{1}{\left(b^2-2a^2b+a^4\right)\left(x+b\right)}-\dfrac{1}{\left(b-a^2\right)\left(x+b\right)^2}+\dfrac{1}{\left(b^2-2a^2b+a^4\right)\left(x+a^2\right)}\right)\mathrm{d}x$

So here the fraction is decomposed into 3 parts. Afterwards the integral is easily solved by applying linearity. How to determine in how many parts the fraction should be decomposed and how to actually decompose it without guessing?

答: This is usual partial fraction decomposition. The referenced Wikipedia article provides algorithms to compute the decomposition.</p>

Title: Partial Fraction Decomposition methods


答:
Here $x = -b$ is a pole of order $2$  and $x = -a^2$ is a pole of order $1$. There is no quadratic or higher order terms here in the denominator.

So the partial fraction will consist of $2+1 = 3$ terms.

In general (considering only linear factors in the denominator)

$\frac{c}{(x-p_1)^{k}(x-p_2)^{h}}$ ($k$ and $h$ are positive integers) can be split into $k+h$ partial fractions as follows.

$$\begin{align}\frac{c}{(x-p_1)^{k}(x-p_2)^{h}} &= \left[\frac{l_1}{x-p_1} +\frac{l_2}{(x-p_1)^2} +\cdots+\frac{l_k}{(x-p_1)^k}\right] \\&+\left[ \frac{m_1}{x-p_2} +\frac{m_2}{(x-p_2)^2} +\cdots\frac{m_h}{(x-p_2)^h} \right]\end{align} $$

($l_i$ and $h_i$ are constants to be determined.)

Title: Partial Fraction Decomposition methods

Body:
I have this problem where I have to solve the following integral.

${\displaystyle\int}\dfrac{1}{\left(x+a^2\right)\left(x+b\right)^2}\,\mathrm{d}x$

I checked the solution on https://www.integral-calculator.com/ and the first step of the solution is a trick.

$={\displaystyle\int}\left(-\dfrac{1}{\left(b^2-2a^2b+a^4\right)\left(x+b\right)}-\dfrac{1}{\left(b-a^2\right)\left(x+b\right)^2}+\dfrac{1}{\left(b^2-2a^2b+a^4\right)\left(x+a^2\right)}\right)\mathrm{d}x$

So here the fraction is decomposed into 3 parts. Afterwards the integral is easily solved by applying linearity. How to determine in how many parts the fraction should be decomposed and how to actually decompose it without guessing?


问:
Solving a surface integral
Given the unit sphere $S = \{(x,y,z)\in \mathbb{R}^3:x^2+y^2+z^2=1\}$. I have to find the value of the surface integral $\iint_S \{ \frac{2x}{\pi} + sin(y^2))\hat{x} + (e^z - \frac{y}{\pi})\hat{y} + (\frac{2z}{\pi} + sin^2y)\hat{z}\}d\sigma$. I used Divergence Theorem to arrive at the answer 3. But the answer is shown to be 4.

答:
You have not shared your working so difficult to say where you made a mistake but $4$ is indeed the correct answer.

$\vec{F} = (\frac{2x}{\pi} + \sin y^2) \hat x + (e^z - \frac{y}{\pi}) \hat y + (\frac{2z}{\pi} + \sin^2 y) \hat z$

$\nabla \cdot \vec F = \frac{2}{\pi} - \frac{1}{\pi} + \frac{2}{\pi} = \frac{3}{\pi}$

$\displaystyle \iint_S \vec F \cdot \hat n \ dS = \iiint \nabla \cdot \vec F \ dV = \frac{3}{\pi} \iiint  dV = \frac{3}{\pi} \times \frac{4 \pi}{3} = 4$.

$\iint dV$ is the volume of the unit sphere which we know is simply $\frac{4\pi}{3}$. If you are integrating to find that, you can write the integral in spherical coordiantes as,

$\displaystyle \int_0^{2\pi} \int_0^{\pi} \int_0^1 \rho^2 \ sin \phi \ d\rho \ d\phi \ d\theta = \frac{4 \pi}{3}$.

Title: Solving a surface integral


问:
Sum of sum of odd terms of Vandermonde's Identity
The Vandermonde's Identity tells us:

$\displaystyle \large {m+n \choose r}=\sum_{k=0}^r{m \choose k}{n \choose r-k}$

However, I want to find out the closed form or the simplest form of the expression:

$\displaystyle \large \sum_{r=1}^{p}\sum_{k=0}^r{m \choose k}{n \choose r-k}$, where $p$ is a natural number.

But I only want to include the terms with odd $k$ to be included in the inner sum (the $\displaystyle \large \sum_{k=0}^r{m \choose k}{n \choose r-k}$ one)

I had previously posted a similar question before, but I couldn't get a simplification for that case. However, with a change in the nature of the sum I believe will render it to have a simplification, at least this time.

答:
Idea. The original vandermonde identity is a consequence of

$$ (1+x)^m (1+x) ^n = (1+x) ^{m+n}$$

(look at the coefficient of $x^r$ !!). Now we do this trick: if we want only odd $k$ to appear, we should consider $((1+x) ^m - (1-x) ^m ) /2$ (why? This is an instructive exercise). In this way we get,

$$ ((1+x) ^m - (1-x) ^m )(1+x) ^n /2 = ((1+x) ^{m+n} - (1-x)^m (1+x) ^n )/2$$

Unfortunately, I don't think this will have a closed form. Nevertheless, for $p=m+n $ in which we sum all terms we have a lucky strike. Summing up all the coefficients of a polynonial is like calculating it in 1, so that the result is $2^{m+n}$. Since you wanted to exclude $r=$0 which contributed with +1, you'll get $2^{m+n}-1$.

These techniques are based on generating functions, which I encourage you to deepen since they are a wonderful subject. Hard equalities like Vandermonde comes from very easy algebraic manipulations.


答:
Since $\overrightarrow{F} = (\frac{2x}{\pi} + sin y^2, e^z - \frac{y}{\pi}, \frac{2z}{\pi} + sin^2 y)$,
$\overrightarrow{\nabla}\cdot\overrightarrow{F} = \frac{2}{\pi} -\frac{1}{\pi} + \frac{2}{\pi} = \frac{3}{\pi}$

Since the volume of the sphere is $\frac{4\pi}{3}$, the answer is 4.

Title: Solving a surface integral

Body: Given the unit sphere $S = \{(x,y,z)\in \mathbb{R}^3:x^2+y^2+z^2=1\}$. I have to find the value of the surface integral $\iint_S \{ \frac{2x}{\pi} + sin(y^2))\hat{x} + (e^z - \frac{y}{\pi})\hat{y} + (\frac{2z}{\pi} + sin^2y)\hat{z}\}d\sigma$. I used Divergence Theorem to arrive at the answer 3. But the answer is shown to be 4.


问:
Does quotient rule and negative exponent property sometimes lead to different answers?
So what should I do is taking the second derivative of:
$$[\frac{1}{1-\lambda t}]^a$$
with respect to t at t = 0.

When I use the quotient rule to find the second derivative at t = 0.

$${\lambda a} * [\frac{1}{1-\lambda t}]^{a-1}$$

Is my first derivative and when I differentiate this expression with respect to t at t = 0. Using the quotient rule I find the answer as:

$$\lambda^2 a^2 - \lambda^2 a$$

Using the negative exponent property makes the expression:

$$ (1- \lambda t)^{-a} $$

and I find its derivative with respect to t at t = 0 as:

$$ \lambda a * (1-\lambda t)^{-a-1} $$

So now I see that it is pretty much different than the derivative of the same expression I have had by quotient rule. When I take its second derivative with respect to t at t = 0

I find the answer correctly as:

$$ \lambda^2 a^2 + \lambda^2 a $$

So where am I mistaken while applying the quotient rule?

答:
You should get the same answer by using the quotient rule as you do by using the negative exponent property. When you take the first derivative using the quotient rule, you should let $f(x)=1$ and let $g(x)=(1-\lambda t)^a$. That gives us $f'(x)=0$ and $g'(x)=-\lambda a(1-\lambda t)^{a-1}$. Then the function that you're differentiating is $f(x)/g(x)$, so its derivative is $\frac{g(x)f'(x)-f(x)g'(x))}{g^2(x)}$. Plugging in the functions we found earlier, we get $$\frac{\lambda a(1-\lambda t)^{a-1}}{(1-\lambda t)^{2a}}=\frac{\lambda a}{(1-\lambda t)^{a+1}}$$

which is equal to what you found using the negative exponent property. If you use the quotient rule on this again, you should get the same second derivative.

As a rule of thumb, if you have a fraction where the numerator is a constant, you should just use the negative exponent property, since the quotient rule will give you the same answer, but in an overly complicated form.


问:
Characterize $\frac{df}{dx}(x)f(x)f(x)=x$
can you help me characterize $\frac{df}{dx}f(x)f(x)=x$? Should be 1st order, not-linear, not elliptic, parabolic, hyperbolic.

Solution with characteristics not possible because not (quasi)linear.

Any ideas for characterization/solution theory?

Thanks!

答:
This is an example of a separable ODE. Note that $f'f^2 = (f^3/3)'$. So just set $g=f^3/3$, then the equation is simply $g'=x$. We can solve by just integrating both sides
$$ g(x) = \frac{x^2}2 + C$$
i.e. the general solution is
$$ f(x)= \left(\frac{3x^2}2 + C\right)^{1/3}$$


问:
Show that a bounded subset of a normed vector space is in a closed ball
Let $E$ be a normed vector space and $A$ be a non-empty bounded subset of $E$.

Let $I = \{R \in \mathbb{R}^+, A \subset \mathcal{B}(x,R), x\in E\}$. Let $r = \inf I$.

The goal is to show that $A \subset \mathcal{B}(x,r)$ where $x\in E$.

I started to show that $I$ is non-empty set and that $r$ does exist. I have shown it using a proof by contradiction. If $I$ is empty then $A$ is not bounded.

I have defined $R_n=r + \frac{1}{n}$. I must show that there exists $x_n \in E$ such that $A \in \mathcal{B}(x_n, R_n)$. I don't know how to prove it since it seems intuitive that a smaller radius of the ball implies that $x_n$ is closer from $A$.

Then I've shown that $(x_n)$ is bounded (again by contradiction). I have to deduce that there exists a convergent subsequence. I immediately thought about Bolzano-Weierstrass theorem, but I need to prove that $E$ is a compact set to use the theorem, and I don't see how to prove it. I also thought about the fact that in a normed vector set, closed balls are compact sets but I couldn't progress any further than that.

Finally I have to show that $A \subset \mathcal{B}(x,r)$ where $x\in E$, where $x$ is the limit of the convergent subsequence. Here is the beginning of my proof for that one but I block at some point:

Let $\epsilon > 0$.

$\exists N_1 \in \mathbb{N}, \forall n \geq N_1, \lvert\lvert x_{\phi{n}} - x \rvert  \rvert < \epsilon$.

$\exists N_2 \in \mathbb{N}, \forall n \geq N_2, \lvert R_{\phi(n)} - r \rvert < \epsilon$.

Let $N_3 = \max(N_1, N_2)$ and $y \in A$.

$$ \lvert\lvert x-y \rvert\rvert \leq \lvert\lvert x - x_{\phi(n)}\rvert\rvert + \lvert\lvert x_{\phi(n)} - y\rvert\rvert $$
$$ \lvert\lvert x-y \rvert\rvert \leq \epsilon + R_{\phi(n)}$$

All in all, I think I have the intuition for the exercise. I can make a drawing of it on a sheet of paper but I can't really formalize the proof.

答:
"$E$ is a compact set". NO. The only compact normed vector space is the trivial space $\{0\}$.

"In a normed vector set, closed balls are compact". NO. This is only true in finite-dimensional normed vector spaces.

The main result may be false if $E$ is not a Banach space. For example let $E$ be the set of continuous piece-wise linear $f:[0,1]\to \Bbb R$ with the $\sup$ norm $\|f-g\|=\sup_{x\in [0,1]}|f(x)-g(x)|.$ Let $A=\{f\in E: \forall x\in [0,1]\,(|f(x)-x^2|\le 1)\}.$ For every $s>1$ there is an open $E$-ball of radius $s$ that covers $A.$ But even a closed $E$-ball $\overline {B(f,1)}$ cannot cover $A,$ else $f(x)=x^2$ for all $x\in [0,1],$ but then $f\not \in E.$


问:
Integral with modified Bessel functions of the first and second kind
Consider the following integral
$$I\left(a, b\right)=\int\limits _0^{+\infty}\cos\left(x\right)I_1\left(ax\right)K_1\left(bx\right)dx,$$
where $a$ and $b$ are parameters such that $b\geq a \geq 0$, and $I_\nu\left(x\right)$ and $K_\nu\left(x\right)$ are the modified Bessel functions of the first and second kind, respectively.

Is it possible to express this integral in terms of elementary and basic special functions?

Similar issue has already been addressed; the result involves complete elliptic integrals.

答:
For $a=b>0$, a CAS produces
$$I(a,a)=\frac{\left(2 a^2+1\right) K\left(-4 a^2\right)-E\left(-4 a^2\right)}{2 a^2}$$


问:
Q: Construct a basis $β$ for $V$ such that $[T]_{\beta}$ = $\begin{pmatrix}0_{2x2} & X\\\ 0_{2x2} & 0_{2x2}\end{pmatrix}$ for some $X \in M_{2x2}(C)$
I've done similar questions regarding finding a basis for vector spaces, however this particular question I'm having difficulty to knowing even where to begin.
Any guidance on where to start would be greatly appreciated. Thank you.

Question:
Consider the vector space $V = C^{2}$ with scalar multiplication over the real numbers $\mathbb{R}$, and let $T : V \rightarrow V$ be the linear operator defined by $$T(\begin{bmatrix}z_{1} \\\ z_{2} \end{bmatrix})=\begin{bmatrix}-1 & i \\\ i & 1 \end{bmatrix} \begin{bmatrix}z_{1} \\\ z_{2} \end{bmatrix}$$

Construct a basis $β$ for $V$ such that $[T]_{\beta}$ = $\begin{pmatrix}0_{2x2} & X\\\ 0_{2x2} & 0_{2x2}\end{pmatrix}$ for some $X \in M_{2x2}(C)$.

答: If $\beta=\{v_1,v_2,v_3,v_4\}$, then, since $[T]_\beta=\left[\begin{smallmatrix}0_{2\times2}&X\\0_{2\times2}&0_{2\times2}\end{smallmatrix}\right]$, $T(v_1)=T(v_2)=0$. Besides, if $z_1=i$ and $z_2=1$, you have $T\left(\left[\begin{smallmatrix}z_1\\z_2\end{smallmatrix}\right]\right)=0$. So, take $v_1(i,1)$. For the same reason, you can take $v_2=(-1,i)$. Now, find $v_3$ and $v_4$ such that both $T(v_3)$ and $T(v_4)$ are linear combinations of $v_1$ and $v_2$.


问:
Probability Chip Tossing Problem with Different Chip Tossing Counts
Mandy and Chuck are flipping are flipping chips. The chip has two sides: head and tails. Each round Mandy flips 3 chips and Chuck flips 2 chips. Each round, they might throw an equal number of heads. In the round they first flip different numbers of heads, what is the probability Mandy threw more heads than Chuck?

My work:

I started off by listing the possible outcomes.

Mandy: HHH, HHT, HTH, THH, TTH, THT, HTT, TTT

Chuck: HH, HT, TH, TT

There are 8*4 = 32 possible outcomes in terms if rounds

I listed possible outcomes where Mandy threw more heads than Chuck.

1.HHH, HH
2.HHH, HT
3.HHH, TH
4.HHH, TT
5.HHT, HT
6.HHT, TH
7.HHT, TT
8.HTH, HT
9.HTH, TH
10. HTH, TT
11. THH, HT
12. THH, TH
13. THH, TT
14. TTH, TT
15. HTT, TT

I got 15 different possibilities. Does this mean the answer is 15/32.

Thank you very much

I'm not really sure how to approach this problem?

答:
Let $A$ = the event "They throw different numbers of heads"

Let $B$ = the event "Mandy threw more heads than Chuck"

$P(B)$ is what you computed, $15/32$.

You want $P(B|A) = \frac{P(B \cap A)}{P(A)}$

Let $M$ = the number of heads Mandy gets

Let $C$ = the number of heads Chuck gets

$P(M=C) = P(M=C=0) + P(M=C=1) + P(M=C=2)$

Use binomials in general to find the probabilities of M throwing k heads and Chuck throwing k heads and then multiply them together because they are independent. Then sum those results.

$P(M≠C) = 1 - P(M=C)$

Similar expansion of possibilities gives the numerator.


问:
Curve fitting with polynomials and inner products
So, I was learning about the curve fitting methods, and the teacher was showing an example where he would have some data where each point was represented with $(x_1, x_2, x_3)$ and we had a measurement $y$ at that point. We wanted to approximate the $y$ with a linear combination of polynomials.

We first defined:

$$
\begin{matrix}
p_1(x_1, x_2, x_3) = 1 & p_2(x_1, x_2, x_3) = x_1 & p_3(x_1, x_2, x_3) = x_2 \\
p_4(x_1, x_2, x_3) = x_3 & p_5(x_1, x_2, x_3) = x_1^2 & p_6(x_1, x_2, x_3) = x_2^2 \\
p_7(x_1, x_2, x_3) = x_3^2 & p_8(x_1, x_2, x_3) = x_1x_2 & p_9(x_1, x_2, x_3) = x_1x_3 \\
p_{10}(x_1, x_2, x_3) = x_2x_3
\end{matrix}
$$

We proceeded to define a matrix $A$ such that:

$$
A = \begin{bmatrix}
\left<p_1, p_1\right> & \left<p_1, p_2\right> & \dots & \left<p_1, p_{10}\right> \\
\left<p_2, p_1\right> & \left<p_2, p_2\right> & \dots & \left<p_2, p_{10}\right> \\
\vdots & \vdots & \ddots & \vdots \\
\left<p_{10}, p_1\right> & \left<p_{10}, p_2\right> & \dots & \left<p_{10}, p_{10}\right> \\
\end{bmatrix}
$$

And $b$ such that:

$$
b = \begin{bmatrix}
\left<y, p_1\right> \\
\left<y, p_2\right> \\
\vdots \\
\left<y, p_{10}\right> \\
\end{bmatrix}
$$

Finally, we proceeded to solve the equation $A\vec{c} = \vec{b}$ where $\vec{c}$ is the vector with the coefficients such that (using $c$ because $x$ was already used for the features):

$$
f(x_1, x_2, x_3) = \sum c_ip_i(x_1, x_2, x_3)
$$

The thing is, I didn't really understand why we need to compute the inner product of those polynomials to create those matrices, my first idea would be to just compute the polynomials and fit using the Moore-Penrose inverse, what is the difference between those two approaches? I didn't really get the teacher's explanation and I can't find the explanation in any book.

Edit 1
So, I wasn't very clear on how I would approach the problem, instead of my professors solution, so I wanted to clarify with an example, let's say I had the data

$$
\begin{array}{|cc|c|}
\hline
x_1 & x_2 & y \\
\hline
1 & 2 & 1\\
1 & 3 & 2\\
2 & 1 & 3\\
2 & 2 & 2\\
\hline
\end{array}
$$

And I wanted to approximate with $f(x, y) = c_1 + c_2x_1 + c_3x_2 + c_4x_1^2 c_5x_2^2 + c_6x_1x_2$ I would create a matrix A like that:

$$
A = \begin{bmatrix}
1 & x_1 & x_2 & x_1^2 & x_2^2 & x_1x_2
\end{bmatrix} = \begin{bmatrix}
1 & 1 & 2 & 1 & 4 & 2 \\
1 & 1 & 3 & 1 & 9 & 3 \\
1 & 2 & 1 & 4 & 1 & 2 \\
1 & 2 & 2 & 4 & 4 & 4
\end{bmatrix}
$$

And have $b$ like that:

$$
b = \begin{bmatrix}
1 \\
2 \\
3 \\
2
\end{bmatrix}
$$

and proceed to solve $A\vec{c} = \vec{b}$ to find the coefficients, without computing the inner product.

Edit 2 So, another clarification would be on how the inner product is defined on my teacher's example, let's say we have the same data as my example before, then:

$$
\left<p_2, p_3\right> = 1\cdot2 + 1\cdot3 + 2\cdot1 + 2\cdot2 = 11
$$

Also, right now it's an exact fit because I have more features than data, which I did to leave the example shorter, but you can imagine that in the real scenario I have much more data. Also, the best-fit is the one that minimizes the squared error.

答:
Per your edit: In your method, each data point gives a row to $A$. If you have a lot more data points, then you have a lot more rows. If $A$ has more rows than columns, then $A\vec c = \vec b$ is unlikely to have any solution, as it is overly constrained. You can solve this by increasing the maximum degree of your polynomial until you have at least as many columns as rows again. But you've given up the ability to limit the degree of your answer, and as mentioned in my original answer, that leads to wildly oscillating polynomials.

Note that in your professor's version, the size of $A$ is not dependent on the number of data points. Only the definition of the inner products change with that. $A$ is always a square matrix. And since the polynomials themselves are chosen to be independent of each other, you are guaranteed that $A\vec c = \vec b$ has at least one solution, even when $A$ is not invertible.

Also, you can prove that the professor's version comes up with a solution that minimizes squared error. Your method almost certainly will not.

Original Post:

How you produce a best-fit polynomial depends on your definition of "best-fit". Your example defines it as "exact match at these four points", and cares not the slightest how the polynomial behaves anywhere else. Such polynomials have a tendency to swing wildly between points. For functions of one variable that are almost linear the interpolating polynomial will generally end up looking sinusoidal over the range of interest instead of the nice flat line you would want.

How your professor's method defines "best-fit" is not clear, because you did not indicate which inner product of polynomials is being used. (And it has been ages since I've dealt with this, so I'd probably have to think hard about what it was doing before I could answer anyway.)


问:
What is a connex relation?.
I've been recently trying to teach myself some topology and in  the book I'm reading there is the definition of an order relation which confuses me a lot
.

I've looked online for more information and they seem to define an order relation differently namely one that is transitive,antisymmetric and reflexive.

So why does the author define it in this way my 2nd confusion has to do with what comparability actually means as there is actually no definition online and I cant quite understand what a connex relation is either and that seems to come up a lot in the definition of an order relation.

Thanks in advance.

答:
The book you’re reading is not helping yourself. It defines an order relation to be a linear order relation, which is really unusual and not good.

Then an order can be strict or not. Wikipedia order theory presents order relations in a more standard way.

Title: What is a connex relation?.


问:
If $m(A_k)\ge (\frac{5}{9})^k$, where $A_1, \dots, A_n$ in $(X,M,m)$, for which $n\, \exists A_i, A_j$ s.t. $A_i \cap A_j \neq \emptyset$?
I am struggling to find a good idea to set up this exercise, could someone give me some hints about it? Thank you.

Let $(X,M,m)$ be a measurable space s.t. $m(X) = 1\,$, let be $n\ge2$ and $A_1, \dots , A_n$ measurable sets s.t. $m(A_k)\ge (\frac{5}{9})^k$, $k \in \{1,\dots,n\}$ . For which $n \in \mathbb{N}$ are we sure that $\exists A_i, A_j$, with $i \neq j$ s.t. $A_i \cap A_j \neq \emptyset$? If it was $m(A_k)\ge (\frac{1}{4})^k$ is it true that $A_i \cap A_j = \emptyset$, $\forall i, j$ s.t. $i \neq j$?

答: By the pigeonhole principle, $n$ just needs to be big enough so that $\sum_{k=1}^{n} (5/9)^k > 1$.  In the second case, $\sum_{k=1}^\infty (1/4)^k < 1$ so there is no such $n$.


问:
Determine the distribution and the expected value of $Z_1=XY_1$
Let $X, Y_1, Y_2$ be independent random variables, all are defined on the same probability space.

Let $X$ be Bernoulli distributed on $\{0,1\}$ with sucessparameter $p_0\in (0,1)$ and let $Y_i,\ i=1,2$ be binomial distributed with parameter $(n,p), \ n\in \mathbb{N}, p\in (0,1)$.

(a) Determine the distribution and the expected value of $Z_1=XY_1$.

(b) Determine the covariance of $Z_1=XY_1$ and $Z_2=XY_2$.

$$$$

For question (a) :

We have that $$P(X=0)=1-p_0 \ \text{ and} \ P(X=1)=p_0$$ and $$P(Y_1=y)=\binom{n}{y}p_0^y(1-p_0)^{n-y}$$

For $Z$ do we calculate the product of these two?

Body:
You have :

$$P(Z_1=k)=P(Z_1=k|X=0)P(X=0)+P(Z_1=k|X=1)P(X=1) \\
= P(X=0)\delta_{k,0}+P(Y_1=k|X=1)P(X=1)\\
=(1-p_0)\delta_{k,0}+\binom{n}{k}p_0^{k+1}(1-p_0)^{n-k}$$


问:
Expected winnings from a game where I accept the money randomly drawn from the interval 0 to 100 if the proposed amount is more than 50 dollars
I am trying to solve this probability puzzle: https://math.stackexchange.com/questions/1977360/expected-value-from-a-game-which-gives-random-numbers-from-1-to-100-with-reducin#:~:text=%22In%20front%20of%20you%20is,a%20new%20proposal%20next%20period.

The puzzle is:

"In front of you is an infinitely-lived machine that proposes amounts of money, which you can either accept or reject. If you accept, the machine hands over the proposed amount, but shuts down and will never give you anything else. If you reject, it'll show you a new proposal next period. Each period's proposal is an independent draw from a uniform distribution on [0, 100]. The time between periods is long — several months, say — and you are impatient: a dollar next period is worth only 0.9 to you today; similarly, a dollar two periods from now is worth 0.9*0.9 = 0.81 today, et cetera. If your strategy were to always accept, you'd expect to make 50 dollars, i.e. the mean of the first draw. If instead you decided to accept any first proposal above 50, and — in case you reject the first — any second proposal whatsoever, your expected discounted payoff would be 60 dollars. But you can do better! If you follow the strategy that maximizes your expected discounted payoff, what is the threshold above which you should accept the machine's first proposal?"

For a start, I am trying to calculate the expected winnings if I choose the threshold to be 50 dollars. I understood the solution mentioned in the above link but I came up with another approach which does not give the correct value when the threshold is 50 dollars. As per the question, for this threshold, the expected discounted payoff will be 60 dollars. But as per the formula derived in the solution, it comes out to be 68.18 dollars.

$$ V = \frac{(50000-5t^2)}{(1000-9t)} = 68.18, \text{for } t = 50$$.

Does the question contain the wrong value?

Now here is my approach:

Let $U_1, U_2, ... , U_N$ be i.i.d. drawn from $U(0,100)$ distribution. Here N is a random variable following geometric distribution with p = 0.5 i.e. N is the first period where we observe the proposed money to be more than 50 dollars. We then need to find $E[0.9^NU_N]$ i.e. the present value of our winnings. Now conditioning of N we can write,

$$E[0.9^NU_N] = E[E(0.9^NU_N|N)] = E[0.9^NE(U_N|N)] = 75E[0.9^N]$$

As 75 will be the expected value of $U_N$ if its more than 50 and less than 100. Now,

$$E[0.9^N] = \sum_{n=1}^\infty \frac{0.9^n}{2^n} = \frac{9}{11}$$

as $P(N = n) = 1/2^n$ i.e the probability of amount less than 50 in first $n-1$ draws and amount more than 50 in the $n^{th}$ draw. So the answer I get is 61.36 dollars which is incorrect. What is wrong with my solution?

答:
If you accept the first offer when it is over $50$ then there is a probability of $\frac12$ of accepting an expected amount of $75$.

If you otherwise accept the second offer unconditionally then there is also a  probability of $\frac12$ of accepting an expected amount of $50$ next time but that is only worth $0.9\times 50=45$ now.  So the overall expected gain is $\frac{75}{2}+\frac{45}{2}=60$ as in the question.

But if you otherwise always accept later offers when they exceed $50$ then expected value now is $$\frac12\times 75 + \frac14 \times 0.9^1\times75  + \frac18 \times 0.9^2\times75+\cdots \\= \frac{75}{2}\sum_{n=0}^\infty \left(\frac{0.9}{2}\right)^2 \\=\frac{75}{2}\frac{1}{1-\frac{9}{20}} =\frac{750}{11} \approx 68.1818$$ as my formula in the linked answer suggested.


问:
How does one figure out the exact equation of the boundary of the area the line sweeps through?
I am thinking of this question, but I am having a hard time solving this, or even easier problems related to this.

There is a graph $y=\sin\frac{\pi}{4}x$ in the domain $0 \leq x\leq 4$. Let A and B two points on the x-axis, with distance 1 apart inside the domain. For example, lets say $A(t, 0)$ and $B(t+1, 0)$. Let C and D be two points the intersection of vertical lines $x=t$ and $x=t+1$ with $y=\sin\frac{\pi}{4}x$, respectively. So this means that $C(t, \sin\frac{\pi}{4}t)$, and $D(t+1, \sin\frac{\pi}{4}(t+1))$. Think of the line segment $CD$. If the points A and B move with distance 1 apart inside the domain, so will C and D, and therefore the line segment, too. What can we say about the boundary of the territory line segment $CD$ sweeps across?

I tried to think as the line segment of CD at a particular $t$ will be the tangent line for the area the line sweeps through. I also tried to simplify the problem, as in circles, or squares. This is what I thought for the square.

Let $0\leq t\leq1$, and let $A(t, 0)$ and $B(0, 1-t)$. What will be the area where the line segment AB sweeps through?

I am not sure about this neither, and I am having a hard time trying to figure this out. Are there any given "structure/ways' for these kind of "sweeping" problems?

答:
Let $t$ range over [0,3]. Let $s$ range over [0,1]. At any values of t and s, the point P directly between C and D has coordinates $$(t+s, (1-s)\sin{(\frac{\pi}{4}t)}-s\cdot \sin{\frac{\pi}{4}(t+1))})$$ Your region is the set of points described.

Now rewrite $t = x-s$, so that $$y =(1-s)\sin{(\frac{\pi}{4}(x-s))}-s\cdot \sin{\frac{\pi}{4}(x-s+1))} $$
For constant x, take $\partial y/\partial s$ to find the extrema. Don't forget the endpoints. Finally integrate over x to get your answer.


问:
In a primitive symmetric association scheme, why does $E_j$ occur in some power of $E_i$ for each $i,j$?
I am having some trouble in the proof of the Absolute Bound Condition for primitive symmetric association Schemes in the book Algebraic Combinatorics I by Bannai and Ito (Chapter 2, Section 4, Theorem 9):

Let $\chi=(X,\lbrace{R_i\rbrace}_{0\leq i \leq d})$ be a primitive symmetric association scheme of class $d$. Let the $\lbrace E_i\rbrace_{0\leq i\leq d}$ be the primitive idempotents of the adjacency algebra of $\chi$. Denote by $E_i^h$ to be the Hadamard product $\circ$ of $E_i$ with itself $h$ times.
Since $\chi$ is primitive, we have for any $j$ and $i$ an $0\leq h \leq d$ such that $E_j$ occurs in the linear decomposition of $E_i^h$ with respect to the basis $E_0,\ldots,E_d.$

I do not understand how the emboldened statement is justified. It says in the book that this is because imprimitivity is a equivalent to there being a suitable rearrangement of indices $1,\ldots,d$ and an index $0<t<d$ such that the Hadamard product $E_i \circ E_j$ is a linear combination of $E_0,E_1,\ldots,E_t$ for all $0\leq i,j \leq t.$

To get an idea, I tried to show this by contradiction on the special case when $d=3$ and the assumption that $E_2$ does not occur in the decomposition of $E_1^h$ for $0\leq h \leq 3.$ With this assumption, we get $q_{11}^2=0$ so
\begin{align*}
E_1^2 = &q_{11}^0 E_0 + q_{11}^1 E_1 + q_{11}^3E_3,\\
E_1^3 =&q_{11}^0(q_{01}^0 E_0 +  q_{01}^1 E_1 + q_{01}^2 E_2 + q_{01}^3 E_3) + q_{11}^1(q_{11}^0 E_0 +  q_{11}^1 E_1 + q_{11}^2 E_2 + q_{11}^3 E_3)\\
      &+q_{11}^3(q_{13}^0 E_0 +  q_{13}^1 E_1 + q_{13}^2 E_2 + q_{13}^3 E_3).
\end{align*}
Since $E_2$ does not occur in $E_1^3$, the latter gives us $q_{11}^0q_{01}^2 + q_{11}^1 q_{11}^2 + q_{11}^3q_{13}^2 = 0.$ Since $q_{01}^2=0=q_{11}^2,$ we obtain $q_{11}^3q_{13}^2=0.$

If $q_{11}^3=0$, then since $q_{11}^2=0$, we have the linear span of $E_0,E_1$ as a subalgebra of the adjacency algebra, violating primitivity. However, I no longer know what to do when $q_{13}^2=0.$ If I could show $q_{33}^2=0$, then the linear span of $E_0,E_1,E_3$ would form a subalgebra of the adjacency algebra and would violate primitivity. However, I don't know how to show $q_{33}^2=0.$

答:
The following supplement is mainly due to Dr. Godsil's excellent answer.

We give a direct way of showing that if for some $0\leq i,j\leq d$ we have that $E_j$ does not occur in the decomposition of any $E_i^h$ for any $0\leq h\leq d$, then it follows that $\chi$ is an imprimitive association scheme. To do this, we will show that there exists a reindexing of the indices $1,\ldots,d$ and an index $0<t<d$ such that the Hadamard product $E_i\circ E_j$ is a linear combination of $E_0,E_1,\ldots,E_t$ for all $0\leq i,j\leq t$.

Without loss of generality, we assume that $E_d$ does not occur in the decomposition of any power $0\leq h\leq d$ of $E_1$. Let $S$ be the set of indices $0\leq j \leq d$ such that $E_j$ occurs in the linear decomposition of some power $0\leq h\leq d$ of $E_1$. Observe that $S$ always contains $0$ and $1$ and never contains $d$. Without loss of generality, we may rearrange the indices such that the indices that occur in $S$ are precisely the first $t<d$ indices $0,1,\ldots,t$ where $t=|S|-1$. It remains to show that if $i,j \in S$ and $k\notin S$, then $q_{ij}^k=0.$ We prove this by contradiction.

Suppose that there exist $i,j \in S$ and $k \notin S$ with $q_{ij}^k>0$. Let $a,b$ be nonnegative integers such that $E_1^a=\sum_r \alpha_rE_r$ and $E_1^b=\sum_r \beta_rE_r$ where $\alpha_i,\beta_j>0$; i.e., such that $E_i$ and $E_j$ respectively occur in $E_1^a$ and $E_1^b$. By multiplying these, we see that $E_1^{a}E_1^b$ contains a term of the form $\alpha_i\beta_j E_i\circ E_j.$ By expanding this, we see that the linear decomposition of $E_1^{a}E_1^{b}$ contains $\alpha_i \beta_j q_{ij}^k E_k$ where $\alpha_i\beta_j q_{ij}^k$ is nonzero. Since the Krein parameters are nonnegative, this term will not be cancelled out by considering the other coefficients of $E_k$ when expanding $E_1^{a}E_1^{b}$. In particular, this implies that $E_k$ occurs in the decomposition of $E^{a+b}.$ If $a+b\leq d$, we are done, for we obtain the contradiction that $k\in S$. If $a+b>d$, note that $E_1^0,\ldots,E_1^{d+1}$ are linearly dependent since the algebra is $d+1$ dimensional. Thus, any power of $E_1$ above $d$ is expressible as a linear combination of powers of $E_1$ at most $d$. Hence, $E_1^{a+b}$ is expressible as a linear combination of the $E_1^h$ where $0\leq h\leq d$. Since $E_k$ occurs in $E_1^{a+b}$, it must occur in some $E_1^h$ where $0\leq h\leq d$. Thus, $k \in S$ and we obtain a contradiction as before.


问:
Dimensionality of datasets in multiple regression
As an example, let's say that a linear regression is performed of the form

$$Y = \beta_0 + \beta_1X_1 + \beta_2X_2 + \cdots + \beta_nX_n +\varepsilon$$

where $Y$ is a vector of $10,000$ measurements of peak acceleration of different car models, and the regressors correspond to different technical features of the cars.

From a linear algebra standpoint $Y$ lives in $\mathbb R^{10000}$, and the coefficients are found by minimizing the sum of the square distances of this vector on a hyperplane.

Now, from the point of view of dimension being the number of linearly independent vectors that span a space, this vector $Y$ is just $1$ dimension.

If it is truly $1$-dimension of a $\mathbb R^{10000}$ ambient space, the Euclidean projection on the hyperplane that underpins the process of finding the coefficients does not have any dimensionality issues (collinearity between the regressors being a separate topic). Otherwise, $L^2$ norms in high dimensions do pose problems.

So is $Y$ (the vector of $10,000$ observations) $1$-mimensional or high dimensional?

答:
The issue of dimensionality is the context of regression analysis is the ratio between $n$, number of observations, and $p$, number of estimated parameters. As closer $n$ to $p$, the less reliable your estimated model is. Assume that your model is
$$
y = \beta_0 + \sum_{j=1}^px_j\beta_j + \epsilon,
$$
hence in order to find the OLS estimators of $\beta = (\beta_0,...,\beta_p)$ you project the vector $y$ on the affine space spanned by $(1,x_1,...,x_p)$, hence it is a $p$ dimensional space. The number of observations, $n$, is not count as dimension. If you have a continuous stochastic process, then you can sample from it infinitely many times, i.e., $n\to \infty$, that is usually a good feature because you can safely use asymptotic results. Notably, in such a case, there is another problem of artificially low p.values, but this is unrelated to the dimension of the model or the embedded space.


问:
How to show that $RH^2$ is closed in $RL^2$?
I have faced difficulty when I try to learn functional analysis myself and to do problem 2.14 in N. Young's "An introduction to Hilbert space".
The question is asking you to show that $RH^2$ is closed in $RL^2$ where $RL^2$ denotes the space of rational functions which are analytic on the unit circle
$$\partial D=\{z\in C:|z|=1\}$$
with the inner product
$$<f,g>=\frac{1}{2\pi i}\int_{\partial D}f(z)\overline{g(z)}\frac{dz}{z}$$
the integral being taken anti-clockwise around $\partial D$. and the $RH^2$ is the subspace of $RL^2$ consisting rational functions which are analytic on the closed unit disc clos $D$, where
$$D=\{z \in C:|z|<1\}$$ with the inner product in $RL^2$.

Could anyone give me a guide on how to do it? I try to start by checking whether the complement of $RH^2$ is open but I'm not sure what is the complement of $RH^2$.

答:
This problem is a bit weird in the sense that neither spaces above ($RH^2, RL^2$) are Hilbert so sequences from $RH^2$ can converge to more general analytic functions in the $L^2$ norm on the unit circle (leading to Hardy spaces etc).

This being said it is easy to see using Cauchy formula that for a compact set $K$ included in the open unit disc, there is $C_K$ st $\sup_{z \in K} |f(z)| \le C_K ||f||_2$ for any analytic function on the unit disc that extends continuously say to the boundary

(using the Cauchy Schwarz inequality and the fact that $|z-\zeta| >A_K>0, z \in K, \zeta \in \partial D$).

This implies that any $||.||_2$ convergent sequence converges locally uniformly hence to an analytic function $F$ on the open unit disc; if now the sequence also converges to a rational function in the $||.||_2$ norm on the unit circle, it immediately follows that $F$ is rational on the unit circle hence inside the unit disc by analytic continuation (two meromorphic functions equal on the unit circle, are equal everywhere), so $F$ is also in $RH^2$.

However I want to emphasize again that in general if $f_n$ is in $RH^2$ and $f_n$ converges into the $||.||_2$ norm it doesn't, in general, follow that the limit is in $RL^2$ (hence in $RH^2$).


答:
We need to find probability of Mandy getting more H than Chuck given they both get different number of H. So our sample space reduces to possibilities where they get different number of H.

Number of possibilities that get same number of heads - either they both get no H, they both get $1$ H or they both get $2$ H.

$ = 1 + 2 \cdot 3 + 3 = 10$

So number of possibilities that they both get different number of H $ = 32 - 10 = 22$

Please note that possibilities of Mandy getting more H than Chuck is a subset of them getting different number of H. Here are the possibilities for Mandy to get more H than Chuck -

Chuck gets no H, then it is $7$ ways (other than Mandy getting zero H)

Chuck gets one H, then it is $2 \cdot 4$ ways (Mandy can get two or three H)

Chuck gets two H, then it is just $ \ 1$ way (Mandy must get three H)

So total number of possibilities for Mandy to get more H than Chuck $ = 7 + 2 \cdot 4 + 1 = 16$.

Desired probability $ = \frac{16}{22} = \frac{8}{11}$.

Title: Probability Chip Tossing Problem with Different Chip Tossing Counts

Body:
Mandy and Chuck are flipping are flipping chips. The chip has two sides: head and tails. Each round Mandy flips 3 chips and Chuck flips 2 chips. Each round, they might throw an equal number of heads. In the round they first flip different numbers of heads, what is the probability Mandy threw more heads than Chuck?

My work:

I started off by listing the possible outcomes.

Mandy: HHH, HHT, HTH, THH, TTH, THT, HTT, TTT

Chuck: HH, HT, TH, TT

There are 8*4 = 32 possible outcomes in terms if rounds

I listed possible outcomes where Mandy threw more heads than Chuck.

1.HHH, HH
2.HHH, HT
3.HHH, TH
4.HHH, TT
5.HHT, HT
6.HHT, TH
7.HHT, TT
8.HTH, HT
9.HTH, TH
10. HTH, TT
11. THH, HT
12. THH, TH
13. THH, TT
14. TTH, TT
15. HTT, TT

I got 15 different possibilities. Does this mean the answer is 15/32.

Thank you very much

I'm not really sure how to approach this problem?


问:
Why don't I understand "What is mathematics?" by Richard Courant
I am 14 years old and can do A level maths happily. However, I want to further my knowledge of mathematics to pave my way for more discrete maths and harder analysis etc. So I decided to pickup "What is mathematics?" by Richard Courant because I heard it was very useful and it would be good for someone of my knowledge.

I am having a lot of trouble understanding it though, and I'm not sure what it is that is stopping me. All I know is that it takes me a long time and some SE questions to understand even a page. It could be because of the way it is written which I am not used to?

If anyone has ever encountered this before then could you please offer me some advice and maybe a new way to approach it.

答:
A couple of suggestions from someone who's been teaching college students for a long time :) First, learn to read math books with pencil and paper and try to figure things out as you go. Don't be passive. And yes, serious mathematics is hard to read casually.

A couple of suggestions of books to look at. A nicely written book that gives an introduction to proofs and mathematical thinking is How to Think Like a Mathematician: A Companion to Undergraduate Mathematics, by Kevin Houston, Cambridge University Press, 2009. Also, check out any of Ian Stewart's books written for the neophyte mathematician; he has great stuff in there. You will see zillions of books on Amazon, but he will give you in many of his books a fascination with current mathematics, not just ancient stuff :). For something older but full of fascinating stuff, look at Hilbert and Cohn-Vossen, Geometry and the Imagination.


问:
$\lim_{n \rightarrow \infty}\left(1+\dfrac{2}{n}\right)^{n^2}e^{-2n}$
$\lim_{n \rightarrow \infty}\left(1+\dfrac{2}{n}\right)^{n^2}e^{-2n}$

$\lim_{n \rightarrow \infty}\left(e^{-2}\left(1+\dfrac{2}{n}\right)^n\right)^{n}$

It is indeterminate form $1^{\infty}$

I solve this like this $e^{\lim_{n \rightarrow \infty}}\frac{\ln(e^{-2}(1+\frac{2}{n})^n)}{\frac{1}{n}}$

I can't apply L'Hôpital's rule now how can I solve this problem quickly?

答:
You have :

$$\left(1+\dfrac{2}{n}\right)^{n^2}=\exp\left(n^2\log\left(1+\frac{2}{n}\right)\right)=\exp\left(n^2\left(\frac{2}{n}-\frac{4}{2n^2}+o\left(\frac{1}{n^2}\right)\right)\right)\\=\exp\left(2n-2+o\left(1\right)\right)$$

So you get :

$$\left(1+\dfrac{2}{n}\right)^{n^2}e^{-2n}=e^{-2+o(1)} $$

So you get the final result :

$$\lim_{n \rightarrow \infty}\left(1+\dfrac{2}{n}\right)^{n^2}e^{-2n}=e^{-2} $$


答:
Without doing a lot of calculation, just observe that $Z_1=XY_1$ can take the values in

$$\{0;1;2;\dots;n\}$$

$Z_1$ will take the value zero when $X=0$, that is with probability $1-p_0$ plus when $X=1$ and $Y_1=0$, thus with probability $p_0(1-p)^n$

Concluding:

$$P(Z_1=0)=1-p_0+p_0(1-p)^n$$

For the rest of support values, the probabilities are the binomial ones multiplied by $p_0=P(X=1)$

As per the expectation is concerned, the text says that the rv are independent thus...

(same reasoning for $Z_2$)

For the covariance, simply expand the definition formula and easy get the solution:

$$Cov(Z_1,Z_2)=\mathbb{V}[X]\cdot\mathbb{E}^2[Y_1]=p_0(1-p_0)\cdot n^2p^2$$

Body:
Let $X, Y_1, Y_2$ be independent random variables, all are defined on the same probability space.

Let $X$ be Bernoulli distributed on $\{0,1\}$ with sucessparameter $p_0\in (0,1)$ and let $Y_i,\ i=1,2$ be binomial distributed with parameter $(n,p), \ n\in \mathbb{N}, p\in (0,1)$.

(a) Determine the distribution and the expected value of $Z_1=XY_1$.

(b) Determine the covariance of $Z_1=XY_1$ and $Z_2=XY_2$.

$$$$

For question (a) :

We have that $$P(X=0)=1-p_0 \ \text{ and} \ P(X=1)=p_0$$ and $$P(Y_1=y)=\binom{n}{y}p_0^y(1-p_0)^{n-y}$$

For $Z$ do we calculate the product of these two?


问:
How many ways can 2 rooks and a bishop be placed on a 4x4 board such that no piece attacks another piece?

How many ways can $2$ rooks and a bishop be placed on a $4\times 4$ board such that no piece attacks another piece?

I stumbled upon this question and I don't know how to find the answer mathematically. I wrote this piece of code to find the answer to it. I think the code is right, but I get the answer as $232$, which doesn't "feel" right. So how do you actually go about solving this question?

答: I'm not sure there is a way to solve the problem without a good deal of brute force.  It's pretty easy to calculate the number of ways to place two rooks that don't attack each other (You can choose the two rows in $4\choose2$ ways and the columns associated with them in another $4\cdot 3$), but then analyzing how to add a bishop would require at least some subcases and probably quite a lot.


问:
Length of triangle edges

X is the middle of AB

Z is the middle of AC

Y is the middle of BC

is AX always equal to ZY?

答:
Yes. Because of Thales theorem

$AZYX$ and $XZYB$ are parallelograms so $ZY=XB$ and $ZY=AX$ implies $ZY=\frac12 AB$

Title: Length of triangle edges


答:
Just a brief explanation about the answer above:

Observe that

$$\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots .$$

Then, it's clear that (for $x = \frac{2}{n}$)

$$\log\left(1 + \frac{2}{n}\right) = \frac{2}{n} - \frac{4}{2n^2} + \ldots .$$

Body:
$\lim_{n \rightarrow \infty}\left(1+\dfrac{2}{n}\right)^{n^2}e^{-2n}$

$\lim_{n \rightarrow \infty}\left(e^{-2}\left(1+\dfrac{2}{n}\right)^n\right)^{n}$

It is indeterminate form $1^{\infty}$

I solve this like this $e^{\lim_{n \rightarrow \infty}}\frac{\ln(e^{-2}(1+\frac{2}{n})^n)}{\frac{1}{n}}$

I can't apply L'Hôpital's rule now how can I solve this problem quickly?


问:
How fast does the sequence converge?
We have the sequence $$a_1=\frac{9}{4} \quad \qquad a_{n+1}=2\frac{a_n^3-a_n^2+1}{3a_n^2-4a_n+1}$$ that converges to $\frac{9}{4}$.

I want to check how fast the convergence is.

Do we have to calculate $\lim_{n\rightarrow \infty}\frac{|a_{n+1}-9/4|}{|a_n-9/4|}$ ?

Title: How fast does the sequence converge?

Body:
Hint: Define
$$f(x)=2\frac{x^3 - x^2 + 1}{3 x^2 - 4 x + 1}$$

Prove the sequence $a_{n+1}=f(a_n)$ converge to $2$ with $a_1 =\frac{9}{4}$ as you said.

\begin{align}
a_{n+1} &= f(a_n) \\
&= 2\frac{(a_n-2)((a_n-2)^2+1)}{((a_n-2)+1)(3(a_n-2)+5)} \tag{1}\\
\end{align}

Denote $b_n = a_n-2$, we have $b_n \rightarrow 0$ when $n\rightarrow +\infty$ or $b_n=\mathcal{o}(1)$. And from (1) we have

$$b_{n+1}+2 = f(b_n+2)$$
or
\begin{align}
b_{n+1} &=f(b_n+2)-2 \\
&= \frac{b_n(b_n^2+1)}{(b_n+1)(3b_n+5)} -2 \\
&= \frac{2}{5}b_n(b_n^2+1)(1-b_n+\mathcal{O}(b_n^2))(1-\frac{3}{5}b_n+\mathcal{O}(b_n^2)) -2\\
&= \frac{4}{5}b_n^2-\frac{22}{5}b_n^3+\mathcal{O}(b_n^4)\\
&= \frac{4}{5}b_n^2(1+\mathcal{O}(b_n)) \tag{2}\\
\end{align}
Then,
$$\ln(b_{n+1})= \ln(\frac{4}{5})+ 2\ln(b_{n}) +\ln(1+\mathcal{O}(b_n))  $$
$$\iff (\ln(b_{n+1})-\ln(\frac{4}{5}))=  2(\ln(b_{n})-\ln(\frac{4}{5})) +\mathcal{O}(b_n)  \tag{3}$$
We notice that from (2), we can deduce also that $b_{n+1}=\mathcal{O}(b_n^2)=\mathcal{o}(b_n)$ or $\mathcal{O}(b_{n+1})=\mathcal{o}(b_n)$. So,
$$(3)\iff (\ln(b_{n+1})-\ln(\frac{4}{5})+\mathcal{O}(b_{n+1}))=  2(\ln(b_{n})-\ln(\frac{4}{5})+\mathcal{O}(b_n)) $$
$$\iff \ln(b_{n})-\ln(\frac{4}{5})+\mathcal{O}(b_{n})=  2^{n-1}(\ln(b_{1})-\ln(\frac{4}{5})) =2^{n-1}\ln(\frac{5}{16})  $$
or
$$b_n=\frac{4}{5} \left(\frac{5}{16} \right)^{2^{n-1}}$$
(because $b_n \rightarrow 0$ when $n \rightarrow +\infty$ then $\exp(\mathcal{O}(b_{n}))=1$)

Conclusion:
$$a_n \approx 2+\frac{4}{5} \left(\frac{5}{16} \right)^{2^{n-1}}$$


问:
Polynomial with two double roots
I need to find $a$, $b$ real numbers such that the polynomial $P(x)=x^{4}-4 x^{3}+10 x^{2}+a x+b$ has 2 double roots. I have tried this $P^{\prime}(x)=4 x^{3}-12 x^{2}+20 x+a=0$ but I still have a so I cant find roots. I know I can write this polynomial as (x-A)^2*(x-B)^2 where A and B  are 2 double roots but I don't know how to proceed from here?

答: Suppose $P(x)$ has two double roots. Then, as you said, $P(x)=(x-A)^2(x-B)^2$. Multiply out the brackets and expand. Then you can compare the coefficients of the polynomials (you will get $4$ simultaneous equations) from which you can immediately solve for $a,b$.

Title: Polynomial with two double roots


问:
Probability a hypercube is non-empty in a uniformly distributed point set in higher dimensions
Before I start, I just want to say I'm actually a computer science student and thus, my maths is not very strong so please feel free to make your explanations more detailed via adding additional links or comments and such.

The Scenario

I am randomly generating $N$ points in $D$ dimensional space (let's call this the data set) and I also generate a test set of points. The values are generated with uniform probability distribution and range from $[0,1]$. So let's say $D$ = 5, then an example point would be $(0.5, 0.2, 0.1, 0.9, 0.7)$.

I want to determine the probability $p$ that there exists at least one point in the data set which lies within a hypercube of side length $2e$ centred at $Q$, where $Q$ is a point in the test set. In other words, I want an equation relating $p$ and $e$.

My thoughts

For a point $Z$ to lie within a hypercube of side $2e$ centred at $Q$, each of its coordinates would need to lie within ${\pm}e$ of each corresponding coordinate of $Q$. For example, if $Q$ =$(Q_1, Q_2, Q_3,...,Q_D)$ and $Z$ = $(Z_1, Z_2, Z_3,...,Z_D)$, for $Z$ to lie within this hypercube, we need:

$Q_1-e{\leq}Z_1{\leq}Q_1+e$

$Q_2-e{\leq}Z_2{\leq}Q_2+e$

... and so on.

Let's rephrase $P(Q_i-e \leq Z_i \leq Q_1+e)$ as $P(|Z_i-Q_i|\leq{e})$. Since $Z_i$ and $Q_i$ are 2 independent uniform variates in $[0,1]$, then:

$P(|Z_i-Q_i|\geq{e}) = (1-e)^2$ (visual depiction of why) and thus:

$P(|Z_i-Q_i|\leq{e}) = 1-(1-e)^2 = 2e-e^2$

Then the probability that $Z$ lies within the hypercube of $Q$ with coordinates bounded by $[0,1]$ in $D$ dimensions is

= (2e-e^2)^D$.

Of course, the probability will vary with each point since this assumes $Z$ and $Q$ are independently uniformly random (e.g. if your test point $Q$ is $(0, 0)$ at the very corner, then the probability of $Z$ being in the hypercube will be much smaller since much of the hypercube is "cut off" example of what I mean in 2 dimensional space. The non-shaded area has no points inside it) but over a large number of test points, it will average out* (I think this is the problem, please read the bottom).

After testing this, it seems to be accurate. For example with, $D = 2, e = 0.2$, we get $(2(0.2)-0.2^2)^2=0.1296$. If we have $N=100 000$ and a test set of 10 000 points and for each test point, we counted the number of data set points which lied within its hypercube, we'd expect a total count of $0.1296...{\times}10 000{\times}100 000=129 600 000$ points. My test gives me 129 393 108 points.

But what is the probability that the hypercube is non-empty? If $P(k)$ = the probability that $k$ points lie inside the hypercube of side $2e$ centred at a test point $Q$, then a simple binomial gives us:

$P(k)=((2e-e^2)^D)^k(1-(2e-e^2)^D)^{N-k}\binom{N}{k}$

Since $p$ is the probability that there is at least one point in this hypercube:

$p=1-P(0)$

$=1-(1-(2e-e^2)^D)^N$

Subjecting $e$ and we get:

$e=1-\sqrt{1-(1-(1-p)^{\frac{1}{N}})^\frac{1}{D}}$.

My Problem

However, this doesn't seem to be correct. As the dimensions go higher, it gets more and more inaccurate. Let's take a data set of $N = 100 000$ and a test set of 10 000 and $p=0.99$. I should be generating around 100 empty hypercubes (1% of 10 000 test points should have no points in the data set within a hypercube of side $2e$ centred at the test point).

For $D=2$, I get $e=0.003399$. This generates 113 empty hypercubes.

For $D=5$, I get $e=0.070334$. This generates 209 empty hypercubes.

For $D=10$, I get $e=0.205269$. This generates 644 empty hypercubes.

What I think is wrong

As pointed out below by David K, maybe the events are not independent because the each test point $Q$ is fixed when testing against all N points in the data set. Therefore, binomial probability would not apply to this analysis since we assumed 2 independently random points Z and Q but in this case, each test point Q would be fixed for multiple Z.

Intuitively, I think I need to also analyse the distribution of the probability that Z lies in the hypercube of Q for a fixed Q or to analyse something about the "cut-off" areas for when Q is closer to the corners/edges.

Any insights are appreciated.

答:
I think the error in the calculated probability is that the cases where
$Q$ is in a corner do not "average out" over a large number of test points.

Let $d(A,B)$ be the $L_\infty$ distance between points $A$ and $B$,
so that $d(A,B)<e$ means $A$ is in the hypercube of side $2e$ centered at $B$.
Then for a large number of test points $Z,$ $Z',$ $Z'', \ldots,$
you are dealing with the events
$d(Z,Q)<e,$ $d(Z',Q)<e,$ $d(Z'',Q)<e,$ and so forth.
The probability of each of these events is $(2e-e^2)^D,$ as you found,
but the events are not independent.
So the rest of your calculations are based on a false assumption.


问:
Prove if det(A) = 0 and det(A -kI)=0 then k=0.
It seems to be, since $detA = 0$, If $det(A-kI)=0$ the kI part should be zero. So we can take k = 0 because I is nonzero. But I haven't any idea to take mathematical representation to prove it.

Can you help me. :-)

答:
Just look the follwing arguments: for clear concept of it.

(1)    if $ lemda$ is a eigen value  of a matrix  $A$  then we get a nonzero  vector $v$ (known as the eigen vector) such that

$A*v=lemda*v$

Then it can be rewritten as   $ (A-  lemda* indentity  matrix)*v=0$
Looking carefully you can see that it is a homogeneous  system of equation. For non trivial solution ( since the trivial solution would not work for us as we have nonzero vector $v$ )of it we should have

$ det (A-lemda * identity matrix)=0$

(2) Also  you should know that the product of the eigen values of a matrix is equal to the determinent of that matirx.

Since det(A) is zero so one of the eigen values of the matrix $A$  is zero.But this does not tell that all the eigen values is zero.
Which can be seen by the early users counterexample.


问:
Find range of $f(x)=\sqrt {2\{x\} -\{x\}^2} -\frac 34$

here {} implies fractional part function

Evaluating $2\{x\}-\{x\}^2\ge 0$

$$\{x\} \in [0,2]$$

Which is weird since $\{x\}\in [0,1)$ is a known range.

Basically I don’t know what to do here, and this was just my attempt

Ans is $[0,\frac 12]$

答:
The answer can't be $[0, 1/2]$ because $f(0) = -3/4$ and $\lim_{x \to 1^-}f(x) = 1/4$. (You showed that the expression under the radical is never negative, which is certainly good to know.)

Indeed, we can show that the range is $[-3/4, 1/4)$ as follows. First, $\{x\}$ has range $[0, 1)$. Second, $\sqrt{2x - x^2}$ has range $[0, 1)$ on $[0, 1)$. Finally, $x - \frac34$ has range $[-3/4, 1/4)$ on $[0, 1)$. Composing these three functions will give you your result.


答:
Let $A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$. Then
$\det(A)=\det(A-kI)=0$ is equivalent to $ad=bc=(a-k)(d-k)$. This yields $k(k-(a+d))=0$. We see that either $k=0$, or $k=a+d$. In the latter case, $k$ need not be zero. So we obtain counter examples.

For example, take
$$
A=\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix},
$$
and $k=2$. Then $\det(A)=\det(A-2I)=0$.

Title: Prove if det(A) = 0 and det(A -kI)=0 then k=0.

Body:
It seems to be, since $detA = 0$, If $det(A-kI)=0$ the kI part should be zero. So we can take k = 0 because I is nonzero. But I haven't any idea to take mathematical representation to prove it.

Can you help me. :-)


问:
Understanding the differential $dx$ when doing $u$-substitution
I just finished taking my first year of calculus in college and I passed with an A.  I don't think, however, that I ever really understood the entire $\frac{dy}{dx}$ notation (so I just focused on using $u'$), and now that I'm going to be starting calculus 2 and learning newer integration techniques, I feel that understanding the differential in an integral is important.

Take this problem for an example:

$$ \int 2x (x^2+4)^{100}dx $$

So solving this...

$$u = x^2 + 4 \implies du = 2x dx \longleftarrow \text{why $dx$ here?}$$

And so now I'd have: $\displaystyle \int (u)^{100}du$ which is $\displaystyle \frac{u^{101}}{101} + C$ and then I'd just substitute back in my $u$.

I'm so confused by all of this.  I know how to do it from practice, but I don't understand what is really happening here.  What happens to the $du$ between rewriting the integral and taking the anti-derivative?  Why is writing the $dx$ so important?  How should I be viewing this in when seeing an integral?

答:
$dx$ is what is known as a differential. It is an infinitesimally small interval of $x$:

$$dx=\lim_{x\to{x_0}}x-x_0$$

Using this definition, it is clear from the definition of the derivative why $\frac{dy}{dx}$ is the derivative of $y$ with respect to $x$:

$$y'=f'(x)=\frac{dy}{dx}=\lim_{x\to{x_0}}\frac{f(x-x_0)-f(x_0)}{x-x_0}$$

When doing u-substitution, you are defining $u$ to be an expression dependent on $x$. $\frac{du}{dx}$ is a fraction like any other, so to get $du$ you must multiply both sides of the equation by $dx$:

$$u=x^2+4$$

$$\frac{du}{dx}=2x$$

$$dx\frac{du}{dx}=2x\space{dx}$$

$$du=2x\space{dx}$$


答:
You just have to determine  $f\bigl([0,1)\bigr)$. This can be done using the basic properties of inequalities:

Rewrite $f(x)$ as $\sqrt{1-(x-1)^2\strut}-\frac 34$ and note that $-1\le x-1<0$, so $0\le 1-(x-1)^2<1$, and ultimately:
$$-\frac34\le f(x)<\frac14.$$

Title: Find range of $f(x)=\sqrt {2\{x\} -\{x\}^2} -\frac 34$

Body:
here {} implies fractional part function

Evaluating $2\{x\}-\{x\}^2\ge 0$

$$\{x\} \in [0,2]$$

Which is weird since $\{x\}\in [0,1)$ is a known range.

Basically I don’t know what to do here, and this was just my attempt

Ans is $[0,\frac 12]$


问:
Where does the 3rd power of the denominator of the curvature of the graph of a function come frome?
According to Wikipedia, the curvature of the graph of a function $f$ is given by the following ratio (assuming second-differentiability).

$$\mathscr{k}_f(x) = \frac{f^{\prime\prime}(x)}{(1 + [f^{\prime}(x)]^2)^{\frac{3}{2}}}$$

The denominator of $\mathscr{k}_f(x)$ is nearly of the form $(a^2 + b^2)^{\frac{1}{2}}$ that makes me think about it as the radius of a circle, but this intuition is cut short by taking that "radius" to the third power. Where does this third power come from in the derivation or definition of this functional?

答:
The curvature $K$ is defined as (the absolute value of) the rate of change of the angle of the curve's tangent with respect to the curve's arc length. That is, for
$$y = f(x)$$
$$K = |\frac{d\theta}{ds}|$$

where $\theta$ is the angle of the tangent and $s$ is the arc length.

Now
$$\tan\theta = f'(x) = y'$$
so
$$\sec^2\theta\frac{d\theta}{dx} = y''$$
$$(1+\tan^2\theta)\frac{d\theta}{dx} = y''$$
$$(1+y'^2)\frac{d\theta}{dx} = y''$$
Therefore
$$\frac{d\theta}{dx} = \frac{y''}{1+y'^2}$$

From Pythagoras' theorem,
$$ds^2 = dx^2 + dy^2$$
so
$$\frac{ds}{dx} = (1 + y'^2)^\frac12$$

And since
$$K = |\frac{d\theta}{ds}| = |\frac{d\theta}{dx} \frac{dx}{ds}|$$
we have
$$K = \frac{|y''|}{(1+y'^2)^\frac32}$$

(The denominator is guaranteed to be positive).

You should verify that a circle has constant curvature, which is the reciprocal of its radius.


答:
$\left[0,\frac12\right]$ is wrong

The fractional part ranges in $[0,1)$  subtract $\frac34$ and you have the result I gave in my comment: $$\left[-\frac34,\frac14\right)$$


问:
How can I derive what is $1\cdot 2\cdot 3\cdot 4 + 2\cdot 3\cdot 4\cdot 5+ 3\cdot 4\cdot 5\cdot 6+\cdots + (n-3)(n-2)(n-1)(n)$ ??
I'd like to evaluate the series $$1\cdot 2\cdot 3\cdot 4 + 2\cdot 3\cdot 4\cdot 5+ 3\cdot 4\cdot 5\cdot 6+\cdots + (n-3)(n-2)(n-1)(n)$$

Since I am a high school student, I only know how to prove such formula's (By principal of mathematical induction). I don't know how to find result of such series.

Please help. I shall be thankful if you guys can provide me general solution (Since I have been told that there exist a general solution by my friend who gave me this question).

答:
As shown in this post,
$$ \sum_{k=1}^n x^k = x \sum_{k=1}^{n}  \binom{n}{k} (x-1)^{k-1}$$

Differentiate five times:

$$ \sum_{k=1}^n k (k-1)(k-2)(k-3) x^{k-4} = \frac{d^5}{dx^5}  \left[ x \sum_{k=1}^{n}  \binom{n}{k} (x-1)^{k-1} \right] \tag{1}$$

For left side,

$$ \sum_{k=1}^n k (k-1)(k-2)(k-3)(k-4) x^{k-4}$$

We can start this sum:

$$ \sum_{k=5}^n (k-1)(k-2)(k-3) x^{k-4}$$

And, I can sub $k-5 =u$, this turns into:

$$ \sum_{u=0}^{n-5} (u+4)(u+3)(u+2)(u+1)x^u$$

And putting this back in(1):

$$ \sum_{u=0}^{n-5} (u+4)(u+3)(u+2)(u+1)x^u= \frac{d^5}{dx^5}  \left[ x \sum_{k=1}^{n}  \binom{n}{k} (x-1)^{k-1} \right] \tag{1}$$

Now shift $ n \to n +5$, this leads to:

$$ \sum_{u=0}^{n} (u+4)(u+3)(u+2)(u+1)x^u= \frac{d^5}{dx^5}  \left[ x \sum_{k=1}^{n+5}  \binom{n+5}{k} (x-1)^{k-1} \right] \tag{1}$$

Evaluate RHS using leibniz product rule and take limit on both side as $x \to 1$, you get the formula


答: This isn't true. Take a look at $A:=\begin{pmatrix}1 & 1 \\ 1 & 1 \end{pmatrix}$. Then you can find solution for $\det{(A-kI)}=0$ for $k_{\pm} = 1 \pm 1$ ($k_+ = 2, k_-=0)$. You see there exists a solution with $k \neq 0$


问:
Derivative of sigmoid function $\sigma (x) = \frac{1}{1+e^{-x}}$
In my AI textbook there is this paragraph, without any explanation.

The sigmoid function is defined as follows

$$\sigma (x) = \frac{1}{1+e^{-x}}.$$

This function is easy to differentiate because

$$\frac{d\sigma (x)}{d(x)} = \sigma (x)\cdot (1-\sigma(x)).$$

It has been a long time since I've taken differential equations, so could anyone tell me how they got from the first equation to the second?

答:
No answer yet involves the logarithm. If $\log$ denotes the natural logarithm then by the chain rule
$$\frac{d}{dx} \log(\sigma(x)) = \frac{1}{\sigma(x)}\frac{d \sigma(x)}{dx}.
$$
Furthermore $\log(\sigma(x))=\log(e^x) -\log(1+e^x)=x - \log(1+e^x)$ so that
$$\frac{d}{dx} \log(\sigma(x)) = 1 - \frac{e^x}{1+e^x} = 1-\sigma(x).$$
Equaling the two displays gives $(d/dx)\sigma(x) = \sigma(x)(1-\sigma(x))$ as desired.


问:
Linear Approximation
I have an exercise, giving this question.

Find the linear approximation $Y$ to $f(x)$ near $x=a$.
  $$
f(x) = x + x^4,\quad a=0
$$

I can see in my result list that it says $Y=x$, however, after multiple tries I can only convince myself that it gives $Y=0$.

Can anyone explain why it gives $1$ and not $0$?

答:
For any differentiable function $f$, the best linear approximation to $f(x)$ at the point $x=a$ is the line defined by
$$y=f'(a)(x-a)+f(a).$$
(Intuitively, this makes sense: $f'(a)$ is like "the slope of $f$ at $a$", and the line with slope $f'(a)$ that goes through the point $(a,f(a))$ is precisely the one specified above).

Can you find the derivative of $f(x)=x+x^4$ with respect to $x$?

Title: Linear Approximation


问:
$\lim_{n\to\infty}ne^{\frac{x}{n}}-n = x$?
Im fairly sure it's $x$. Here's what I've got so far.
$\lim_{n\to\infty}ne^{\frac{x}{n}}-n = \lim_{n\to\infty}n(e^{\frac{x}{n}}-1) = \lim_{n\to\infty}n(-1+\sum_{i=0}^\infty\frac{x^i}{i!n^i}) = \lim_{n\to\infty}x+\sum_{i=0}^\infty\frac{x^i}{i!n^{i-1}}$

答:
You have :

$$ne^{\frac{x}{n}}-n=n\left(1+\frac{x}{n}+o\left(\frac{1}{n}\right)\right)-n=x+o(1) $$

So you get :

$$\lim_{n\to\infty}ne^{\frac{x}{n}}-n = x $$


问:
upper bound of n-derivative of sigmoid function
Let $\sigma(x)=\frac{1}{1+e^{-x}}$ be the sigmoid function, and $\sigma^{(n)}(x)$ be its n-derivative, for any $x\in [-1/2,1/2]$, any $n\geq 1$, is it true that $|\frac{\sigma^{(n)}(x)}{n!}|\leq C$? C is a constant.

Thanks!

答: I got a way to solve this problem, which also uses Leibniz rule, but in a tricky way. Note that $\sigma^{(1)}=\sigma(1-\sigma)$, therefore, $\sigma^{(n+1)}=(\sigma(1-\sigma))^{(n)}=\sum_{k}\frac{n!}{k!(n-k)!}\sigma^{(k)}(1-\sigma)^{(n-k)}$. Now do induction, suppose $|\sigma^{(k)}|\leq k!$, we can easily get the final result.


问:
Basis of vector fields on manifold
For a typical real manifold of $n$ dimensions, the basis of a tangent vector space $T$ at point $p$ is $$\frac{\partial}{\partial x_i},\ldots,\frac{\partial}{\partial x_j}$$
So is the general basis of the $k$-vector

$$\frac{\partial}{\partial x_{i}}\otimes\frac{\partial}{\partial x_j}\otimes\frac{\partial}{\partial x_{l}}\otimes\cdots\otimes\frac{\partial}{\partial x_t}$$

where $i,j,l,\ldots,t$ are $k$ indices, run from $1$ to $n$ and $\otimes$ is tensor multiplication?

Also for tensor field at $p$ of valence $(a,b)$ is the basis

$$\frac{\partial}{\partial x_i}\otimes\cdots\otimes\frac{\partial}{\partial x_t}\otimes dx_q\otimes dx_r\otimes\cdots\otimes dx_y$$

where $i,...,t$ are $a$ indices and $q,r,...,y$ are $b$ indices?

答:
Yes, although most modern books will use subscripts with indices (which look scary) but make the counting clearer than writing letters. For example, a $(2,1)$ tensor should look like
$$\sum a^i_{j_1j_2}\frac{\partial}{\partial x_i}\otimes dx_{j_1}\otimes dx_{j_2}\,.$$
Most of us think of a $k$-vector as dual to a $k$-form (so it will be an alternating $(0,k)$-tensor), so I'd write it as linear combinations of
$$\frac{\partial}{\partial x_{i_1}}\wedge\frac{\partial}{\partial x_{i_2}}\wedge \dots \wedge\frac{\partial}{\partial x_{i_k}}\,.$$

Title: Basis of vector fields on manifold


答:
Here is my Basic code (17 lines, including comments). It prints $116$, so it looks like you are treating the two rooks as different.

```<code>Private R1X, R1Y, R2X, R2Y, BX, BY
Private Counter = 0

For R1X = 1 To 3 : For R2X = R1X + 1 To 4
  For R1Y = 1 To 4 : For R2Y = 1 To 4
    If R2Y <> R1Y Then
      For BX = 1 To 4 : For BY = 1 To 4
        If BX <> R1X And BX <> R2X And BY <> R1Y And BY <> R2Y Then
          Rem  A bishop at (BX,BY) attacks a piece at (X,Y) if
          Rem  Abs(BX-X) = Abs(BY-Y):
          If Abs(BX - R1X) <> Abs(BY - R1Y) And Abs(BX - R2X) <> Abs(BY - R2Y) Then
            Counter = Counter + 1
          End If
        End If
      Next BX, BY
    End If
Next R1X, R1Y, R2X, R2Y

Print Counter
</code>```

Title: How many ways can 2 rooks and a bishop be placed on a 4x4 board such that no piece attacks another piece?

Body:
How many ways can $2$ rooks and a bishop be placed on a $4\times 4$ board such that no piece attacks another piece?

I stumbled upon this question and I don't know how to find the answer mathematically. I wrote this piece of code to find the answer to it. I think the code is right, but I get the answer as $232$, which doesn't "feel" right. So how do you actually go about solving this question?

Body:
Here is my Basic code (17 lines, including comments). It prints $116$, so it looks like you are treating the two rooks as different.

Private R1X, R1Y, R2X, R2Y, BX, BY
Private Counter = 0

For R1X = 1 To 3 : For R2X = R1X + 1 To 4
  For R1Y = 1 To 4 : For R2Y = 1 To 4
    If R2Y <> R1Y Then
      For BX = 1 To 4 : For BY = 1 To 4
        If BX <> R1X And BX <> R2X And BY <> R1Y And BY <> R2Y Then
          Rem  A bishop at (BX,BY) attacks a piece at (X,Y) if
          Rem  Abs(BX-X) = Abs(BY-Y):
          If Abs(BX - R1X) <> Abs(BY - R1Y) And Abs(BX - R2X) <> Abs(BY - R2Y) Then
            Counter = Counter + 1
          End If
        End If
      Next BX, BY
    End If
Next R1X, R1Y, R2X, R2Y

Print Counter


问:
The crumpled paper problem
I saw a theorem recently (and have seen similarly phrased problems before), which I found quite surprising. Unfortunately, I'm not sure what its name is (or if it has one). I'll refer to it as the "crumpled paper problem". It goes something like this:

Suppose you were to take two pieces of paper, identical in size and shape, and place one on top of another such that all 4 corners and all 4 edges line up. Now, every point along the top sheet can be "mapped" to its corresponding point on the bottom sheet. If you were to then crumple up the top sheet of paper, and toss it on top of the bottom sheet, there necessarily exists a point on the top sheet that lies precisely above its "mapping point" on the bottom sheet, no matter how it was crumpled or where it was placed.

This result, to me, at least, is very surprising. Intuitively, it seems to me that it must be that there is some way to crumple the paper, or location to toss it, such that no points line up. I can sort've imagined progressively folding the paper (analogous to crumpling), and kind of see how this might be, however, it's still quite challenging.

My question is, how is this proven, does the problem have a name, and to what field of mathematics does this belong? I'm very curious how this can be shown.

I've added those tags which I thought may be relevant. Unfortunately, I have no clue to what precise area of mathematics this belongs to. Please do correct tags as necessary.

答:
Let's assume that when you crumple a sheet of paper, the fibers don't stretch, so any two points on the crumpled paper will be no further apart than they were on the uncrumpled sheet. We also assume that crumpling is a continuous function.

Now let $P_0$ be a point on the uncrumpled sheet that is as close as possible to the shadow $Q_0$ of its location on the crumpled version. (Such a point exists because crumpling is continuous.) Let $P_1$ be the midpoint of the line segment $P_0Q_0$ (on the uncrumpled sheet), and let $Q_1$ be the shadow of the location of $P_1$ on the crumpled sheet.

If $d$ is the distance from $P_0$ to $Q_0$, then $Q_1$ must be inside (or on) a circle of radius $d/2$ centered at $Q_0$ but outside (or on) a circle of radius $d$ centered at $P_1$. There is only one such point: $Q_1$ must be on the extension of the line segment $P_0Q_0$, with $Q_0$ as the midpoint of $P_1Q_1$.  But we can now replace $P_0$ and $Q_0$ with $P_1$ and $Q_1$ (because they are the same minimal distance apart). Iterating this procedure produces a sequence of points $P_0,P_1,P_2,\ldots$, all in a straight line at distance $d$ apart. If $d\gt0$, this is a contradiction.

Title: The crumpled paper problem


问:
Prove that $f(x)>f(y)$ if $x>y$
I am working on Spivaks calculus book and have the following question:

We have a function with properties (Chapter 3, Ex. 17, (d)):
$$f(x+y)=f(x)+f(y)$$
$$f(xy)=f(x)f(y)$$
and we are trying to prove that $f(x)=x$ for $x$ a real number.
In this step (d) we want to prove that $f(x)>f(y)$ if $x>y$ and here I am struggling.

My attempt was to do the following:
$x>y$ is the same as $x-y>0$.
Then
$$f(x-y) = f(x)+f(-y) = f(x)+f(-1)f(y)=f(x)+f(y)$$
because from (a) we know that
$$f(1)=1=f((-1)(-1))=f(-1)^2$$
therefore
$$f(-1)=1$$
From (c) we know that $f(x)>0 $ if $x>0$ so
$$f(x-y)=f(x)+f(y)>0$$
but i want $f(x)-f(y)>0$.
The solution manual says instantly $f(x-y)=f(x)-f(y)$, but why is that? And what is wrong in my reasoning?

答:
Note that $f(x+y)=f(x)+f(y)$ implies that:

$f(0)=0$ because $f(0)=f(0+0)=f(0)+f(0)$

$f(-x)=-f(x)$ because $0=f(0)=f(x-x)=f(x)+f(-x)$

$f(x-y)=f(x)-f(y)$ because $f(x-y)=f(x)+f(-y)=f(x)-f(y)$

These are classical properties of group homomorphisms.

There's something wrong with your reasoning though. $f(-1)^2=1$ implies that $f(-1)=\pm 1$ not $f(-1)=1$. And indeed, in reality it is $-1$, because ultimately $f(x)=x$ is supposed to be a solution, so $f(-1)=-1$. So either $f(x)=x$ is not a solution (spoiler alert: it actually is) or you've made a mistake.

Here's a different idea. First of all if $f(x_0)=0$ for some $x_0\neq 0$ then $f(x)=0$ for all $x$. That's because

$$f(x)=f(x_0x_0^{-1}x)=f(x_0)f(x_0^{-1}x)=0\cdot f(x_0^{-1}x)=0$$

Note that constant $0$ is a valid solution.

Now assume that $f$ is not constant $0$.

For any $x> 0$ we have

$$f(x)=f(\sqrt{x}\sqrt{x})=f(\sqrt{x})\cdot f(\sqrt{x})=f(\sqrt{x})^2\geq 0$$

But as we've said $f(x)$ cannot be equal $0$ because $f$ is not constant $0$. And so $f(x)>0$.

Therefore if $x> y$ then $x-y> 0$ and so $f(x-y)> 0$ meaning $f(x)-f(y)> 0$ which finally leads to $f(x)> f(y)$.

Title: Prove that $f(x)>f(y)$ if $x>y$


问:
What is the relation between vectors in physics and algebra?
Vector math is something I find very interesting. However, we have never been told the link between vectors in physics (usually represented as arrows, e.g. a force vector) and in algebra (e.g. represented like a column matrix). It was really never explained well in classes.

Here are the things I can't wrap my head around:

How can a vector (starting from the algebraic definition) be represented as an arrow? Is it correct to assume that a vector (in a 2-dimensional space) $v = [1,1]$ could be represented as an arrow from the origin $[0,0]$ to the point $[1,1]$?

If the above assumption is correct, what does it mean in the physics representation to normalize a vector?

If I have a vector $[1,1]$, would the vector $[-1,1]$ be orthogonal to that first vector? (Because if you draw the arrows they are perpendicular).

How can one translate an object along a vector? Is that simply scalar addition?

These questions probably sound really odd, but they come from a lack of decent explanation in both physics and algebra.

答:
Actually, I think your intuition is very good here. An "arrow" has two bits of information: its length and direction. When you have a vector $v = [a, b]$, it's describing an arrow that moves $a$ in the horizontal direction and $b$ in the vertical direction. Since this defines a right triangle, this can be converted into length and direction information.

Normalizing a vector simply means adjusting the length of a vector to be $1$ without changing its direction.

To address your orthogonality question, yes, as you mentioned, $[-1, 1]$ and $[1,1]$ are indeed orthogonal, because of the geometric reasoning you correctly described. However, they are only orthogonal with respect to the standard inner product. This is imply the dot product, which has a lot of geometric connections. However, in mathematics, we can define many other types of inner products that don't have as nice geometric connections.


问:
What is a connex relation?.
I've been recently trying to teach myself some topology and in  the book I'm reading there is the definition of an order relation which confuses me a lot
.

I've looked online for more information and they seem to define an order relation differently namely one that is transitive,antisymmetric and reflexive.

So why does the author define it in this way my 2nd confusion has to do with what comparability actually means as there is actually no definition online and I cant quite understand what a connex relation is either and that seems to come up a lot in the definition of an order relation.

Thanks in advance.

答:
Traditonally, we first define a partial order (sometimes even a preorder before that) to be reflexive, antisymmmetric ($xRy$ and $yRx$ implies $x=y$, this a pre-order need not be) and transitive. Lots of examples of this: order on $\Bbb Q$ adn $\Bbb Z$, inclusion relations etc. etc.

Then among those we single out linear orders (the orders we grow up with on "numbers" are mostly linear): we want comparability: $x=y$ or otherwise $xRy$ or $yRx$ must hold (not both as this would imply $x=y$ again). But then peope wanted to axiomatise $x < y$ instead of $x \le y$ and the distinguishing things there are: no reflexivity (never $x< x$ so $xRx$ is always forbidden) and asymmetry: if $xRy$ then never $yRx$), so the relation always has to decide on one of them.

A connex relation is sort of a generalisation of the linear order: we always want either $xRy$ or $yRx$. We then don't demand transivity or asymmetry or antisymmetry necessarily. So it can be a very different thing. E.g. After a round-robin tournament in a game without draws, the players will have constructed a connex relation at the end ($xRy$ : $x$ has beaten $y$), which is the also nonreflexive as you cannot beat yourself.

Linear orders I can see the importance of, and they have a rich theory (order types etc) but connex relations don't add much in my opinion. I've never seen them in any of the papers I studied.

Title: What is a connex relation?.