问:
Motivation for vector spaces
For instance I can make a narrative for defining some familiar sets, leading up to complex numbers, as follows:

Natural numbers are obvious. They are "ordered sets" useful for counting things, and we can define addition and multiplication on them in a useful way.

Integers were defined because we wanted additive inverses, e.g. $5+x=3$. This also an "ordered set"

Rationals were defined because we wanted multiplicative inverses, e.g. $3x=2$, and now we have a "field" and since they are ordered, an "ordered field."

Reals were defined because we wanted to solve $x\cdot x = 2$, these are also an "ordered field."

Complex numbers were defined because we wanted $x^2=-1$. We defined multiplication and addition specifically so that they satisfy the field axioms, which makes complex numbers also a "field." (We lose "ordered" because we can't define it in a consistent way, but the reals are a subfield which we can order so we're still ok.)

Ok, so now we have ordered sets and fields and ordered fields. Is there a similar narrative for defining "vector spaces?"

Most textbooks simply start with "here are fields and here are the axioms for vector spaces over fields." e.g. what is the sequence of logic that led us to invent $\mathbb{R}^N$ or $\mathbb{C}^N$, along with the addition/scaling/identity/closure axioms that go with it? Why didn't we introduce the concept of multiplication (I don't mean inner/outer/cross product) into the definition?

答:
(Disclaimer: This is not at all a historical account - it is a path of intuition that leads to vector spaces, not the only path)

Vector spaces! (a.k.a. stuff you can add and scale, but not multiply)

Vector spaces are intentionally capturing the notion of quantities that can be sensibly added together or scaled, but not necessarily anything else. This situation arises naturally in various phenomenon - I'll discuss two simple examples: durations and displacements. Then, we can discuss why a weaker set of axioms, extracted only from the goal of having addition and scaling, is a useful thing - especially when this is a suitable model of natural objects we care about. Finally, we get to $\mathbb R^n$ (and its rules for addition and multiplication) which arises through the use of coordinates on (finite dimensional) vector spaces.

Example 1: The space of durations

For durations, we can clearly have some idea that it ought to be possible to take two lengths of time and sum them together - certainly expressions such as
$$1\text{ minute} + 17\text{ seconds}$$
$$1\text{ hour} + 34\text{ minutes}$$
make sense, where we can just imagine placing such intervals of time end-to-end and will naturally run into such things. It also makes sense to scale a duration - I can say that an hour is sixty times as long as a minute, and generally multiplications such as
$$2\cdot( 51\text{ seconds})$$
$$\frac{1}7\cdot (1 \text{ hour})$$
clearly refer to meaningful durations - where I can take a duration and a positive scaling factor and get a new duration. What doesn't make sense*, however, is to try to multiply two durations:
$$(1\text{ hour})\cdot (1\text{ hour}) =\,\,???$$
$$(60\text{ minutes})\cdot (60\text{ minutes}) =\,\,???$$
Sure, you could come up with arbitrary definitions for multiplication of durations, but it wouldn't really mean anything tangible. We could get around this sort of issue if we all agreed "duration will always be represented as a real number their length in minutes," but this would be a completely arbitrary choice and wouldn't tell us anything new about durations. If we want to stick to the natural phenomenon, we're left with only a couple operations on durations: we can add and we can scale. We might envision a duration as something that lives on a timeline - just being some segment of that line that we can lay end to end to add or dilate to scale - but that's about all you can do with line segments without adding any extra information.

You might generalize a bit to add some sign information to this - maybe we want to discuss how 5:00 PM is one hour before 6:00 PM, but 7:00 PM is one hour after. Okay, fine, we can revise our model to imagine that durations are more like arrows between points on a timeline - they have a length and a direction. This is much like how one gets from natural numbers to integers.

At this point all we really know is that we can add and subtract signed durations, as well as scale them by real factors. There's our first example of a vector space, albeit a one-dimensional one.

Example 2: The space of displacements

Displacements (changes in position) are a more canonical example of the same phenomenon - and are absolutely fundamental in any pursuit of physics. For the purposes of this post, let's imagine** that Earth is flat to avoid clever people poking holes in things. A displacement is just a relationship between points - for instance, I could say "this road is 5 miles north of that road" or "that road is 5 miles south of this road" or "that satellite is 254 miles above me." I could geometrically imagine them as just arrows in space, but where I'm not interested in where the arrow starts and ends, but only in the size and direction of the arrow itself, so that "5 miles north" is a displacement regardless of where I base it.

It should be clear that we can add these things together - certainly, "move 10 feet North, then 50 kilometers downwards" is a sensible way to describe the spatial relationship of two points, and can visually be thought of laying two arrows end to end - and this is needed to say, for instance, "my net change in position of two days is the sum of the change in the first day with the change in the second day," or to, more sophisticatedly, develop calculus to deal with the motion of objects through space and say "the net change in position is the integral of velocity (rates of displacement) over time."

Similarly, it makes sense to scale these arrows - saying "it's thrice three furlongs east" is sensible as is "it's half an angstrom above." We can also negate displacements by reversing their direction and even define scaling by negative factors to be scaling by a positive, then reversing direction.

Again, multiplication does not make sense: what is "two inches east times five meters north"? In fact, it makes even less sense than with durations, because now we have the issue of direction - where exactly is east times north? How's it relate to up times up? That's all utter nonsense!

All we really know is that we can add and subtract displacements, as well as scale them by real factors. Hm, if only we had a name for stuff that we can add and scale...

Vector spaces, abstractly

A vector space is meant to model this kind of behavior where it's possible to add and scale things - and we can sort of work out what axioms we expect to hold. There's certainly plenty of work there, but once we've decided that we want to talk about addition and scaling as two operations, the vector space axioms are all fairly reasonable to assume, even if their necessity may not immediately be apparent. It's not so objectionable to ask that there be a zero vector, or that addition be associative and commutative, or that additive inverses exist, nor should statements like "scaling by $a$ then $b$ is the same as scaling by $ab$" or "scaling $x+y$ by $a$ should give $a\cdot x + a\cdot y$" or "scaling $x$ by $a+b$ should give $a\cdot x + b\cdot x$" or "scaling by $1$ shouldn't do anything" - and that's literally all that the vector space axioms say.

The trick of this sort of axiomatization is this: a lot of things satisfy these axioms, be it durations, displacements, weights, velocities, or accelerations - or even more exotic things such as waves or abstract ones such as $\mathbb R^n$ or "the space of solutions to a recurrence relation" or "the space of continuous functions $\mathbb R\rightarrow\mathbb R$." This is to say: the axioms aren't too strong to exclude having lots of examples. However, the axioms are strong enough to prove useful facts - such as bringing ideas of matrices and bases and dimension to bear on problems in any of these instances - and we know we've done something right when we find a set of axioms that gives us abstract theorems that apply to loads of situations that we cared about before we had the axioms.

We would lose the broad applicability of the theory if we started insisting upon multiplication or anything else - yeah, there are things such as algebras, which are vector spaces with a multiplication rule, or inner product spaces or Banach spaces - and sometimes you do need this extra structure - but not everything is like that, so we can get more general results by not including anything we don't need. This is a bit different from the situation with complex numbers where it sometimes doesn't hurt to have extra stuff in your space - we're defining a class of things rather than a single thing, and we actually benefit from not requiring anything we're not going to use.

Coordinates (a.k.a. why $\mathbb R^n$ is so ubiquitous)

Let me finish this discussion by bringing coordinates into play. The final piece of the puzzle is that vector spaces are defined (up to isomorphism) by their dimension - which is to say, they can all be given coordinates.

In the two explicit examples, it should be fairly clear that a signed duration is just any expression of the form $x\text{ years}$ for any real $x$. So, durations can be represented by a single real number, and adding them together adds this number, whereas scaling them multiplies by that real number. This isn't inherent to the space of durations - you can choose any non-zero duration to base your coordinate system on and then say every other is some factor times that one - but it is possible.

Similarly, any displacement can be written uniquely as
$$x\text{ meters north} + y\text{ meters east} + z\text{ meters up}$$
where we have three coordinates - and the sum of
$$x_1\text{ meters north} + y_1\text{ meters east} + z_1\text{ meters up}$$
$$x_2\text{ meters north} + y_2\text{ meters east} + z_2\text{ meters up}$$
just turns out to be
$$(x_1+x_2)\text{ meters north} + (y_1+y_2)\text{ meters east} + (z_1+z_2)\text{ meters up}$$
and scaling by a factor $c$ would give
$$(cx)\text{ meters north} + (cy)\text{ meters east} + (cz)\text{ meters up}.$$
These are facts with real geometrical significance - they give us a way to talk about displacements mathematically. We might try to abbreviate these notations by just writing a tuple of $(x,y,z)$ instead to start being able to tersely write:
$$(x_1,y_1,z_1)+(x_2,y_2,z_2)=(x_1+x_2,y_1+y_2,z_1+z_2)$$
$$c\cdot (x_1,y_1,z_1)=(cx_1,cy_1,cz_1)$$
and then, oh look, we just invented the vector space $\mathbb R^3$ consisting of tuples of three real numbers, with addition and scaling defined per-component - and this turns out to be a perfectly good representation of displacements. Of course, we might immediately start to generalize about other things we care about - a velocity can be written as
$$x\text{ meters/second north} + y\text{ meters/second east} + z\text{ meters/second up}$$
and it's going to have the same rules for addition and multiplication of the tuples $(x,y,z)$ as before - suggesting that $\mathbb R^3$ can be used to represent these too.

Again, these notions aren't intrinsic to the space - I could just as well say any displacement can be written as
$$x'\text{ lightyears northeast} + y'\text{ nanometers west} + z'\text{ smoots up-south}$$
where a "smoot" is any unit of distance. Sure, each displacement would have to be written using a different tuple $(x',y',z')$ than it was in the previous form - and we might find this form less intuitive - but it's equally correct, and gives another way we could represent the space using $\mathbb R^3$.

This process is essentially just picking a basis for our vector space - as long as we all agree which displacement is represented by $(x,y,z)$, we can happily use that representation pretty much everywhere. Even when our end goal isn't to talk about $\mathbb R^3$ (and it hardly ever is), the fact is that a lot of spaces we do care about can easily be manipulated through coordinates, which makes the space of tuples $\mathbb R^3$ rather fundamental - and, of course, it should bring us no obstacle to realize that we can have $\mathbb R^n$ for any $n$ that we like, with similar rules for addition and multiplication.

(*Okay, okay, you could get $1\text{ hour}^2$, whatever that means, and I'm sure physicists sometimes are happy to do this sort of thing - but clearly that's not a duration - it's transformed into something else)

(**Or, if you already believe this, don't stop believing it)

Title: Motivation for vector spaces


问:
How should one explain the conventional formula for relative variation of signed variables?
The conventional formula for the relative variation between two signed values is $\displaystyle \frac{b-a}{|a|}$. Unfortunately, I could not find any clear justification for it, especially when considering examples like this one: according to the conventional formula, the relative variation between -10 and +20 is +300%, while the relative variation between -20 and +20 is +200%. This is puzzling, because the absolute variation for the former is +30 while the absolute variation for the latter is +40 and both datasets end at the same +20 value, yet the relative variation for the former is greater than the relative variation for the latter.

How can one justify this counter-intuitive scenario?

答:
To better understand what is going on, it usually helps to think of the relative growth between two values of different signs as being composed of two separate parts: the relative growth from the first value to zero, plus the relative growth from zero to the second value. For example, the relative growth between $-10$ and $20$ is equal to the relative growth between $-10$ and $0$, plus the relative growth between $0$ and $20$.

The relative growth between $-10$ and $0$ is $\displaystyle \frac{0-(-10)}{|-10|} = \frac{10}{10} = 1 = 100\%$ according to our previous formula.

In fact, the relative growth between any negative value and $0$ is always equal to $100\%$, which actually makes sense when one comes to think about it. But things become a bit more tricky when we need to compute the relative growth between $0$ and a positive value, like $20$ in our previous example. There, we cannot simply use our formula, because it would lead to a division by zero. Instead, we have to compute this relative growth in relation to the previous one, by computing the ratio of two absolute growths.

For example, when going from $-10$ to $20$, we gain $10$ in absolute terms between $-10$ and $0$, then $20$ in absolute terms again between $0$ and $20$. Therefore, we gained twice as much going from $0$ to $20$ than we did when going from $-10$ to $0$. And as we have seen earlier, the relative growth between $-10$ and $0$ is $100\%$, therefore the relative growth between $0$ and $20$ should be twice that amount, or $200\%$, and the relative growth between $-10$ and $20$ should be the sum of our two relative growths, or $100\% + 200\% = 300\%%$, which matches what our adjusted formula gave us in the first place. In other words, we found two ways of getting to the same result, but this should not come as a surprise if you look at the original equation.

Indeed, if $a$ is negative, we can rewrite $\displaystyle \frac{b-a}{|a|}$ as $\displaystyle \frac{b-a}{-a}$. Then, we can split the fraction as $\displaystyle \frac{b}{-a} + \frac{-a}{-a}$ and further simplify it into $\displaystyle \frac{b}{-a} + 1$, or even $\displaystyle 1 + \frac{b}{-a}$. This addition has two addends, $1$ and $\displaystyle \frac{b}{-a}$. The former is the relative growth from the first value to zero (always equal to $1$), and the latter is the ratio of the absolute growth between the first value and $0$ and the absolute growth between $0$ and the second value, which we can write as $\displaystyle \frac{b - 0}{0 -a}$.

When we look at the formula through that angle, we realize that a relative growth between two values of different signs is only affected by the ratio between the two values $\displaystyle \frac{b}{-a}$, unlike the relative growth between two values of the same sign, which is the ratio between the difference of the two values and the first value $\displaystyle \frac{b-a}{a}$. But this is where the adjusted formula becomes misleading, because we should not look at the former as a ratio between two values. Instead, we should look at it as a ratio between two absolute growths $\displaystyle \frac{b - 0}{0 -a}$.

Make no mistake though: this is not a trivial arithmetic rewriting. Instead, by writing $\displaystyle \frac{b}{-a}$ as $\displaystyle \frac{b - 0}{0 -a}$, we clearly communicate the fact that our fraction is a ratio between two absolute growths, instead of a ratio between two values. And this is what explains that the relative growth between $-10$ and $20$ ($300\%$) is greater than the relative growth between $-20$ and $20$ ($200\%$).

As explained earlier, to grow from $-10$ to $20$, you first need to grow from $-10$ to $0$, then grow from $0$ to $20$, and the first growth component to $0$ is always equal to $100\%$. But when you grow from $0$ to $20$, while this absolute growth ($20$) is the same whether you started from $-10$ or $-20$, the relative growth is twice as much if you started from $-10$ than if you started from $-20$.

In other words, what the relative growth of values of different signs really measures is the relative growth after you have crossed the zero line, because the relative growth before you cross the zero line is always the same ($100\%$).

ContentLicense: CC BY-SA 4.0


问:
Find $\lim_{k\rightarrow \infty}\binom{2k+1}{3}\frac{1}{2^{2(2k-1)}}\left[\binom{2k-2}{k-2}-\binom{2k-2}{k-1}\right]^2$.
While trying to compute the limit as $k\rightarrow \infty$ of the level 3 Energy Spectrum of the Majority Boolean Function, the following expression arises

$$\binom{2k+1}{3}\frac{1}{2^{2(2k-1)}}\left[\binom{2k-2}{k-2}-\binom{2k-2}{k-1}\right]^2.$$

Are there any tricks\bounds (non brute force) to compute this limit?

As $\sum_{j=0}^{k-1}\binom{2k-2}{2j}=\sum_{j=1}^{k-1}\binom{2k-2}{2j-1}=2^{2k-3}$, my guess is that this limit is zero, but ignoring the term $-2\binom{2k-2}{k-2}\binom{2k-2}{k-1}$ and using that $\binom{2k-2}{j}\leq 2^{2k-3}$ is not sharp enough.

答: I am not quite sure, if this qualifies as brute force, so you be the judge of that. Try to prove the identity $\binom{2n}{n-1}-\binom{2n}{n}=-\frac1{n+1}\binom{2n}{n}$ to simplify the expression between the parantheses. Then using that $\binom{2n}{n}\sim\frac{2^{2n}}{\sqrt{\pi n}}$ for n large enough (hint: Stirling's approximation) you can calculate the limit.


问:
Equation for orthocentre in argand plane
I would like to know the general equation for orthocentre (and other centres of the triangle as well, excluding the centroid, if possible) in the argand plane. In my case, I faced some difficulty in a particular problem where all vertices of the triangle lie on a circle. So, please provide me with the solving methodology, if a separate one exists for this case.

答:
Lemma. Let $O$ be the circumcentre of $\triangle ABC$. Then $\overrightarrow{OH}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}$.

Proof. You can do this yourself -- just verify that $\overrightarrow{AH}\cdot \overrightarrow{BC}=0$ etc.

So in complex numbers, if your vertices are $a$, $b$, $c$, circumcentre is $o$ and orthocentre $h$, then
$$(h-o)=(a-o)+(b-o)+(c-o)\implies h=a+b+c-2o.$$
In general, the circumcentre of $a$, $b$, $c$ is given by the fearsome determinant formula:
$$\begin{vmatrix}
a&a\overline{a}&1\\
b&b\overline{b}&1\\
c&c\overline{c}&1
\end{vmatrix}\div
\begin{vmatrix}
a&\overline{a}&1\\
b&\overline{b}&1\\
c&\overline{c}&1
\end{vmatrix}.$$
The proof of this formula is to consider the system
\begin{align*}
\lvert a-o\rvert^2&=R^2\\
\lvert b-o\rvert^2&=R^2\\
\lvert c-o\rvert^2&=R^2.
\end{align*}
Expand the moduli in terms of conjugates, then apply Cramer's rule where the unknowns are $o$, $\overline{o}$ and $R^2-o\overline{o}$.

This circumcentre formula becomes much more tractable when $c=0$ -- we just get $$\frac{ab\overline{(a-b)}}{\overline{a}b-a\overline{b}}.$$
So if you want to compute an arbitrary circumcentre, shift one of the vertices to the origin, apply the above (simpler) formula, then shift back.


问:
$A_{\mathfrak{p}}\otimes_AB\cong B_\mathfrak{q}$
Question: Let $f: A\rightarrow B$ be a homomorphism of rings which makes $B$ an $A$-algebra. Suppose $\mathfrak{q}$ is a prime ideal of $B$ and $f^{-1}(\mathfrak{q})=\mathfrak{p}$, then I want to know if we have $A_{\mathfrak{p}}\otimes_AB\cong B_\mathfrak{q}$ as rings?

I don't assume that $f$ is surjctive. I can consturct a map $A_{\mathfrak{p}}\otimes_AB\rightarrow B_\mathfrak{q}$ by $\frac{a}{s}\otimes b\rightarrow \frac{f(a)b}{f(s)}$. To prove the isomorphism, I want to construct the inverse map, but it seems to me there is no obvious map, since $f$ is not assumed to be surjcetive.

Motivation: This result is used to prove $\mathcal{O}_{X, x}/\mathfrak{m}y \, \mathcal{O}_{X,x} \simeq \mathcal{O}_{X_y, x}$ in this question, User "nowhere dense" considered the map $B\xrightarrow{g} B\otimes_A A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$ in that question, and he got $(B\otimes_A A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p})_\tilde{\mathfrak{q}}\cong
( A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}\otimes_A B)\otimes_B B_\mathfrak{q}$. I don't know why this is true.

Edit. I don't think this question is just $S^{-1}A\otimes_A B\cong f(S)^{-1}B$, since we don't know $A-\mathfrak{p}$ doesn't map to $B-\mathfrak{q}$.

Any help is appreciated.

答:
This is not true. Suppose $A= \mathbb{C}, p=0$ and $B=\mathbb{C}[t], q=(t) $. The map $f: \mathbb{C} \to \mathbb{C}[t] $ is the inclusion. Note that $f^{-1}(q) = q \cap \mathbb{C} = 0$.

You have that $A_p = A$ because $\mathbb{C}$ is a field and everything beside zero is already invertible. Now
$$ A_p \otimes_A B = A \otimes_A B \simeq B$$

but
$$ B_q = \mathbb{C}[t]_{(t) }$$

is not isomorphic to $B$, because it is a local ring while $B$ has infinitely many maximal ideals.


问:
Algebra help with distance, two runners on circular track. How far before they meet.
Two runners race around a circular track, with a 300 ft radius, in opposite directions, from the same starting point. Tom averages 8 mph and Dick averages 12 mph. How far has Dick run before he meets Tom?</p>

答:
The track length is $\quad300\times 2\pi\approx 1885'.\quad $
The two runners are converging at $$\quad12+8=20\text{ mph}=
\frac{20\times 5280}{3600}=\frac{88}{3}\text{ fps}.$$

This means they will meet in $$\frac{3}{88}\cdot 1885
\approx 64.26 \text{ seconds}$$

Dick is running at $$\frac{12}{20}\cdot \frac{88}{3}=17.6 \text{ fps}$$

The distance Dick has run when he meets Tom is
$17.6\times 64.26\approx 1131'$

We can compare this to the easy way where Dick's distance is
$$\frac{12}{20}\cdot 1885=1131'$$


问:
Can we have a subgroup of order $20$ for a group of order $30$?
Here is the problem: we have to show that there does not exist a subgroup of order $20$ when the whole group itself has the order $30$.

Is using Lagrange's theorem sufficient i.e. $|G|/|H|=30/20=3/2$ is not an integer, thus the group of order $30$ cannot have a subgroup of order $20$?

答: No.  Because by Lagrange's theorem, the order of any subgroup divides the order of the group.  And $20\nmid 30$.


问:
Result on subspace Borel $\sigma$-algebra: Sorgenfrey plane provides counterexample?
Let $X\supseteq Y$ be sets, let $\tau$ be a topology on $X,$ and let $\tau_Y$ be the subspace topology on $Y;$ then
$$\sigma_Y(\tau_Y)=\{A\cap Y:A\in\sigma(\tau)\},$$
where for $Z\subseteq Y$ we define $\sigma_Y(Z)$ to be the $\sigma$-algebra generated, on $Y,$ by $Z.$ (This result is proved in e.g. user642796's Math.Stackexchange post.)

However, doesn't Example 4.12 in Wise & Hall's Counterexamples in Probability and Real Analysis contradict this result?

(In the notation of the above example, our result says $\sigma_B(\tau_B)=\{A\cap B:A\in\sigma(\tau)\},$ so that every $C\in\sigma_B(\tau_B)$ is of the form $A\cap B$ with $A\in\sigma(\tau);$ but $B$ is closed and hence is in $\sigma(\tau),$ so $A\cap B\in\sigma(\tau),$ contradicting the statement of the Example.)

Maybe Wise & Hall made a mistake somewhere...

答: There is indeed an error: The Borel $\sigma$-algebra of $H$ properly contains the Borel $\sigma$-algebra of $\Bbb R^2$. This is clear from the fact that every subset of the reverse diagonal $y=-x$ is closed in $H$ and therefore a Borel set in $H$, but not every subset of the reverse diagonal is a Borel set in $\Bbb R^2$. For further explanation see the answer to this MathOverflow question.


问:
Prove an entire function is a constant under an inequality
f is an entire function, suppose $|f(z^{2})| \leq  2|f(z)|$ for all C, then f is a constant.

I 'm trying to use Liouville's theorem, but it seems that it isn't helpful.

答:
Let $M(r)=\max_{|z| \le r} |f(z)|$. The hypothesis yields that $M(e^{2^k}) \le 2^k M(e)$ for all integer $k \ge 0$, so $f$ has sublinear growth. Since analytic functions are harmonic, we conclude that $f$ is constant by [1].

[] A harmonic function with sublinear growth at infinity is constant


问:
An art dealer receives five old paintings, given that a painting chosen at random is a forgery what is the probability that all are forgeries?
An art dealer receives a shipment of five old paintings from abroad, and, on the basis of past experience,
she feels that the probabilities are, respectively, 0.76, 0.09,
0.02, 0.01, 0.02, and 0.10 that 0, 1, 2, 3, 4, or all 5 of them
are forgeries. Since the cost of authentication is fairly
high, she decides to select one of the five paintings at random and send it away for authentication. If it turns out
that this painting is a forgery, what probability should she
now assign to the possibility that all the other paintings
are also forgeries?

So far I tried assigning A as the event all paintings are forgeries and N that none of the paintings are forgeries so the required probability becomes P(A|N*) = $$\left(\frac{P(A \cap N*)}{P(N*)}\right)$$

which gives $\left(\frac{0.10}{0.24}\right)$ cause $P( A \cap N*) = $ P (A) = 0.10 and P(N*) = 1-0.76 = 0.24
and my answer is o.4166 but the textbook gives the answer as 0.6757

答:
Intuitively, the art dealer has more information than just "at least one is a forgery" when she discovers that a particular one is a forgery.  This is the reason your answer doesn't match the correct answer.

Let $F_i$ denote the event that $i$ are forgeries and $C$ denote the event that she chooses a forgery.  Then

$$\mathbb{P}(F_5 | C) = \frac{\mathbb{P}(F_5)\mathbb{P}(C|F_5)}{\mathbb{P}(C)}$$

A couple of these are trivial: $\mathbb{P}(F_5) = 0.10$ and $\mathbb{P}(C|F_5) = 1$, but the last one requires a little bit of work:

$$\begin{align*}\mathbb{P}(C) &= \sum_{i=0}^5\mathbb{P}(F_i)\mathbb{P}(C|F_i) \\&= \sum_{i=0}^5\mathbb{P}(F_i)\cdot \frac{i}{5} \\&= 0.09\cdot \frac{1}{5}+0.02\cdot\frac{2}{5}+0.01\cdot\frac{3}{5} + 0.02\cdot\frac{4}{5}+0.10 \\&= 0.148\end{align*}$$
(here, we also can note that the probability she chooses a forgery is the expected number of forgeries divided by the number of paintings)

Therefore $$\mathbb{P}(F_5|C) = \frac{0.10}{0.148} \approx 0.675675676$$


问:
maximum volume?

I know I have to derive the formula for the volume $V=A_{base}*h$, with $h=600cm$, set it equal to zero and the result would be the maximum, but I don't know what parameters to take to get the area of the triangle.
I hope you can help me, thank you!

答: From your diagram, we can see that $(x/2)^2 + y^2 = 0.5^2$ by the Pythagorean theorem on the triangle face. From your diagram, we can see that the volume is equal to $\frac{xy}2$ for the face area times $6$ for the height, so we get $V=3xy$, which is the quantity we want to maximize. Use this equation $(x/2)^2 + y^2 = 0.5^2$ and solve for $y$. We get $ y = \sqrt{0.25-x^2/4}$. Substitute this into the volume equation to get $V=3x\sqrt{0.25-x^2/4}$. Find $V'$ and set it to $0$, and find the $x$ that makes $V' = 0$. Use $x$ to then find $y$. I hope this helps.

Title: maximum volume?


问:
Are all diagonalizaitions necessarily eigendecompositions?
If a matrix $A \in \mathbb{F^{NxN}}$ can be diagonalized, i.e. factored into the form:

$A = V \Lambda V^{-1}$, where V is a basis for $\mathbb{F^N}$ and $\Lambda$ is a diagonal matrix, do $V$ and $\Lambda$ necessarily contain the eigenvectors and eigenvalues of $A$ respectively, or could a diagonalization be in terms of other vectors/diagonal values?

Is the answer to this question different between asymmetric square matrices on the one hand and symmetric/Hermitian or normal matrices which decompose as $Q\Lambda Q'$ for unitary $Q$ on the other hand? And if so, or if all diagonalizations are necessarily in terms of eigenvectors/values, why?

答:
Rewrite
$$
A = V \Lambda V^{-1}
$$
as
$$
AV = V \Lambda
$$
and let $v_1$ be the first column of $V$.

Now look at the first column of $AV$ --- it's just $Av_1$.

What about the first column of $V \Lambda$? It's just $\lambda_1 v_1$.

The same reasoning applies to all other columns, hence each column of $V$ is an eigenvector of $A$.

I leave you to think about the complex case on your own...


问:
In any tree, what is the maximum distance between a vertex of high degree and a vertex of low degree?
In any undirected tree $T$, what is the maximum distance from any vertex $v$ with $\text{deg}(v) \geq 3$ to the closest (in a shortest path sense) vertex $y$ with $\text{deg}(y) \leq 2$? That is, $y$ can be leaf.

It seems to me that this distance can be at most $\dfrac{\text{diam}(T)}{2}$, and furthermore that the maximum distance will be attained from a graph center. Is this true? There's probably simple argument for it somewhere.

答:
In answer to the first version of the question (between any $v$ and any $y$):

------------------------------------------------------<

y is at the left, v is at the vee

The answer to the second version of the question is yes.  Suppose that $v$ were more than $\frac{D}{2}$ away from all leaves, where $D$ denotes the graph diameter.  Pick two of the branches leading from $v$.  Each of them contains at least one leaf.  The shortest path between these leaves passes through $v$, and is of distance more than $D$, being the sum of two distances each more than $\frac{D}{2}$.  This is a contradiction.

The bound given is tight, by considering the Y graph on four vertices.

ContentLicense: CC BY-SA 3.0


答:
Since $A=V\Lambda V^{-1}$ the eigenvalues of $A$ and $\Lambda$ are the same, since $\Lambda$ is diagonal, we can read the eigenvalues immediately off from $\Lambda$. Writing the equation as $AV=V\Lambda$, then if $\Lambda=\text{diag}(\lambda_1,\ldots,\lambda_n)$ and the column vectors of $V$ are $v_1,\ldots,v_n$, we see that $Av_i=\lambda_i v_i$ holds, so we can read off from $V$ an eigenvector $v_i$ corresponding to the eigenvalue $\lambda_i$ for every $i$.

A diagonalisation in terms of different $V$ and $\Lambda$ is possible. For instance, we could permute the entries of $\Lambda$. Also, if $v$ is an eigenvector, so is $\lambda v$ for a non-zero scalar $\lambda$. Therefore, we could multiply every column of $V$ with a different non-zero scalar and still get a diagonalisation.

Title: Are all diagonalizaitions necessarily eigendecompositions?

Body:
If a matrix $A \in \mathbb{F^{NxN}}$ can be diagonalized, i.e. factored into the form:

$A = V \Lambda V^{-1}$, where V is a basis for $\mathbb{F^N}$ and $\Lambda$ is a diagonal matrix, do $V$ and $\Lambda$ necessarily contain the eigenvectors and eigenvalues of $A$ respectively, or could a diagonalization be in terms of other vectors/diagonal values?

Is the answer to this question different between asymmetric square matrices on the one hand and symmetric/Hermitian or normal matrices which decompose as $Q\Lambda Q'$ for unitary $Q$ on the other hand? And if so, or if all diagonalizations are necessarily in terms of eigenvectors/values, why?


问:
How to find angle in between two right triangles when one side is twice of other?
The problem is as follows:

$\begin{array}{ll}
1.&10^\circ\\
2.&12^\circ\\
3.&15^\circ\\
4.&16^\circ\\
\end{array}$

I'm not sure which sort of construction can be used here to solve this problem?.

I've attempted to draw a perpendicular line from $B$ to segment $AC$. But this did not yielded good results.

Then I've attempted tracing a perpendicular segment to $BC$ intersecting $AC$, from looking on this possibility. I got still stuck.

I don't know how to use the given angles, they suspiciously add up to $3\alpha$.

I still don't know how to use $QC=2HC$. Can someone give me some ideas on what to do?. Should congruence be used here?.

I'm assuming that the intended approach is try to spot triangles such as:

$3-4-5$ or $1-\sqrt{3}-2$ or something along those special right triangles.

I do hope someone could help me on how to solve this problem relying on euclidean geometry can this be done?.

答: Hint: Using trigonometry and $|QC|=2|HC|$, express the lengths of $BQ,BP,PC$, and $BC$ in terms of $|HC|$. Now observe that $B,P,C$ are collinear.


问:
Is a thick Möbius strip genus 2?
A cylinder has genus 0, but a thick cylinder has genus 1. This gets me into wondering, since a Möbius strip has genus 1, would a thick Möbius strip have genus 2? I can't quite comprehend if the cuttings along a thick Möbius strip would be a closed curve or not.</p>

答: A "thick" Möbius strip may be smoothed out into a torus, so it still has genus $1$.

Title: Is a thick Möbius strip genus 2?


问:
Prove that if $z,w\in\Bbb{C}$, then $|z+w|^2+|z-w|^2=2|z|^2+2|w|^2$
Prove that if $z, w\in \mathbb{C}$ then $|z+w|^2+|z-w|^2=2|z|^2+2|w|^2$.
I'm just taking this course for the first time any hint or process is welcomed, thank you.

答: $|z+w|^2=(z+w)\overline{(z+w)}=(z+w)(\bar z+\bar w)=z\bar z+z\bar w+w\bar z+w\bar w=|z|^2+z\bar w+w\bar z+|w|^2$ and similarly $|z-w|^2=|z|^2-z\bar w-w\bar z+|w|^2$.  Thus...


答:
Hint:

Note that $|z|^2=z\overline{z}$. Now you can expand both sides using this property.

Notes.

This is called the parallelogram law.

Title: Prove that if $z,w\in\Bbb{C}$, then $|z+w|^2+|z-w|^2=2|z|^2+2|w|^2$

Body:
Prove that if $z, w\in \mathbb{C}$ then $|z+w|^2+|z-w|^2=2|z|^2+2|w|^2$.
I'm just taking this course for the first time any hint or process is welcomed, thank you.


问:
Can you implicitly construct a non-Borel subset $X \subset \mathbb{R}$ such that $\overline{X} = \mathbb{R}$?
I'm just now reading into measure theory and the idea of a $\sigma$ algebra. My textbook says implicit constructions of non-Borel subset of $\mathbb{R}$ with AOC are possible.  Can you construct such a non-Borel subset $X \subset \mathbb{R}$ such that $\overline{X} = \mathbb{R}$?

If so, how would you go about such a construction?

答:
It is possible to construct such a set even without the axiom of choice, but the axiom of choice certainly makes it easier:

First construct any non-Borel set $A\subseteq\mathbb{R}$. For example, the Vitali set is a construction by first partitioning $\mathbb{R}$ into the subsets $\{a+\mathbb{Q}:a\in\mathbb{R}\}$, then from each set $a+\mathbb{Q}$ choosing a distinct member to form the set $A$. This $A$ is not even in the Lebesgue $\sigma$-algebra, so it's not in the Borel $\sigma$-algebra. There are other wats to make such $A$ (and even ways without invoking the axiom of choice), so I'll assume that you've seen at least one.

Then the key observation is, we can make $A$ into a dense set without losing its non-Borel-ness! Consider the set $B=A\cup\mathbb{Q}$, which is certainly dense in $\mathbb{R}$. If $B$ were Borel, then note that $A\cap\mathbb{Q}$ is a countable set and hence Borel, and $A=(B\setminus\mathbb{Q})\cup(A\cap\mathbb{Q})$ is the union of two Borel sets and hence Borel, a contradiction.

This $B\subseteq\mathbb{R}$ is what you wanted.


问:
How can I describe a linear equation that becomes sinusoidal after a certain point?
For example, take $y = x$, but at $x \geq 5$, $y$ now equals $\sin x + 5$.
So for all $x$ values $\lt 5$, $y$ has linear relationship, but at $\geq 5$, $y$ now has sinusoidal relationship.
Is there a concise way to express this, perhaps a single equation?

答:
This is called a PIECEWISE function. It is a function that behave differently depending on which parts of the domain you are on. It is basically made out of PIECES of other functions, and they are tied together by taking sections of them and combining them into one function. For example, $|x|$ is a piecewise function because it is equal to $x$ for $x \ge 0$, and it is equal to $-x$ for $x \le 0$.
A piecewise function might look like this:
$
f(n) =
\begin{cases}
n/2,  & \text{if $n$ is even} \\
3n+1, & \text{if $n$ is odd}
\end{cases}
$

You can read more about piecewise functions here: https://en.wikipedia.org/wiki/Piecewise


问:
Counting permutations $\sigma \in S_n$ such that $\sigma$ is a product of $k$ disjoint cycles of length $r$.
Problem: If $kr \leq n$, with $1<r\leq n$, then the number of permutations that are products of $k$ disjoint cycles of length $r$ is
$$\frac{1}{k!}\frac{1}{r^k}[ n(n-1)\cdots(n-(kr-1)]$$

I already found that the number of $r$-cycles in $S_n$ is
$$A=\frac{1}{r}[n(n-1)\cdots(n-(r-1)]$$

I can't seem to find a way to get the result. I think I understand what I need to do, but I don't know how to do the counting.

Given the set $C_r$ of all $r$-cycles, $|C_r|=A$, I choose one of them, say $\sigma_1$, for which I have $A$ choices. Once I make a choice, I need to remove all the $r$-cycles in $C_r$ that are not disjoint with $\sigma_1=(i_1\cdots i_r)$

How many of those are there? How do I count them? I need to remove all the cycles where any of the $i_j$ occur, but I don't know how to do that.

Once I do that, I can choose one of those, say $\sigma_2$, and repeat the same thing , until I have chosen $k$.

Any help would be appreciated

答: Imagine $k$ subsequences of length $r$ representing the disjoint cycles which you fill up one element at a time with elements of the set $S_n$ acts on. There are $n(n-1)\cdots(n-kr+1)$ ways to do so. The division by $k!$ is because the subsequences may be permuted among themselves; the division by $r^k$ is because cyclic rotations do not change a cycle.


答:
A function like this is said to be a piecewise function. The typical way to notate them is to enumerate the subdomains and the functions defined on such:

$$f(x) = \left\{\begin{array}{ll} x, & \text{if }x<5\\ x+\sin x, & \text{if }x\geq 5\\\end{array}\right.$$

Since any real $x$ satisfies one of these two conditions, we could also write this as

$$f(x) = \left\{\begin{array}{ll} x+\sin x, & \text{if }x\geq 5\\ x, & \text{else }\end{array}\right.$$

Another, less common notation is that of the Iverson bracket. The Iverson bracket $[P]$ for a conditional $P$ is defined as $$[P]=\left\{\begin{array}{ll} 1, & \text{if }P\text{ is true}\\ 0, & \text{else }\end{array}\right.$$

This allows us to express piecewise functions with one-line notation, e.g., $$f(x)=x [x<5]+(x\sin x)[x\geq 5].$$ Since $x\leq 5$ is the negation of $x<5$, we may eliminate the first bracket as $[x<5]=1-[x\geq 5]$ and therefore also have
\begin{align}
f(x)&=x(1-[x\geq 5])+(x+\sin x)[x\geq 5]\\
&=x+(\sin x)[x\geq 5].
\end{align}
This captures the intuition that, whether or not $x\geq 5$, we start "at minimum" with $f(x)=x$; if we $x\geq 5$, we add on $\sin x$ as well. Of course, we could also have done this with our original notation:

\begin{align}
f(x)
&= x+ \left\{\begin{array}{ll} \sin x, & \text{if }x\geq 5\\ 0, & \text{else }\end{array}\right.\\
&= x+ (\sin x)\left\{\begin{array}{ll} 1, & \text{if }x\geq 5\\ 0, & \text{else }\end{array}\right.
\end{align}

So these are ultimately just variations in notation, all of which present the same concept.

Title: How can I describe a linear equation that becomes sinusoidal after a certain point?

Body:
For example, take $y = x$, but at $x \geq 5$, $y$ now equals $\sin x + 5$.
So for all $x$ values $\lt 5$, $y$ has linear relationship, but at $\geq 5$, $y$ now has sinusoidal relationship.
Is there a concise way to express this, perhaps a single equation?


问:
Volume of a solid of revolution - Getting two different results.
The region between $y=x^2+1$ and $y=-x+3$ is rotated about the $x$-axis.

I have to compute the volume. The intersection between these two curves are at $x=-2$ and $x=1$.

At first, I thought about $V(x)=\pi\displaystyle\int_{-2}^1(3-x-(x^2+1))^2dx=\frac{81\pi}{10}$, where the radius is the difference between the two curves.

Another approach I made was $V(x)=\pi\displaystyle\int_{-2}^1[(3-x)^2-(x^2+1)^2]dx=\frac{117\pi}{5}$. I'm not sure why both results are different, and which one is correct.

答:
It might be easier to think about it in steps. So, we know the formula for the volume of a solid of revolution about the $x$ axis from $a$ to $b$, right?

$$V = \pi \int_a^b (f(x))^2 dx$$

Think of the $f^2$ as serving the role of a radius, and the integral as contributing the length by summing up over all of the super-small bits of length $dx$ on the interval.

If you have the area between two curves $f$ and $g$, though - let $f \ge g$ here - you can think of it not as revolving the area between $f$ and $g$, but rather as just revolving the outer area $f$ and then cutting a hole from it based on $g$, and you want the volume left over. That is, for instance,

$$V = \underbrace{\pi \int_a^b (f(x))^2 dx}_{\text{original volume}} - \underbrace{\pi \int_a^b (g(x))^2 dx}_{\text{hole you cut out}}$$

Simplifying, we get

$$V = \pi \int_a^b \Big( (f(x))^2 - (g(x))^2 \Big) dx$$

or, if you understand that $f^2(x)$ means $(f(x))^2$, you can make this a little neater to write:

$$V = \pi \int_a^b \Big( f^2(x) - g^2(x) \Big) dx$$

Thus your second approach would be the correct one.


问:
Row reduction and how to find unknown
So I have been struggling with this problem for about two hours now. I have these equations:

And I'm asked to find out for which values of k do we have

a) No soultions
b) A unique solution
c) Infinitely many solutions

I have seen a similar problem where they first find the reduced row echelon form of the expanded matrix, but I'm stuck.

I have tried a lot of different things now, for example change the order of the equations. But I always get stuck when I come to the last equation where I'm trying to change the second "element" in the row to a $0$. Does anyone have some good tips?

答:
We have this, hopefully it is helpful:

Case (I) $k\in \{3,-5\}$, in which case I am getting a free variable (you can check this for $k = 3$ and $k = -5$ individually via standard gaussian elimination), so infinitely many solutions.

Case (II) $k = -\frac{1}{2}$, in which case I am getting a unique solution (you can check this via standard gaussian elimination).

Case (III) $k\in \mathbb{R}\setminus \{3,-5, -\frac{1}{2}\}$. I used substitution to solve for $x_1,x_2$, and $x_3$ in terms of k as follows:

We have

\begin{equation}
x_1 = 2 -kx_2 + x_3 \text{ (by Eq.1)}\\
4 - (2k+ 1)x_2+ (2+k)x_3 = 5 \text{ (by the above line and Eq.2)}\\
- (2k+ 1)x_2+ (2+k)x_3 = 1\\
x_2 = \frac{1 - (2+k)x_3}{-(2k+1)}
\end{equation}
substitute this into Eq.3 to solve for $x_3$, viz:

\begin{equation}
[2 -kx_2 + x_3] + 10[\frac{1 - (2+k)x_3}{-(2k+1)}] -6x_3 = 1\\
[2 -k\frac{1 - (2+k)x_3}{-(2k+1)} + x_3] + 10[\frac{1 - (2+k)x_3}{-(2k+1)}] -6x_3 = 1\\
(-k+10)\frac{1 - (2+k)x_3}{-(2k+1)}-5x_3 = -1\\
\frac{-k+10}{-(2k+1)} - \frac{(2+k)(-k+10)}{-(2k+1)}x_3 - 5x_3 = -1\\
\bigg[\frac{(2+k)(-k+10)}{2k+1} - 5\bigg]x_3 = -1 - \frac{10-k}{-(2k+1)}\\
x_3 = \frac{-1 - \frac{10-k}{-(2k+1)}}{\bigg[\frac{(2+k)(-k+10)}{2k+1} - 5\bigg]}
\end{equation}

Plugging this into the last line of the first block of equations obtains:

\begin{equation}
x_2 = \frac{1 - (2+k)}{-(2k+1)}\frac{-1 - \frac{10-k}{-(2k+1)}}{\bigg[\frac{(2+k)(-k+10)}{2k+1} - 5\bigg]}
\end{equation}

Plugging our expressions for $x_2$ and $x_3$ into $x_1 = 2 -kx_2 + x_3$ gives:

\begin{equation}
x_1 = 2 - k\frac{1 - (2+k)}{-(2k+1)}\frac{-1 - \frac{10-k}{-(2k+1)}}{\bigg[\frac{(2+k)(-k+10)}{2k+1} - 5\bigg]} + \frac{-1 - \frac{10-k}{-(2k+1)}}{\bigg[\frac{(2+k)(-k+10)}{2k+1} - 5\bigg]}
\end{equation}

We can do this since $k \notin \{3,-5, -\frac{1}{2}\}$ (the roots of all of the denominators, when thought of as polynomials in k). So we have a unique solution in this case.

Title: Row reduction and how to find unknown


问:
Characterization of disks of constant curvature and whose boundaries have constant geodesic curvature
Is it true that if $D$ is a Riemannian $2$-disk having constant Gaussian curvature equal to $1$ and whose boundary has constant geodesic curvature, then $D$ is isometric to some geodesic ball of the unit sphere $\mathbb{S}^2 \subset \mathbb{R}^3$? I strongly suspect so, but I couldn't find a complete argument.

答:
A Riemannian $2$-manifold $D$ with constant Gaussian curvature $1$ is locally isometric to the standard unit sphere $\mathbb{S}^2$. This known as Minding's Theorem, and a proof can be found in section 4-6 of Do Carmo's book, Differential geometry of curves and surfaces. It follows that there is an isometric immersion $D \rightarrow \mathbb{S}^2$.

Its boundary $C$ is therefore an immersed curve in $\mathbb{S}^2 \subset \mathbb{R}^3$. We want to show that if $C$ has constant geodesic curvature, then it is the intersection of $\mathbb{S}^2$ with a plane. Let $p \in C$ and $v$ be a unit vector tangent to $C$ at $p$.

Here is one way: On one hand, the existence and uniqueness theorem of ODEs, given $\kappa \ge 0$, there exists a unique unit speed immersed curve with constant geodesic curvature $\kappa$ passing through $p$ with velocity $v$. Therefore, this curve must be $C$. On the other hand, it is easy to show that there exists a unique plane $P$ containing $p$ and parallel to $v$ such that the curve $P\cap\mathbb{S}^2$ has constant geodesic curvature $\kappa$. It follows that $C = P\cap\mathbb{S}^2$.

Here is a second way: Parameterize $C$ by arclength. Let $(e_1,e_2,e_3)$ be an orthonormal frame along $C$ such that $e_1$ is tangent to $C$ and $e_3$ is normal to $\mathbb{S}^2$. It's straightforward to check that
\begin{equation}
\begin{bmatrix} e_1' \\ e_2' \\ e_3' \end{bmatrix} =
\begin{bmatrix} 0 & \kappa & -1 \\ -\kappa & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix}
\begin{bmatrix} e_1 \\ e_2 \\ e_3 \end{bmatrix},
\end{equation}
where $\kappa$ is the geodesic curvature of $C$.
If $\kappa$ is constant, then it is easy to verify that $(e_2 + \kappa e_3)' = 0$. It follows that $C$ lies in a plane normal to $e_2 + \kappa e_3$ and has curvature equal to $\sqrt{1+\kappa^2}$.

Either way, $C = P \cap \mathbb{S}^2$ and $D$ is one of the two spherical caps with boundary $C$.


问:
Find $\lim_{k\rightarrow \infty}\binom{2k+1}{3}\frac{1}{2^{2(2k-1)}}\left[\binom{2k-2}{k-2}-\binom{2k-2}{k-1}\right]^2$.
While trying to compute the limit as $k\rightarrow \infty$ of the level 3 Energy Spectrum of the Majority Boolean Function, the following expression arises

$$\binom{2k+1}{3}\frac{1}{2^{2(2k-1)}}\left[\binom{2k-2}{k-2}-\binom{2k-2}{k-1}\right]^2.$$

Are there any tricks\bounds (non brute force) to compute this limit?

As $\sum_{j=0}^{k-1}\binom{2k-2}{2j}=\sum_{j=1}^{k-1}\binom{2k-2}{2j-1}=2^{2k-3}$, my guess is that this limit is zero, but ignoring the term $-2\binom{2k-2}{k-2}\binom{2k-2}{k-1}$ and using that $\binom{2k-2}{j}\leq 2^{2k-3}$ is not sharp enough.

Body:
Without a full simplification

Start simplifying the expression $$y_k=\binom{2k+1}{3}\frac{1}{2^{2(2k-1)}}\left[\binom{2k-2}{k-2}-\binom{2k-2}{k-1}\right]^2$$ using
$$\binom{2k+1}{3}=\frac{1}{3} k (2 k-1) (2 k+1)$$
$$\binom{2k-2}{k-2}-\binom{2k-2}{k-1}=\frac{\Gamma (2 k-1)}{\Gamma (k-1) \Gamma (k+1)}-\frac{2^{2 k-2} \Gamma
   \left(k-\frac{1}{2}\right)}{\sqrt{\pi } \Gamma (k)}$$ Now, start series expansions
$$A=\frac{\Gamma (2 k-1)}{\Gamma (k-1) \Gamma (k+1)}$$
$$\log(A)=k \log (4)-\frac{1}{2} \log (16 \pi  k)-\frac{5}{8 k}-\frac{3}{8 k^2}-\frac{55}{192
   k^3}-\frac{15}{64 k^4}+O\left(\frac{1}{k^5}\right)$$

$$A=e^{\log(A)}=\frac{4^{k-1}}{\sqrt{k\pi } }\Bigg[1-\frac{5}{8 k}-\frac{23}{128 k^2}-\frac{95}{1024 k^3}-\frac{1701}{32768
   k^4}+O\left(\frac{1}{k^5}\right) \Bigg]$$

$$B=\frac{2^{2 k-2} \Gamma
   \left(k-\frac{1}{2}\right)}{\sqrt{\pi } \Gamma (k)}$$
$$\log(B)=k \log (4)-\frac{1}{2} \log (16 \pi  k)+\frac{3}{8 k}+\frac{1}{8 k^2}+\frac{3}{64
   k^3}+\frac{1}{64 k^4}+O\left(\frac{1}{k^5}\right)$$
$$B=e^{\log(B)}=\frac{4^{k-1}}{\sqrt{k\pi }}\Bigg[1+\frac{3}{8 k}+\frac{25}{128 k^2}+\frac{105}{1024 k^3}+\frac{1659}{32768
   k^4}+O\left(\frac{1}{k^5}\right) \Bigg]$$
$$(A-B)^2=4^{2 k} \left(\frac{1}{16 \pi  k^3}+\frac{3}{64 \pi  k^4}+O\left(\frac{1}{k^5}\right)\right)$$
All of the above make
$$y_k=\frac{1}{3 \pi }\left(1+\frac{3}{4 k}+\frac{9}{32 k^2}+\frac{21}{128
   k^3}+O\left(\frac{1}{k^4}\right) \right)$$

Much faster !

For a quick check
$$y_{10}=\frac{3929991065}{34359738368}\sim 0.1143778$$ while the truncated series expansion gives
$$y_{10}\sim \frac{137981}{384000 \pi }\sim 0.1143769$$

With a full simplification

Using the hint @user299843 gave in comments
$$A-B=-\frac{\Gamma (2 k-1)}{\Gamma (k) \Gamma (k+1)}$$ gives
$$y_k=\frac{1}{3} k (2 k-1) (2 k+1)\frac{1}{2^{2(2k-1)}}\Bigg[\frac{\Gamma (2 k-1)}{\Gamma (k) \Gamma (k+1)} \Bigg]^2$$ Take logarithms and use Stirling approximation and continue with Taylor
$$\log(y_k)=\log \left(\frac{1}{3 \pi }\right)+\frac{3}{4 k}+\frac{3}{32 k^3}+O\left(\frac{1}{k^5}\right)$$
$$y_k=e^{\log(y_k)}=\frac{1}{3 \pi }\left(1+\frac{3}{4 k}+\frac{9}{32 k^2}+\frac{21}{128
   k^3}+O\left(\frac{1}{k^4}\right) \right)$$


问:
Probabilty Density as a Quadratic?
I'm trying to express the natural log of a probability density function.

$$\displaystyle p(x)=\frac{1}{\sqrt{2\pi\sigma^2}}exp\left(\frac{-(x-\mu)^2}{2\sigma^2}\right)$$
$$h(x) = ln(p(x))$$
$$\displaystyle h(x)=ln\left(\frac{1}{\sqrt{2\pi\sigma^2}}exp\left(\frac{-(x-\mu)^2}{2\sigma^2}\right)\right)$$

attempting h(x) as a quadratic in the the form $ax^2+bx+c$

I think I may be a bit lost.

$$\displaystyle h(x)=ln\left(\frac{1}{\sqrt{2\pi\sigma^2}}\right)+ln\left(exp\left(\frac{-(x-\mu)^2}{2\sigma^2}\right)\right)$$
$$\displaystyle h(x)=ln\left(\frac{1}{\sqrt{2\pi\sigma^2}}\right)+\left(\frac{-(x-\mu)^2}{2\sigma^2}\right)$$
$$\displaystyle h(x)=exp\left(ln\left(\frac{1}{\sqrt{2\pi\sigma^2}}\right)\right)+\left(\frac{-(x-\mu)^2}{2\sigma^2}\right)$$
$$\displaystyle h(x)=\frac{1}{\sqrt{2\pi\sigma^2}}+\frac{-(x-\mu)^2}{2\sigma^2}$$
$$\displaystyle h(x)=\left(\frac{1}{\sqrt{2\pi\sigma^2}}\right)^2+\frac{-(x-\mu)^2}{2\sigma^2}$$
$$\displaystyle h(x)=\frac{1}{2\pi\sigma^2}+\frac{-(x-\mu)^2}{2\sigma^2}$$

答:
You're right up until the fifth line, I'm not sure how you added an $\exp(\cdot)$ around your $\ln(\cdot)$. But starting from your fifth line, we have the following:

$$ h(x) = \ln\left(\frac{1}{\sqrt{2\pi\sigma^2}}\right) + \left(\frac{-(x-\mu)^2}{2\sigma^2} \right) $$

$$ = \ln\left(\frac{1}{\sqrt{2\pi\sigma^2}}\right) - \left(\frac{x^2-2\mu x + \mu^2}{2\sigma^2} \right) $$

$$ = - \frac{1}{2\sigma^2}x^2 + \frac{\mu}{\sigma^2}x - \frac{\mu^2}{2\sigma^2} + \ln\left(\frac{1}{\sqrt{2\pi\sigma^2}}\right) .$$

So, you have

$$ a = - \frac{1}{2\sigma^2} , $$

$$ b = \frac{\mu}{\sigma^2} , $$

$$ c = - \frac{\mu^2}{2\sigma^2} + \ln\left(\frac{1}{\sqrt{2\pi\sigma^2}}\right) . $$

Title: Probabilty Density as a Quadratic?


问:
Is it legitimate to quantify a variable twice?
I am wondering if it is legitimate in first-order logic to quantify one variable twice in a single formula, for instance, as in the following proposition:
\begin{equation}
\left(\forall s\right) \left(\exists t\right) \left[P\left(t\right) \wedge R\left(s\right) \wedge \left(\exists s\right) Q\left(s\right)\right].
\end{equation}
It seems to me that there is no problem in $s$ appearing twice as quantifiers. But it looks a little weird.

P.S. I am having this question as I read the following statement in set theory about variable substitution:

If $\Psi$ is $\left(\forall v_{k}\right)\Theta$ for some formula $\Theta$, and if $k \neq i$, then $\Psi^{*}$, which is achieved from substitution of $v_{j}$ for $v_{i}$ in $\Psi$, is just $\left(\forall v_{k}\right)\Theta^{*}$, where $\Theta^{*}$ is achieved from substitution of $v_{j}$ for $v_{i}$ in $\Theta$. If $k = i$, then $\Psi^{*}$ is $\left(\forall v_{j}\right)\Gamma$, where $\Gamma$ is achieved from substitution of $v_{j}$ for $v_{i}$ in $\Theta$.

It seems that in set theory, variables are quantified only once. Otherwise, $\Gamma$ should be achieved from substitution of $v_{j}$ for FREE occurrences of $v_{i}$ in $\Theta$, which means that the substition should only take place the first time $v_{i}$ is met as a quantifier, and not for others included inside. Is it that in set theory, laws from first-order logic are relaxed so that each variable is assumed to be quantified only once?

答:
Your formula is a perfectly acceptable and grammatical formula in first-order logic, and it is exactly as you say: the $s$ in $Q(s)$ is quantified by the the $\exists s$, while the $s$ in $R(s)$ is quantified by the $\forall s$.

Note that you can even have something like $\forall s \ \exists s \ P(s)$. In this case, the $s$ in $P(s)$ gets quantified by the $\exists s$, meaning that there are no free variables $s$ to quantify for the $\forall s$. This is why in this formula the $\forall s$ is called a 'null quantifier': while still following the grammatical rules, it effectively serves no purpose. Indeed, the formula is equivalent to simply $\exists s \ P(s)$

But yes, you can use quantifiers with the same variable any number of times. But as you also say, it looks weird, and can certainly invite confusion, so I would recommend never to do this in practice. And if you need a lot of variables, you can always use $x_1, x_2, x_3, ...$


答:
Yes, it is okay to reuse the same symbol for a variable in different "scopes". A quantifier binds the variable that it introduces within the subexpression headed by that quantifier.

Note that it's always possible to rewrite an expression into an equivalent one that does not reuse variable symbols, unless there are only finitely many variable symbols.

A variable may appear free in a well-formed formula, meaning that it isn't bound

$$ x \;\;\text{is free in}\;\; P(x) \land Q(x) $$

We can be more precise and use $\text{FV}$ to denote the set of free variables.

$$ \text{FV}(P(x) \land Q(x)) = \{ x \} $$

If a variable is bound by a quantifier then it is no longer free.

$$ \text{FV}\!\left(\forall x \mathop. P(x)\right) = \emptyset $$

We can think about the meaning of $\forall$ as well:

$$ \Gamma \models \forall x \mathop. \varphi \;\;\text{if and only if} \;\; \text{$\Gamma \models \varphi$ for every $x$ in the domain of discourse } $$

So, in the course of determining the truth of a well-formed formula headed by a quantifier, we can remove the quantifier and then consider the result of replacing the newly free variable with all the elements of the domain.

However, instances of $x$ that are hidden behind another quantifier won't actually be free in $\varphi$.

More concretely, the following two statements are equivalent

$$ \forall x \mathop. (P(x) \to \exists x \mathop. Q(x)) \;\;\text{is equivalent to}\;\; \forall x \mathop. (P(x) \to \exists y \mathop. Q(y)) $$

Title: Is it legitimate to quantify a variable twice?

Body:
I am wondering if it is legitimate in first-order logic to quantify one variable twice in a single formula, for instance, as in the following proposition:
\begin{equation}
\left(\forall s\right) \left(\exists t\right) \left[P\left(t\right) \wedge R\left(s\right) \wedge \left(\exists s\right) Q\left(s\right)\right].
\end{equation}
It seems to me that there is no problem in $s$ appearing twice as quantifiers. But it looks a little weird.

P.S. I am having this question as I read the following statement in set theory about variable substitution:

If $\Psi$ is $\left(\forall v_{k}\right)\Theta$ for some formula $\Theta$, and if $k \neq i$, then $\Psi^{*}$, which is achieved from substitution of $v_{j}$ for $v_{i}$ in $\Psi$, is just $\left(\forall v_{k}\right)\Theta^{*}$, where $\Theta^{*}$ is achieved from substitution of $v_{j}$ for $v_{i}$ in $\Theta$. If $k = i$, then $\Psi^{*}$ is $\left(\forall v_{j}\right)\Gamma$, where $\Gamma$ is achieved from substitution of $v_{j}$ for $v_{i}$ in $\Theta$.

It seems that in set theory, variables are quantified only once. Otherwise, $\Gamma$ should be achieved from substitution of $v_{j}$ for FREE occurrences of $v_{i}$ in $\Theta$, which means that the substition should only take place the first time $v_{i}$ is met as a quantifier, and not for others included inside. Is it that in set theory, laws from first-order logic are relaxed so that each variable is assumed to be quantified only once?


问:
If a random variable converge in distribution then scalar multiplication also converges in distribution.
Let c be a constant and suppose that $X_n\xrightarrow{d}X$, prove that

i)  $cX_n\xrightarrow{d}cX$

ii) $X_n+c\xrightarrow{d}X+c$

I've tried to use that:

$lim_{n\rightarrow\infty} F_{cX_n}(x) = lim_{n\rightarrow\infty} P(cX_n\leq x) = lim_{n\rightarrow\infty} P\left(X_n\leq \dfrac{x}{c}\right) = lim_{n\rightarrow\infty} F_{X_n}\left(\dfrac{x}{c}\right) $

But that's not exactly what I was looking for.

答:
The sequence $X_n$ converges in distribution to $X$ as $n \rightarrow \infty$ (denoted $X_n \Rightarrow X$) if and only if, for any bounded continuous function $f$ (denoted $f \in C_b$), we have
$$\lim_{n\rightarrow\infty} E[f(X_n)] = E[f(X)].$$
Let $c \in \mathbb{R}$ be given. The maps $g_c^\times(x) = cx$ and $g_c^+(x) = x + c$ are continuous, so for any $f \in C_b$, $f\circ g_c^\times$ and $f\circ g_c^+$ are also in $C_b$. Therefore, because $X_n \Rightarrow X$ we have
$$\lim_{n\rightarrow\infty} E[f(X_n^{c,\times})] = \lim_{n\rightarrow\infty} E[(f\circ g_c^\times)(X_n)] = E[(f\circ g_c^\times)(X)] = E[f(X_c^\times)],$$
and ditto for $X_n^{c,+}$.


问:
Fastest way to implement Secant method on calculator
I am looking for the fastest way to implement the secant method (the one which is used for finding roots to a given single-variable equation) on a fx-991es calculator (scientific, non programmable calculator). What I have done till now:

Assign initial values to A,B,C,D
(Later the variables will serve as A = $x_{n-1}$, B = $x_n$, C = $x_{n+1}$, $D = f(x_{n-1})$, $E = f(x_n)$)

Perform the operation $\frac{AE-BD}{E-D} \rightarrow C$, followed by the operations $B \rightarrow A, E \rightarrow D, C \rightarrow B$. Then, perform $f(C)\rightarrow E$. For example: ($cos(c)-c e^c \rightarrow E$)

Repeat step 2 till the desired accuracy is reached.

I found this pretty useful in implementing Regular-Falsi(False position) method/ bisection method, as only one set of variables is replaced there (A,D or B,E).

However, navigating history after each operation 2 takes some time (to find the operation, then implement $\frac{AE-BD}{E-D} \rightarrow C$). Also, there are multiple replacement operations. Please post a shorter solution, if possible. A general comment on implementing numerical solutions using calculator (for other methods) would be helpful too.

About the calculator: It has 9 variables for storage option (including M), operations supported in Vectors, Matrices too. Maybe a 2-d vector solution to this problem can be found.

Note: I am aware that on the computer(Matlab)/programmable calculator, the process can be automated. However, I require a solution which is doable using a non-programmable calculator as these options are not available in the exam. So, please don't post such comments.

答:
Not only does the form

$$x_{n+1}=x_n-\frac{f(x_n)}{f(x_n)-f(x_{n-1})}(x_n-x_{n-1})$$

provide more numerical stability, it can also be implemented with fewer variables (and hence less updates per iterations) since it can be rewritten as

\begin{cases}x_{n+1}=x_n+\Delta x_n\\\Delta x_n=\dfrac{f_n}{f_{n-1}-f_n}\Delta x_{n-1}\\f_n=f(x_n)\end{cases}

which requires only one value of $x$, one value of $\Delta x$, and two values of $f$ per iteration:

Initialize $(A,B,C,D)=(x_1,\Delta x_0,f_1,f_0)=(x_1,x_1-x_0,f(x_1),f(x_0))$.

Update $B:=\dfrac{BC}{D-C}$, then $A:=A+B$, then $D:=C$, then $C:=f(A)$.

Repeat 2. until desired accuracy.


问:
Morse Theory proof of Fundamental Theorem of Algebra
Suppose that p(z) is a nonconstant polynomial with no roots.  The complex plane with additional point ∞ is homeomorphic to the 2-sphere.  At each z in the plane, let the vector at z be 1/p(z), which is defined since p(z) is nonzero everywhere.  As |z| goes to infinity, p(z) goes to 0, since p(z) is not constant, and hence the vector field is 0 at ∞.  This gives a vector field on the 2-sphere with only one critical point, whereas Morse Theory tells us there must be at least two critical points.

I think this is correct.  What perplexes me is, what prevents defining a nonzero constant vector field likewise, which would have no critical points?

Thanks for your help.

答:
Morse function says that for any Morse function on $S^2$, it has at least two critical point. However, you constructed a vector field instead of a Morse function. The right machinery to use here should be Poincare-Hopf index theorem, which says the index of a vector field is equal to the Euler characteristic of the space. In this case, $\chi(S^2)=2$.

Consider your vector field given by $1/p(z)$, which has only one isolated zero at infinity. If you invert that, i.e. taking $z\mapsto 1/z$, your vector field becomes $1/p(1/z)$ with an isolated zero at the origin.

Let $p(x)=a_0+a_1x+\cdots+a_nx^n$ with nonzero $a_n$. Then
$$\frac{1}{p(1/z)}=\frac{z^n}{a_0z^n+a_1z^{n-1}+\cdots+a_n}$$
which is a degree-$n$ function. So the vector field has index $n$.

According to Poincare-Hopf, we have $n=2$. But we know that the Fundamental Theorem of Algebra is true for $n=2$ because we have the quardratic formula.


问:
Is it inuitively correct to say that $\lim_{n \to \infty} \frac{x}{n} = \mathrm{d}x$?
We know that from the Riemann sum that $\Delta x = \mathrm{d}x$ where $n \to \infty$ and $\Delta x = \dfrac{b - a}{n}$. If, however, $x$ represents some length of interval, can we also say that $\displaystyle\lim_{n \to \infty}\dfrac{x}{n} = \mathrm{d}x$?

答:
I think this is quite a tricky topic to talk about because of how people interpret things vs how they are actually defined. I find it best to think of $dx$ as a separate value, rather than a function or scaling of the variable $x$. whilst we are talking about a small change in $x$ the key word here is change, so rather than $x/n\to0$ really we want the change in $x$, or the interval that we are looking at to be very small rather than the initial or end values.

One nice way to think of this is by looking at your average integral:
$$I_{a,b}[f]=\int_a^bf(x)dx$$
now think of the integral symbol as a stretched out $S$, which is the first letter of sum. Now inside the integral (the part we are summing) is $f(x)\times dx$. In other words we can think of an integral as a sum over our domain $[a.b]$ where the height of each one is $f(x)$ and the width is $dx$. However for our error to tend to zero we require the rectangles to fit the function $f(x)$ well, and actually tend to it exactly. The best way for us to do this is to reduce the width $dx$ to a very small value close to zero, but it cannot actually equal zero or be "infinitely small". Whilst this may seem strange it does not actually matter as we do not need a numerical method for $dx$, unless using some methods for approximating and in that case the statement is not strictly true. Hope this helps


问:
Prove $ |\frac{\text{Cov}\left(X,Y\right)}{\sqrt{\text{Var}[X]\text{Var[Y]}}}|=1\iff\exists a,b\in\mathbb{R}:(\mathbb{P}[Y=aX+b]=1) $

Let $ \rho\left(X,Y\right)=\dfrac{\text{Cov}\left(X,Y\right)}{\sqrt{\text{Var}[X]\text{Var[Y]}}}$. Prove that $$ |\rho\left(X,Y\right)|=1\iff\exists a,b\in\mathbb{R}:(\mathbb{P}[Y=aX+b]=1). $$

I prove the $ \Leftarrow $ direction. But I'm struggeling to prove the other direction.

Here's what I've tried:

Assume that $ |\rho\left(X,Y\right)|=1 $

Then

$ \left(\text{Cov}\left(X,Y\right)\right)^{2}=\text{Var}[X]\text{Var[Y]} $.

Now :

$ \require{cancel}
\mathbb{E}^{2}[XY]+2\mathbb{E}[XY]\mathbb{E}[Y]\mathbb{E}[X]+\cancel{\left(\mathbb{\mathbb{E}}[X]\mathbb{E}[Y]\right)^{2}}=\mathbb{E}[X^{2}]\mathbb{E}[Y^{2}]-\mathbb{E}[X^{2}]\left(\mathbb{E}[Y]\right)^{2}-\left(\mathbb{E}[X]\right)^{2}\mathbb{E}[Y^{2}]+\cancel{\left(\mathbb{E}[X]\mathbb{E}[Y]\right)^{2}}
$

So

$ \mathbb{E}^{2}[XY]+2\mathbb{E}[XY]\mathbb{E}[Y]\mathbb{E}[X]=\mathbb{E}[Y^{2}]\text{Var}[X]-\mathbb{E}[X^{2}]\left(\mathbb{E}[Y]\right)^{2} $

$ \mathbb{E}[Y^{2}]\text{Var}[X]=\mathbb{E}[\mathbb{E}^{2}[XY]+2\mathbb{E}[XY]\mathbb{E}[Y]X+\left(\mathbb{E}[Y]\right)^{2}X^{2}] $

$ \mathbb{E}[Y^{2}]=\mathbb{E}[\frac{\left(\mathbb{E}[XY]+\mathbb{E}[Y]X\right)^{2}}{\text{Var}[X]}] $

Define: $ b:=\frac{\mathbb{E}[XY]}{\sqrt{\text{Var}[X]}},\thinspace\thinspace\thinspace a:=\frac{\mathbb{E}[Y]}{\sqrt{\text{Var}[X]}} $

$ \mathbb{E}[Y^{2}]=\mathbb{E}[\left(aX+b\right)^{2}] $

$ \mathbb{E}\left(Y^{2}-\left(aX+b\right)^{2}\right)=0 $

I feel close because if the random variable in the expectation would have been non-negative I guess I could proceed from there, but it is not necessarily true.

Im looking for the most direct way to prove it, preferable without any tricks that I couldnt think of really fast (this is a question from an exam), is there a way to solve it using what I started?

Thanks in advance.

答:
Hint: We may standardize the variables $X,Y$ via the substitutions

$$\hat{X}:=\frac{X-\mathbb{E}[X]}{\sqrt{\text{Var}(X)}},\quad \hat{Y}:=\frac{Y-\mathbb{E}[Y]}{\sqrt{\text{Var}(Y)}}.$$
What are the mean and variance of these new variables? What is their covariance? What do you conclude about the quantity $E[(\hat{X}\pm \hat{Y})^2]$?


问:
What does $p+q=k$ mean in the index of summation?
I need help solving something I don't understand. OK so the problem is this:

$$H^k(X,C)=\bigoplus_{p+q=k} H^{p,q}(X),$$

What does the $\;p+q=k\;$ mean?

Thank you anybody that helps! :)

答:
There are multiple ways to indicate a summation (similarly with a product, direct summation, etc.)  In all cases, the sum is evaluated in all ways such that the condition is met, subject to natural restrictions that are implicit in the problem.  Typical restrictions are that the index is an integer, and sometimes that the index is nonnegative.

Examples: $$\sum_{i\ge 0} f(i)$$
and $$\sum_{i\in[0,6]}f(i)$$
and $$\sum_{j|n} f(j)$$
and $$\sum_{j|n, j\neq 3} f(j)$$
and $$\sum_{jk=n} f(j)f(k)$$
and $$\sum_{j+k=n} f(j)$$


问:
Prove inequality with $a;b;c \in R$
Let $a;b;c \in  \mathbb{R} $ such that $a+b+c=0$
Prove that: $P=\dfrac{a-1}{a^2+8}+\dfrac{b-1}{b^2+8}+\dfrac{c-1}{c^2+8} \geq -\dfrac{3}{8}$

I tried to do this:
$\dfrac{8a-8}{a^2+8}+2+\dfrac{8b-8}{b^2+8}+2+\dfrac{8c-8}{c^2+8}+2 \geq 3$

$\Leftrightarrow \dfrac{(a+2)^2}{a^2+8}+\dfrac{(b+2)^2}{b^2+8}+\dfrac{(c+2)^2}{c^2+8} \geq \dfrac{3}{2}$

Then WLOG $c=max\{a;b;c\} \Rightarrow c \geq 0$

I hope I can prove: $\dfrac{(a+2)^2}{a^2+8}+\dfrac{(b+2)^2}{b^2+8} \geq f(c)$

So I only need to find minimum of $f(c)+\dfrac{(c+2)^2}{c^2+8}$

But I can't find $f(c)$, I stuck here.

答:
Your idea is very good and we will use that :)

As $a+b+c=0$ we can find a pair such that they have the same sign hence WLOG $ab\ge 0$. Now from your attempt we have to prove  $$\Leftrightarrow \dfrac{(a+2)^2}{a^2+8}+\dfrac{(b+2)^2}{b^2+8}+\dfrac{(c+2)^2}{c^2+8} \geq \dfrac{3}{2}$$
by C-S $$\dfrac{(a+2)^2}{a^2+8}+\dfrac{(b+2)^2}{b^2+8}\ge \frac{{(a+b+4)}^2}{a^2+b^2+16}=\frac{{(4-c)}^2}{a^2+b^2+16}\ge \frac{{(4-c)}^2}{c^2+16} $$ here we used $a^2+b^2\le {(a+b)}^2=c^2\iff 2ab\ge 0$

it remains to prove $$\frac{{(4-c)}^2}{c^2+16}+\frac{{(c+2)}^2}{c^2+8}\ge \frac{3}{2}$$ $$\iff \frac{c^2{(c-4)}^2}{2(c^2+16)(c^2+8)}\ge 0$$ which is true

Equality when $a=b=c=0$,

$c=4,a=-2,b=-2$(upto permutations)

Title: Prove inequality with $a;b;c \in R$


问:
Soft question- The Bashing Technique and Other powerful techniques for Olympiads
I am relatively new to the Olympiad style maths problems and I find the word "bashing" being thrown around a lot in the community and I haven't really understood it. I do know about "co-ordinate bashing" using Barycentric co-ordinates and complex numbers but I've also seen people use the term "bashing" in Number Theory, Combinatorial and Algebraic questions and don't know why.

My Question- What does the term "bashing" generally mean in Mathematics(in the broader sense)?

Also, if anyone could provide me some other powerful problem solving tools like Bashing which are useful for Maths Olympiads I would be highly obliged. By "Problem Solving Tools" I refer to concepts like Inversive, Projective and Homothety in geometry;LTE and Chinese Remainder Theorem in Number Theory; Vieta Root Jumping in Algebra. I only know a few of these and would like to see a comprehensive list of all(or most) of such important techniques. Even a link would suffice.

Thanks in advance.(First question on Stack Exchange).

答:
Bashing is usually a way of using brute force to answer a question. In most cases, it doesn't require any mathematical knowledge of theorems and techniques.

The way to go in basing is usually just trial and error but for higher efficiency, you can use theorems to set upper and lower bounds for your answer. This will increase efficiency while bashing.

The most common bashing is done with induction. Where you are trying to prove a theorem so you substitute 1 and then 2 and then we know that if it satisfies for n, it should satisfy for n+1

I must warn you that in olympiads like RMO, INMO, EGMO, USMO, AIME, IMO, etc, Basing is looked down upon and will have less weightage because in these olympiads they check your method of answering as well as your answer.


问:
Is it true that if the complex function $\bar{f}$ is differentiable, then $f$ is differentiable?
If $\bar{f}=u(x,y)-iv(x,y)$ is differentiable, then the first-order partial derivatives of $u$ and $-v$ exist and they are continuous. Also, the Cauchy-Riemann equations hold, i.e., $u_x=-v_y$ and $u_y=v_x$, but from these relationships, how can we know $f$ is differentiable?

答:
In my text, by Saff &Snyder Complex Math. For Scientists , Engineers and Scientists on analyticity,pg 51,

“Notice that we will have to ban the congugate function $\bar z$ because if we admit we will open the gate to x[=(z+$\bar z$/2)] and y[=(z-$\bar z/2i)]$

$\bar z $=x-iy (inadmissible)


问:
Prove that $\int 1/x\, \text{d}x=\ln |x|$
What is the proof that $$\int \frac{1}{x}\, \text{d}x=\ln |x|?$$

Please note that similar questions refer to $\ln x$, and not $\ln |x|$ as I have seen in Calculus textbooks.

Given the formula that $$\frac{\text{d}}{\text{d}x}\ln x=\frac{1}{x},$$ it is obvious that for $x>0$, $\int 1/x\, \text{d}x=\ln x$; however I am unsure how the above definition is proven over all $x\neq 0$.

答:
$$y=\ln|x|$$
$$e^y=|x|$$
$$e^y\frac{dy}{dx}=\operatorname{sgn}(x)$$
$$\frac{dy}{dx}=\frac{\operatorname{sgn}(x)}{|x|}=\frac1x$$
where the exponential function:
$$e^y=\exp(y)$$
is defined as the function satisfying the following statements:
$$f(x)=\exp(x),\,f^{-1}(x)=\ln(x)$$
$$f'=f$$


答:
$p+q = k$ indicates that the sum of indexes $p, q$ is equal to $k$, where $k$ is constant, for the given $k$.  If $k=4$ you can take $p=1, q=3$  or $p = q = 2,$ etc.

So every possible pair $(p, q)$ for which $p+q = k$; where $p, q$ are the indices which indicate which $H^{p, q}(X)$ to sum.

From comment

Indeed: to "spell it out the notation" you've posted, for a particular value of $k$, let's let $k = 4.\;$ Then:
$$H^4(X,C)=\bigoplus_{p+q=k} H^{p,q}(X) = H^{0, 4}(X) \oplus H^{1, 3}(X) \oplus H^{2,2}(X) \oplus H^{3,1}(X)\oplus H^{4,0}(X)$$

Title: What does $p+q=k$ mean in the index of summation?

Body:
I need help solving something I don't understand. OK so the problem is this:

$$H^k(X,C)=\bigoplus_{p+q=k} H^{p,q}(X),$$

What does the $\;p+q=k\;$ mean?

Thank you anybody that helps! :)


答:
The existing answers are correct, but let me add the following:

In some contexts we find it convenient to restrict attention to formulas which do not do this. For example, it makes some arguments around substitutions easier. This is never truly a big deal as far as I'm aware, but the translation of an arbitrary first-order formula into a "no-redundant-quantification" formula is simple enough that it often (annoyingly!) goes unstated. While elementary logic texts will generally be careful about this, advanced texts and papers often make claims about arbitrary first-order formulas with the implicit assumption that they have no multiply-quantified variables.

It's also worth noting that there are two drawbacks to the translation above. First, it introduces new variables, so if you care about how many variables you're using you lose information. This is actually a meaningful issue - see e.g. chapter $11$ of Libkin's wonderful book Elements of finite model theory. Second, it impacts the complexity of the set of well-formed formulas (e.g. regular, context-free, ...). This isn't a topic I know much about so I don't have much to say here, but I'm sure there are times when this is an issue.


问:
Euclids Algorithm for polynomials and a greatest common divisor
I have question about a problem I've encountered while attempting to solve an exercise (it's from an exercise in a homework series).

Suppose we have two polynomials $f$ and $g$ (presumably over the reals, although the field over which these polynomials are defined isn't mentioned explicitly). Assume that deg($g$) > 1. Then there exist polynomals $q_1,r_1$ such that $f = q_1 g + r_1$. We can now repeat the polynomial divison, which gives $g = q_2 r_1 + r_2$ for polynomials $q_3,r_3$ and $r_1 = q_4 r_2 + r_4$ for polynomials $q_4,r_4$, etcetera. At some point we get $r_{p-2} = q_{p}r_{p-1} + r_p$ and $r_{p-1} = q_{p+1} r_p$. I now want to show that $r_p$ is the greatest common divisor of $f$ and $g$.

I have already showed (I think) that $r_p$ divides both $f$ and $g$ by writing the equations in matrixform, giving $$\left(\begin{array}{c}f\\g\end{array}\right) = \Pi_{i=1}^{p+2}\left(\begin{array}{cc} q_{i} & 1 \\ 1 & 0 \end{array}\right) \left(\begin{array}{c}r_p\\0\end{array}\right)$$ But now I want to show that $r_p$ is the greatest divisor.  I think the best way to do it is to assume that there is some $\tilde{r}_p$ which also divides $f$ and $g$ but which has a degree larger than $r_p$. At this moment I'm not sure how to proceed (I'm a bit rusty on my abstract algebra). I feel that the solution is obvious but I'm not seeing it. How is this sort of problem best handled?

答:
That $\rm\:r_p\:$ is a greatest common divisor, i.e. that it is divisible by every common divisor $\rm\:d\:$ of $\rm\:f,g\:$, follows immediately from the fact that, by construction, it is an $\rm\:\mathbb R[x]$-linear combination of $\rm\:f,g\:,\ $ i.e. for some $\rm\: h,k\in \mathbb R[x]\:,\:$ we have $\rm\  r_p =\: h\ f + k\ g\:,\ $ so if $\rm\ d\:|\:f,g\ $ then $\rm\ d\ |\ h\ f + k\ g = r_p\:,\ $ i.e. more explicitly $\rm\ f/d,\:g/d\in\mathbb R[x]\ \Rightarrow\ (h\ f + k\ g)/d = h\: (f/d) + k\: (g/d)\in \mathbb R[x]\:.$

That $\rm\:r_p\:$ is in fact such a linear combination is easy to prove inductively. The base case is trivially true, i.e. both $\rm\:r_{-1} = f\:$ and $\rm\: r_0 = g\:$ are trivially linear combinations of $\rm\:f,\:g\:.\:$ Suppose, as inductive hypothesis, for all $\rm\:k\le i\:$ that $\rm\:r_k\:$ is a linear combination of $\rm\:f,\:g\:.\:$  Then so to is $\rm\ r_{i+i} = r_{i-1} - q_i\ r_i\:$ since $\rm\:\mathbb R[x]$-linear combinations of $\rm\ f,\:g\ $ are preserved under multiplication by elements of $\rm\:\mathbb R[x]\ $ (here $\rm\:q_i\:$), and they are also preserved under subtraction. Thus the inductive proof is complete.

If you know about ideals, then the above inductive proof may be interpreted as saying that each step of the Euclidean algorithm simply changes the basis of the ideal, so preserves the ideal, i.e.

$$\rm\:I_{\:i+1} =\: (r_{i+1},\:r_i)\ =\ (r_{i-1}-q_i\ r_i,\:r_i)\ =\ (r_{i-1},\:r_i)\: =\ I_{\:i}$$

So $\rm\ I_i = I_0 = (f,g)\ \Rightarrow\ r_i\in I_i = (f,g) = f\ \mathbb R[x]+g\ \mathbb R[x] = \mathbb R[x]$-linear combinations of $\rm\:f,g\:.$

Generally in any Bezout domain, i.e. any domain where finitely generated ideals are principal, the gcd is given by any generator of an equivalent principal ideal, i.e.

$$\rm (a,b)\ = (c)\ \ \iff\ \ gcd(a,b)\ =\ c $$

$\ $ since $\rm\quad\quad\quad\  (a,b)\ =\ (c)$

$\rm\quad\quad\quad \iff\ [\ (d)\supset (a,b)\ \iff\ (d)\supset (c)\ ] $

$\rm\quad\quad\quad \iff\ [\ d\ |\ a,\:b\ \iff\ d\ |\ c \ ] $

$\rm\quad\quad\quad \iff\ gcd(a,b)\ =\ c $


问:
Inverse fourier transform of a function which is a fundamental solution
Let $f:\mathbb{R}^3 \to \mathbb{R}$ be $f(x)=(1+|x|^2)^{-1}$.

I need to calculate $\mathcal{F}^{-1}(f)$.

I've proven that $f\in L^2(\mathbb{R}^3)$ and I know that the fourier transform is an isometry of $\mathcal{S}'$ where $L^2$ is happily embedded.

I tried using the direct formula of the inverse transform but that is an insance calculation.

I suspect I am supposed to use the fact that $\mathcal{F}^{-1}f$ is a fundamental solution to $I-\Delta$ but I'm not sure how.

Thanks in advance, Shai.

答:
One can do this almost straightforwardly. Define
\begin{align}
\hat{f}(\mathbf{k})=\mathcal{F}^{-1}(f)=\frac{1}{(2\pi)^3}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{e^{-i(k_x x+k_yy+k_zz)}}{1+x^2+y^2+z^2}dxdydz
\end{align}
Now rotate the coordinate system in $x,y,z$-space so that the direction of $\mathbf{k}=(k_x,k_y,k_z)$ coincides with $z$-direction in new coordinates. Then, passing to spherical coordinates, we get
\begin{align}
\hat{f}(\mathbf{k})&=\frac{1}{(2\pi)^3}\int_0^{\infty}\int_0^\pi\int_0^{2\pi}\frac{e^{-ikr\cos\theta}}{1+r^2}r^2\sin\theta dr d\theta d\varphi=\\
&=\frac{1}{4\pi^2}\int_0^{\infty}\int_0^\pi\frac{e^{-ikr\cos\theta}}{1+r^2}r^2\sin\theta dr d\theta=\\
&=\frac{1}{2\pi^2k}\int_0^{\infty}\frac{\sin kr}{1+r^2}rdr=\\
&=-\frac{1}{2\pi^2k}\cdot\frac{\pi\sinh k}{2}=\\
&=-\frac{\sinh k}{4\pi k}.\tag{1}
\end{align}
Here $k=\sqrt{k_x^2+k_y^2+k_z^2}$. It is easy to check that (1) solves $(-\Delta_k+1)\hat{f}=0$ for $k\ne0$, as you expected.</p>


问:
Distributing 10 books to 5 schools
The problem asks:

```<code>10 copies of a book are distributed in 5 schools.
1. In how many ways can these books be distributed among schools?
2. What if at least one book is assigned to each school?
</code>```

What I'm thinking is "I have 10 books to distribute in 5 schools, no matter how, so perhaps I can even give all the books to a single school"
So my answer to the first question is $C'_{10,5}=C_{14,5}$.
For the second question I was thinking "I have to give at least one book at school (so $C_{10,5}$), the rest (5) to any school, no matter how, so $C'_{5,5}$ my final answer is $C_{10,5} C'_{5,5}$".
Actually,  I was wrong on both answers, the answers are: $$1)\ C_{14,10}\\2)\ C_{9,5}$$
Why?

答:
This is a simple combinatorics problem where, as there is no restriction on having at least 1 book per school or such we can just plug in this formula.

(n+r-1) Choose r , which in this case is 14 choose 5 which is 2002.

Now if you need at least 1 book per school, This means that we can control the distribution of 5 books in any order. So again buy the same formula we get 9 choose 5 which is 126.

This theorem can be proved by the stars and bars method ([Stars and bars method discription])1. Please note that if your training for olympiads this theorem is bound to come in handy.

Title: Distributing 10 books to 5 schools


问:
Questions about Absoluteness (Set Theory)
I have a few questions about absoluteness in set theory; I think I'm largely fine with absoluteness regarding formulas, but then I see absoluteness applied to non-formulas, and that's confusing me.

Fix some transitive classes $M, N$ with $M \subseteq N$ and maybe $N = V$ for convenience. So absoluteness of a formula $\varphi(x_1, \cdots, x_n)$ with $a_1, \cdots, a_n \in M$, simply means that $\varphi^M(a_1, \cdots, a_n) \iff \varphi^N(a_1, \cdots, a_n)$. Basically it's trying to say $M \prec_{\varphi} N$ except since it's a proper class you can't quite say that.

But then, given sets $A$, I've also seen people talk about the set $A^M \in M$. $A$ is a set, not a formula, so what does $A^M$ mean?

My interpretation has been that if $A$ is defined by a formula $\varphi$ in $V$, then $A^M$ is the unique set that satisfies $\varphi^M$. Is that correct? Couching it in model theory terms, $A$ is the unique thing such that $V \vDash \varphi(A)$ and $A^M$ is the unique thing such that $M \vDash \varphi(A^M)$. (Ignoring issues with defining $\vDash$ when talking about classes due to Tarski Indefinability shenanigans).

So if $\varphi$ is absolute (and uses parameters from $M$), does this mean that $A^M = A \cap M$?

I feel like this is largely the right interpretation. I'm even more confused about the absoluteness of functions. Given a (possibly class) function $F : V \rightarrow V$, what does the absoluteness of $F$ mean exactly, and what does $F^M$ refer to?

Is it right to say that if $\varphi$ defines $F$ (with parameters from $M$), meaning that for every $x$, $\varphi(F(x), x)$, then $\varphi$ is absolute? I believe there is also an additional restriction that $F(x) \in M$ for every $x \in M$, which makes sense. So is it fine to say that in this case, $F^M = F \cap M$ as sets ($F$ is conflated with the graph of $F$)? Appealing to model theory, can I imagine expanding the language/model with $F$, and that $M$ forms a submodel in this new language (i.e, basically $F^M = F |_{M})$?

Absoluteness of functions is something I'm confused with, and in particular how to relate absoluteness of the set with the absoluteness of the (graph of) the function.

Finally, I would just like to cement some general intuition about absoluteness. Initially I was thinking about $\varphi$ being absolute to $M, N$ as simply saying "$\varphi$ is true or false in both $M$ and $N$". But now I'm thinking that perhaps it's more accurate to say that $M$ and $N$ "think" that $\varphi$ is true or false at the same time? that is, even if $\varphi$ is absolute to $M$ and $V$, this doesn't necessarily mean $\varphi$ is 'followed' in $M$, it's just that $M$ thinks it is followed. Is that an accurate assessment?

答:
Your confusion is a good one: when people say e.g. "$\omega$ is absolute between transitive models of $\mathsf{ZFC}$," what they really mean is "the formula $\varphi_\omega$ is absolute between transitive models of $\mathsf{ZFC}$" where $\varphi_\omega$ is the usual formula defining $\omega$ in $V$.

Note the bolded clause, however - there are multiple ways to define $\omega$ in $V$, and not all of them are absolute! For example, suppose $\mathsf{CH}$ holds in $V$, and consider the formula $$\psi(x)\equiv (\mathsf{CH}\rightarrow \varphi_\omega(x))\wedge(\neg\mathsf{CH}\rightarrow \forall y(\neg y\in x)).$$ This still defines $\omega$ in $V$, but is clearly not absolute between transitive models of $\mathsf{ZFC}$: if $M\models\mathsf{ZFC+\neg CH}$ is transitive then $\psi^M=\emptyset$.

So actually there's a huge potential for nonsense here, and this conflation of an object with some fixed definition of it is a really annoying abuse of notation. Fortunately it's rarely an issue, since generally there's only one reasonable choice of definition and it's clear from context, but it's still very much an abuse. (And every so often it does matter - less in terms of proving results than in developing intuitions, but still.)

The same is true for (class) functions: we often conflate an object with some fixed formula defining it. And again, this depends crucially on what definition we pick.

Meanwhile I don't really understand your last paragraph: it sounds like you're asserting a difference between "$\varphi$ is true in $M$" and "$M$ thinks $\varphi$ is followed," but I don't see what that difference would be.

(Note that, following the above abuse of notation, we do have $V^M=M$: the "standard" formula defining $V$ is $x=x$, and $M\models\forall x(x=x)$. So in fact if $M\models\theta$ then $M$ thinks $\theta$ is true in $V$ - since $M$ thinks it is $V$. Maybe that's what you're getting at?)


答:
In short: No. The limit of that sort of expression is, generally speaking, going to be a real expression (provided the limit exists and is finite). A "real expression" in this context means a mathematical expression in terms of $x$, where you can plug in real numbers as possible values of $x$ and get out real numbers as a result (like a function). On the other hand, $\mathrm{d}x$ is not a real expression; it is a different kind of object called a differential form, which is (to greatly oversimplify) "a thing that you can put under an integral sign."

The defining property of the real numbers is that you can take limits on them, so much so that a limit is usually assumed to be taken over the reals unless context implies some other set of numbers (e.g. the complex numbers, if an $i$ is present). But another important property is the archimedean property, which states that there are no infinities or infinitesimals. In other words, no real number is infinitely larger or smaller than any other (except for zero, which is infinitely smaller than all non-zero values). More formally, if you have two real numbers $x$ and $y$, you can add $x$ (or $-x$) to itself repeatedly, and the sum will eventually be greater than $y$ (or $-y$).

If you want to think of $\mathrm{d}x$ as an infinitesimal, then it immediately follows that $\mathrm{d}x$ cannot be a real number. However, in standard analysis, differentials are actually a much stranger sort of object, which arises out of abstract algebra (see link above). You don't need to know this algebra in order to do simple manipulations like the chain rule (cancelling differentials when multiplying two derivatives), but it is important to understand if you want to start talking about things like $\mathrm{d}x$ by itself (and not under an integral or part of a derivative).

Nonstandard analysis, which actually does use infinitesimals, most commonly uses the hyperreal numbers rather than the reals, and (usually) doesn't bother much with limits as such at all. In that context, it might make sense to write something like "Let $\mathrm{d}x = \frac{x}{\infty}$" (although we must take care not to confuse this $\mathrm{d}x$ with the previous paragraph's definition!). But this is a somewhat misleading thing to write, because the hyperreals do not just have one infinity - they have infinitely many. So you need to specify which infinity you want, or at least clarify that you're always using the same infinity. For simple cases, we might write $\omega$ to mean an arbitrary infinity, and thus have $\mathrm{d}x = \frac{x}{\omega}$. Then, if we write $\omega$ in another place, it will be clear that we mean the same infinity, and not $2\omega$ or something else.

Title: Is it inuitively correct to say that $\lim_{n \to \infty} \frac{x}{n} = \mathrm{d}x$?

Body: We know that from the Riemann sum that $\Delta x = \mathrm{d}x$ where $n \to \infty$ and $\Delta x = \dfrac{b - a}{n}$. If, however, $x$ represents some length of interval, can we also say that $\displaystyle\lim_{n \to \infty}\dfrac{x}{n} = \mathrm{d}x$?


问:
Question on convolution problem
I got a question in which I was asked to:
Find the convolution of two time continuous functions- 3cos(2t) and a piece wise function in which the value of the function for t<0 is exp(t) and exp(-t) for t>0.

I know how to do the integral but I'm a bit confused on the limits. I usually integrate from 0 to t but they were not piece-wise functions. In this case the there is a part of the integral for t<0 but again we are integrating u. I have sawn some places where they have integrated the integral from positive to negative infinity. I'm confused on the limits and that's all. Please tell me what limits should be taken and if it is possible what exactly is u in basic terms.

Thank you for reading the question.

答: In general the integration is from $-\infty$ to $+\infty$. If you are convoluting two functions which vanish on $(-\infty,0)$ then the integral would be from from $0$ to $t$ because the integrand vanishes  over the remaing part. In this case the integral is from $-\infty$ to $+\infty$.

Title: Question on convolution problem


问:
$\pi$ & $\phi$ (Golden ratio), Pentagon inscribed in unit circle
Everyone is aware that square inscribed in unit circle and infinite product giving rise to $\pi$.
One of the simplest way to represent $\pi$ with the help of nested radical as follows
$$\pi = \lim_{n\to \infty} 2^n\cdot \sqrt{2-{\sqrt{2+\sqrt{2+\sqrt{2+...}}}}}\tag1$$

where number of 2 inside the nested radicals is n. Viete's formula for $pi$

Approach to get similar result with inscribed initial polygon as PENTAGON will be as follows..

According to Chord in Trigonometry

A perpendicular line from the centre of a circle bisects the chord. Or otherwise can be stated as, 2. The line segment through the centre bisecting a chord is perpendicular to the chord.

The length of the chord inside the unit circle for Pentagon will be $2\sin36^\circ$. Therefore circumference of Regular pentagon will be $5\times2\sin36^\circ$.

For the sake of simplicity let us apply rules which are basic

$$2\sin\frac{\pi}{5} = \sqrt{2 - 2\cos\frac{2\pi}{5}}$$

Now doubling the sides the length of the each chord inside the unit circle will be $2\sin(\frac{\pi}{2\cdot5})$ and therefore circumference of decagon is $5\cdot2^2\sin18^\circ$. where $2\sin18^\circ = \sqrt{2-2\cos\frac{\pi}{5}}$

Therefore circumference can be represented as $C_{10} = 5\cdot2^1\sqrt{2-2\cos\frac{\pi}{5}} = 5\cdot2^1\sqrt{2-\phi}$

(Here we get $2\cos\frac{\pi}{5} = \phi $, the Golden ratio)

As $\pi = \frac{C}{D}$

Iterating the sequence we get the circumference of infinitely increasing polygons as follows and simplifying for $\pi$

Assuming doubling from Pentagon as initial polygon

For Pentagon, calculating $\pi$ as $\frac{C}{D}$

Circumference of Pentagon is $5\times 2\sin\frac{\pi}{5} = 5\cdot\sqrt{2-2\cos\frac{2\pi}{5}}$ (by Half angle cosine formula). As the diameter is 2 in unit circle. Value of $\pi_{n}$ where n represents the number of sides of the polygon of increasing order, i.e. doubling everytime from Pentagon

From Regular Pentagon
$\pi_{5}$ = $\frac{1}{2}\cdot 5\cdot\sqrt{2-2\cos\frac{2\pi}{5}} = \frac{5}{2}\cdot\sqrt{3-\phi} = 2.938926...$

From decagon
$\pi_{5\cdot2^1} = 5\cdot2\sin\frac{\pi}{5\cdot2} = 5\cdot\sqrt{2-2\cos\frac{\pi}{5}} = 5\cdot\sqrt{2-\phi} = 3.0901699...$

From icosagon (20 sided polygon)
$\pi_{5\cdot2^2} = 5\cdot2^2\sin\frac{\pi}{5\cdot2^2} = 5\cdot2^1\cdot\sqrt{2-2\cos\frac{\pi}{5\cdot2^1}} = 5\cdot2^1\cdot\sqrt{2-\sqrt{2+2\cos\frac{\pi}{5}}} = 5\cdot2^1\sqrt{2-\sqrt{2+\phi}} = 3.128689...$

From 40gon

$\pi_{5\cdot2^3} = 5\cdot2^2\sqrt{2-\sqrt{2+\sqrt{2+\phi}}} = 3.138363...$

From 80gon

$\pi_{5\cdot2^4} = 5\cdot2^3\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\phi}}}} = 3.140785...$

And so on

The value of $\pi$ for $5\cdot2^{10}$gon is $\pi_{5\cdot2^{10}}=3.14159245...$

$$\pi = \lim_{n \to \infty} 5\cdot2^n\sqrt{2-\sqrt{2+\sqrt{2+...\text{(n terms)}+\phi}}} \tag2$$ by applying limits $2^n$ becomes infinitely large and nested radical becomes infinitely small. The resulting value of $\pi$ is increasingly accurate and converges little faster compared to $\pi$ from incribed polygon as square.

The question is

$Equation1 \over Equation2$ $= 5$ explained as follows

$\frac{\pi}{\frac{\pi}{5}} = 5 = \frac{\lim_{n\to\infty}2^n\cdot\sqrt{2-\sqrt{2+\sqrt{2+...\text{n times}}}}}{\lim_{n\to\infty}2^n\cdot\sqrt{2-\sqrt{2+\sqrt{2+...\text{n times}+\phi}}}}$ (The $2^n$ gets cancelled out)

How to explain the fact that $\sqrt{2-\sqrt{2+\sqrt{2+...+\phi}}}$ is an infinitely small number different from $\sqrt{2-\sqrt{2+\sqrt{2+...+\sqrt{2}}}}$ another infinitely small number which is 5 times larger than the the first one? (By the geometric method we are deriving the value of $\pi$ from right side proceeding towards left to get the value of nested square root. (Computing also does the same). It seems that we should change the conventional thinking of representing infinite nested radicals in this respect. I need comments on this)

Infinite expansion happens between both ends. Can we call these as INFINITELY EXPANDING BALLOON NESTED RADICALS? (Bear with me for poor image quality)

When trigonometry is more concerned with unit circle practically, is $\epsilon = \frac{\pi}{\infty}$(all operations as in real number systems, here also it is applicable) different from traditional $\epsilon = \frac{1}{\infty}$? (Even though it looks trivial) (Of course there are different ways to get the infinitesimals in surreal/hyperreal number systems)

By deriving $\pi$ from polygons other than square, equilateral triangle and pentagon, probably it may be possible to get infinite number of infinitesimals by the way of representing as nested square roots. Please comment or clarify on these.

答:
George Orwell would say that all infinitely small numbers are small, but some of them are more small than others.

An infinitely small number $\varepsilon$ is an informal understanding of a sequence $(\varepsilon_n)$ of real numbers converging to zero, as your nested radicals. The smallness of $\varepsilon$ can be thought as a convergence speed of $(\varepsilon_n)$. This understanding provides a huge hierarchy of infinitely small numbers.

For each infinitely small number $\varepsilon=(\varepsilon_n)$ which is non eventually zero and each positive real $\lambda$, a sequence $(\lambda\varepsilon_n)$ converges “in $\lambda$-times” slower, and can be thought as an infinitely small number $\lambda\varepsilon$. So, for instance $\tfrac 12\varepsilon$ is smaller than $\varepsilon$.
But an infinitely small number can be even smaller than $\lambda\varepsilon$ for each $\lambda>0$. For instance, a sequence $(\varepsilon^2_n)$ eventually converges quicker than $(\lambda \varepsilon_n)$.

I guess that in a non-standard analysis infinitely small numbers are explicitly provided in an extension of a field of real numbers. So we can add and multiply them, but when we divide by an infinitely small numbers then as a result we can obtain an infinitely large number.

PS. Let $\mathcal P$ be a set of infinitely small numbers $(\varepsilon_n)$, such that all members $\varepsilon_n$ are positive. We can endow $\mathcal P$ with a preorder $\le$, putting $(\varepsilon_n)\le (\delta_n)$, if $(\varepsilon_n)\le (\delta_n)$ for all sufficiently large $n$. It turns out that this preorder has interesting set-theoretical properties.

For any sequence $(\varepsilon_n)_1, (\varepsilon_n)_2,\dots$ of elements of $\mathcal P$ there exists $\delta=(\delta_n)\in\mathcal P$ such that $\delta\le (\varepsilon_n)_i$ for each $i$. Indeed, it suffices to put $\delta_n=\min_{i\le n} (\varepsilon_n)_i$ for each $n$. Moreover, we can define $\mathfrak b$ be the smallest cardinality of a subset $B$ of $\mathcal P$ such that there exists no $\delta\in \mathcal P$ such that $\delta\le\varepsilon$ for each $\varepsilon\in B$ and $\mathfrak d$ be the smallest cardinality of a subset $D$ of $\mathcal P$ such that for
each $\varepsilon\in \mathcal P$ there exists no $\delta\in D$ such that $\delta\le\varepsilon$. Clearly, $\omega_1\le \frak b\le\frak d\le \frak c$. It turns out, that values of $\frak b$ and $\frak d$ depend on additional axiomatic assumptions. For instance, Continuum Hypothesis implies $\omega_1=\frak b=\frak d=\frak c$, but each of claims $\omega_1=\frak b<\frak c$,
$\omega_1=\frak d<\frak c$, $\omega_1<\frak b<\frak c$, $\omega_1<\frak d<\frak c$,  $\omega_1<\frak b=\frak c$, and $\omega_1<\frak d=\frak c$ is consistent with ZFC, see, for instance, Section 5 of [vDou].

References

[vDou] Eric K van Douwen, The integers and Topology, in: K.Kunen, J.E.Vaughan (eds.) Handbook of Set-theoretic
Topology, Elsevier, 1984, 111–167.


问:
How to calculate the square area in coordinate system given only y coordinates?
I'm given a square in the coordinate plane, the y co-ordinates of the square vertices are $y1$, $y2$, $y3$, and $y4$. How should I calculate the area of the square?

I tried to have $x1$, $x2$, $x3$, and $x4$ set, and then

$(x1-x2)^2 + (y1-y2)^2 = (x2-x3)^2 + (y2-y3)^2$

$(x2-x3)^2 + (y2-y3)^2 = (x3-x4)^2 + (y3-y4)^2$

$(x3-x4)^2 + (y3-y4)^2 = (x4-x1)^2 + (y4-y1)^2$

$(x1-x2)^2 + (y1-y2)^2 = (x4-x1)^2 + (y4-y1)^2$

$(x1-x3)^2 + (y1-y3)^2 = 2(x2-x3)^2 + 2(y2-y3)^2$

$2(x1-x2)^2 + 2(y1-y2)^2 = (x2-x4)^2 + (y2-y4)^2$

Trying to solve x is difficult. Is there a better way?

答: Order the y-axis coordinates in increasing order so that $y_1 \leq y_2 \leq y_3 \leq y_4$. Let $a = y_2 - y_1$ and $b = y_3 - y_1$, and notice that in following figure the inscribed square is the square you are trying to find the area. By the Pythagorean theorem, the area of that square is $a^2 + b^2$.