答:
The reason the author says this is that it's a kind of motivation for the definition of a linear mapping. Just saying a linear map is any $f : V \rightarrow W$ for vector spaces $V$ and $W$ which satisfies:

$$f(u+v) = f(u) + f(v)$$

and

$$f(\lambda u) = \lambda f(u)$$

is indeed the definition, but it is not enlightening in the slightest, so the author is attempting to give some insight into why we have this definition.

Generally, when vectors are added together, or multiplied by a scalar, the result is also a vector (the same type of object).
How are the above properties of a linear transformation preserving this?

Let's go over what vector properties mean. First, we have that adding vectors produces a vector, so, for vectors in boldface, we have:

$$\mathbf{u + v} = \mathbf{u} + \mathbf{v}$$

For example:

$$\begin{bmatrix} 1 + 2 \\ 4+5 \\ -1 + 15 \end{bmatrix} = \begin{bmatrix} 1  \\ 4 \\ -1 \end{bmatrix} + \begin{bmatrix}2 \\ 5 \\ 15 \end{bmatrix}$$

We also have the ability to pull out scalars:

$$\mathbf{au} = a\mathbf{u}$$

For example:

$$\begin{bmatrix} 10  \\ 40 \\ -10 \end{bmatrix} = 10\begin{bmatrix} 1  \\ 4 \\ -1 \end{bmatrix}$$

What this tells us is that the linear map is compatible with the rules of vector arithmetic. This is one of the author's intentions behind saying that it maintains structure.

Indeed, if $\mathbf{w} = \lambda \mathbf{u} + \delta \mathbf{v}$, if $f$ is a linear mapping, then $f(\mathbf{w}) = \lambda f(\mathbf{u}) + \delta f(\mathbf{v})$ (can you prove this?). So a linear map uniquely preserves the relationship between different vectors. Can you find three vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$ such that $\mathbf{a} = \mathbf{b} + \mathbf{c}$, but for $g(x) = x^2$, we have $g(\mathbf{a}) \neq g(\mathbf{b}) + g(\mathbf{c})$? Why does this tell us that $g$ is not a linear map? Can you find a counterexample to prove that $g$ is not a linear map?

This idea that maps "preserve the decomposition of a vector into a sum of scaled vectors" will be vital to defining things like a basis of a vector space, and other useful properties.

As long as the LHS and RHS of both given properties belong to the same vector space, can I not consider the vector properties preserved? Why do they have to be equal?

In this case, once again consider $g(x) = x^2$. This function runs from $\mathbb{R}$ to $\mathbb{R}$, both of which are vector spaces, but it does not satisfy our properties. Hence while the output is a vector, the way that the output relates to its component parts is not the same as the way that the input relates to its component parts.

Once again, this is related to the idea of a basis, in which we find we may write a vector as a unique linear combination of a finite set of vectors (if the vector space is finite dimensional), and linear maps preserve that combination, but I do not know if you have been exposed to a basis yet.

Title: How do linear transformation properties preserve vector space structure?

Body:
Currently I'm reading the Mathematics for Machine Learning book, and in the beginning of section 2.7 of the book, the following is given:

2.7 Linear Mappings

In the following, we will study mappings on vector spaces that preserve their structure, which will allow us to
deﬁne the concept of a coordinate. In the beginning of the chapter, we
said that vectors are objects that can be added together and
multiplied by a scalar, and the resulting object is still a vector. We
wish to preserve this property when applying the mapping:  Consider
two real vector spaces V,W. A mapping Φ : V → W preserves the
structure of the vector space if

Φ(x + y) = Φ(x) + Φ(y) (2.85)

Φ(λx) = λΦ(x) (2.86)

for all x, y ∈ V and λ ∈ R.

Generally, when vectors are added together, or multiplied by a scalar, the result is also a vector (the same type of object).

How are the above properties of a linear transformation preserving this?

As long as the LHS and RHS of both given properties belong to the same vector space, can I not consider the vector properties preserved? Why do they have to be equal?

What does it mean by they preserve the structure of a vector space?

Is there a particular reason Φ(λx + ψy) = λΦ(x) + ψΦ(y) has a form similar to a linear combination (y = a1.x1 + a2.x2 +...)?


问:
Clarification on higher dimensional unit sphere/unit cube properties
I've been learning probability lately and I have come across this neat result (in Rick Durrett's book):

Most of the volume of the unit cube in $\mathbb{R}^n$ comes from the set $A_{n,\epsilon} := \{x \in \mathbb{R}^n \, : \,(1-\epsilon)\sqrt{\frac{n}{3}} < |x| < (1+\epsilon)\sqrt{\frac{n}{3}} \},$ which is almost the sphere of radius $\frac{n}{3}.$

At the same time, we know the two basic facts that the volume of the unit ball in $\mathbb{R}^n$ goes to zero as $n$ grows and the volume of the unit cube remains the same, but most of the volume get concentrated in the corners of the cube.

I realize that these two points are pretty different ideas altogether, but it seems strange to me that most of the volume in a high dimensional cube can be contained within a sphere and yet also be concentrated in the corners.  Could someone clarify what is really going on? Or perhaps my understanding is off.. I am not very experienced in geometry at all.  Thanks!

答:
After leaving my comments to your post asking about the meaning of "concentrated in the corners" I gave the question some more thought and now feel it potentially useful to leave this crude attempt at an answer. I want to make clear that the mention of $\sqrt{n/2}$ in my comment was referring to the cube $[0,1]^n$ and sphere centered at the origin. In this answer it will be convenient to talk about the cube $[-1,1]^n$, with sphere still centered at the origin. The former cube is one of $2^n$ cubes making up the latter cube, and the spherical shell of radius $\sqrt{\frac{n}{3}}$ intersects all $2^n$ constituent cubes equally, so that the concentration phenomenon holds whether you consider $[0,1]^n$ or $[-1,1]^n$.

Note that I am not going to address the derivation of the quoted concentration result, which is done elsewhere, for example here and here. My focus will be on intuitive understanding of how the result can be consistent with the other facts you mention.

To address your remark about the volume of the $n$-dimensional ball of unit radius going to zero, the point, of course, is that the spherical shell in the quoted result has radius that grows with the dimension, making it at least possible that the volume doesn't go to zero. Indeed, we can check that this is the case. The volume of the $n$-ball of radius $R$ is given by
$$
V_n=\frac{\pi^{n/2}}{\Gamma\left(\frac{n}{2}+1\right)}R^n,
$$
which does go to $0$ for any fixed $R$. By Stirling's approximation, $\Gamma\left(\frac{n}{2}+1\right)\sim\sqrt{\pi n}\left(\sqrt{\frac{n}{2e}}\right)^n$, so if we take $R=\sqrt{n/3}$ we get
$$
V_n\sim\frac{1}{\sqrt{\pi n}}\left(\sqrt{\frac{2\pi e}{3}}\right)^n.
$$
Since $\sqrt{\frac{2\pi e}{3}}\approx2.39$, this grows faster than $2^n$, which it needs to do to keep pace with the volume of the cube of side $2$. So there is no contradiction there.

Now for the intuitive, but crude part. If the statement that the volume of the cube is "concentrated in the corners" is just shorthand for "away from the center", then I think it's fine. But to go farther we need a more precise understanding of how the volume is really distributed. The crude model I will use is taken from Rubik's cube, where the cube of side $2$ is divided into $3^3$ cubies of side $\frac{2}{3}$. Generalize to $n$ dimensions, so that each edge of the $n$-cube is divided into equal thirds, $\left[-1,-\frac{1}{3}\right]$, $\left[-\frac{1}{3},\frac{1}{3}\right]$, $\left[\frac{1}{3},1\right]$, making a total of $3^n$ cubies. Conceptually (although if you open it up you see it's not that way inside) the Rubik's cube has four types of cubie: $8$ corner cubies, $12$ edge cubies, $6$ face cubies, and $1$ body cubie. In $n$ dimensions there will be $n+1$ different types of cubie, each type associated with a particular dimension of facet. So corner cubies are associated with $0$-dimensional facets and edge cubies with $1$-dimensional facets, but there will also be cubies associated with, say, $10$-dimensional facets. Now we can ask: which types of cubie contain most of the volume?

The answer comes from combinatorics. A $d$-dimensional facet of the $n$-cube is specified by fixing each of $n-d$ coordinates to $\pm1$ while allowing the other $d$ coordinates to vary continuously between $-1$ and $1$. The number of $d$-dimensional facets is therefore $2^{n-d}\binom{n}{d}$, which we may express as $3^n\left(\frac{1}{3}\right)^d\left(\frac{2}{3}\right)^{n-d}\binom{n}{d}$. In this expression we see the $3^n$ cubies, multiplied by a Bernoulli distribution function with mean $\frac{n}{3}$ and standard deviation $\frac{\sqrt{2n}}{3}$. For large $n$ this is close to a normal distribution and has a sharp peak around $d=\frac{n}{3}$.

Taking $n=100$ as an example, the mean is about $33.3$ and the standard deviation is about $4.7$. Taking the $3$ standard deviation rule, more than $99.7\%$ of the volume of the $100$-dimensional cube is in the cubies associated with facets of dimensions $19$ through $48$. There is a vanishingly small amount of volume associated with corner cubies. Although $2^{100}$ might seem like a big number, there's a sense in which the $100$-dimensional cube has very few corners: $33$-dimensional facets outnumber corners by a factor of $3.4\times10^{16}$. I won't go so far as to say that the statement about volume being concentrated in corners is wrong. My crude measure of proximity to corners is just one possibility, and other measures may give different results. But I am still curious about what was meant by that statement.

Now here's a key point: cubies associated with the same dimension of facet lie at the same distance from the center, and cubies associated with facets that are close in dimension are of similar distance from the center. Furthermore, the distance of a cubie depends rather strongly on the dimension of the associated facet. If we use the centers of cubies as reference points for measuring distance (so center coordinates will all be $\pm\frac{2}{3}$ or $0$), then cubies associated with $(n-1)$-dimensional facets lie at distance $\frac{2}{3}$ from the center, while corner cubies lie at distance $\frac{2}{3}\sqrt{n}$ from the center. Since the number of cubies associated with a given dimension of facet is sharply peaked, the number of cubies as a function of distance from the center is also sharply peaked, and cubies whose distance from the center is far from the peak value don't contribute much to the volume.

To make a more concrete, but crude, estimate of the distance at which most cubies lie from the center, we focus only on the peak of the distribution at dimension $n/3$ and ignore the fact that the volume of a cubie isn't all at its center. For cubies associated with facets of dimension $n/3$ the center will have $2n/3$ coordinates equal to $\pm\frac{2}{3}$ and $n/3$ coordinates equal to $0$, giving a distance
$$
\sqrt{\frac{2n}{3}\cdot\left(\frac{2}{3}\right)^2+\frac{n}{3}\cdot0^2}=\sqrt{\frac{8}{9}}\sqrt{\frac{n}{3}}\approx0.94\sqrt{\frac{n}{3}}.
$$
So even this very crude picture produces an answer close to the quoted value.

A final remark: it may seem puzzling that $33$-dimensional facets can outnumber corners by the large factor $3.4\times10^{16}$ when each $33$-dimensional facet contains $2^{33}$ corners. The explanation for this is that each corner is shared by a huge number of other $33$-dimensional facets. For $d\le e$, an $e$-dimensional facet contains $2^{e-d}\binom{e}{d}$ $d$-dimensional facets. The number of $e$-dimensional facets incident on a given $d$-dimensional facet is $\binom{n-d}{e-d}$, which is the number of ways of choosing which of the $n-d$ fixed coordinates should be allowed to vary. The ratio of the number of $d$-dimensional facets to the number of $e$-dimensional facets is therefore
$$
\frac{2^{e-d}\binom{e}{d}}{\binom{n-d}{n-e}}=\frac{2^{n-d}\binom{n}{d}}{2^{n-e}\binom{n}{e}}.
$$
When $n=100$, $d=0$, and $e=33$ the factor $2^{33}\approx8.6\times10^9$ in the numerator is completely overcome by the factor $\binom{100}{67}\approx2.9\times10^{26}$.

Added: Perhaps one should be troubled by the roughness of the approximations in the penultimate paragraph and maybe should wonder what the unwanted $\sqrt{\frac{8}{9}}$ is trying to tell us. The approximation of considering only the peak of the distribution seems unlikely to pose any problem, as the peak gets sharper and sharper as $n$ grows. But using the centers of cubies in the distance calculation is a pretty poor approximation. For one thing, cubies, like all cubes, are surprisingly large when the dimension gets big. The distance from the center of a cubie to a corner, for example, is $\frac{1}{3}\sqrt{n}$. Furthermore, there is almost no volume at the center of a cubie. Dividing each cubie into sub-cubies, the center is located in the body sub-cubie, which is unique among the $3^n$ sub-cubies, and the least representative of all sub-cubies.

Fortunately, any intra-cubie distances we might need to account for are corrections to a distance which is itself large: $\sqrt{n/3}$, so our approximation can't be completely off. Furthermore, we know where most of the volume of a cubie lies: it's in the sub-cubies associated with facets of dimension $n/3$ and the centers of these sub-cubies are at distance $\frac{1}{3}\sqrt{\frac{8}{9}}\sqrt{\frac{n}{3}}$ from the center of the cubie. Now the direction along which some of these sub-cubies lie will be aligned or anti-aligned with the direction along which the cubie itself lies, but this will happen a vanishingly small fraction of the time. In almost all cases, the sub-cubie direction will be uncorrelated with the cubie direction and will be nearly orthogonal to it. Therefore we may use the Pythagorean theorem to estimate the correction to the distance:
$$
\left[\left(\sqrt{\frac{8}{9}}\sqrt{\frac{n}{3}}\right)^2+\left(\frac{1}{3}\sqrt{\frac{8}{9}}\sqrt{\frac{n}{3}}\right)^2\right]^{1/2}.
$$
Of course we have just used the some rough center-point approximation on the sub-cubies, so we should correct that as well by iterating:
$$
\left[\left(\sqrt{\frac{8}{9}}\sqrt{\frac{n}{3}}\right)^2+\left(\frac{1}{3}\sqrt{\frac{8}{9}}\sqrt{\frac{n}{3}}\right)^2+\left(\frac{1}{3^2}\sqrt{\frac{8}{9}}\sqrt{\frac{n}{3}}\right)^2+\ldots\right]^{1/2}.
$$
The geometric sum $1+1/9+1/9^2+\ldots$ equals $\frac{9}{8}$, which exactly compensates for the factor $\frac{8}{9}$, giving the refined distance estimate $\sqrt{\frac{n}{3}}$, a gratifying result!

Of course I'm not suggesting that these heuristic arguments are the way one should actually do the computation: the usual rigorous derivation of the concentration result is actually much simpler, and can be found at the links I gave above. But this picture does provide a way of understanding the concentration phenomenon that is simultaneously combinatorial and geometric and, at least for me, makes things much more concrete.

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 4.0


问:
Interchange order of integral
Can someone tell me which step in the process did wrong? My result seems wrong
$$D = \{(x, y) \mid \sqrt{2x - x^2} \leq y \leq \sqrt{2x}, 0 \leq x \leq 2 \}$$
$$\iint_D f(x,y) \, dy \, dx$$
Then I can draw the Graph

According to the graph, we can know that this is the area enclosed by $x = \frac{y^2}{2}$ and $x = 2$ minus the semicircle area
$$\iint_D f(x,y)\,dy\,dx = -\int_0^4 \int_{\frac{y^2}{2}}^2 f(x,y)\,dx\,dy - 2\int_0^1 \int_1^{1 + \sqrt{1 - y^2}} f(x,y)\,dx\,dy $$

答:
The natural way to write the iterated integral is of course $$\iint_D f \, dA = \int_{x=0}^2 \int_{y=\sqrt{2x-x^2}}^{\sqrt{2x}} f \, dy \, dx.$$  The first alternative where the order of integration is switched, is as you attempted, the area of the parabolic sector minus the semicircle.  Note that the parabolic segment satisfies $$y = \sqrt{2x} \iff x = \frac{y^2}{2},$$ so when $x = 2$, $y = \sqrt{4} = 2$, not $y = 4$ as shown in your diagram.  Then the semicircular arc is a portion of the circle $y^2 = 2x - x^2 = 1 - (x-1)^2$, hence $$x = 1 \pm \sqrt{1 - y^2}.$$  Thus the semicircular region is, for $y \in [0,1]$, bounded above by $1 + \sqrt{1 - y^2}$ and below by $1 - \sqrt{1 - y^2}$, and we have
$$\iint_D f \, dA = \int_{y=0}^2 \int_{x=y^2/2}^2 f \, dx \, dy - \int_{y=0}^1 \int_{x=1 - \sqrt{1 - y^2}}^{1 + \sqrt{1 - y^2}} f \, dx \, dy.$$  A second alternative is to compute the area from $y = 0$ to $1$ as the sum of two sub-areas, plus a third sub-area from $y = 1$ to $2$:  $$\iint_D f \, dA = \int_{y=0}^1 \left( \int_{x=y^2/2}^{1-\sqrt{1-y^2}} f \, dx + \int_{x=1+\sqrt{1-y^2}}^2 f \, dx \right) dy + \int_{y=1}^2 \int_{x=y^2/2}^2 f \, dx \, dy.$$

Title: Interchange order of integral


问:
Why are the order-of-operations conventions good?
Children are sometimes taught silly mnemonics like "PEMDAS" to remember conventions on order of operations.  (I never heard of "PEMDAS" until long after graduating from college, as far as I can recall.  I think it means (1) parentheses, (2) exponentiation, (3) multiplication and division, and (4) addition and subtraction.)

I think it would be better to help them understand why those particular conventions, rather than some others, are a good thing.  Maybe even demonstrably optimal by some precisely definable desiderata?

How would one make the case for the usual conventions against possible alternatives?

I think I might say that if one operation distributes over another, it should come first, and we go from left to right because we normally read that way.  Maybe I'll post my own answer to this if I'm so inspired at some point.

答:
Well, since parentheses exist precisely to specify the intended order of operations in case the usual default rules don't cut it, it makes sense that they come first

As for exponentation, I'd say that this is a consequence of using superscripts to indicate exponentation, since those (via font size) provide a natural grouping. It'd certainly be very weird if $a^b + c$ meant $a^{(b+c)}$ instead of $(a^b) + c$, since the different font sizes of $b$ and $c$ indicate that they're somehow on different levels.

As MJD pointed out though, this arguments only applies to the exponent. Font size alone doesn't explain why $a + b^c$ means $a + (b^c)$ and not $(a + b)^c$ and the same for $a\cdot b^c$ vs. $a\cdot(b^c)$ respectively $(a\cdot b)^c$. For these, I'd argue that it's also a matter of visual grouping. In both $a\cdot b^c$ and $a + b^c$, the exponent is written extremely close to the $b$, without a symbol which'd separate the two. On the other hand $a$ and $b$ are separated by either a $+$ or a $\cdot$. Now, for multiplication the dot may be omitted, but it doesn't have to be omitted, i.e. since $ab$ and $a\cdot b$ are equivalent, one naturally wants $ab^c$ and $a\cdot b^c$ to be equivalent too.

For multiplication, division, addition subtraction, I always felt that the choice is somewhat arbitrary. Having said that, one reason that does speak in favour of having multiplication take precedence over addition is that one is allowed to leave out the dot and simply write $ab$ instead of $a\cdot b$. Since this isn't allowed for addition, in a lot of cases the terms which are multiplied will be closer together than those which are added, so most people will probably recognize them as "belonging together".

You may then ask "how come we're allowed to leave out the dot, but not the plus sign". This, I believe is a leftover from times when equations where stated in natural language. In most langues, you say something like "three apples" to indicate, well, three apples. In other words, you simply prefix a thing by a number to indicate multiple instances of that thing. This property of natural languages is mimicked in equations by allowing one to write $3x$ with the understanding that it means "3 of whatever $x$ is".


问:
$E(1/(1+x^2)) $under a normal distribution
I want to know as mentioned in topic $E(1/(1+x^2))$ under a normal distribution $N(0,1)$. If it's not analytical, are there any bounds that are possible?

So basically,
$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \frac{e^{-x^2}}{1+x^2}  dx $$

Thanks in advance,
Sachin

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{{1 \over \root{2\pi}}
\int_{-\infty}^{\infty}{\expo{-x^{2}} \over 1 + x^{2}}\,\dd x} =
{\expo{} \over \root{2\pi}}
\int_{-\infty}^{\infty}{\expo{-\pars{x^{2} + 1}} \over
1 + x^{2}}\,\dd x
\\[5mm] = &\
{\expo{} \over \root{2\pi}}
\int_{-\infty}^{\infty}
\int_{1}^{\infty}\expo{-\pars{x^{2} + 1}y}\,\,\dd y\,\dd x
\\[5mm] = &\
{\expo{} \over \root{2\pi}}
\int_{1}^{\infty}\expo{-y}\int_{-\infty}^{\infty}
\expo{-yx^{2}}\,\,\dd x\,\dd y =
{\expo{} \over \root{2\pi}}\int_{1}^{\infty}\expo{-y}
\pars{\root{\pi}y^{-1/2}}\dd y
\\[5mm] = &\
\root{2}\expo{}\int_{1}^{\infty}\expo{-y^{2}}\,\dd y =
\root{2}\expo{}\bracks{{\root{\pi} \over 2}\on{erfc}\pars{1}}
\\[5mm] & = \bbx{{\root{2\pi} \over 2}\,\expo{}\on{erfc}\pars{1}}
\approx 0.5359\\ &
\end{align}
See this DLMF link.


问:
Closed subsets of compact spaces are compact.
I'm having trouble understanding a point in the following standard argument to show that closed subsets of compact spaces are compact.

The proof usually goes as follows.

Let $X$ be compact and let $C$ be closed. Let $\{U_\alpha \}$ be some open cover of $C$. Now, $\{U_\alpha \}$ with $X \setminus C$ is an open cover of $X$. As such, we get a finite open cover of $X$ from this collection, from which we get finite subcover of $C$.

The point I am having trouble understanding is that if $\{U_\alpha \}$ is an open cover of closed set $C$, it does not seem necessary that all $U_\alpha$ is open in $X$. The example I am thinking of is $[0,1]$ in $\mathbb{R}$. I believe there is an open cover(with respect to subspace topology of $[0,1]$) of $[0,1]$ with $[0,0.6)$ and $(0.4,1]$. However, both of these are not open sets in $\mathbb{R}$. So I am troubled by the point that $\{U_\alpha \}$ with $X \setminus C$ forms an open cover of $X$.

Thanks for your help!

答: You are right, if $U$ is open in $C$ then it is not necessarily open in $X$. However there will be an open subset $U'$ of $X$ such that $U=C\cap U'$. Then you can simply replace the $U_\alpha$ with the $U_\alpha'$.


问:
Convex optimization books, except with a focus on problem solving and formalism later?
Let me start off by saying that I'm well aware of these books:

Convex Optimization, Boyd and Vandenberghe

Lectures on Convex Optimization, Nesterov

The books by Bertsekas

Convex Analysis by Rockafellar

Convex Analysis and Linear Optimization by Borwein and Lewis

I'd also like to mention that my math background is by no means deficient: I have a math bachelor's degree, a master's in statistics, and am doing a two-semester sequence in measure theory. The advice seems to consistently be that if you know linear algebra and multivariable calculus, you should be able to handle this stuff.

I personally can't stand these books. Here's why:

There is too much emphasis on formalism. Boyd and Vandenberghe, especially, seem insistent on defining term after term after term at you from the very beginning. Unless people who read this book are walking dictionaries, I don't see how one can reasonably learn from this book.

There is too little emphasis on problem solving. Why do I want to learn optimization? So that I understand how to minimize/maximize an objective function with constraints. I understand the need for formalism, but the fact that this is front and center in every one of these books makes them extremely difficult for self study. Furthermore, I do not want to just learn how to formulate a problem as a convex optimization problem; I would like to at least know how to solve such problems by hand (or enough theory so that I can program such solutions).

Now that all of that is out of the way, to my main question:

Does there exist a convex optimization book which

has a problems-first emphasis, and if it provides formal results, does provide them but primarily in the context of actually solving a convex optimization problem?

is suitable for self-study, without supplementary material necessary? (I am not going to watch videos on top of a textbook; the book should be able to stand on its own.)

is not just a definition-theorem-proof-type of book?

Please note that I am not looking for a linear optimization book, but a convex optimization one.

答:
Given your background in statistics perhaps you could look at Online Convex Optimisation, which is an area that combines convex optimization and regret minimisation in sequential learning problems. The resources in this area usually cover a lot of convex optimisation theory that is useful for solving actual problems in that area. Online learning is an area with a lot of open problems and has been an active area for research, but also covers a lot of practical problems you'd face in an industry setting, e.g. portfolio optimisation, recommender systems.

Some links:

Online Convex Optimization Tutorial and list of resources

Elad Hazan's book on online convex optimization

Online Learning notes by Francesco Orabona


问:
What does "$\binom n k$" mean?
\begin{align}
& \binom n 0 + \binom n 1 + \binom n 2 + \cdots + \binom n n = \\[8pt]
& \text{(A)}\,\,\, 2^{n-1} \qquad \text{(B)} \,\,\, {}^{2n}C_n \qquad \text{(C)} \,\,\,2^n \qquad \text{(D)}\,\,\,2^{n+1}
\end{align}

Why is there nothing inside the bracket (between the $n$ and the number)?

答:
The numbers inside the brackets don't represent fractions. They are known as binomial coefficients and that is how we write them. For example,
$$\binom{3}{2}=\frac{3!}{2!1!}$$

Very simply, the binomial coefficent $\binom{n}{k}$ (pronounced $n$ choose $k$) represents the number combinations that can be formed from $k$ items out of $n$ items.

Title: What does "$\binom n k$" mean?


Title: What does "$\binom n k$" mean?

Body:
\begin{align}
& \binom n 0 + \binom n 1 + \binom n 2 + \cdots + \binom n n = \\[8pt]
& \text{(A)}\,\,\, 2^{n-1} \qquad \text{(B)} \,\,\, {}^{2n}C_n \qquad \text{(C)} \,\,\,2^n \qquad \text{(D)}\,\,\,2^{n+1}
\end{align}

Why is there nothing inside the bracket (between the $n$ and the number)?

Body:
Simply,

$$\binom{n}{k}=\frac{n!}{k!(n-k)!}$$


问:
Fixed point result for a differentiable function.
Let $f$ be a differentiable function on $[0, 1]$ with $0$ and $1$ be its fixed points. Also, it is given that $$f'(0)>1, f'(1)>1.$$ Then prove that

(i) $f$ has at least one fixed point in $(0, 1)$.

(ii) $f$ has a fixed point $\xi\in (0, 1)$ such that $f'(\xi)\leq 1$.

My effort: I started by considering $g(x)=f(x)-x$. Clearly, $g$ is a differentiable function which means $g'(x)=f'(x)-1$. This means $g'(0)>0, g'(1)>0$. But I am not able to conclude anything. Please help.

答:
As no one else has given an answer addressing part (ii), let me prove both parts at once (and also prove the intermediate value theorem as a corollary).

As you've reduced, what we need to do is to prove:

Suppose $g$ is a differentiable function with $g'(0)>0$ and $g'(1)>0$. Prove that there exists some $z\in (0,1)$ so that $g(z)=0$ and $g'(z)\leq 0$.

As others have noting, using the derivatives at either end tells us that least that there is some $x_0$ so that $\frac{g(x_0)}{x_0} > 0$ and some $x_1$ so that $\frac{-g(x_1)}{1-x_1} > 0$ which in particular implies $g(x_0)>0$ and $g(x_1)<0$ - or, in fact, that these inequalities hold for all $x_0$ in a neighborhood of $0$ and $x_1$ in a neighborhood of $1$, which we subtly need to ensure that we choose $x_0$ and $x_1$ so that $x_0 < x_1$.

Now, intuitively, we are looking for some place where the graph of $g$ crosses downwards - from positive to negative. A nice way to do that is to consider the last spot in $[x_0,x_1]$ where $g$ is non-negative. In particular, let $$S=\{x\in [x_0,x_1] : g(x) \geq 0\}$$
Note that $S$ is a closed (bounded) set and thus has some maximum $z\in S$. Moreover, note that $x_1$ is not in $S$ thus $x_0\leq z<x_1$. We claim that $g(z)=0$ and $g'(z)\leq 0$.

We know that the image of every interval $(z-h,z+h)$ under $g$ contains a negative number, since every input greater than $z$ gives a negative output. We also know that $g(z)\geq 0$. By continuity, we need that every neighborhood of $g(z)$ includes negative numbers. Together, this implies $g(z)=0$.

To see that the derivative is negative at this $z$, consider the quotient
$$\frac{g(z+h) - g(z)}{h}=\frac{g(z+h)}h.$$
Observe that this is negative for every $0 < h \leq x_1 - z$, which implies that the limit of this as $h$ goes to $0$ must be non-positive - or, otherwise put, $g'(z)\leq 0$.

The proof that $g$ has a zero crossing suffices as a proof of the intermediate value theorem - and is a proof that I particularly like because it captures some extra geometric intuition about the crossing, which is exactly what is needed for this problem.


问:
Prove that if $A_{n}$ belongs to a $\sigma$-algebra $\mathcal{F}$ for each $n \geq 1$, then $\bigcap^\infty_{n = 1} A_{n} \in \mathcal{F}$
I am going over properties of sigma-algebras but am struggling how to go about proving certain properties or conditions. I would be able to do this if it was talking about the union, but the intersection I am unsure of.</p>

答:
Here it is worth emphasizing the definition of $\sigma$-algebra.

Definition

Let $\Omega$ be a non-empty set, where $2^{\Omega}$ indicates its power set. We say that a class of subsets $\mathcal{F}$ of $\Omega$ is a $\sigma$-algebra iff

$\Omega\in\mathcal{F}$,

If $A\in\mathcal{F}$, then $A^{c}\in\mathcal{F}$,

If $A_{n}\in\mathcal{F}$ is a countable collection of sets, then one has that
\begin{align*}
\bigcup_{n=1}^{\infty}A_{n}\in\mathcal{F}
\end{align*}

Solution

Thus, if $A_{n}$ belongs to $\mathcal{F}$, then $A^{c}_{n}$ also belongs to $\mathcal{F}$.

According to the countable union property, one concludes that
\begin{align*}
\bigcup_{n=1}^{\infty}A^{c}_{n}\in\mathcal{F} \Rightarrow\left[\bigcup_{n=1}^{\infty}A^{c}_{n}\right]^{c} = \bigcap_{n=1}^{\infty}A_{n}\in\mathcal{F}
\end{align*}

Hopefully this is what you are looking for.


问:
Is the same to add two vectors and then compute the norm of the resulting one to first compute the norm of both vectors and then adding those values?
This may be a dumb question but I pretty new in vector math and I couldn't find the answer anywhere.

What I'm asking is basically this:

$$\|\vec{v}\| + \|\vec{u}\| \stackrel{?}{=} \|\vec{v} + \vec{u}\| $$

答:
In fact $||\vec{u}||+||\vec{v}||\ge||\vec{u}+\vec{v}||$, with the equality reached only in special cases ($u=0$, $v=0$ or $v=\lambda u$ for $\lambda>0$). This is called a "triangle inequality" for reasons obvious from the following picture:

(Every side in a triangle is shorter than the sum of the other two sides.)


问:
Reference Needed: $m$-faced dice with $n$ options
I would like to investigate the following question and I am not sure if similar ideas have been investigated. I cannot really find a detail discussion about this matter on the Internet. It will be great if you can tell me what domain I should look into, or you may also give your view on the following questions.

Suppose we have one fair dice. We would like to choose an option randomly from 6 possible options. We roll the dice once then we can decide which option we should choose.

Suppose we now have 8 options. To choose an option randomly, different algorithms can be used.
For example, roll the dice once first.

If the result is odd, we focus on option 1 to 4. Then roll the dice again, if we get 1 to 4, we know which option to use. If it is a 5 or 6, repeat the second step again.

If the result is even, we focus on option 5 to 8. Then roll the dice again, if we get $n$, which $1\leq n\leq 4$, we choose option $n+4$. If it is a 5 or 6, repeat the second step again.

One can check the the probability of choosing each option is exactly $\frac{1}{8}$.

My question now is to investigate if it is possible that for whatever number of options we have, we can always use a dice to choose an option randomly. And is there a general algorithm to follow in order to choose an option?

My second question is more advance. Suppose we have a fair $m$-faced dice and we have $n$ total options. Can we choose an option randomly by using the dice for any $m$ and $n$?

Once again, you don't have to answer the above two questions. I am looking for hints or domain that I should look into. Thank you.

答:
We are given a die of $m$ faces, and $n$ options to choose from.  Like in jlammy's answer, let $\ell$ be the number of die rolls.  There are then $m^\ell$ possible outcomes, so we certainly need $m^\ell \geq n$ so that every option has a chance of being chosen.

This answer will tweak jlammy's answer, to be more efficient.  For example, suppose $m=6$ and $n=7$.  If we use $\ell= 2$, there will be $6^2 = 36$ outcomes.  In jlammy's answer, only $7$ of these are used, one for each of the $n=7$ options.  So there is a $\frac{29}{36}$ chance of needing to re-roll.  Instead, we'd like to assign $5$ outcomes to each of the $7$ options.  That way we use $5\cdot 7 = 35$ of the $36$ outcomes, so there is only a $\frac{1}{36}$ chance of needing to re-roll.

To do this, divide $m^\ell$ by $n$, using integer division, so we get a quotient $q$ and a remainder $r$.  Thus $m^\ell = qn+r$, where $0\leq r < n$.  Also, remember that we choose $\ell$ such that $m^\ell \geq n$, which implies $q\geq 1$ (for, if $q\leq 0$ then $m^\ell = qn+r \leq r < n$).  In particular, we have $0\leq r < qn$.

The idea is that each of the $n$ options will be assigned $q$ separate outcomes, with $r$ outcomes left over so the probability of a re-roll will be $\frac{r}{m^\ell}$.  To assign the outcomes, write the die faces as $0, 1, \ldots, m-1$, and the options as $0, 1, \ldots, n-1$.  Now the sequence of $\ell$ die rolls corresponds to an integer written in base-$m$.  Call this number $Z$, so $0\leq Z \leq m^\ell-1$.

If $Z\geq qn$, then re-roll.  Otherwise, assume $0\leq Z\leq qn-1$.  Then divide $Z$ by $q$, and record the quotient (discard the remainder).  That new quotient is the option we select.

Let $p = \frac{r}{m^\ell}$ be the probability of re-rolling, so $0\leq p < 1$.  The expected number of die rolls is
$$\begin{eqnarray*}
E & =& (1-p)(\ell) + p(1-p)(2\ell) + p^2(1-p)(3\ell) + \ldots\\
& =&  \ell (1-p) \left( \sum_{k=1}^\infty k p^{k-1} \right)\\
& =& \ell (1-p) \frac{d}{dp} \sum_{k=1}^\infty  p^k\\
& =& \ell (1-p) \frac{d}{dp} \left(\frac{1}{1-p} - 1 \right) \\
& =& \frac{\ell}{1-p},
\end{eqnarray*}
$$
since $0\leq p < 1$.  Indeed,
$$
E = \frac{\ell}{1-\frac{r}{m^\ell}} = \ell \left(\frac{ m^\ell}{m^\ell - r} \right)  = \ell \left( \frac{qn+r}{qn}\right) = \ell\left(1 + \frac{r}{qn}\right).
$$
Since we know $r< qn$, we get $E< 2\ell$.  But there are $\ell$ die rolls per round of rolling, so the expected number of rounds is less than $2$.  (Compare with the bound of $m$ rounds in jlammy's answer.)

Also, notice that, in terms of minimizing expected number of die rolls, choosing $\ell = \left\lceil \log_m (n)\right\rceil$ may not be optimal.  Above, we have only assumed that $m^\ell \geq n$, so $\ell \geq \left\lceil \log_m(n)\right\rceil$.  We are allowed to choose larger $\ell$ if we want.

For example, suppose $m=6$ and $n=19$.  Then choosing $\ell=2$ gives us $6^2 = 36 = 1\cdot 19 + 17$ outcomes, with a $\frac{17}{36}$ chance of a re-roll.  The expected number of rolls is
$$
2 \left(  \frac{36}{19} \right) = \frac{72}{19}\approx 3.789.
$$
But with $\ell=3$, we have $m^\ell=216 =11\cdot 19 + 7$. So we have expectation
$$
3\left(\frac{216}{209} \right) = \frac{648}{209} \approx 3.100.
$$
In fact $\ell=3$ is optimal for this example.  We must choose $\ell\geq 2$, and notice that we always have $E\geq \ell$.  Thus since we have already found an expectation less than $4$, there is no need to try $\ell$-values that are $4$ or more.  So the only possibilities (for an optimal $\ell$-value) are $\ell=2$ and $\ell=3$, and we have seen that $\ell=3$ is better.


问:
prove that the product of 4 consecutive positive integers is divisible by 24
How can I prove that prove that the product of $4$ consecutive positive integers is divisible by $24$, ie for any positive integer $n$ : $n(n+1)(n+2)(n+3)$
is divisible by $24$.
I've noticed that:
$24$ = $2^3 * 3$

$n(n+1)(n+2)(n+3)$ is divisible by $2*2$ so by $4$ (as there are at least 2 even numbers, obvious)

$n(n+1)(n+2)(n+3)$ is divisible by $3$ (as there is the product of 3 consecutive integers, easily provable using congruences)

so I've proved that you can divide it by $12$ until now. What am I missing here?

答:
Notice that among $4$ consecutive numbers exactly one is divisible by 4 and one more divisible by 2 (say $n$ and $n+2$ or $n+1$ and $n+3$).

Notice also that among $3$ consecutive numbers exactly one is divisible by 3.

So the number is divisible by $3\cdot 8$.


问:
Can we write $a \leq a$
I am proving some inequality and one of the steps i took is that i wrote $n+1 \leq n+1$ is it correct mathematically to write such thing ?

Note :
i was try to prove that $2^{n-1} \leq n! \leq n^{n} \space \space \forall n \in \mathbb{N^*}...P_n $

so i did it by induction and i got to $2^{n-1} \leq n! \leq (n+1)^n$

then i wrote that $2 \leq n+1 \leq n+1$ and i multiplied the inequalities side by side to get to $P_{n+1}$

答:
The binary relation $\leq$ is a partial order over $\mathbb{R}$, thus is satisfies the reflexivity, i.e. :

$$\forall a \in \mathbb{R}, \; a\leq a$$

Title: Can we write $a \leq a$


答:
Consider the diagram of what it would look like. Below is the two dimensional case: from the cosine rule, we know that $$||u+v||^2=||u||^2+||v||^2-2||u||||v||\cos(x)$$
which directly implies that $$||u+v||\leq ||u||+||v||$$

which is the general triangle inequality.

We have equality if either vector is zero or their directions are the same.

Title: Is the same to add two vectors and then compute the norm of the resulting one to first compute the norm of both vectors and then adding those values?

Body:
This may be a dumb question but I pretty new in vector math and I couldn't find the answer anywhere.

What I'm asking is basically this:

$$\|\vec{v}\| + \|\vec{u}\| \stackrel{?}{=} \|\vec{v} + \vec{u}\| $$


问:
How do I find the x and y position given a latitude and longitude coordinate?
I am building an app that allows me to track the user's position inside a building. For this I use GPS and an image of the floor plan.

I have the latitude and longitude of each of the four corners and I now want to work out what the x and y position is in the image based on the GPS coordinate I receive.

I have provided an image to make this a lot clearer:

Click here - don't have enough rep yet :(

I've tried many things so far, but this one came the closest:

Find the range between top left (min lat/lon) and bottom right (max lat/lon):

lonrange = |minLon - maxLon|

latrange = |minLat - maxLat|

Calculate the scale factors for pixel per degree:

scalex = width / lonrange

scaley = height / latrange

Now get pixel positions:

x = (currentLon - minLon) * scalex

y = (currentLat - minLat) * scaly

What I've done above is kinda accurate when the lat/lon point I'm trying to convert is somewhere directly between the topLeft and bottomRight of  the image but gets wildly inaccurate the more I deviate towards the other corners.

I think it's probably because of how rotated my image is compared to the map but I don't know how to account for it (I don't even know how much it's rotated angle wise).

Any help would be appreciated!

Note: also I apologise if my tags are inaccurate! I'm not very well versed in terms of what is what in maths.

答:
Using $x$ and $y$ coordinates is a method to describe a position in a plain. However, the earth is not a plane. Therefore, you need to use a projection to project the points, described as lat-lng-values on a reference ellipsoid, to points on a map. Many different map projections are known, each of them with different properties. The first step is to choose one projection that suits your need. For example, you could use UTM projections.

After you projected your coordinates onto the map, all you have to do is to perform a rotation (which can be achieved by applying a $2\times 2$ rotation matrix to a point) and translation (which can be achieved by adding a vector).


问:
Conceptual Question About the Nature of PDE Boundary Conditions
In this $1D$ boundary value explanation, it is stated that "Taking the example of a straight line, whose slope at the boundary points is decidedly not $0$ [the animation shows a mostly-unlabeled graph of temperature as a function of time, for the sake of concreteness we'll call it the line segment $T = .8t$ from $t = 0$ to $t = L$], as soon as the clock starts, those boundary values will shift infinitesimally, such that the slope there suddenly becomes $0$, and remains that way through the remainder of the evolution. In other word, finding a function satisfying the heat equation itself is not enough; it must also the property that it's flat at each of those endpoints for all times greater than $0$."

The idea seems to be that the slope of the line at the boundaries is not equal to $0$ at time $t = 0$, but becomes $0$ immediately after time begins progressing, as a consequence of the Laplacian seeing only one neighboring point at each boundary point instead of two.

However, when an example is worked, it seems like the PDE solution satisfying the boundary condition is engineered precisely AT $t = 0$, the one value of $t$ where the boundary condition DOESN'T need to be satisfied.

Is this "flattening of the boundaries at the first nonzero instant" feature incidental to the first linked animation, or is it a general feature of boundary conditions? Is it hiding somewhere in the worked example of the heat equation?

答:
The infinitesimal boundary shift is caused by the additional assumption that is the boundary condition, namely that we require
$$ \frac{\partial u}{\partial x}(0,t) = \frac{\partial u}{\partial x}(L,t) = 0 $$
for all $t>0.$ As the video explains this is an additional constraint that comes from the model (corresponding to no heat flow on the rod ends), and something we need to specify to obtain the "correct" solution when we solve the equation.

What is happening with the example where $u(x,0) = x$ is an incompatibility between the initial and boundary conditions; if the partial derivatives of $u$ were to be continuous up to the boundary, then by the above boundary conditions we would expect that
$$ \lim_{x \to 0} \frac{\partial u}{\partial x}(x,0) = \frac{\partial u}{\partial x}(0,0) = \lim_{t \to 0} \frac{\partial u}{\partial x}(0,t) = 0 $$
and similarly as $x \to L.$ However our initial condition $u(x,0)=x$ evidently does not satisfy this, so we get this jump discontinuity in $\frac{\partial u}{\partial x}$ at the corner point - this is the instantaneous smoothing condition. In general this will always occur if the initial condition is not compatible with the boundary conditions in the above sense.

How exactly this jump occurs when solving the equation is what you will see in the next part - you will write the solution as an infinite series, where it will not be a-priori obvious if this converges (i.e. if the infinite sum evaluates to a reasonable limit), or if you can differentiate it. It will turn out this subtlety with the discontinuous derivative is caused by delicate issues relating to differentiating this infinite sum of functions - however from the above discussion you see that this is inevitable due to the incompatibility.

As for the worked example, the family of solutions
$$ u_n(x,t) = \cos\left(\frac{n\pi}L x\right) \exp\left(-\frac{\alpha n^2\pi^2}{L^2}t\right) $$
can be derived from a general technique known as separation of variables, where we seek solutions of the form $u(x,t) = X(x)T(t).$ We then require that $u$ solves the heat equation subject to the boundary condition, which in this case reduces to requiring $X'(0)=X'(L)=0.$ Hence any solution of the above form must necessarily satisfy the boundary condition for all $t \geq 0.$

It turns out that the above $u_n$ are the only solutions that can be written as a product $X(x)T(t),$ and this is also why we can obtain series solutions using these, but the full story requires some more involved theory.


问:
How do you express this as a product of two 1-forms?
$dx\wedge dy+3\,dz\wedge dy+4\,dx\wedge dz$

I tried to do this and I reached

$dx\wedge (dy+4\,dz)+3\,dz\wedge dy$

I do not know how to proceed from here. Any hints will be appreciated.

答:
You can always do this the "stupid" way which is guaranteed to work. A short calculation shows that
$$ (a dx+b dy+cdz)\wedge (a'dx+b'dy+c'dz)=(ab'-a'b)dx\wedge dy+(ac'-a'c)dx\wedge dz+(bc'-b'c)dy\wedge dz$$
One solution set to these equations is $a=1,b=-\frac{3}{4},c=0, a'=0, b'=1, c'=4$. Indeed,
$$ (dx-\frac{3}{4}dy)\wedge (dy+4dz)$$
works.


问:
Finding the angle between $x-5=0$ and $2x+2y-7=0$
I have 2 line equations: $x-5=0$ and $2x+2y-7=0$

I believe the angle is $45^\circ$, but I don't know how to calculate the second equation when there's a $y$ as well.

Thanks!

答:
A more fundamental/visual approach to solving this problem is as follows:

Recognize that there will be a point of intersection between the two lines. We know this must occur at $(x,y)=(5,\frac{3}{2})$, as concluded via substitution:

$$x = 5$$
$$\Rightarrow 2(5) + 2y - 7 = 2y + 3 = 0 $$
$$\Rightarrow y = -\frac{3}{2} $$

Now, we need side lengths of a triangle in order to calculate the angle we desire. We know that the $y$-component will be from the line $x-5=0$. In other words, $3/2$ is the length of the triangle's vertical.

What about the horizontal? If we take $x=5$, then our angle measurement would be towards the origin. However, we want our angle to be towards the line described by $2x+2y-7=0$. To do this, we need to know this line's intersection with the x-axis, which (by substituting $y=0$) must be at: $$(x,y)=(\frac{7}{2},0)$$ The final $x$-component/horizontal is thus: $$\Delta x=5-\frac{7}{2}=\frac{3}{2}$$ To help you understand why, I graphed this on Desmos. In this image, the purple line represents the point of 1intersection. If you don't calculate $\Delta x$ and use $x=5$, you can see that you'll get the wrong angle.

Since the horizontal and vertical are both $3/2$, and we have a right-angle triangle since we're working off the axes, we conclude the angle measured from $x=5$ to $2x+2y-7=0$ must be $45^\circ$, or: $$\theta=atan(\frac{3/2}{3/2})=atan(1)=45^\circ$$


问:
Probability of withdrawing a card with or without replacement
The traditional deck features suits in red (diamonds and hearts) and black (clubs and
spades), what is the approximate probability

a. that the card drawn is a seven or a red suit card?

b. of drawing a seven, returning it to the deck and right after pulling a red suit card?

c. of drawing a seven and then a red card without replacing the first card?

a. The probability is given by $\dfrac{4}{52} + \dfrac{26}{52} - \dfrac{2}{52} = 0,53$

b. $\dfrac{1}{13} \cdot \dfrac{1}{2} = \dfrac{1}{26}$

Are my calculations in a and b correct? I can't see any difference between calculating b and c.

答:
b & c are different scenarios, but you get the same answer, still. b states that you need to find the probability of 'drawing a seven, returning it to the deck and right after pulling a red suit card?'. That means you need to find the probability of drawing a 7 from 52 cards, which is $\frac4{52}$. Now, you need to put the card you drew back into the pile, leaving it with 52 cards again. Now find the probability of drawing a red card. There are 2 colors, red and black, each with equal probability, so this is equal to $\frac12$. Now multiply the probabilities you got. $\frac4{52}$ * $\frac12$ = $\frac1{26}$. So that is the answer to b.

c is very similar, but after you draw the seven, you don't put the 7 back in the pile, so there are only 51 cards left. Now we split this up into 2 cases. If the 7 is red, there is a $\frac2{52}$ chance to draw it. Since we drew a red seven, there are 25 reds to draw out of the remaining 51 cards, so the chance of drawing a red card after drawing a red 7 is $\frac{25}{51}$. Now multiply the probabilities $\frac2{52}$ * $\frac{25}{51}$ = $\frac{25}{1326}$. The second case is that the 7 we drew was black, a chance of $\frac2{52}$. This leaves 26 red cards out of 51 left to draw, so the chance of drawing a red is now $\frac{26}{51}$. Multiply the probabilities to get $\frac2{52}$ * $\frac{26}{51}$ = $\frac1{51}$. Now add the probabilities of both cases to get $\frac{25}{1326}$ + $\frac1{51}$ = $\frac{1}{26}$.


问:
Conversion of second order ode into integral equation
The second order differential equation is
$$-\phi''(x,\psi)+g(x)\phi(x,\psi)=\lambda^2\phi(x,\psi) $$
where $\lambda\in\mathbb{C}$ and $x>0$. with conditions $$\phi(0,\psi)=0,\;\;\phi'(0,\psi)=1$$ can be converted into following Voltera equation
$$\phi(x,\psi)=\lambda^{-1}\sin(\lambda x)+\int_0^x\lambda^{-1}\sin(\lambda(x-y)g(y)\phi(y,\psi))dy$$
I don't know how this ode is converted in the above equation: any help will be highly appreciated. In particular I don't know how trigonometry involved and how to tackle $g(x)$, while converting. I don't know much about integral equations. Thanks in advance.

答:
I will consider the simplified case for the IVP $$u''(t)+\lambda^2u(t)=f(t,u(t)),\qquad u(0)=0,\quad u'(0)=1, \qquad\lambda>0.$$ You can apply these methods to your equation, I hope it helps.

1st way:
Apply the Laplace transform to this equation, $$s^2\mathcal{L}\{u(t)\}(s)-su(0)-u'(0)+\lambda^2\mathcal{L}\{u(t)\}(s)=\mathcal{L}\{f(t,u(t)\}(s)$$ then rearrange the equality, $$\mathcal{L}\{u(t)\}(s)=\frac{1}{\lambda}\frac{\lambda}{s^2+\lambda^2}+\frac{1}{\lambda}\frac{\lambda}{s^2+\lambda^2}\mathcal{L}\{f(t,u(t)\}(s),$$ finally apply inverse Laplace transform and use the convolution operation.

2nd way:
Multiply the equation with $\sin\lambda(t-s)$ and integrate, use integration by parts and do some calculation (as mentioned in this question).

3rd way:
Use variations of parameters formula see [F. Brauer, J. A. Nohel, Qualitaive Theory of Differential Equations: An Introduction, Dover Publ., pp:110].


问:
Efficient algorithm to find the permutation that reduces distance between two lists
Two lists $A$ and $B$ contain equal number of elements. List $A$ is in ascending order. List $B$ is randomly ordered.

$$
A = \{a_1, ..., a_N\}, a_i \leq a_{i+1}
$$
$$
B = \{b_1, ..., b_N\}
$$

What is an efficient algorithm to find the permutation of $B$ for which $|B-A|_1$ or $|B-A|_2$ is minimum?

If the solution to the above problem is simply the ascending order of $B$, then it is trivial to use any sorting algorithm to find the solution. But I am stuck at establishing whether it is true for lists of any arbitrary length.

答:
Longer than a comment. In this question  it was about the $l^1$ distance.
The Hungarian algorithm was suggested. I was thinking that naively one could try to minimize
$$\sum c_{ij} a_{ij}$$
where $c_{ij}$ positive are the given costs and $(a_{ij})$ vary over the doubly-stochastic matrices. Then we could do linear programming and get a solution at an extreme point of the polytope of doubly stochastic matrices, which will be a permutation matrix. But it seems the H alg beats this naive approach.

$\bf{Added:}$ As Rob Pratt indicated, for $l^2$ the best matching is order preserving. Same for $l^1$, I sketched a proof in the comments to my posting. Interesting for other $l^q$ norms.


问:
prove $\prod_{i=0}^{n}x-\frac{i}{n}\le\frac{n!}{4n^{n+1}}$ for any $x\in[0,1]$ and $n>2$
prove $\prod_{i=0}^{n}x-\frac{i}{n}\le\frac{n!}{4n^{n+1}}$ for any $x\in[0,1]$ and natural $n>2$.

I tried using AM-GM, but found nothing interesting.

I was looking also for a Taylor series. no luck

答:
Let $(\sigma_0,\ldots,\sigma_n)$ be a permutation of $(0,\ldots,n)$ such that $$|x-\tfrac{\sigma_0}{n}| \le |x-\tfrac{\sigma_1}{n}| \le \cdots \le |x-\tfrac{\sigma_{n-1}}{n}| \le |x-\tfrac{\sigma_n}{n}|.$$ In other words, of the points $\{\tfrac{0}{n},\tfrac{1}{n},\ldots,\tfrac{n-1}{n},\tfrac{n}{n}\}$, $\tfrac{\sigma_0}{n}$ is the closest to $x$, $\tfrac{\sigma_1}{n}$ is the 2nd closest to $x$, ..., and $\tfrac{\sigma_n}{n}$ is the farthest from $x$, with ties broken arbitrarily.

If $x = \tfrac{k}{n}$ for some $k = 0,\ldots,n$, then $\prod_{i = 0}^{n}|x-\tfrac{i}{n}| = 0 \le \tfrac{n!}{4n^{n+1}}$ holds trivially.

Now, suppose $x \in (\tfrac{k-1}{n},\tfrac{k}{n})$ for some $k = 1,\ldots,n$. Then $\{\sigma_0,\sigma_1\} = \{k-1,k\}$, and thus, $$\left|x-\tfrac{\sigma_0}{n}\right| \cdot \left|x-\tfrac{\sigma_1}{n}\right| = (x-\tfrac{k-1}{n})(\tfrac{k}{n}-x) \le \left(\dfrac{(x-\tfrac{k-1}{n})+(\tfrac{k}{n}-x)}{2}\right)^2 = \dfrac{1}{4n^2}.$$

Also, for $i = 2,\ldots,n$, we have $|x-\tfrac{\sigma_i}{n}| \le \tfrac{i}{n}$ since at least $i+1$ of the points $\{\tfrac{0}{n},\tfrac{1}{n},\ldots,\tfrac{n-1}{n},\tfrac{n}{n}\}$ are within $\tfrac{i}{n}$ of the point $x$. Hence, $$\prod_{i = 0}^{n}|x-\tfrac{i}{n}| = \prod_{i = 0}^{n}|x-\tfrac{\sigma_i}{n}| = \left|x-\tfrac{\sigma_0}{n}\right| \cdot \left|x-\tfrac{\sigma_1}{n}\right| \cdot \prod_{i = 2}^{n}\left|x-\tfrac{\sigma_i}{n}\right| \le \dfrac{1}{4n^2}\prod_{i = 2}^{n}\dfrac{i}{n} = \dfrac{n!}{4n^{n+1}}.$$


答:
Let $f(x)=\prod_{i=0}^n\left|x-\frac{i}{n}\right|$. Since $f$ is continuous on the compact interval $[0,1]$, it attains its max at some $y\in[0,1]$. As $f(x)=f(1-x)$, we may assume WLOG that $y\in[0,1/2]$.

If $y>1/n$, then
$$f\left(y-\frac{1}{n}\right)=\left(1-\frac{1}{n}-y\right)\cdot\prod_{i=0}^{n-1}\left|y-\frac{1}{n}-\frac{i}{n}\right| > y\cdot\prod_{i=1}^n\left|y-\frac{i}{n}\right|=f(y).$$
So we must have $0\leq y\leq\frac{1}{n}$. Then by AM-GM, $y\left(\frac{1}{n}-y\right)\leq\frac{1}{4n^2}$, so
$$f(y)=y\left(\frac{1}{n}-y\right)\cdot\prod_{i=2}\left(\frac{i}{n}-y\right)\leq\frac{1}{4n^2}\cdot\prod_{i=2}^n\frac{i}{n}=\frac{n!}{4n^{n+1}}.$$

Title: prove $\prod_{i=0}^{n}x-\frac{i}{n}\le\frac{n!}{4n^{n+1}}$ for any $x\in[0,1]$ and $n>2$

Body:
prove $\prod_{i=0}^{n}x-\frac{i}{n}\le\frac{n!}{4n^{n+1}}$ for any $x\in[0,1]$ and natural $n>2$.

I tried using AM-GM, but found nothing interesting.

I was looking also for a Taylor series. no luck


问:
Absolute value of difference of squares
I am working through the Stewart Calculus text independently and am stuck on one of the practice problems (edition 7e - problem 29 in section 1.7).

I am confused by particularly by hint #3 which says:

Why is $\lvert x^2 -4x + 4 \rvert = \lvert x + 2 \rvert \lvert x - 2 \rvert$

I am quite confused by this and I cannot figure out how these two sides of the equation are equal. For full context of the homework hints please see here (I would have attached the image but I cannot since I have not posted enough).

Left to my own device I would have figured that $\lvert x^2 -4x + 4\rvert \lt  \epsilon $ where the left hand sides could be reworked as $ \lvert  (x -2)^2 \rvert \lt \epsilon $ and furthermore that $ x - 2 = \sqrt{\epsilon}$

Not sure where I am going wrong here, can somebody please help?

答:
Solution

Suppose that $|x - 2| \leq \delta$. Then one has that
\begin{align*}
|f(x) - f(2)| & = |x^{2} - 4x + 5 - 1|\\\\
& = |x^{2} - 4x + 4|\\\\
& = |x-2|^{2} \leq \delta^{2} = \varepsilon
\end{align*}

Thus we can conclude that for every $\varepsilon > 0$, there corresponds $\delta = \sqrt{\varepsilon}$ such that
\begin{align*}
|x - 2| \leq \delta \Rightarrow |f(x) - f(2)| \leq \varepsilon
\end{align*}

and we are done.

If you still have any doubts, please let me know.

Comments

The proposed identity is not correct. Indeed, $|x^{2} - 4x + 4| \neq |x-2||x+2|$.

Title: Absolute value of difference of squares


答:
$${n \choose k}$$ means "for a set of size $n$, how many subsets have size $k$" ?  For example, the set $\{a, b, c, d\}$ of size 4 has 6 subsets of size 2: $\{a, b\}, \{a, c\}, \{a, d\}, \{b, c\}, \{b, d\}, \{c, d\}$
, so
$${4 \choose 2} = 6$$

$$\begin{align}
& \binom n 0 + \binom n 1 + \binom n 2 + \cdots + \binom n n = \\[8pt]
& \text{(A)}\,\,\, 2^{n-1} \qquad \text{(B)} \,\,\, {}^{2n}C_n \qquad \text{(C)} \,\,\,2^n \qquad \text{(D)}\,\,\,2^{n+1}
\end{align}$$

This is asking "how many subsets of size 0 plus subsets of size 1 plus...." which can become the question "how many subsets are there (from a set of size n)?"

Title: What does "$\binom n k$" mean?


答: Regarding your question, I think there was a typo in that hint #3; that should be $$|x^2-4x+4| = |x\color{red}-2||x-2|$$ instead.

Title: Absolute value of difference of squares

Body:
I am working through the Stewart Calculus text independently and am stuck on one of the practice problems (edition 7e - problem 29 in section 1.7).

I am confused by particularly by hint #3 which says:

Why is $\lvert x^2 -4x + 4 \rvert = \lvert x + 2 \rvert \lvert x - 2 \rvert$

I am quite confused by this and I cannot figure out how these two sides of the equation are equal. For full context of the homework hints please see here (I would have attached the image but I cannot since I have not posted enough).

Left to my own device I would have figured that $\lvert x^2 -4x + 4\rvert \lt  \epsilon $ where the left hand sides could be reworked as $ \lvert  (x -2)^2 \rvert \lt \epsilon $ and furthermore that $ x - 2 = \sqrt{\epsilon}$

Not sure where I am going wrong here, can somebody please help?


问:
Separation of variables on a second order ODE
I am attempting to grasp the basics of separation of variables for a second order separable differential equation, and am failing to do so:

Given the equation:

$$ x=\frac{d^2y}{dx^2}$$

I know from calculus class intuitively that the solution is:

$$ y=\frac{x^3}{6}$$

But if I am to use separation of variables, why don't I get:

$$ \frac{y^2}{2}=\frac{x^3}{6} $$

Since I should have had to integrate the $y$ side twice as well.

答:
You have treated $\frac{d^2y}{dx^2}$ as $\frac{d^2y}{dx^2} = (\frac{dy}{dx})^2 = \frac{dydy}{dxdx}$. This of course isn't true. The second derivative of $y$ wrt $x$ is $\frac{d}{dx}(\frac{dy}{dx})$ which if you want to use the fraction analogy can be thought of as $\frac{dy}{dxdx}$. Now, using separation of variable on the equation you've supplied would yield $$\iint  xdxdx = \int  dy$$

Solving this would produce the correct solution. This isn't a mathematically rigorous explanation, but it works to show the difference between $\frac{d^2y}{dx^2}$ and the first derivative squared.


问:
An unbounded sequence s.t. $\forall t>0 |[a_nt]-a_nt|\rightarrow 0$
While studying probability I accidentally proved the result below by what I think is an overly complicated proof (a sketch of this proof below). Is there an easy way to prove it?

The result is the following.

Let $\{a_n\}_{n\in\mathbb{N}}$ be a sequence over the reals such that for all $t>0$ we have $\lim_{n} |[a_nt]-a_nt|=0$, where we denote with $[x]$ the nearest integer to x.

Then $\{a_n\}_{n\in\mathbb{N}}$ converges to $0$.

This is equivalent to another result.

Let $\{a_n\}_{n\in\mathbb{N}}$ be a sequence over the reals and $a$ a real number such that for all $t>0$ we have $\lim_{n} \text{e}^{\text{i}a_nt}=e^{\text{i}at}$

Then $\{a_n\}_{n\in\mathbb{N}}$ converges.

Proof. Let us suppose that $\{a_n\}_{n\in\mathbb{N}}$ doesn't converge. We will study two possible scenarios, both leading to contradiction.

First case. The sequence $\{a_n\}_{n\in\mathbb{N}}$ is bounded.
Then, choosing a small enough $t$, we get that $\forall n |a_nt|<\pi$, and we conclude, using the continuity of the complex logarithm, that the sequence converges to $a+2k\pi$ for some $k\in\mathbb{Z}$.

Second case. The sequence $\{a_n\}_{n\in\mathbb{N}}$ is unbounded. Then, we consider a sequence $\{X_n\}_{n\in\mathbb{N}}$ of random variables such that $\forall n X_n\sim N(a_n,1)$ and their respective characteristic functions $\{\text{exp}(\text{i}a_nt-t^2/2)\}_{n\in\mathbb{N}}$. Our initial assumptions imply that the sequence of characteristic functions is pointwise convergent to $\text{exp}(\text{i}at-t^2/2)$, so, by Paul-Levy's theorem, the sequence of random variables $\{X_n\}_{n\in\mathbb{N}}$ converges in distribution to a random variable $X$ such that $X\sim N(a,1)$. However, it can be proved via Chebishev inequality that if the sequence of numbers $\{a_n\}_{n\in\mathbb{N}}$ is unbounded then the sequence of random variables doesn't converge in distribution to any random variable.

答:
Notation: I use $d(r, \Bbb Z)$ for the (non-negative) distance from $r\in \Bbb R$ to the nearest integer.

We show that $[1]$ if there were a counter-example, that is, if $d(ta_n, \Bbb Z)\to 0$ for all $t>0$ but $\neg (a_n\to 0)$ then there is a strictly monotonic sequence $(z_n)_n$ of positive integers such that $d(tz_n,\Bbb Z)\to 0$ for all $t>0.$

We also show that $[2]$ if $(z_n)_n$ is a strictly monotonic sequence of positive integers then there exists $t>0$ such that $d(tz_n,\Bbb z)\in [1/4,3/4]$ for infinitely many $n.$

Hence no counter-example to the main Q can exist.

Proof of $[1]$. Suppose $a=(a_n)_n$ is a counter-example. Let $b=(b_n)_n$ be a sub-sequence of $a$ with $\inf_n|b_n|>0.$

Now b cannot have any convergent sub-sequence $c=(c_n)_n$ with a limit $l. $ Otherwise, for any $t>0$ we would have $tc_n\to tl,$ but $d(tc_n,\Bbb Z)\to 0$, so $tl\in \Bbb Z,$ but $l\ne 0$ (as $|l|\ge \inf_n|b_n|>0$), so every $t\in \Bbb R^+$ would belong to the countable set $\{z/l:z\in \Bbb Z\},$ which is absurd.

So $b$ has a strictly monotonic sub-sequence $d=(d_n)_n$ diverging to $\infty$ or to $-\infty$ such that $|[d_{n+1}]|-|[d_n]|$ is strictly increasing. And $d$ has a sub-sequence $e=(e_n)_n$ such that $d(e_n,\Bbb Z)$ converges to a limit $u.$

Let $e_n=[e_n]+u+\delta_n,$ with $\delta_n\to 0.$ Finally, let $z_n=[e_{n+1}]-[e_n].$ We have $$tz_n=t([e_{n+1}]-[e_n])=$$ $$=t(\,(e_{n+1}-u-\delta_{n+1})-(e_n-u-\delta_n)\,)=$$ $$=(te_{n+1}-[te_{n+1}])-(te_n-[te_n])-t(\delta_{n+1}-\delta_n)+([te_{n+1}]-[te_n]).$$ From this, since $([te_{n+1}]-[te_n])\in\Bbb Z,$ we have $$ d(tz_n,\Bbb Z)\le   |te_{n+1}-[te_{n+1}]|+|te_n-[te_n]|+|t|\cdot |\delta_{n+1}|+|t|\cdot |\delta_n|,$$ which $\to 0$ as $n\to\infty.$

Proof of $[2]$.Let $(z_n)_n$ be a strictly increasing sequence in $\Bbb N$. We obtain $t>0$  such that $d(tz_n,\Bbb N)\in [1/4,3/4]$ for infinitely many $n$ by an approximating sequence $(t_n). $

The idea is that

(i) $\,(t_n)$ is a positive increasing sequence with limit $t$, and

(ii) $\,f:\Bbb N\to \Bbb N$ is strictly increasing, and

(iii) for each $n$ and for $1\le j\le n$ we have $1/4< t_n\cdot z_{f(j)}-v_{n,j}<3/4$ where each $v_{n,j}\in \Bbb N_0$.

If we can do this then we will have $d(tz_{f(n)},\Bbb Z)\in [1/4,3/4]$ for every $n$.

So let $f(1)=1$ and let $t_1=1/2z_1$ .  Inductively on $n$, suppose  $1/4< t_n\cdot z_{f(j)}-v_{n,j}<3/4$ when $1\le j\le n$ with each  $v_{n,j}\in \Bbb N_0.$ We find $t_{n+1}$ and $f(n+1)$  as follows:

(a) If there exists $m>f(n)$ such that $d(t_nz_m,\Bbb Z)\in (1/4,3/4)$ then let this $m=f(n+1)$ and let $t_{n+1}=t_n.$

(b) If (a) does not apply, but if there are infinitely many $m>f(n)$ such that $t_nz_m=k_{n,m}+p_{n,m}$ with $k_{n,m}\in \Bbb N_0$ and $p_{n,m}\in [0,1/4]:$ Take such an $m,$ large enough that $1/(3z_m)<2^{-n}$ and also large enough that $(t_n\cdot z_{f(j)}-v_{n,j})+1/(3z_m)<3/4$ for $1\le j\le n.$ Now let $ f(n+1)=m$ and let $t_{n+1}=t_n+1/(3z_m)$.

(c) If (a) and (b) do not apply then there are infinitely many $m>f(n)$ such that $t_nz_m=k_{n,m}+p_{n,m}$ with $k_{n,m}\in \Bbb N_0$ and $p_{n,m}\in [3/4,1).$ Take such an $m,$ large enough that $7/(12z_m)<2^{-n}$ and also large enough that $(t_n\cdot z_{f(j)}-v_{n,j})+7/(12z_m)<3/4 $ for $1\le j\le n.$ Now let $ f(n+1)=m$ and let $t_{n+1}=t_n+7/(12z_m)$.


答:
Set up the problem to find the unknown coefficients $a,b,c,p,q,r$ such that:

$$\begin{array}{rcll}(adx+bdy+cdz)\wedge(pdx+qdy+rdz)&=&&(aq-bp)(dx\wedge dy)\\&&+&(br-cq)(dy\wedge dz)\\&&+&(cp-ar)(dz\wedge dx)\\&=&&dx\wedge dy+3dz\wedge dy+4dx\wedge dz\end{array}$$

so we are trying to find $a,b,c,p,q,r$ such that:

$$\begin{array}{rcr}aq&-bp&&=&1\\&br&-cq&=&-3\\-ar&&+cp&=&-4\end{array}$$

The rest is just a routine system of linear equations, say, in unknowns $a,b,c$ with parameters $p,q,r$. Distinguish the cases and do the elimination until you find some (or all) solutions.

Title: How do you express this as a product of two 1-forms?

Body:
$dx\wedge dy+3\,dz\wedge dy+4\,dx\wedge dz$

I tried to do this and I reached

$dx\wedge (dy+4\,dz)+3\,dz\wedge dy$

I do not know how to proceed from here. Any hints will be appreciated.


问:
Are $S^1$ and $\mathbb{R}/{\sim}$ the "same thing"?
I am reading quotient space of topology and I am a little bit confused. I am looking at the relationship between $\mathbb{R},S^1$ and the quotient space $\mathbb{R}/{\sim}$, where the relation $\sim$ corresponds to the partition $\mathbb{R}=\mathbb{Z}\cup(\mathbb{R}-\mathbb{Z})$.

Function $f:\mathbb{R}\to S^1$ give by $f(t)=(\cos(2\pi t),\sin(2\pi t))$ is continuous, onto but not one-to-one. We can also define function $g:\mathbb{R}\to \mathbb{R}/{\sim}$ where $g(t)=[t]$.

The circle $S^1$ and the quotient space $\mathbb{R}/{\sim}$ is not the same mathematical object, I suppose. But intuitively speaking they should be the "same thing". So there should be some relationship (bijective function, I guess) between these two spaces. But I was considering the function $\pi\circ f^{-1}$ and it's not injective. Maybe I should define the equivalence relation in a different way? Like "$x\sim y$ if $x\equiv y\pmod {2\pi}$" Also, suppose we can find such a function, can it be a homeomorphism?

答:
We could also show that $f\colon\mathbb R\to S^1$ given by $$f(x)=e^{ix}$$ is an open map. To check this, it is enough to check that image of any open interval $(a,b)\subseteq\mathbb R$ is an open set in $S^1$. (If image of every set from a basis is open, then the map is open.)

The we could check continuity of $f$, which can be done in various ways. For example, we can use $f(x)=\cos x+i\sin x$ and the fact that cosine and sine are continuous functions.

It is also clear that $f$ is surjective.

Once we get this, we can use the fact that every continuous surjective open map is a quotient map.


问:
Comparing two standard distributions and getting best outcome
Let's say I have 2 normal distributions

A - Mean: $\$1000,$ Variance: $100^2$

B - Mean: $\$1500,$ Variance: $300^2$

and I want to know the probability that I get more money selecting B than I will A. I know you take the difference of the mean and add the variances, make a new standard distribution using the difference of the means and the new variance, and then see the percentage greater than zero on that distribution.

Now let's say I get 20 random selections with A and only 10 random selections on B. How would I calculate the probability that the highest/most profitable selection would be from subset B?

What formulas would I use for this situation which can be used for any number of random of selections of either subset?

Body: Let $\bar{A}=\max\{A_1,\ldots,A_{20}\}$ and $\bar{B}=\max\{B_1,\ldots,B_{10}\}$. By independence $\Big(\bar{A},\bar{B}\Big)\sim f_{\bar{A}\bar{B}}$ where $$f_{\bar{A}\bar{B}}(a,b)=200\Big(F_{A}(a)\Big)^{19}\Big(F_{B}(b)\Big)^{9}f_{A}(a)f_{B}(b)$$ Here $A\sim N(1000,100^2)$ while $B\sim N(1500,300^2)$. You need to compute $P(\bar{B}>\bar{A})$. This equals $$\int_{-\infty}^{\infty} \int_\alpha^{\infty} f_{\bar{A}\bar{B}}(\alpha,\beta)d\beta d\alpha$$ FYI: $$f_{A}(\alpha)=\frac{1}{100\sqrt{2\pi}}e^{-\frac{1}{2}\Big(\frac{\alpha-1000}{100}\Big)^2}$$ $$f_{B}(\beta)=\frac{1}{300\sqrt{2\pi}}e^{-\frac{1}{2}\Big(\frac{\beta-1500}{300}\Big)^2}$$ $$F_{A}(\alpha)=\int_{-\infty}^{\alpha}f_{A}(x)dx=\frac{1}{2}+\frac{1}{2}\text{erf}\Bigg(\frac{\alpha-1000}{100\sqrt{2}}\Bigg)$$ $$F_{B}(\beta)=\int_{-\infty}^{\beta}f_{B}(x)dx=\frac{1}{2}+\frac{1}{2}\text{erf}\Bigg(\frac{\beta-1500}{300\sqrt{2}}\Bigg)$$


问:
Indefinite integral $\displaystyle \int \dfrac {\mathrm d x} {p^2 - q^2 \sinh^2 a x}$
Evaluating this integral:

$$\displaystyle \int \dfrac {\mathrm d x} {p^2 - q^2 \sinh^2 a x}$$

Spiegel's "Mathematical Handbook of Formulas and Tables" (Schaum, 1968), item $14.556$ gives:

$$\frac 1 {2 a p \sqrt {p^2 + q^2} } \ln \left({\frac {p + \sqrt {p^2 + q^2} \tanh a x} {p - \sqrt {p^2 + q^2} \tanh a x} }\right) + C$$

(As usual he glosses over negative argument to the $\ln$, but all you need to do is modulus it.)

First I multiplied top and bottom by $\operatorname {sech}^2 a x$, then used $\operatorname {sech}^2 a x = 1 - \tanh^2 a x$ for the bottom, then substituted $u = \tanh a x$ to convert it into the form:

$$\frac 1 {a (p^2 + q^2) } \int \frac {\mathrm d u} {u^2 - \left({\frac p {\sqrt {p^2 + q^2} } }\right)^2}$$

This is a standard integral, giving:

$$\frac 1 {2 a p \sqrt {p^2 + q^2} } \ln \left|{\frac {\sqrt {p^2 + q^2} u - p} {\sqrt {p^2 + q^2} u + p} }\right| + C$$

which leads to:

$$\frac 1 {2 a p \sqrt {p^2 + q^2} } \ln \left|{\frac {p - \sqrt {p^2 + q^2} \tanh a x} {p + \sqrt {p^2 + q^2} \tanh a x} }\right| + C$$

As can be seen, the fraction is upside down in the argument of the $\ln$. This means there's a negative sign gone astray somewhere. I can't see where.

答:
Wow you gave me such a hard time on this, I have never used these functions before but the mistake was simpler than I thought. This is just a simple sign mistake in the first step of your calculation.

$$\int \dfrac{\mathrm{sech}^2(ax) \mathrm{d}x}{\mathrm{sech}^2(ax) (p^2-q^2 \sinh^2(ax))} = \int \dfrac{\mathrm{sech}^2(ax) \mathrm{d}x}{p^2\mathrm{sech}^2(ax)-q^2 \tanh^2(ax)}$$

$$=\int \dfrac{(1-\tanh^2(ax)) \mathrm{d}x}{p^2(1-\tanh^2(ax))-q^2 \tanh^2(ax)}$$

$$= \int \dfrac{(1/a) \tanh'(ax) \mathrm{d}x}{p^2-(p^2+q^2) \tanh^2(ax)}$$

$$= \dfrac{1}{a(p^2+q^2)} \int \dfrac{ \tanh'(ax) \mathrm{d}x}{\left(\dfrac{p}{\sqrt{p^2+q^2}}\right)^2- \tanh^2(ax)} = - \dfrac{1}{a(p^2+q^2)}\int \dfrac{\mathrm{d}u}{u^2-\left(\dfrac{p}{\sqrt{p^2+q^2}}\right)^2}$$


问:
Compute $\int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx$
How does one compute the following integral?
$$\int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx$$

I have tried extending $x$ to the complex plane then evaluating the following contour integral
$$\oint_C \frac{\sqrt{x}e^{ix}}{1+x^2} dx$$
with the contour $C$ running along the whole real axis and then upper semicircle. I obtain
$$\int_{0}^{\infty} \frac{\sqrt{x}\big(\cos(x)+\sin(x)\big)}{1+x^2}\,\mathrm dx=\frac\pi{e\sqrt2}$$
but not the original integral.

答:
Using the Laplace transform of the sine function, we get $$ \begin{align}  \int_{0}^{\infty} \frac{\sqrt{x}\sin x}{1+x^{2}} \, \mathrm dx &= \int_{0}^{\infty} \sqrt{x} \sin(x) \int_{0}^{\infty} \sin(t) e^{-xt} \, \mathrm dt \, \mathrm dx  \\ &= \int_{0}^{\infty}  \sin(t) \int_{0}^{\infty} \sqrt{x} \sin(x) e^{-tx} \, \mathrm dx \, \mathrm dx  \\ &= -\Im\int_{0}^{\infty} \sin(t) \int_{0}^{\infty} \sqrt{x} e^{-(t+i)x} \, dx \, \mathrm dt \\ &= -\frac{\sqrt{\pi}}{2} \, \Im  \int_{0}^{\infty} \frac{\sin(t)}{(t+i)^{3/2}} \, \mathrm dt \tag{1}\\ &= - \sqrt{\pi} \,  \Im \int_{0}^{\infty} \frac{\cos (t)}{\sqrt{t+i}} \, \mathrm dt \tag{2}\\ &=- \frac{\sqrt{\pi}}{2} \,  \Im \left(\int_{0}^{\infty} \frac{e^{it}}{\sqrt{t+i}} \, \mathrm dt + \int_{0}^{\infty}\frac{e^{-it}}{\sqrt{t+i}} \, \mathrm dt \right). \end{align} $$

$(1)$ $\mathcal{L} \{\sqrt{x} \} (s) = \frac{\sqrt{\pi}}{2 s^{3/2}}$ if $\Re(s) >0$

$(2)$ Integrate by parts.

Next, using the principal branch of the square root, let's integrate the function $$f(z) = \frac{e^{iz}}{\sqrt{z+i}}, $$ around an infinitely large closed quarter-circle in the first quadrant of the complex plane.

Applying Jordan's lemma, we get

$$ \begin{align} \int_{0}^{\infty} \frac{e^{it}}{\sqrt{t+i}} \, \mathrm dt&=e^{i \pi/4} \int_{0}^{\infty} \frac{e^{-t}}{\sqrt{t+1}} \, \mathrm dt \\ &= 2 e^{i \pi/4} \, e \int_{1}^{\infty} e^{-u^{2}} \, \mathrm du \tag{3} \\ &= 2 e^{i \pi/4} e \, \frac{\sqrt{\pi}}{2} \operatorname{erfc} (1) \\ &= e^{i \pi/4} e \sqrt{\pi}\operatorname{erfc} (1). \end{align}$$

And by integrating the function $$g(z) = \frac{e^{-iz}}{\sqrt{z+i}} $$ around an infinitely large closed quarter-circle in the fourth quadrant of the complex plane (indented at $z=-i$), we get $$ \begin{align} \int_{0}^{\infty}\frac{e^{-it}}{\sqrt{t+i}} \, \mathrm dt &= \int_{0}^{1}  \frac{e^{-t}}{\sqrt{(1-t)e^{ i \pi /2}}} \, (-i \, \mathrm dt)  + \int_{1}^{\infty} \frac{e^{-t}}{\sqrt{(t-1)e^{-i \pi /2} }} \, (-i \mathrm \, dt) \\ &=  -e^{i \pi/4}\int_{0}^{1} \frac{e^{-t}}{\sqrt{1-t}} \, \mathrm dt +  e^{ - i \pi /4} \int_{1}^{\infty} \frac{e^{-t}}{\sqrt{t-1}} \, \mathrm dt \\ &= -\frac{2 e^{ i \pi /4}}{e} \int_{0}^{1} e^{v^{2}} \, \mathrm dv + \frac{2e^{ - i \pi/4}}{e} \int_{0}^{\infty} e^{-w^{2}} \, \mathrm dw \tag{4} \\ &= -\frac{2e^{ i \pi /4}}{e} \frac{\sqrt{\pi}}{2} \operatorname{erfi}(1) +  \frac{2 e^{ - i \pi /4}}{e}\frac{\sqrt{\pi}}{2}  \\ &= \frac{\sqrt{\pi}}{e} \left(-e^{ i \pi /4} \operatorname{erfi}(1) + e^{-i \pi /4} \right). \end{align}$$

$(3)$ Let $u^{2} = t+1$.

$(4)$ Let $v^{2} = 1-t$ and let $w^{2} = t-1$.

Putting everything together, we have

$$ \begin{align} \int_{0}^{\infty} \frac{\sqrt{x}\sin x}{1+x^{2}} \, \mathrm dx  &= - \frac{\sqrt{\pi}}{2} \Im \left(e^{i \pi/4} e \sqrt{\pi}\operatorname{erfc}(1) + \frac{\sqrt{\pi}}{e} \left(-e^{ i \pi /4} \operatorname{erfi}(1) + e^{-i \pi /4} \right) \right)\\ &= - \frac{\sqrt{\pi}}{2} \left(\frac{e \sqrt{\pi}}{\sqrt{2}} \operatorname{erfc}(1) - \frac{\sqrt{\pi} }{\sqrt{2}e} \operatorname{erfi}(1) - \frac{\sqrt{\pi}}{\sqrt{2}e} \right)  \\ &= \frac{\pi}{2 \sqrt{2}e} \left(1+ \operatorname{\color{red}{erfi}}(1) - e^{2} \operatorname{erfc}(1) \right). \end{align}$$


问:
Understanding weighted inner product and weighted norms
I am reading  this book where at page 27 following definitions about weighted inner product and weighted  norms are given.

Let $M$ and $N$ be Hermitian positive definite matrices of order $m$ and $n$ respectively. The weighted inner products in $\mathbb{C}^{m}$ and $\mathbb{C}^{n}$ are

$(x, y)_{M} = y^{*}Mx$ , $x, y \in \mathbb{C}^{m}$ and $(x, y)_{N} = y^{*}Nx$ , $x, y \in \mathbb{C}^{n}$ ....$(1)$

The definitions of weighted vector norms are

$\|x\|_{M} = (x, x)^\frac{1}{2}_{M} = (x^{*}Mx)^\frac{1}{2} = \|M^\frac{1}{2} x\|_{2}$, $x\in\mathbb{C}^{m}$ ....$(2)$

$\|x\|_{N} = (x, x)^\frac{1}{2}_{N} = (x^{*}Nx)^\frac{1}{2} = \|N^\frac{1}{2} x\|_{2}$, $x\in\mathbb{C}^{n}$ ....$(3)$

The definitions of weighted matrix norm are

$\|A\|_{MN} = \max_{\|x\|_{N} = 1}{\|Ax\|_{M}},\;  x \in\mathbb{C}^n and ~~A\in \mathbb{C}^{m\times n}$

$\|B\|_{NM} = \max_{\|x\|_{M} = 1}{\|Bx\|_{N}},\;  x \in\mathbb{C}^n and ~~B\in \mathbb{C}^{n\times m}$

Such a norm is sometimes called an operator norm subordinate to vector norm. It is easy to verify that

$\|A\|_{MN} =  \|M^\frac{1}{2} A  N^\frac{-1}{2} \|_{2}$ ....$(4)$

$\|B\|_{NM} =  \|N^\frac{1}{2} B  M^\frac{-1}{2} \|_{2}$ ....$(5)$

Could anybody explain me about the significance of weighted norms? Why we need weighted norm? In $(2)$ how we got $\|M^\frac{1}{2} x\|_{2}$? How could we find square root of matrix $M$?  How did we got equation $(4)$ and $(5)$.

I would be very much thankful for the help and suggestions.

答:
Weighted norms have a variety of uses. Suppose you're measuring the size of vectors that are coming out of some random or physical process, and they look like this:
$$
\begin{bmatrix}
+5.4\times 10^{-10} \\
-1.3\times 10^{+6} \\
\end{bmatrix}
\begin{bmatrix}
+1.8\times 10^{-9} \\
-4.3\times 10^{+5} \\
\end{bmatrix}
\begin{bmatrix}
-2.3\times 10^{-9} \\
+3.4\times 10^{+5} \\
\end{bmatrix}
\begin{bmatrix}
+8.6\times 10^{-10} \\
+3.6\times 10^{+6} \\
\end{bmatrix}
\begin{bmatrix}
-3.2\times 10^{-10} \\
+2.7\times 10^{+6} \\
\end{bmatrix}
$$
Would it make sense to use the standard Euclidean norm $\|\cdot\|_2$ to measure the size of these vectors? I say no. The values of $x_1$ hover around $10^{-9}$, $x_2$ around $10^6$. Since $x_1$ is so much smaller than $x_2$, $\|x\|_2\approx |x_2|$. You're losing information about $x_1$ with this measurement.

What you might choose to do in this circumstance is select a diagonally weighted norm $\|x\|_D\triangleq\sqrt{x^*Dx}$, with the values of $D_{ii}>0$ chosen to "normalize" each entry. For instance, I might choose $D_{11}=10^{18}$ and $D_{22}=10^{-12}$. The values of $D^{1/2} x$ are
$$
\begin{bmatrix}
+0.54 \\
-1.3
\end{bmatrix}
\begin{bmatrix}
+1.8 \\
-0.43
\end{bmatrix}
\begin{bmatrix}
-2.3 \\
+0.34
\end{bmatrix}
\begin{bmatrix}
+0.86 \\
+3.6
\end{bmatrix}
\begin{bmatrix}
-0.32 \\
+2.7
\end{bmatrix}
$$
Now small relative changes in $x_1$ will have approximately the same impact on the norm $\|x\|_D=\sqrt{x^*Dx}=\|D^{1/2}x\|_2$ as small relative changes in $x_2$. This is probably a more informative norm for this set of vectors than a standard Euclidean norm.

Diagonally weighted norms are probably the easiest to justify intuitively, but in fact more general weighted norms have their uses. For instance, they come up often in proofs about  Newton's method.

For information about matrix square roots, Wikipedia really is not a bad place to start, or any reasonably good linear algebra text. Square roots exist for any Hermitian positive semidefinite matrix---that is, any Hermitian matrix with nonnegative real eigenvalues.

Two types of square roots are typically considered for a real symmetric/complex Hermitian PSD matrix $M$. The lower triangular Cholesky factor $L$ satisfying $M=LL^*$ is simpler to compute in practice. But the symmetric/Hermitian square root $Q=M^{1/2}$ satisfying $M=Q^2$ is often preferred in proofs, because then you don't have to keep track of transposes, and because sometimes it is helpful for $Q$ and $M$ to share eigenvectors.

With the symmetric square root defined, the derivations for (2) are straightforward:
$$\|M^{1/2}x\|_2 = \left(x^*M^{*/2}M^{1/2}x\right)^{1/2} =
\left(x^*M^{1/2}M^{1/2}x\right)^{1/2} =
\left(x^*Mx\right)^{1/2} = \|x\|_M.$$
Here is a derivation for (4). First, we convert the numerator:
$$\|M^{1/2}AN^{-1/2}\|_2 = \max_{\|x\|_2=1} \|M^{1/2}(AN^{-1/2}x)\|_2 =
\max_{\|x\|_2=1} \|AN^{-1/2}x\|_M$$
Now we define $y=N^{-1/2} x$, or $x=N^{1/2} y$:
$$\max_{\|x\|_2=1} \|AN^{-1/2}x\|_M = \max_{\|N^{1/2} y\|_2=1} \|Ay\|_M = \max_{\|y\|_N=1}\|Ay\|_M.$$


问:
Does the reduced column echelon form of a matrix $A$ have the same null space as $A$?
Row reduction never changes the null space of a matrix $A$. But can column reduction change the null space?

I do not think this is true, my attempt at a counterexample is

$$ A =
\begin{pmatrix}
1&0&1&0\\
0&1&0&1\\
\end{pmatrix}
$$

The null space of $A$ equals the span of $[1, 1, -1, -1]$. The column reduced form of $A$ is

$$
R=\begin{pmatrix}
1&0&0&0\\
0&1&0&0\\
\end{pmatrix}
$$

and the null space of $R$ is the span of $[0,0,c_1, c_2]$ where $c_1, c_2 \in \mathbf{R}$.

So $N(A) \neq N(R)$. But I may have made a mistake as these concepts are still new to me.

As a side note, I suspect that for any invertible matrix, column reduction would not change the nullspace, but I'm not sure how to prove this.

答:
Column operations can change the nullspace.

Your counterexample is good, except the nullspaces are not quite right. They should be $\text{span}\{[1,0,-1,0], [0,1,0,-1]\}$ and $\text{span}\{[0,0,1,0], [0,0,0,1]\}$ respectively. (I guess "span of $[0,0,c_1, c_2]$ where $c_1,c_2 \in \mathbb{R}$" is technically fine, although the word "span" could be dropped".)

For invertible matrices, check that each column operation does not change the fact that the columns are linearly independent.


问:
How many meetings would it take for 12 people to meet in 4 groups of 3 until they met everyone?
I have a group of 12 people that I would like to meet in four groups of three each month.  How many minimum months would it take such that each person has been in at least one group with every other person? Below is the brute force method I used to get it to seven months:

答:
You can solve this set covering problem via integer linear programming as follows.  For each of the $\binom{12}{3,3,3,3}/4!=15400$ possible groupings $g\in G$ of 12 people into 4 groups of 3, let binary decision variable $x_g$ indicate whether that grouping is used.  For each pair $(i,j)$ of people with $i<j$, let $G_{i,j} \subset G$ be the set of groupings that cover that pair.  The problem is to minimize $\sum_{g \in G} x_g$ subject to
$$\sum_{g \in G_{i,j}} x_g \ge 1 \quad \text{for all $i<j$}$$
The minimum turns out to be 7.


问:
Completion of nonstandard numbers allowed and isomorphic to non-negative, non-infinitesimal, integer hyperreals?
I recently learned about nonstandard number systems satisfying Peano's axioms (when they use only first order logic). And this gives us what is often described by "$\mathbb N$ with densely many copies of $\mathbb Z$."

So we have that $\mathbb N=\{0,1,2,3,\ldots\}$ (all of the standard natural numbers) and (probably being quite loose and informal) let us define addition with $\mathbb Z$ (the standard integers) according to successor/predecessor rules, e.g. $n+1$ is the successor of number $n$ and $n-2$ is the predecessor of the predecessor of number $n$, etc. Let $\mathbb Z_1$ be a nonstandard copy of $\mathbb Z$ that is "beyond $\mathbb N$" and $\mathbf c$ be some (privileged) nonstandard number in $\mathbb Z_1$. Now let $\mathbb Z_q$ be the nonstandard copy of $\mathbb Z$ corresponding to $q\in\mathbb Q$, i.e. that $\mathbb Z_q=\{q\mathbf{c}+z \mid z\in\mathbb Z\}$. So the full collection of nonstandard numbers is
$\mathbb N \cup \left(\bigcup_{q\in\mathbb Q_{>0}}\mathbb Z_q\right)$ with the appropriate ordering.

My questions: Can we "complete" this set of nonstandard numbers into something like $\mathbb N \cup \left(\bigcup_{r\in\mathbb R_{>0}}\mathbb Z_r\right)$? And if so, then is this completion simply a subset of the hyperreals or isomorphic to a subset (e.g. the non-negative, non-infinitesimal, integer hyperreals)?

We could just define equivalence classes of sets of nonstandard numbers or dedekind-like cuts of the set of sets of nonstandard numbers (with an order relation on the sets $\mathbb Z_q$ and whatever other structures are necessary). In this way we create a semi-infinite "continuum" of sets of nonstandard numbers where each point on the continuum corresponds to a nonstandard copy of $\mathbb Z$ (except the origin corresponding to the standard $\mathbb N$). I envision the structure being very much like the surreal numbers except that each point in $\mathbb R$ gets a copy of $\mathbb Z$ (a discrete set of "large numbers") instead of a copy of a continuum of infinitesimals.

The structure of such a completion seems like it would be similar to part of the set of hyperreal numbers (i.e. take only the non-negative hyperreals and no infinitesimals). Is the completion of the nonstandard natural numbers isomorphic to the non-negative, non-infinitesimal, integer hyperreals, i.e. $\{x \mid x\geq0, x\in\mathbb N \text{ or } x=r\omega+z \text{ for some } r\in\mathbb R_{>0}, z\in\mathbb Z\}$?

I'm mostly looking to confirm if this intuition is correct. I always welcome any insight whether prose or rigorous. Of course, let me know if I am making some grave error in anything above. I know that the notation and language above is not rigorous at all, but my hope is that someone sufficiently experienced here can understand what I am trying to get at.

答:
First of all, there are lots of nonstandard models of $\mathsf{PA}$, including uncountable ones. Of course the ordertype of an uncountable model of $\mathsf{PA}$ can't be $\mathbb{N}+\mathbb{Z}\cdot\mathbb{Q}$ just for cardinality reasons.$^1$

The following is true however:

If $M\models\mathsf{PA}$ is countable and nonstandard, then the ordertype of $M$ is $\mathbb{N}+\mathbb{Z}\cdot \mathbb{Q}$.

So the "start" of your question is basically fine, you just need to restrict attention to the countable models.

Now on to the actual question. Surprisingly, it turns out that the class of possible ordertypes of uncountable models of $\mathsf{PA}$ is extremely complicated, and in particular we have:

There is no model of $\mathsf{PA}$ which has ordertype $\mathbb{N}+\mathbb{Z}\cdot\mathbb{R}$.

See e.g. here. So any completion process of the type you describe has to result in a structure quite different from what you start with, and indeed the argument shows that even very simple algebraic properties must be lost. In particular, you won't find an "algebraically nice" subset of the hyperreals (or more accurately, a hyperreal field - there is no such thing as "the" hyperreals) which will do the desired job.

$^1$Note that the ordertype alone doesn't determine the algebraic structure - we can see this e.g. by noting that the linear order $\mathbb{N}+\mathbb{Z}\cdot\mathbb{Q}$ is "computably describable" and then contrasting that with Tennenbaum's theorem. And the gap between order-theoretic and semiring-theoretic structure is exactly the obstacle to defining a "completion" process of the type you have in mind.

And in fact this points to an issue with your proposed construction of a nonstandard model of $\mathsf{PA}$. It's true that we can give a system of "names" to the elements of such a model which is relatively simple; however, that alone doesn't determine the algebraic structure (think about multiplying elements) and indeed there is no simple way to whip up an appropriate multiplication notion.


问:
$\mathcal{P}(\omega \times \omega)$ contains a copy of every countable ordinal
I am trying to understand this proof of the existence of an uncountable ordinal. I don't see why $\mathcal{P}(\omega \times \omega)$ contains a copy of every countable ordinal as it is said.

For example, what element of $\mathcal{P}(\omega \times \omega)$ would correspond to $\omega\cdot 2$ ?

答:
Let $\alpha$ be a countable ordinal, and let $f\colon\omega\to\alpha$ be a bijection. Then the relation $\{\langle m,n\rangle\mid f(m)\in f(n)\}$ is a well-ordering of $\omega$ with order type $\alpha$ (with $f$ being the isomorphism), and it is an element of $\mathcal P(\omega\times\omega)$.

If you want a particular relation which is isomorphic to $\omega\cdot2$, take the following: $$\{\langle m,n\rangle\mid (m<n\text{ and }m\equiv n\pmod 2)\lor m\text{ is odd, and }n\text{ is even}\}.$$


问:
How to prove the sum of 2 Gaussian distribution is also a Gaussian distribution using the characteristic function
Let X and Y be two $ \mathcal{N}(0, 1) $ distributions. I have to prove that for $(a,b)\in \mathbb{R}^2 $, $ aX + bY $ is equal to $\mathcal{N}(0, a^2 + b^2)$.

I'm trying to do this using the characteristic function of a Gaussian distribution.
$$ \phi_{aX + bY}(t) = \int_{\mathbb{R}}{ \mathbb{e}^{it(ax+by)}{\frac{1}{2} \mathbb{e}^{-\frac{x^2}{2}}} dx} $$

I don't really know what to do since by changing the variable I can't replace both x and y. Any sugestions ?

答:
Let $Z=aX+bY$. The characteristic function of $Z$ is:

$\phi_Z(t)=E\{e^{itZ}\}=E\{e^{it(aX+bY)}\}=E\{e^{i(at)X}e^{i(bt)Y)}\}$

EDIT (Sloppy mistake...) If X and Y are independent:

$\phi_Z(t)=E\{e^{i(at)X}\}E\{e^{i(bt)Y)}\}=\phi (at) \phi (bt)$,

where $\phi(w)=e^{-\frac{w^2}{2}}$ is the characteristic function of the normal distribution. So,

$\phi_Z(t)=e^{-\frac{1}{2}(at)^2}e^{-\frac{1}{2}(bt)^2}=e^{-\frac{1}{2}(a^2+b^2)t^2}$,

which is the characteristic function of the normal distribution $\mathcal{N}(0,a^2+b^2)$.


问:
Asymptotic expansion of the inverse of $x\mapsto x+x^{\small\sqrt2}+x^2$ near zero
This is a follow-up to my previous question "Asymptotic expansion of the inverse of $x\mapsto x+x^\phi$ near zero".

Consider a continuous real-valued monotone increasing function $f:\mathbb R^+\to\mathbb R^+$ satisfying $f\big(x+x^{\small\sqrt2}+x^2\big)=x.$

I am interested in an asymptotic expansion of $f(z)$ for $z\to0^+$ in terms of powers of $z$. I was able to find a few initial terms by manually balancing coefficients:
$$f(z)=z-z^{\small\sqrt2}+\sqrt2\;z^{\small\unicode{x202f}2\unicode{x202f}\sqrt2-1}-z^2+\mathcal O\big(z^{\small\unicode{x202f}3\unicode{x202f}\sqrt2-2}\big), \quad z\to0^+.\tag{$\diamond$}$$
Computing next terms in an ad hoc fashion quickly becomes tedious, so I am looking for a more systematic approach that would allow to obtain a general formula for the terms of this series. I expect it to be a mix of integer powers of $z$ and irrational powers involving $\sqrt2$. Also, I would like to know the radius of convergence of that series.

More generally, I am looking for a uniform approach for inverting generalized polynomials of a single variable that may contain both rational and irrational powers.

答:
A naive approach:

We have to solve ( with $y=f$)

$$y+ y^{\sqrt{2}} + y^2 = x$$

Let's take a simpler equation $y - y^2 = x$, equivalent to $y = x + F(y)$,
so should have
$$y = x + F(x+F(y)) = x+F(x+F(x+F(y)))= x+F(x+F(x+\ldots )))$$
(note that $F$ is not a linear operator, if it were, we would have the Neumann series for $(I-F)^{-1}$), still we have a fixed point problem that is solved with iterations.
Using this iteration approach we get the Taylor series for the solution  $y = \frac{1 - \sqrt{1 - 4 x}}{2}$.

We might use this approach for
$$y = x - y^{\sqrt{2}}- y^2 = x - ( x- y^{\sqrt{2}}- y^2)^{\sqrt{2}} -( x- y^{\sqrt{2}}- y^2)^2= \cdots $$

$\bf{Added:}$

The equation $y = T(y)= x - y^{\sqrt{2}} - y^2$ can be solve by successive approximations
$y_{n+1} = T(y_n)$. For instance, we can consider $y_0=x$, $y_1 = x - x^{\sqrt{2}} - x^2$, $y_2 = x - (x - x^{\sqrt{2}} - x^2)^{\sqrt{2}} - (x - x^{\sqrt{2}} - x^2)^2$, etc. One should figure out (easy) how to expand

$$( x+ \sum_{\alpha>1} a_{\alpha}x^{\alpha} )^{\beta} = x^{\beta} ( 1 + \sum_{\gamma >0} c_{\gamma} x^{\gamma})^{\beta}$$
We use the expansion
$$(1+ \sum_{\gamma} t_{\gamma})^{\beta}  = \sum_{\beta_1 < \ldots \beta_r}\sum_l \frac{\beta(\beta-1)\cdots (\beta- |l|+1)}{l!} t_{\gamma_1}^{l_1}\cdots t_{\gamma_r}^{l_r} $$
We are dealing with formal series with exponents positive and $\to \infty$.

To be continued...


问:
Can I use divergence theorem right after using Stokes' theorem?
Stokes’ theorem converts a line integral to a surface integral. Divergence theorem relates a surface integral to a triple integral. So is it possible to take the result from Stokes’ theorem, and apply the divergence theorem to it? I’m not good at using latex, else I would’ve written down what I meant.</p>

答:
Short Answer

Arguably "yes, you can", but certainly not in a useful way. Any application of the combination of the Divergence Theorem and (Kelvin–)Stokes' Theorem results in $0$ for a trivial reason.

Explanation

Using $\mathop{\rm bd}R$ to mean "the boundary of $R$", and ignoring some of the technical conditions of the theorems, there's no immediate reason why we can't chain them together to get something like the following (notations vary a lot) as long as all the conditions are satisfied:
\begin{equation}\int_{\mathop{\rm bd}(\mathop{\rm bd}R)} \mathbf F\cdot\widehat{\mathbf{r}}\,\mathrm ds=\iint_{\mathop{\rm bd}R} (\nabla\times\mathbf F)\cdot\widehat{\mathbf{n}}\,\mathrm dS=\iiint_R\nabla\cdot(\nabla\times\mathbf F)\,\mathrm dV\tag{$\star$}\end{equation}

However, calculus puts some severe restrictions on what $\nabla\cdot(\nabla\times\mathbf F)$ can be, and geometry puts some severe restrictions on what $\mathop{\rm bd}(\mathop{\rm bd}R)$ can be. Either of these restrictions alone turn out to force the triple/single integral to be $0$.

Integrand Issue

In "nice situations" where the order of partial derivatives doesn't matter, it turns out that $\nabla\cdot(\nabla\times\mathbf F)=0$. You can see an algebraic proof in this answer to Why is $\operatorname{Div}\big(\operatorname{Curl} F\big) = 0$? Intuition? and get some intuition for it among the answers to What is an intuitive explanation for $\operatorname{div} \operatorname{curl} F = 0$?.

For reasons studied in real analysis, if the triple integral makes sense, we're guaranteed to be in a "nice situation" at almost all the points we're integrating over, so the triple integral must come out to be $0$. The main details are a version of Clairaut's Theorem and a generalization of Lebesgue's criterion for Riemann integrability (see, for example, the comments on this question).

This means that if you had a triple integral that you wanted to calculate via $\star$, you'd know it was $0$ without looking at the region $R$ because the integrand would just be $0$.

Domain Issue

Just like the integrand of the triple integral forced a value of zero, it turns out that the domain of integration of the single integral forces a value of zero, too.

This means that if you had a single integral that you wanted to calculate via $\star$, you'd know it was $0$ without looking at the integrand $\mathbf F$ because of the nature of the curve(s) you're integrating over.

Geometric Intuition

If you start with a nice round 3D region $R$ for the triple integral, like a ball or  the region bounded by an ellipsoid or a bumpy variant*, then $\mathop{\rm bd}R$ would be a smooth surface with no edges. In that case, $\mathop{\rm bd}\mathop{\rm bd}R$, if it exists at all, would geometrically seem to be something trivial/without length — like the empty set, or a single point. Whatever it is, it would make the single integral equal to zero.

And if we instead have a region that's not so smooth, like a solid cube, then we could fix it by pulling in the edges and rounding them slightly. This would lose an arbitrarily tiny fraction of the volume (so the triple integral is not significantly affected), and then we'd again be in a situation where the single integral is forced to be zero.

*like those displayed in Samantha Driskill's Footenote 18 for Naala Brewer's Spring 2008 section of "Math 267 Calculus for Engineers III" at Arizona State University

Careful Consideration

We can understand this a bit more precisely by thinking about the orientation relationship between the surface normal(s) we use in the double integral, and the directions of any curves used in the single integral. For a refresher, the relationship required by (Kelvin–)Stokes' is described on Paul's Online Notes and in OpenStax Calculus Volume 3.

If we have a closed surface $\mathop{\rm bd}R$ that bounds a 3d region $R$, then there must be some of the surface on two sides of any curves making up the edges (in $\mathop{\rm bd}\mathop{\rm bd}R$) of the surface. For example, one of the $12$ edges of a cube bounds two faces. The equator of a sphere bounds two hemispheres, etc.

Note that the Divergence Theorem gives the surface an outward normal everywhere. But this forces the normal vectors for the pieces of the surface on two sides of an edge to go in "roughly opposite" directions — directions that require following the edge in opposite directions in order to satisfy the orientation condition of (Kelvin–)Stokes'. For example, in the cube below, the middle horizontal edge must be followed rightwards for the top blue face, but leftwards for the front yellow face:

What this means is that any curve pieces in the single integral must be followed in both directions (an equal number of times), so that the integrals along each directed/oriented piece of the curve cancel out because switching the direction/orientation changes the sign. Since everything cancels out because of the nature of $\mathop{\rm bd}\mathop{\rm bd}R$, the singe integral must be $0$.

Deeper Context

The equations and patterns seen here generalize to higher dimensions.

Divergence and Stokes' Theorems

These theorems (and others like the fundamental theorem for line integrals) generalize to (generalized) Stokes' Theorem and the fundamental theorem of geometric calculus.

Getting zero

Zero Integrand

That the two "derivative" operations of divergence and curl end up giving you zero is generalized in the theory of differential forms (or closely-related things like "geometric calculus"): the exterior derivative gives you $0$ when applied twice, in any dimension. This is used as the basis of things like de Rham cohomology, where we can detect topological holes in a space by noting things like "on this region, the divergence of this field is zero even though it's not the curl of another field" (see the answer to Problem understanding a solenoidal vector field that is not a curl. for an example that shows a hole at the origin of 3D space).

"Zero" curves

That the boundary of the boundary of $R$ seems to give curves that "cancel out" comes up in discussions of simplicial homology and singular homology, which can also be used to measure topological holes.

There are deep connections between these sorts of things in algebraic topology, but they're not directly related to your question.


问:
The algebraic closure of an algebraic field extension of a field is also an algebraic extension of the field

If $K\subseteq E$ is an algebraic field extension and $\overline{E}$ is the algebraic closure of $E$, then the field extension $K\subseteq\overline{E}$ is also algebraic.

This is simply stated in my notes but I don't know how to prove it. I have seen this answer by egreg but it just says "by well-known results", which I don't get.

If $a\in\overline{E}$ then it is a root of some polynomial with coefficients in $E$, and every element of $E$ is a root of a polynomial with coefficients in $K$, but I don't know how to use this last statement.

答:
Fact. Let $L/K$ be a field extension, and let $\alpha \in L$. $\alpha$ is algebraic over $K$ if and only if $K(\alpha)/K$ is a finite extension.

You should make sure you can prove this! Let me know if you're stuck.

Now let $\alpha \in \overline{E}$ be arbitrary. Then there is some polynomial $p \in E[x]$ such that $p(\alpha) = 0$. Let $p = \sum_{i=0}^n a_i x^i$. Letting $F = K(a_0, \dots, a_n)$, we see that $p \in F[x]$. Now (exercise!) $F/K$ is finite, and we have that $\alpha$ is algebraic over $F$ so $F(\alpha)/F$ is finite. Finally,
$$[K(\alpha) : K] \leq [F(\alpha) : K] = [F(\alpha) : F] [F : K]$$
shows that $K(\alpha)/K$ is finite. Thus, $\alpha$ is algebraic over $K$.