问:
Integral mean value theorem on the sphere.
Let $\mathbb{S}^d$ be an unit sphere with dimension $d$.

Do we have an integral mean value theorem on the unit sphere? That is we can find
a point $t\in\mathbb{S}^d$, such that $$f(t)=\frac{\int_{\mathbb{S}^d}f(y)d\sigma(y)}{\int_{\mathbb{S}^d} d\sigma(y)}?$$

Since we have this result on the interval case: on the interval $[a,b]$, we can find a point $t\in(a,b)$, s.t. $$\int_a^bf(x)dx=f(t)(b-a).$$ Not quite sure this also works on the sphere.

Thank you for any suggestions!

答:
If $f$ is a continuous function $S^d\to \mathbb R$, then the answer is yes, for the same reason as the usual integral mean value theorem (essentially the intermediate value theorem). Namely, the value on the right of your equation is some real number between the maximum and minimum value of $f$. Since $f$ is continuous and $S^d$ is connected (assuming $d>0$), the image of $f$ is an interval and hence contains all numbers between its maximum and minimum.

If $f$ maps to a higher-dimensional space, the answer may be no, for example the inclusion $f: S^1 \to \mathbb R^2$ has average value $0\in \mathbb R^2$ but does not attain that value.

If $f$ is not continuous, then certainly the answer is no in general.


问:
Give an example of a bounded function $f:R\to R$ that has a limit at every point except 2
Give an example of a bounded function $f:R\to R$ that has a limit at every point except 2

This question is very open ended and I'm not sure where to go with this. I know that if we have a denomator of x-2, then a limit of 2 would not work. Am I on the right track?

答:
Hello，here is a function you can consider
y=sin（1/（x-2））。</p>

ContentLicense: CC BY-SA 4.0


问:
Predator-prey dynamical system. Find the vector $x_k$ for a certain $k$

Consider a predator-pray model with the matrix $A= \begin{bmatrix}
 0.5 & 0.6\\
 -0.2 & 1.4 \end{bmatrix}$ ($A$ is the population distribution) and the initial state is $\vec{x}_0 = \begin{bmatrix} 1\\ 10 \end{bmatrix}$.
Find the vector $\vec{x}_k$ for $k = 10$.

How can I find $\vec{x}_{k=10}$? And could I find $\vec{x}_{k=11}$ from $\vec{x}_{k=10}$?

答:
The brute force way is simply to multiply it out; you need $A^{10}$, so calculate $A^2$, then $A^4$, then $A^8$, then multiply by $A^2$.

Another way is to find the eigenvalues and eigenvectors of the matrix. If you can diagonalize A, finding its powers is easy.

If $A = PDP^{-1},$ then $A^{10} = PD^{10}P^{-1}$.


问:
The Lebesgue Integeral of $f(x) = \operatorname{int}(x^{2})$ over $[0,2]$ (integer part)
I just finished working on this Lebesgue integration problem and I am having doubts regarding my work. Note that I will use the notation $\int_{a}^{b}f(x)dx$ to denote the Lebesgue integral in this problem.

$ex$: Find the Lebesgue integral of $f$ over $E$ where $f(x) := \operatorname{int}(x^{2})$ and $E = [0,2]$. Note that $m$ is the Lebesgue measure on $\mathbb{R}$ and $\operatorname{int}$ is the integer part of the real number.

Here is my work:

$$\begin{align*}
\int_{0}^{2}\operatorname{int}(x^{2})dx &= 0m([0,1])+1m([1,\sqrt2)]+4m([\sqrt2,\sqrt3]) \\
 &= \sqrt2 - 1 +4\sqrt3-4\sqrt2.  \\
\end{align*}$$

For some reason in my textbook, the solution has the integral going from $0$ to $4$ so i'm not sure if this is correct or not (since the textbook solution looks wrong).

Any feedback is welcome.

答:
$$\begin{align*}
\int_{0}^{2}\operatorname{int}(x^{2})dx &= 0m([0,1])+1m([1,\sqrt2)]+2m([\sqrt2,\sqrt3]) \\ +3(2-\sqrt 3)
 &= \sqrt2 - 1 +2\sqrt3-2\sqrt2+3(2-\sqrt 3).  \\
\end{align*}$$ Integer part of $x^{2}$ when $x$ is between $\sqrt 2 $ and $\sqrt 3$ is $2$, not $4$. (This should agree with the answer in the bbok (which uses susbtitution to evalute the integral).


问:
What is the difference between a natural transformation and a 2-morphism?
Is one notion more general than the other? Are they just completely distinct ideas? I know very little category theory, but the intuitive definition often given for natural transformations is that they are morphisms between morphisms, and that seems to be exactly what $2$-morphisms are.

If they are the same idea in some contexts, when if at all is the terminology exchangeable, and when would you use one or the other?

答: 2-categories are a generalization of categories. They do not only possess objects and morphisms between those objects, they also possess "morphisms between the morphisms", so-called 2-morphisms (all obeying various axioms, of course). A specific exaple of a 2-category is the 2-category of small categories, whose objects are categories, whose morphisms are functors between those categories and whose 2-morphisms are natural transformations between those functors. Thus, 2-morphism is the general notion and natural transformations are a specific example thereof.


问:
Issue solving a linear differential equation.
I am currently working through a chapter in my textbook, and it is on solving a form of linear differential equation. I have been trying to figure out what is happening for about 3 hours as of now.

I don't understand how $\frac{d}{dt}e^{k(t))}x = 0$, since it says that $\frac{d}{dt}e^{k(t)} = p(t)e^{k(t)}$. Surely this would mean that$\frac{d}{dt}e^{k(t)}x = x*p(t)e^{k(t)}$.

答:
Start from $e^{k(t)}\frac{dx}{dt}+p(t)e^{k(t)}x=0$

Note that $p(t)e^{k(t)}x=\frac{d}{dt}e^{k(t)}$

Therefore, by the product rule

$\frac{d}{dt}(e^{k(t)}x)=\frac{d}{dt}(e^{k(t)})*x+\frac{dx}{dt}e^{k(t)} = p(t)e^{k(t)}x+e^{k(t)}\frac{dx}{dt}=0$

Let me know if this clears up the confusion


答:
Note that
\begin{align*}
\frac{d}{dt}\left(e^{k(t)}x\right) &= e^{k(t)}\frac{dx}{dt} + \frac{d}{dt}\left(e^{k(t)}\right)x\\
&= e^{k(t)}\frac{dx}{dt} + p(t)e^{k(t)}x
\end{align*}

according to the product rule. So if we run through the lines in reverse order, we have the derivation.

Title: Issue solving a linear differential equation.

Body:
I am currently working through a chapter in my textbook, and it is on solving a form of linear differential equation. I have been trying to figure out what is happening for about 3 hours as of now.

I don't understand how $\frac{d}{dt}e^{k(t))}x = 0$, since it says that $\frac{d}{dt}e^{k(t)} = p(t)e^{k(t)}$. Surely this would mean that$\frac{d}{dt}e^{k(t)}x = x*p(t)e^{k(t)}$.


问:
Factoring ideal into prime ideals in $\mathbb{Z}[\sqrt{-5}]$
I would like to write the ideal $(9)$ as a product of prime ideals in $\mathbb{Z}[\sqrt{-5}]$, which is a Dedekind domain. We have
$$9=3 \cdot 3=(2+\sqrt{-5})\cdot (2-\sqrt{-5}) $$
and I have shown that $3,2+\sqrt{-5},2-\sqrt{-5}$ are irreducible and not prime.

I can see that $(3)$ is not a prime ideal because $(2+\sqrt{-5}) (2-\sqrt{-5})=9 \in (3)$, but $2+\sqrt{-5},2-\sqrt{-5}\notin (3)$.
I am not sure what to do next.

答:
This is exactly the situation that the Dedekind-Kummer theorem treats. This is given in the following two theorems from Keith Conrad's Factoring after Dedekind.

Theorem 1 (Dedekind). Let $K$ be a number field and $\alpha \in O_K$ such  that $K= \mathbb{Q}(\alpha)$. Let $f(T)$ be the minimal polynomial of $\alpha$ in $\mathbb{Z}[T]$. For any prime $p$ not dividing $[O_K:\mathbb{Z}[\alpha]]$, write
$$
f(T) \equiv \pi_1(T)^{e_1}\cdots \pi_g(T)^{e_g} \pmod{p}
$$
where the $\pi_i(T)$ are distinct monic irreducibles in $\mathbb{F}_p[T]$. Then $ (p) = p O_K$ factors into prime ideals as
$$
\newcommand{\p}{\mathfrak{p}}
(p) = \p_1^{e_1} \cdots \p_g^{e_g}
$$
where there is a bijection between the $\p_i$ and $\pi_i(T)$ such that $N(\p_i) = p^{\deg(\pi_i)}$. In particular, this applies for all $p$ if $O_K=\mathbb{Z}[\alpha]$.

He further describes how to obtain generators for the $\p_i$ in Theorem 8.

Theorem 8. In the notation of Theorem 1, when $\p_i$ is the prime ideal corresponding to $\pi_i(T)$ we have the formula $\p_i = (p,\Pi_i(\alpha))$ where $\Pi_i(T)$ is any polynomial in $\mathbb{Z}[T]$ that reduces mod $p$ to $\pi_i(T)$ mod $p$.

Returning to your problem, the minimal polynomial of $\sqrt{-5}$ is $T^2 + 5$. How does this factor mod $3$? What does this tell you about the primes ideals in the factorization of $(3)$ and their generators?


问:
Median and Mode of a probability density function
Let X be the random variable with probability density function

$$
f_X (x) = \begin{cases}
    \frac{3}{32} (-x^2+2x+3), & x\in(-1,3) \\
    0, & \text{otherwise} \\
\end{cases}.
$$

Let $M_X$ be the median of $X$ and $m_X$ be the mode. Determine $M_X+m_X$.

To calculate the mode of $f_X$, it is the value of x at which $f_X$ attains its maximum. In this case, it is $1$.

I know that generally to calculate the median,

$$
\int_{-\infty}^{M_X} f_X (x) = \frac{1}{2}
$$

But as we have an interval on $x$, should the lower limit be $-1$?

答:
Yes, you can integrate from $x = -1$ instead of $x = -\infty$. The relevant concept here is the support of the distribution. When working with probability densities, you can generally ignore the parts of the domain where $f_X(x) = 0$. In these portions of the domain, the CDF is unchanged, so they have no effect on probabilities.

In this particular problem, you can split up the integral used to calculate the median into two parts, only one of which is over the support of the distribution:

$$ \int_{-\infty}^{M(x)} {f_X(x) dx} = \int_{-\infty}^{-1} {f_X(x) dx} + \int_{-1}^{M(x)} {f_X(x) dx} . $$

Then, since $f_X(x) = 0$ for $x \in (-\infty, -1)$, we have

$$ \int_{-\infty}^{M(x)} {f_X(x) dx} $$
$$ = \int_{-\infty}^{-1} {0 dx} + \int_{-1}^{M(x)} {f_X(x) dx} $$
$$ = 0 + \int_{-1}^{M(x)} {f_X(x) dx} $$
$$ = \int_{-1}^{M(x)} {f_X(x) dx} $$


问:
Various proofs: pairs of points in a circle = Möbius strip
It is well-known that the space of "pairs of points in a circle" (also called the symmetric square of $S^1$) can be identified with a Möbius strip. I don't know where the idea originates, but it is a neat elementary result in topology, which has a beautiful application in Vaughan's proof of the rectangular peg problem (3blue1brown gives a really nice account of this). Inspired by a talk I saw on related results, I've been trying to compile a list of proofs that the symmetric square of $S^1$ is a Möbius strip. To be specific, the space $Sym^2(S^1)$ is constructed from $S^1\times S^1$ by identifying $(x,y)\sim (y,x)$ for every $x,y\in S^1$ and taking the quotient topology.

Hence, I am asking this question as community wiki, to try and learn of other peoples' favorite ways to prove this result. A quick ground rule: if someone beats you to the chase on your favorite proof, maybe comment to show your preference for it, rather than posting the same proof again? But if you have a proof that seems like a close variant of one already posted, feel free to post it! Since these are all about the same result, there are bound to be variations on the same idea. Also, if you know where your proof comes from, please give proper credit to its originator. If the proof is your own, be explicit about that fact, so that I can make sure to give you proper credit if I ever cite it elsewhere.

Here are summaries of some proofs that I know of, to get the ball rolling:

View the torus $S^1\times S^1$ as square with opposite sides identified. The identification $(x,y)\sim (y,x)$ corresponds to folding the square along its diagonal, so we get a triangle with gluing between two adjacent sides. A brief cut-and-paste argument turns this into the usual presentation of the Möbius strip. (I don't know where this argument originates.)

Charles Livingston: To any pair of points $x,y\in S^1$ in the unit circle of $\mathbb R^2$, we associate their perpendicular bisector, which is a line through the origin and thus an element of $\mathbb R\mathbb P^1$ (if $x=y$, we take the line through this point and the origin). This defines a map $Sym^2(S^1)\rightarrow \mathbb R\mathbb P^1$, which is a fiber bundle with the interval as its fiber. By checking that this bundle is not orientable (i.e. going around $\mathbb R\mathbb P^1$ once flips the interval), we can see that it is a Möbius strip. (This argument—and a nice picture of the previous one—are included as §2.2 of this paper by Tuffley.)

Hugh Morton (summarized on nLab): This is a case of Morton's more general description of $Sym^n(S^1)$. Consider $S^1\subset \mathbb C$ as the unit circle and define a map $\text{Sym}^2(S^1)\rightarrow S^1$ by sending a pair $(x,y)$ to their product $xy.$ As in the previous proof, we can check that this is a fiber bundle with the interval as its fiber, and that this bundle is not orientable (i.e. going around $S^1$ once flips the interval), so it is a Möbius strip.

Étienne Ghys: We can view $\mathbb R\mathbb P^2$ as the plane $\mathbb R^2$ with a copy of $\mathbb R\mathbb P^1$ at infinity. Then the Möbius strip $M$ is formed by cutting the open unit disk out of our projective plane. For any two points $p,q$ on the unit circle $S^1\subset \mathbb R^2$, we can take the tangents $\ell_p$ and $\ell_q$ to the circle at these points. The map which sends $(p,q)\mapsto \ell_p\cap \ell_q$ is a homeomorphism $Sym^2(S^1)\rightarrow M$, where the $\mathbb R\mathbb P^1$ at infinity corresponds to intersections of parallel lines. (I believe this is from “Prolongements des diffeomorphisms de la sphère,” but I am not certain.)

Karanbir Sarkaria: We can view $\mathbb R\mathbb P^n$ as the space of non-zero, homogeneous polynomials of degree $n$ in $\mathbb R[x,y]$, up to multiplication by a non-zero scalar. Then we have a map $\mathbb R\mathbb P^1\times \mathbb R\mathbb P^1\rightarrow \mathbb R\mathbb P^2$, given by $$\big((rx+sy),(ux+vy)\big)\mapsto (rx+sy)(ux+vy).$$ This descends to a map $\text{Sym}^2(\mathbb R\mathbb P^1)\rightarrow \mathbb R\mathbb P^2$, which is a homeomorphism onto its image. But $ax^2+bxy+cy^2$ lies in the image of this map if and only $b^2-4ac\geq 0$. Hence, we must cut out from $\mathbb R\mathbb P^2$ the set of quadratics with complex roots, which is homeomorphic to the open disk. But removing a disk from $\mathbb R\mathbb P^2$ yields the Möbius strip, as desired. (See Sarkaria's note for a nice illustration of this fact.)

Map a pair of points $\{p,q\}$ on the unit circle to their midpoint $(p+q)/2$. This is a bijection between the punctured disk $D^2-{0}$ and pairs $\{p,q\}$ of non-antipodal points, but every pair of antipodal points gets mapped to $0$. In other words, we can form $Sym^2(S^1)$ from $D^2$ by replacing the origin with a copy of $\mathbb R\mathbb P^1$. This blowup of the disk is precisely a Möbius strip. (Again, I don't know where this argument originates.)

The fact in question also relates to a fascinating result of Bott and Shchepin. Let $\exp_n(S^1)$ consist of all subsets of $S^1$ containing between 1 and $n$ points. This is topologized via the Hausdorff metric on compact subspaces of $S^1$. Bott showed that $\exp_3(S^1)=S^3$ ("On the third symmetric potency of $S^1$") and Schepin showed that the natural embedding $S^1=\exp_1(S^1)\hookrightarrow \exp_3(S^1)=S^3$ is a trefoil knot (unpublished). Identifying pairs of equal elements with singleton sets, we get $Sym^2(S^1)=\exp_2(S^1)$, so the Möbius band $\exp_2(S^1)$ fits into this picture as a (non-orientable) Seifert surface of the trefoil. This theorem is refined in the aforementioned article by Tuffley, and has also been proven by: Nakandakari and Tsukuda via elementary pasting techniques (they animate their construction here), Mostovoy via lattices in $\mathbb C$, and Rose via isometries of the hyperbolic plane. I am not describing these in detail because they are lengthier and I haven't taken the time to fully understand them yet, but if you know of another description of $Sym^2(S^1)$ inside $\exp_3(S^1)=S^3$, that would also be a great thing to contribute.

答:
Lemma 1: The set of lines on the plane is the open Möbius band.

Proof: The set of lines on $\rm\Bbb RP^2\cong S^2/{-1}$ is is continuous bijection with the points of $\rm\Bbb RP^2$: associate each great circle with pair of points located perpendicular to that circle. Deleting the line at infinity (the equator) gets us to the lines on the plane, through the gnomonic projection. This leaves us with $\rm\Bbb RP^2\setminus\{pt\}$, which is the open Möbius band.

(To see that the projective plane minus a point is the Möbius band, relate them both to a cylinder mod antipodes.)

Lemma 2: The set of lines on the plane that intersect the unit circle is the closed Möbius band.

Proof: This is the previous construction minus an open neighborhood of the line at infinity.

Finally, to relate this construction to the set of pairs of points on $S^1$, intersect these lines with the unit circle.

(Incidentally, the set of infinite rows of unit-spaced parallel lines (example element) is topologically a Klein bottle. We can decompose it into two Möbius bands by drawing a diameter $\frac12$ circle and asking whether a given element intersects it or not.)


问:
Can these connections be considered as permutations or combinations?

Any help would be highly appreciated.

答:
No, what you are counting is nothing like a permutation. A permutation refers to an ordered list of objects without repeats. Just because order matters, that does not make it a permutation.

The things you are counting are called matchings in the complete graph. If you do not like that name, you can make up a new one, as long as you define this new term for your audience. I think a descriptive name would be a "pairing," or a "partial pairing."

In general, if you are counting some combinatoric object you made up and do not have a name for it, you can generally refer to the things you are counting as "arrangements," and say "the number of arrangements is...".

Body: Any help would be highly appreciated.


问:
Counting numbers smaller than $N$ with exactly $k$ *distinct* prime factors
Using common notation, $\omega(n)$ is the number of distinct prime factors on $n$. Similiarly, $\Omega(n)$ is the number of prime factors of $n$, not necessarily distinct: $120=2^{3}\cdot 3 \cdot 5$ , therefore $\omega(120)=3$ , $\Omega(120) = 5$.

If we let $W_k(N)$ be the count of numbers not exceeding $N$ with exactly $k$ distinct prime factors {$i | i \leq N , \omega (i) = k$}, is there any way or formulation to calculate it? Or at least a relatively coincise algorithm for it, which does not include factoring each number and checking for $\omega$ individually?

For example, one can easily find that $W_3(100)=8$ , $W_3(1000000)=379720$ , $W_4(1000000)=208034$.

Please note here I am not looking for an asymptotic approximation but rather an exact count. And $N$ can be large, with order of magnitude of $10^{12}$ and over.

I have found that a solution exists for the $\Omega$ counterpart, known as k-Almost Primes, which admits closed-form formulas, but as for $\omega$, I have found non.

答:
I think Karl above has a good start to an algorithmic answer here. But I don't think you even look at divisibility or inclusion-exclusion. I think you actually find every value of $i$ but in a faster way, without checking every $i < N$--just enumerate the possible values of $\{i \mid \omega(i) = k\}$ via multiplication. For the sake of an algorithm, I think you treat the factorizations in their exponent array/vector form; for instance

$$1260 = 2^2\cdot3^2\cdot 5^1\cdot7^1 = [2,2,1,1] \\43472 = 2^4\cdot3^0\cdot 5^0\cdot7^0\cdot11^1\cdot13^1\cdot 17^0\cdot19^1 = [4,0,0,0,1,1,0,1] $$

I'm not an algorithm-optimization person. But here's what I'd do (for $k=4$), because I think it gets you to stop points in the shortest time.

Start with the initial array $[a_2, a_3, a_5, a_7] = [1,1,1,1] = 2\cdot3\cdot5\cdot7 =  210$. Determine the number of different lists of values where all of the elements are $1 \leq a_i \leq 2$ and at least one value is $a_i = 2$. These are: one $2$, two $2$s, three $2$s, and four $2$s.

For each of those possible lists, generate all the permutations, in reverse lexicographical order. This would get you (eliminating brackets and commas for space reasons):

$$2111, 1211, 1121, 1112; 2211, 2121, 2112, 1221, 1212, 1122; 2221, 2212, 2122, 1222; 2222$$

At each step, of course, generate the actual numbers:

$$420, 630, 1050, 1470;\\ 1260, 2100, 2940, 3150, 4410, 7350;\\ 6300, 8820, 14700, 22050, 44100$$

and add to a counter if the number is less than $N$. In each group you can stop when you hit a number that's too high.

Now do the same thing for $1 \leq a_i \leq 3$, with at least one value $a_i = 3$, and so on. You'll want some recursion, and also some recursion. For $n = 10^{16}$, for instance, at the very least $[46,1,1,1]$ and $[44,2,1,1]$ are possibilities.

But eventually you'll hit a point where none of the possibilities is less than $N$. I suggested reverse lexicographical order, as you should hit the smaller numbers earlier each time around. You'll never even try, for instance, $[10,10,10,10]$

Now you add another element to the array to represent $11$, and do the same thing, with the constraint that only four elements can be non-zero. Then add another element for $13$, etc. Eventually you should hit a point where nothing is under $N$, probably when your array looks like $[0,0,....,0,1,1,1,1]$ and the last four elements are the last four primes before the fourth root of $N$.

I may come back later and try to pseudocode this.

Edit: some code available here for lex order, there's code at the bottom for anti-lex order but only in Java, but I'm sure you can generate similar code in your personal favorite language.


问:
Specialization and separatedness
I was reading a proof on the Stacks project about a morphism of schemes $Y\to X$ (where $X$ has the property that every point of $X$ specializes to a unique closed point) when I came across the following assertion:

As $Y\to X$ is a finite morphism, we see that the set of closed points $Y_0$
of $Y$ is the inverse image of the set of closed points $X_0$ of $X$.
Moreover, every point of $Y$ specializes to a unique point of $Y_0$ as (a)
this is true for $X$ and (b) the map $X\to Y$ is separated.

The last point seems to be asserting that

If $y_1\rightsquigarrow y_0$ and $y_1\rightarrow y_0’$ are specializations of $Y$ that are sent to the same specialization $x_1\rightsquigarrow x_0$ of $X$ under a separated morphism of schemes, then $y_0=y_0’$.

This is very close to the valuative criterion of separatedness, but in order to conclude, I need something like ‘I can represent two specializations of the same point by two maps from the spectrum of the same valuation ring.’

Can you help me prove or disprove the italicized assertion, or provide me with a reference? Thanks!

答: The italicized statement is false. Take $X=\mathrm{Spec}\,\mathbb{Z}_5$, and $Y=\mathbb{A}^1_X$. Let $x_1,y_1$ be the generic points of $X,Y$; let $x_0$ be the maximal ideal of $X$ (the unique closed point), but let $y_0,y’_0 \in Y$ correspond to the distinct maximal ideals $(5,x)$ and $(5,x^2+x+1)$.

Title: Specialization and separatedness


问:
Are the field reducts of real closed fields first-order axiomatizable?
We know that the class of real closed fields $(F,+,-,*,0,1,<)$ is first-order axiomatizable. Is the $\{+,-,*,0,1\}$ class of reducts of real closed fields axiomatizable? This is different from my previous question on the axiomatizability of ordered fields, because I am asking about real closed fields, rather than merely ordered fields.

答:
The ordering on a real closed field is determined by the field structure alone: we set $a\le b$ iff the equation $a+x^2=b$ has a solution. This automatically gives axiomatizability, since it says that the reducts of real closed fields are exactly the fields in which the relation $$R(a,b):\equiv\exists x(a+x^2=b)$$ satisfies a particular property.

More generally, suppose we have languages $\Sigma_0\subseteq\Sigma$, an axiomatizable class $\mathfrak{K}$ of $\Sigma$-structures, and for each symbol $R\in\Sigma\setminus\Sigma_0$ a $\Sigma_0$-formula $\varphi_R$ of appropriate arity such that for each $\mathcal{A}\in\mathfrak{K}$, the structure $\mathcal{A}'$ gotten by first taking the reduct of $\mathcal{A}$ to $\Sigma_0$ and then expanding it to $\Sigma$ by interpreting $R\in\Sigma\setminus R_0$ as $\varphi_R$ is also in $\mathfrak{K}$. Then the set of reducts to $\Sigma_0$ of elements of $\mathfrak{K}$ is again axiomatizable: letting $\Theta$ axiomatize $\mathfrak{K}$, consider the $\Sigma_0$-theory $\Theta'$ gotten by replacing each $R\in\Sigma\setminus\Sigma_0$ with $\varphi_R$ in every sentence in $\Theta$.

In the case of real closed fields things are particularly nice: real closed fields are uniquely orderable. But this uniqueness (that is, $\mathcal{A}=\mathcal{A}'$) isn't a necessary condition for the argument above: for example, let $\Sigma=\{U\}$ with $U$ a unary relation symbol, $\Sigma_0=\emptyset$, and $\mathfrak{K}$ be the class of $\Sigma$-structures where $U$ holds of at most one element. While we cannot recover $\mathcal{A}\in\mathfrak{K}$ from its underlying set (= $\mathcal{A}\upharpoonright\Sigma_0$), the formula $\varphi_U:\equiv x\not=x$ satisfies the condition of the above paragraph.


答: The $(+, -, \cdot, 0, 1)$ reducts of real closed fields are precisely those in which the relation defined by $x < y$ iff $\exists z \neq 0(y = x + z^2)$ gives a structure for the signature $(+, -, \cdot, 0, 1, <)$ that satisfies the axioms of a real closed field. This gives the axiomatization that you are looking for.

Title: Are the field reducts of real closed fields first-order axiomatizable?

Body: We know that the class of real closed fields $(F,+,-,*,0,1,<)$ is first-order axiomatizable. Is the $\{+,-,*,0,1\}$ class of reducts of real closed fields axiomatizable? This is different from my previous question on the axiomatizability of ordered fields, because I am asking about real closed fields, rather than merely ordered fields.


问:
Is it possible to solve this quadratic equation without calculating discriminant?
I have the quadratic equation $\;\;5x^2+96x-576=0\;\;$. I wonder can we solve it without using formula $x=\frac{-b\pm\sqrt{\Delta}}{2a}$ ? I suspect there is some way to do it because we have a lot of $24$s , ( $96=24\times4$ and $576=24^2)$ but I can't find it.

答:
On recognizing that $96=24\cdot4$ and $576=24^2$, it makes sense to let $x=24u$, factor out the $24^2$ from all three terms, and reduce the quadratic to

$$5u^2+4u-1=0$$

This factors easily into $(5u-1)(u+1)=0$, at which point you can let $u=x/24$, multiply the $24^2$ back in, and get

$$(5x-24)(x+24)=0$$


问:
Alternative Quadratic Formula
Well the formula for solving a Quadratic equation is :

$$\text{If }\space ax^2+bx+c=0$$

then

$$x=\dfrac{-b \pm \sqrt{b^2 -4ac} }{2a}$$

But looking at this : [Wolfram Mathworld] (And also in other places)

They give An Alternate Formula:

$$x=\dfrac{2c}{-b \pm \sqrt{b^2 -4ac}}$$

How does one get this?

Also in the first formula (the one we know) , $a \neq 0$ ... but here is it still the case?

Please help, Thanks!

答:
The alternative formula can be obtained from the conventional formula quite easily...

(This assumes $a$,$c$ both non-zero...)

The quadratic with roots $\alpha,\beta$:
$$
     a x^2 + b x + c  =  0
$$
can also be written in terms of $a$ and its roots:
$$
     a (x-\alpha)(x-\beta) = 0
$$
Expanding...
$$ ax^2 - a(\alpha+\beta)x + a\alpha\beta = 0 $$

And so $c = a\alpha\beta$, since this is an identity in $x$;

Thus if one root is $\alpha$, the other is
$$\beta = \dfrac {c} {a \alpha} $$

And if

$$\alpha=\dfrac{-b \pm \sqrt{b^2 -4ac} }{2a}$$

Then

$$
\beta = \dfrac {c} {a} \dfrac{2a} {-b \pm \sqrt{b^2 -4ac} } \\
  = \dfrac{2c} {-b \pm \sqrt{b^2 -4ac} }
$$

... which is the 'alternative formula', giving the other root. Change $\pm$ to $\mp$ to get the corresponding root.

As alluded to in my other answer, in a floating point calculation, you can choose the radical with the opposite sign from $b$ when evaluating $-b \pm \sqrt{b^2 -4ac}$, and then use that to obtain one root from each of the two formulae; this can reduce numerical precision issues when $\lvert {ac}\rvert$ is much smaller than $b^2$.

ContentLicense: CC BY-SA 3.0

Title: Alternative Quadratic Formula


问:
Lower bound of $(m,k)$-perfect numbers
Let $\sigma(n)$ represent the divisor sum of $n$ and $\sigma^m(n)$ represent the repeated application of the divisor function $m$ times.

$(m,k)$-perfect numbers are numbers such that $\sigma^m(n) = kn$. For example, $\sigma^3(12) = 120 = 12\times10$, so $12$ is a $(3,10)$-perfect number.

It's easy to see that $\sigma(n) \ge n, \forall n \in \mathbb N$, and it follows that $\sigma^{m+1}(n) \ge \sigma^m(n), \forall n,m \in \mathbb N$.

Now, let $f(m, k)$ represent the smallest $(m,k)$-perfect number, i.e. the minimal solution to $\sigma^m(n) = kn$. For all $m,k$ that actually have a solution, is it possible to show that $f(m, k) \ge k, \forall k \in \mathbb N$?

Context

答:
No. If we think about $n$ as being fixed and allowing $m$ and (particularly $k$) to vary, we have that $n$ is an $(m,k)$-perfect number for some $k$ given any $n$ for which
$$n\ \big|\ \sigma^m(n).$$
In particular,
$$\sigma^4(2)=8,$$
so $2$ is $(4,4)$-perfect. In fact, it seems likely that the sequence $x_i=\sigma^i(2)$ contains infinitely many terms, which would imply that $2$ is $(m,k)$ perfect for some $k$ above any given threshold.

Title: Lower bound of $(m,k)$-perfect numbers


问:
Let $E$ be a Banach space, $F=\mathcal{l}^1$, and let $T\in\mathcal{L}(E,F)$ be surjective, prove that $T$ has a right inverse
this is an exercise from Brezis Book, 2.11. My attempt:
T is surjective so thanks to the open mapping theorem, for every $n\in\mathbb{N}$ there exists  $x_n\in E$ such that $\|x_n\|\leq c$ ($c$ is the constant from the open mapping theorem) and  $T(x_n)=e_n$, where $(e_n)_{n\in\mathbb{N}}$ is the canonical basis of $\mathcal{l}^1$, and define the next function:
$S:\mathcal{l}^1\rightarrow E$ where if $y=(y_n)_{n\in\mathbb{N}}\in\mathcal{l}^1$ then $$S(y)= \displaystyle{\sum_{n=1}^\infty y_n(x_n)}.$$
This "function" is  near to be the right inverse of $T$ because $T\circ S=Id$, $S(y)$ is a convergent serie for every $y\in\mathcal{l}^1$ and $S$ is linear and bounded but I can't  prove that is well defined, I mean, we don't know if $T$ is injective and maybe could exists an $n\in\mathbb{N}$ such that $e_n=T(x_1)=T(x_2)$ with $x_1\neq x_2$ so the definition of $S$ is ambiguous. Please could you explain me what is happening?

答:
Let $T_n(y)=\sum_{i=1}^{i=n}y_nx_n$, $T_n$ is continuous and $\|T_n(y)\|\leq \|y\|c$.

We have $lim_nT_n(x)$ exists since $T_n(x)$ is a Cauchy sequence, the uniform bounded principle implies that there existsa bounded operator $T$ such that $T(x)=lim_nT_n(x)$.
See the corrolaries

https://en.wikipedia.org/wiki/Uniform_boundedness_principle


Title: Is it possible to solve this quadratic equation without calculating discriminant?

Body: I have the quadratic equation $\;\;5x^2+96x-576=0\;\;$. I wonder can we solve it without using formula $x=\frac{-b\pm\sqrt{\Delta}}{2a}$ ? I suspect there is some way to do it because we have a lot of $24$s , ( $96=24\times4$ and $576=24^2)$ but I can't find it.

Body:
An easy method is to start with factoring out $x$ and $24$

$$5x^2+96x-576=0$$
$$(5x^2-24)+(120x-576)\tag{Break into groups}$$
$$x(5x-24)+24(5x-24)\tag{Factor out x and 24}$$
$$(5x-24)(x+24)=0$$


问:
Proof is wrong, but I don't understand why.
I recently got this question wrong on a homework and after looking at my professors explanation I'm still failing to understand what's wrong with my proof. Any help would be much appreciated.

Suppose that the function
$$f: [ 0, 1]  \rightarrow \mathbb{R}$$

is continuous and that
$$f( x) \geq 0,\; \forall x \in [ 0, 1]$$
Prove that
$$\int ^{1}_{0}  f >0$$
if and only if there is a point $x_{0} \in [ 0, 1]$  at which $f( x_{0})   > 0.$

\begin{array}{l}
Proof\ by\ contradictin:Suppose\ that\ the\ function\ f:\ [ 0,\ 1] \ \rightarrow \mathbb{R} .\ is\ continuous\ and\ that\ \\
f( x) \geq 0\ \ for\ all\ x\ in\ [ 0,\ 1] .\ prove\ that\ \int ^{1}_{0} \ f >0\ iff\ \forall x_{0} \ in\ [ 0,\ 1] \ f( x_{0}) \leq 0\\
\\
\forall x_{0} \ in\ [ 0,\ 1] \ f( x_{0}) \leq 0\ \Longrightarrow \ \int ^{1}_{0} \ f >0\\
We\ know\ that\ the\ function\ is\ always\ greater\ than\ or\ equal\ to\ zero,\ so\ for\ f( x_{0}) \ to\ be\ less\ than\ \\
or\ equal\ to\ 0,\ \forall x_{0} \ in\ [ 0,\ 1] \ \ f( x) \ must\ be\ 0\ on\ [ 0,1] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( 0\leq f( x_{0}) \leq 0)\\
\\
this\ means\ f\ is\ a\ constant\ ( c=0) \ on\ [ 0,1] .\\
\\
\int ^{1}_{0} \ f=0( b-a) =0\ \ \ \ \ \ \ \ \ ( example\ 6.5)\\
\\
This\ contradicts\ the\ above\ statement.\ \\
\\
\\
\int ^{1}_{0} \ f >0\ \Longrightarrow \forall x_{0} \ in\ [ 0,\ 1] \ f( x_{0}) \leq 0\\
\\
0< \int ^{1}_{0} \ f\leq U( f,\ P) =\sum ^{n}_{i=1} M_{i}( x_{i} -x_{i-1}) .\\
\\
0< \sum ^{n}_{i=1} M_{i}( x_{i} -x_{i-1}) \ \\
\therefore \ becuase\ ( x_{i} -x_{i-1}) \ is\ positive\ there\ must\ exist\ some\ M_{i}  >0\ \ \ \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( the\ sum\ of\ all\ negative\ or\ zero\ numbers\ is\ \leq 0)\\
\ \ \ \ \ \ \ \ \ \ \ \\
\therefore \ \exists x_{0} \ in\ [ 0,\ 1] \ s.t.f( x_{0}) \  >\ 0\ \ \ \ \ \ \ \ \ \ \ [ M_{i} \ =\ sup\{f( x) \ |\ x\ in\ [ x_{i-1} ,\ x_{i}]\}]\\
\\
This\ contradicts\ the\ above\ statement.
\end{array}

答:
When you have an if and only if (iff) statement, you must prove it both ways.

If $\int_0^1 f(x) > 0$ then there is an $x_0\in [0,1]$ such that $f(x_{0})>0$

This direction you can prove with proof by contradiction:

Suppose $g(x) = 0$ and $f(x)\le g(x)$ for all $x\in[0,1]$ and $\int_0^1 f(x)\ dx > 0$

If $f(x) \le g(x)$ for all $x\in [a,b],$ with $f(x), g(x)$ integrable and $a<b,$  then $\int_a^b f(x)\ dx \leq \int_a^b g(x)\ dx$

$\int_0^1 f(x)\ dx \le \int_0^1 g(x)\ dx = 0$

Contradiction

However, you still must prove the other direction.

If there is an $x_{0}\in[0,1]$ such that $f(x_{0})>0,$ and $f$ is continuous on $[0,1]$, then $\int_0^1 f(x) > 0$

Since $f$ is continuous, there must be some neighborhood around $x_0$ such that $f(x) > 0$ for all $x$ in this neighborhood.

This creates a sub-interval where the integral is strictly positive.

For the remaining sub-intervals

$f(x)\ge 0$ for all $x \in [0,1]$ implies the integral is greater than or equal to zero for all sub intervals.

This makes the integral strictly positive.


问:
How many digits of $\pi$ are needed to approximate $\pi^n$ correctly to the nearest whole number?
In his piece on $\pi^{\pi^{\pi^\pi}}$, Matt Parker plots a graph showing that if $\pi$ is approximated to $n$ digits, then whenever $m \leq 2 n$ (roughly), $\pi^m$ is correct to the nearest whole number.

Is there any theoretical underpinning for this?

答:
Suppose $p>0$ is an approximation of $\pi$ with error $\epsilon$ (say $\epsilon=|p-\pi|$). The first digit where $p$ and $\pi$ disagree is in the $10^{\lceil\log_{10}\epsilon\rceil}$'s column.

Now we can trivially bound the error between $\pi^n$ and $p^n$ since
$$|p^n-\pi^n|=\epsilon \cdot \bigg(\sum_{k=0}^{n-1}p^{k}\pi^{n-k-1} \bigg) < \epsilon \cdot n \cdot \max(\pi,p)^{n-1}.$$
Thus they definitely agree up to the $10^{(n-1)\lceil\log_{10}(\epsilon n\max(\pi,p))\rceil}$'s column.


答:
The difference between our approximation to $\pi^m$ and the true result is
$$\left(\pi + \epsilon\right)^m - \pi^m  \approx \pi^m m \epsilon \sim \frac{\pi^mm}{10^n}$$
where we have used $\epsilon \sim 10^{-n}$ since the error in $\pi$ is in the $n$th digit so $\epsilon \sim 10^{-n}$ and the approximation $(1+x)^m \approx 1+ mx$. Now $\pi^2 \approx 10$ so we will have $\frac{\pi^mm}{10^n} > 1$ when $m \gtrsim 2n$. If you want to be more precise then its $m \gtrsim \log(10) n/\log(\pi)  \approx 2.011n$ for large $n$.

Title: How many digits of $\pi$ are needed to approximate $\pi^n$ correctly to the nearest whole number?

Body:
In his piece on $\pi^{\pi^{\pi^\pi}}$, Matt Parker plots a graph showing that if $\pi$ is approximated to $n$ digits, then whenever $m \leq 2 n$ (roughly), $\pi^m$ is correct to the nearest whole number.

Is there any theoretical underpinning for this?


答:
Let $y = 24$ and you have

$$5x^2+4xy - y^2$$

which easily factors as $(5x-y)(x+y)$.


问:
How to represent a series reversion like this
Consider the sequence: $$\sum_{k=0}^{0}\alpha^{k}\;,\;\sum_{k=0}^{1}\alpha^{k}\;,\;\cdots\;,\;\sum_{k=0}^{4}\alpha^{k}$$
This can be represented as:
$$
x_{n}=\sum_{k=0}^{n}\alpha^{k}\quad \text{if $0\leq n\leq 4$}
$$
My question is for a sequence :
$$
\sum_{k=2}^{6}\alpha^{k}\;,\;\cdots\;,\;\sum_{k=5}^{6}\alpha^{k}\;,\;\sum_{k=6}^{6}\alpha^{k}
$$
How should I represent using sum notation the same way I did for the previous one? It seems this requires a double summation but I actually am just assuming. Any help would be appreciated.

答:
You can write
$$x_n=\sum_{k=2+n}^6\alpha^k;$$
then $x_0$ will be the first sum you indicated, and $x_4$ will be the last sum, consisting only of $\alpha^6$. In general, modifying the indices of a summation is a very useful tool.


问:
Related rates +implicit differentiation cone problem
I'm stuck at the following related rates problem:
As shown in the diagram the apex angle of a cone is $2\theta$ and the slope length is 10cm. The rate of increase of $\theta$ is 0.01 radians per second. The cone starts with $\theta=\frac \pi 6$.
Find the initial rate of change of the volume of the cone.

What I've tried:

The volume $V$ of a cone is given by $V = \frac 1 3 \pi r^2h$. We can replace $r$ and $h$ with $\theta$ using the law of sines if we look at the cross section of the cone:

$\frac {10}{\sin \frac \pi2} = \frac r {\sin \theta}$ => $r = 10\sin \theta$

Similarly

$\frac {10}{\sin \frac \pi2} = \frac h {\sin (\frac \pi 2 -\theta)}$ => $ h = 10 \sin (\frac \pi 2 -\theta) = 10 \cos \theta$

There's probably a better way since this is a right angled triangle but this should work.

From this follows:

$V=\frac 1 3 \pi r^2h = \frac 1 3 \pi (10 \sin \theta)^2  (10 \cos \theta) = \frac {1000} 3 \pi \sin^2 \theta \cos\theta $

So: $\frac {dV} {dt} = \frac {d}{dt} (\frac {1000} 3 \pi \sin^2 \theta \cos\theta)$

$\frac {dV} {dt} = \frac {1000} 3 \pi  \frac {d}{dt} (\sin^2 \theta \cos\theta)$
$\frac {dV} {dt} = \frac {1000} 3 \pi  ( \cos\theta \frac {d}{dt} \sin^2 \theta + \sin^2 \theta \frac {d}{dt} \cos\theta)$

$\frac {dV} {dt} = \frac {1000} 3 \pi  ( 2\cos\theta \sin\theta *  \frac {d}{dt} \sin \theta- \sin^2 \theta \sin\theta\frac {d \theta}{dt})$

$\frac {dV} {dt} = \frac {1000} 3 \pi  ( 2\cos\theta \sin\theta *   \cos\theta\frac {d \theta}{dt}- \sin^2 \theta \sin\theta\frac {d \theta}{dt})$

$\frac {dV} {dt} = \frac {1000} 3 \pi  ( 2\cos^2\theta \sin \theta \frac {d \theta}{dt} - \sin^3 \theta \frac {d \theta}{dt})$

Then I plugged in the given values:
$\frac {dV} {dt} = \frac {1000} 3 \pi  ( 2 \cos^2\frac \pi 6 \sin \frac \pi 6 (0.01) - \sin^3 \frac \pi 6 (0.01)) = 6.545 \frac {cm^3}{s}$.

However my textbook says that it should be $0.654 \frac {cm^3}s$ which seems pretty close to what I have found. Did I make a blunder somewhere or is my reasoning incorrect?

答:
Your answer is correct. The book is incorrect by a factor of $10$.

To simplify your calculation, we have $$V(t) = \frac{1000\pi}{3} \sin^2 \theta(t) \cos \theta(t) = \frac{1000\pi}{3} (\cos \theta(t) - \cos^3 \theta(t)).$$  At time $t = 0$, $\theta(0) = \pi/6$ and $\theta'(0) = \frac{1}{100}$, so $$\begin{align}
V'(0) &= \frac{1000\pi}{3} \left(-\theta'(0) \sin \theta(0) + 3 \theta'(0) \cos^2 \theta(0) \sin \theta(0) \right) \\
&= \frac{1000\pi}{3} \theta'(0) \sin \theta(0)(3 \cos^2 \theta(0) - 1) \\
&= \frac{5\pi}{3} \left(3 \cos^2 \frac{\pi}{6} - 1 \right) \\
&= \frac{5\pi}{3} \left(\frac{9}{4} - 1\right) \\
&= \frac{25\pi}{12}.
\end{align}$$


问:
Solving Integration Problem
What is the value for the following integration :
$\int_{-\infty}^1 (e^{-2|x-2|} dx)$ + $\int_1^\infty (0.5e^{-|x|} dx)$ ? The absolute value is confusing me .

Title: Solving Integration Problem

Body:
When $x$ ranges from $−\infty$ to $1$, we have $|x−2|=2−x$ and, when $x$ ranges from $1$ to $\infty$, we have $|x|=x$.

Now,

$$\int_{-\infty}^1 e^{-2|x-2|} = \int_{-\infty}^1 e^{-2(2-x)} = \int_{-\infty}^1 e^{2x - 4} = \frac{1}{e^4}\int_{-\infty}^1 e^{2x} = \frac{1}{2e^4}e^{2x}]_{-\infty}^1 = \frac{1}{2e^2}$$

Similarly,

$$\frac{1}{2}\int_{1}^\infty e^{-|x|} = \frac{1}{2}\int_{1}^{\infty} e^{-x} = \frac{1}{2e}$$


问:
$n^5-n$ is divisible by $10$?
I was trying to prove this, and I realized that this is essentially a statement that $n^5$ has the same last digit as $n$, and to prove this it is sufficient to calculate $n^5$ for $0-9$ and see that the respective last digits match. Another approach I tried is this: I factored $n^5-n$ to $n(n^2+1)(n+1)(n-1)$. If $n$ is even, a factor of $2$ is guaranteed by the factor $n$. If $n$ is odd, the factor of $2$ is guaranteed by $(n^2+1)$. The factor of $5$ is guaranteed if the last digit of $n$ is $1, 4, 5, 6,$ $or$ $9$ by the factors $n(n+1)(n-1)$, so I only have to check for $n$ ending in digits $0, 2, 3, 7,$ $and$ $8$.  However, I'm sure that there has to be a much better proof (and without modular arithmetic). Do you guys know one? Thanks!</p>

答:
To check if $n^5-n$ is congruent to $0$ modulo $5$, it suffices to check the five cases $n = 0, \pm 1, \pm 2$. Since the polynomial $n^5-n$ is odd, the sign is not important, and we need check only $0$, $1$, and $2$. The first two cases are trivial, and for the last case we calculate manually $32 - 2$ which is divisible by $5$.

Congruence modulo $3$ or modulo $2$ is similar (only involves the "trivial" cases).

Title: $n^5-n$ is divisible by $10$?


问:
Irreducible polynomials of degree $n>3$
I don't quite understand how I know if a polynomial is irreducible, and I would like to know if someone could help me understand that.

For example, if I have the following polynomial: $x^5+x^2+x^3+1$, is this polynomial irreducible in $\mathbb{Z}_{2}[x]$? I know that $\mathbb Z_{2}[x]=\{x^2, x^2+1, x^2+x, x^2+x+1\}$ but I will not continue from there to answer that question

答:
In general, showing that a polynomial is irreducible is hard. There are a few irreducibility tests that can be used, but there are few general methods that don't require lots of computation or case-checking. One special case that you might have seen is that a cubic is irreducible over a field $k$ if and only if it has no roots in $k$; this is because any nontrivial factorization of a cubic must involve a term of degree at most $1$, and thus give a root. You can generalize this to higher degrees -- for example, for a degree $5$ polynomial, you only need to check all possible linear and quadratic factors -- but if your polynomial or field is very large this can get tricky.

However, showing that a polynomial is reducible is "easy" (easy in the sense that the data needed is short, not that it's necessarily easy to find). All you need to do is find a nontrivial factorization of your polynomial, and then you're done.

In particular, for the degree-$5$ polynomial you've written down, $x=1$ is a root in $\mathbb Z/2\mathbb Z$, so you can write it as $(x-1)$ times a degree-$4$ polynomial. This is a nontrivial factorization, and so the polynomial is reducible.

Title: Irreducible polynomials of degree $n>3$


问:
Proof in FOL with no CQ rule

I had the first completed but when doing its converse, since there is no CQ rule, I do not know where to proceed. Please help

答:
As usual the key is to work backwards.

To introduce the universal, assume an arbitrary variable, $a$, then derive $P(a)\to\lnot Q(a)$.

To introduce that conditional, assume $P(a)$ then derive $\lnot Q(a)$.

To introduce that negation, assume $Q(a)$, then derive a contradiction.

To derive that contradiction under the assumptions of $\lnot\exists x~(P(x)\land Q(x))$, $P(a)$, and $Q(a)$, well...

Title: Proof in FOL with no CQ rule


问:
Why are these sums equal
I have to find the value of $$\mathcal L^{-1}\left(\frac{1}{s(1+e^{as})}\right)$$ where $a \gt 0$ and $s$ is a complex variable and $\mathcal L^{-1}$ is the inverse Laplace transform.

In the book, this exercise is solved using The Cauchy residue theorem.
The function $$F(s)=\frac{1}{s(1+e^{as})}$$ has a simple pole at $s=0$ and at $s_n=\frac{2n-1}{a}\pi i$ for $n=0, \pm1,\pm2,...$.
Then the sum of residues becomes $$\sum _{n=-\infty}^{+\infty}\mathcal{Res}(s_n)=\frac12-\sum _{n=-\infty}^{+\infty}\frac1{(2n-1)\pi i}e^{t\frac{2n-1}{a}\pi i}=\frac12-\frac2\pi\sum _{n=1}^{+\infty}\frac1{(2n-1)}\sin{\left(t\frac{2n-1}{a}\pi\right)}$$

This is where I'm stuck. I don't understand why these sums are equal $$\sum _{n=-\infty}^{+\infty}\frac1{(2n-1)\pi i}e^{t\frac{2n-1}{a}\pi i}=\frac2\pi\sum _{n=1}^{+\infty}\frac1{(2n-1)}\sin{\left(\frac{2n-1}{a}t\pi\right)}$$

答: Hint: The series of interest consists of two parts, one with $n\geq 0$ and the other with $n<0$. Split the series up accordingly, and relabel the sum from $n=0$ to $\infty$ using $n'=1-n$. Now forget the label on $n'$ and combine the series.

Title: Why are these sums equal


问:
Double counting volume of revolution
I want to understand why is there no double-counting when using the volume of revolution formula on a circle. Say we want to find the volume of revolution in the first and second quadrants about the $y$ axis. Say the equation of the circle is $x^2+y^2=R^2$. Using the volume of revolution formula, we get
$$\int_{-R}^{R}\pi (R^2-x^2)dx=\frac{4}{3}\pi R^3$$
which is correct. However, aren't we rotating around the $y$ axis basically twice, as we are doing it once with the quarter circle from $x=0$ to $x=R$ and then again with the quarter circle from $x=-R$ to $x=0$?
There is some error with this logic that I can't find. Can someone please explain to me why the formula holds even though we are apparently doing the operation twice?

EDIT: I just had a thought. Is it because $x^2+y^2=R^2$ is not a function (as it fails the vertical line test)?

答:
$$\int_{x=-R}^R \pi (R^2 - x^2) \, dx = \pi \left[R^2 x - \frac{x^3}{3}\right]_{x=-R}^R = \pi \left(R^3 - \frac{R^3}{3} - \left(-R^3 - \frac{(-R)^3}{3}\right)\right) = \frac{\color{red}{4}}{3}\pi R^3.$$

Now that the original question was edited to correct the arithmetic error, the follow-up answer is this:

There is no double counting because the integral is being calculated with respect to $x$, not $y$.  Therefore, the axis of revolution is the $x$-axis, not the $y$-axis, and the differential volume for a representative slice corresponds to a disk of thickness $dx$, and radius equal to the distance from the point $(x,0)$ to $(x, \sqrt{R^2 - x^2})$, i.e. $\sqrt{R^2 - x^2}$.  This gives the full sphere volume as I have shown above.

If you want to calculate the volume of the hemisphere for a volume of revolution above the $x$-axis by rotating about the $y$-axis, then you would write the integral as follows:

$$\int_{y=0}^R \pi (R^2 - y^2) \, dy.$$  For a given $y \in [0,R]$, the differential volume element is a disk of thickness $dy$, with radius equal to the distance from $(0,y)$ to $(\sqrt{R^2 - y^2}, y)$.

Title: Double counting volume of revolution


问:
Geometric similarities between points in an algebraic variety
If $f : \mathbb{R} \to \mathbb{R}$ is a univariate irreducible polynomial, Galois theory says that all roots are equivalent up to field automorphism (specifically, an automorphism of the field extension fixing the base field).

Can anything similar be said for the multivariate case?  Specifically, if $f : \mathbb{R}^d \to \mathbb{R}$ is an irreducible multivariate polynomial, is the local geometry of the zero set $V = f^{-1}(0)$ similar at different points in some sense?

Clearly this isn't true at all points of $V$; the easiest counterexample is the lemniscate (http://mathworld.wolfram.com/Lemniscate.html), where $V$ is manifold except at one point.  However, $V$ does have some self similarity except at this lower dimensional set: in particular, the curvature elsewhere is nonzero.  As another example, I would expect that $V$ is locally a ruled surface either almost everywhere or almost nowhere.  Is there a general statement that would encompass these and related similarities between different points of $V$?

A hint at a possible answer: I believe the statement is true for any geometric property that can be expressed algebraically in terms of the differential structure near $x \in V$, since otherwise adding these equations would separate $V$ into multiple components, contradicting the irreducibility of $V$.

答:
In general, it is not true that points on an algebraic variety "look the same". If $V$ is a smooth variety, and $p, q \in V$, it is not true that there are necessarily neighborhoods $U,V$ of $p,q$ which are isomorphic by an isomorphism taking $p$ to $q$ (a property which is obvious in differential geometry). If $V$ is smooth and projective, such an isomorphism would actually extend to an automorphism of $V$ taking $p$ to $q$. So in fact, on a sufficiently generic variety, no two points look the same!

What I've said amounts to the following thing: for different points, the local rings $\mathcal O_p$ are generally different. However, the process of completion allows us to make sense of the intuition from differential geometry. The local ring $\mathcal O_p$ is a local ring with maximal ideal $\mathfrak m_p$. Its dimension is equal to the dimension of $V$. We can take the formal completion $\hat{\mathcal O_p}$ along the maximal ideal $\mathfrak m_p$; this has essentially the effect of replacing rational functions with power series. Taking Taylor expansion of a rational function allows us to embed the ring $\mathcal O_p$ into $\hat{\mathcal O_p}$. If $V$ is smooth at $p$, the local ring $\mathcal O_p$ is a regular local ring (this is what it means to be smooth); and we have the following theorem characterizing the completion of a regular local ring of dimension $n$ over a field:

Theorem: If $(A, \mathfrak m)$ is a regular local $k$-algebra of dimension $n$, then $\hat A \cong k[[x_1, \dots, x_n]]$, the ring of formal power series in $n$ indeterminates.

Thus, all of the complete local rings $\hat{\mathcal O_p}$ are isomorphic, if $V$ is smooth. However, there is generally no preferred isomorphism from $\hat{\mathcal O_p}$ to $\hat{\mathcal O_q}$.


问:
Show for $1>a>0$ that $(e^{x}-a)^{-1}\sin(x)$ is integrable over $[0,\infty)$
Show  for $1>a>0$ that $(e^{x}-a)^{-1}\sin(x)$ is integrable over $[0,\infty)$

Any help would be super helpful!

答:
Hint: split up the region of integration as $[0,2]\cup[2,\infty)$. In the first interval, you can trivially bound the function. In the second interval, you can use the boundedness of $\sin(x)$ and bound $e^x-a$ in a nice way.

In this way, you can prove the integral converges absolutely and so it converges.


问:
ODE Non - Homogeneous
The equation is:
$$y''-2y'+1=0$$

I am trying to find the general solution and have tried a few different tactics. I can't use the char equation because it won't work. Does anyone have any ideas?

答:
$$y''-2y'+1=0$$
$$y''-2y'=-1$$
Just integrate both sides:
$$\int (y''-2y')dx=-\int dx$$
$$\int dy'-2\int dy=-\int dx$$
$$y'-2y=-x+C_1$$
The DE is of first order apply any method you know to integrate the DE.

Note that :
$$y''-2y'+1=0$$
becomes homogeneous:
$$(y-\dfrac x2)''-2(y-\dfrac x2)' =0$$
$$u''-2u'=0$$
With $u=y-\dfrac x2$.
$$\implies r^2-2r=r(r-2)=0 $$
$$r=0, r=2$$
Therefore
$$u=c_1e^{2x}+c_2$$
$$y=c_1e^{2x}+c_2+\dfrac x2$$

Title: ODE Non - Homogeneous


问:
Probability question on sample space
Recently, I encountered a probability question which can be phrased differently:

Q1: A letter is chosen at random from the word MISSISSIPPI. What is the sample space.

Q2: The letters from the word MISSISSIPPI are put into a bag. What is the sample space.

Q3: Picking a letter at random from a box containing identical cards with letters that spell the word MISSISSIPPI. What is the sample space.

I can understand that the S's, I's and P's are indistinguishable for Q1, and hence the sample space is {M, I, S, P}.

However, my teacher told me that the sample space for Q2 is also {M, I, S, P} which I do not understand. Since each letter is now distinct and distinguished from one another, i.e., the four S's are distinguishable, then why shouldn't the sample space be {M, I1, S1, S2, I2, S3, S4, I3, P1, P2, I4} ?

To make things even more confusing, she said that the sample space for Q3 is {M, I1, S1, S2, I2, S3, S4, I3, P1, P2, I4}.

Thank you in advance.

答:
For Q2, it's still the same sample space because the letters are still indistinguishable. If you pull an S at random from the bag (I'm imagining Scrabble tiles here), you still don't know whether it was the first S in MISSISSIPPI or one of the others.

If we let the random variable $X$ represent the letter drawn, the sample space is still $X\in \{M,I,S,P\}$, with probabilities $P(X=M) = \frac{1}{11}$, $P(X=I) = \frac{4}{11}$, $P(X=S) = \frac{4}{11}$, and $P(X=P) = \frac{2}{11}$.

For Q3, the distinction comes from the cards presumably being ordered. Then, once you've drawn a card, you can tell what position it was in. Each card would have a distinguishable letter and position, so each element of the sample space would consist of the letter and its position.

Title: Probability question on sample space


问:
GNS constructions of two faithful positive linear functional
If $\rho_1$ and $\rho_2$ are two faithful positive linear functionals on a unital $C^*$-algebra $A$. Then $\rho_i(x)=\langle x\omega_i,\omega_i \rangle$,
where $\omega_i=1+N_{\rho_i}$,
$N_{\rho_i}=\{x\in A: \rho_i(x^*x)=0\}(i=1, 2).$

Since each $\rho_i$ is faithful, we have $N_{\rho_i}=0$. Then each cycic vector $\omega_i=1(i=1, 2),$ $\rho_1$ and $\rho_2$ have the same GNS construction.

Does the above conclusion correct?

答:
Martin's example says it all but here is another, more pedestrian counter-example.

Let $A=C([0, 1])$ and let $D_1$ and $D_2$ be  two disjoint, dense subsets of $[0, 1]$.  For example,  take
$$
  D_1=\mathbb Q\cap [0, 1],  \quad  \text{and} \quad   D_2=(\mathbb Q+\pi )\cap [0, 1].
  $$

Fix an enumaration $\{d_n^i\}_{n\in {\bf N}}$ for each $D_i$, and define a state $\rho _i$ on
$C([0, 1])$ by
$$
  \rho _i(f) = \sum_{n\in {\bf N}} 2^{-n}f(d^i_n), \quad\forall f\in C([0, 1]).
  $$

It is not hard to  see that the GNS representation $\pi _i$ associated to $\rho _i$ is equivalent to the representation of
$C([0, 1])$ on $\ell ^2$, where each $f$ in $C([0, 1])$ is mapped to the diagonal operator with diagonal entries
$f(d^i_n)$.

This implies that $\pi _i$ is faithful,  so one might wonder whether or not  $\pi _1$ and $\pi _2$ are equivalent.    To see that
they are not,  it is enough to observe that,  for any  given $f$, the point spectrum (eigenvalues) of the operator
$\pi _i(f)$ is
$$
  \sigma _p(\pi _i(f))= \{f(d^i_n): n\in {\bf N}\},
  $$
and if $f$ is any injective function,  this imlies that
$$
  \sigma _p(\pi _1(f))\neq   \sigma _p(\pi _2(f)).
  $$

This shows that the operators $\pi _1(f)$ and $\pi _2(f)$ are not conjugate to each other,  and hence $\pi _1$ cannot be equivalent to $\pi _2$.


问:
Proving that $SU(n)$ is a smooth manifold
Considering this post: Show that $SL(n, \mathbb{R})$ is a $(n^2 -1)$ smooth submanifold of $M(n,\mathbb{R})$ I dont understand how the manipulations in the limit are done, for instance:

$$ \det(A+tA)=(1+t)^n \det(A)$$

and how the limit overall evaluates to: $n \det(A)$

答:
It's best to write out what you have a question about on this post explicitly, but anyway:
$$
\det(A+tA)=\det((1+t)A)=(1+t)^n\det(A)
$$
by the property of determinants of $n\times n$ matrices that says that $\det(\lambda A)=\lambda^n \det(A)$. Here we just set $\lambda=1+t$. As for the limit in the post:
$$
\lim_{t\to 0}\frac{\det(A+tA)-\det(A)}{t}=\lim_{t\to 0}\frac{(1+t)^n\det(A)-\det(A)}{t}=\det(A)\lim_{t\to 0}\frac{(1+t)^n-1}{t}$$
$$=\det(A)\lim_{t\to 0}\frac{\sum_{k=0}^n{n\choose k}t^k-1}{t}=\det(A)\lim_{t\to 0}\sum_{k=1}^n{n\choose k}t^{k-1}=\det(A){n\choose 1}=n\det(A).
$$


问:
Keyhole integration
Ok so my lecturer gave us this powerful lemma for doing contour integrals over a semi-circle. The lemma is:

Let $C_{R}$ be the contour defined as $\{z \in \mathbb{C} \mid z = R  e^{i \theta } , \theta \in [0, \pi] \}$.

If $|f(z)|  \le \frac{k}{ |z| ^2} $ for some $ k \in \mathbb{R} $ , $\Im(z) > 0$ and $|z|$ large enough, then:

$$\lim_{R \to 0} \int_{C_{R}} f(z) dz = 0$$

So in a lecture we did this problem:

$$ \int_0^{\infty} \dfrac{dx}{ \sqrt{x} (1 + x) }$$

He then defined $\Gamma _{R} = C_R \cup [\epsilon,R] \cup (-C_{\epsilon}) \cup [R,\epsilon]$

Where $C_R = \{z \in \mathbb{C} \mid |z| = R\}$, $C_{\epsilon} = \{z \in \mathbb{C} \mid |z| = \epsilon \}$

He somehow used the lemma to show that:

$$\lim_{R \to 0} \int_{C_{R}} f(z) dz = 0$$

But he can't do that can he?

答: The statment of the powerful lemma has $|f(z)|\le k|z|^{-2}$ as a condition, but it holds if $|f(z)|\le k|z|^{-\alpha}$ with $\alpha>1$. The proof is the same. He is using this with $\alpha=3/2$.

Title: Keyhole integration


问:
Image of closed unit ball under a compact operator
Let $X,Y$ be Banach spaces and $A\in\mathcal L(X,Y)$ . The task is to prove the following:

$A$ is compact if and only if the image of the closed unit ball in $X$ is compact in $Y$.

I have proven this when $X$ is a reflexive space.

Proof. Let $X$ be a reflexive space, $\bar B$ the closed unit ball in $X$, and $A$ a compact operator. Let further $y_n=Ax_n$ be a sequence in $A(\bar B)$.

In reflexive spaces $\bar B$ is weakly compact, so there exists a subsequence $x_{n_j} \to x$ weakly.

Because $A$ is compact, $Ax_{n_j}\to Ax$ strongly.

On the other side, $A(\bar B)$ is relatively compact, so there exists $z_k=Ax_{n_{j_k}}$ that converges strongly to $y\in Y$.

But $z_k\to Ax$ strongly. So by unicity of the limit $y=Ax$ and the image is compact.

It's easily proved that if the image is compact, the operator is also compact.

But I don't know what to do in case of nonreflexive spaces. Is there any counterexample or proof in such case?

答: This is much easier. Use the definition of compact operator. An operator is compact if and only if the image of a bounded set is precompact. The unit ball is bounded and any other bounded set is inside some $\lambda$ times the unit ball.</p>


问:
Polynomial $f$ over $\mathbb{C}$.
I came across a question which asks to find all polynomials $f$ over $\mathbb{C}$.

Does this mean the coefficients of the polynomial can be complex or the polynomial can take complex values?

答:
The  polynomials over  $\Bbb C$ (or more generally any field $\Bbb K$) are expressions of the form

$p_0+p_1X+p_2X^2+\cdots+p_{m-1}X^{m-1}+p_mX^m$,

where $X$ is an indeterminate and $p_0,p_1,p_2,...,p_{m-1},p_{m}$ are in $\Bbb C$ (or more generally $\Bbb K$).

Title: Polynomial $f$ over $\mathbb{C}$.


问:
Finding the sum- $x+x^{2}+x^{4}+x^{8}+x^{16}\cdots$

If $S = x+x^{2}+x^{4}+x^{8}+x^{16}\cdots$

Find S.

Note:This is not a GP series.The powers are in GP.

My Attempts so far:

1)If $S(x)=x+x^{2}+x^{4}+x^{8}+x^{16}\cdots$

Then $$S(x)-S(x^{2})=x$$

2)I tried finding $S^{2}$ and higher powers of S to find some kind of recursive relation.

3)When all failed I even tried differentiating and integrating S.Obviously that was of no good either.

Could anyone give me a hint to solve this?Thanks!

答:
I have worked on this series before.

There is no simple closed form as a geometric series has.

(such as $x+x^2+x^3+x^4+...=\frac{x}{1-x}$ where $|x|<1$).

You can see more information in the link about Lacunary function.

The series can be expressed in closed form of double integral. I shared my result below.

$x+x^{2}+x^{4}+x^{8}+\dots=F(x)$

Let's transform
$x=e^{-2^t} \tag{1}$

Where $-\infty<t<\infty$
Thus x will be in $ (0,1)$ and  $F(x)$ is not divergent in this range.

$e^{-2^t}+e^{-2^{t+1}}+e^{-2^{t+2}}+\dots=F(e^{-2^t})=H(t)$

$e^{-2^t}+H(t+1)=H(t)$

$H(t+1)-H(t)=-e^{-2^t}$

The Fourier transform of both sides

$$\int_{-\infty}^{+\infty} H(t+1)e^{-2πift} \mathrm{d}t-\int_{-\infty}^{+\infty} H(t)e^{-2πift} \mathrm{d}t=-\int_{-\infty}^{+\infty} e^{-2^{t}}e^{-2πift} \mathrm{d}t$$

$$V(f)= \int_{-\infty}^{+\infty} H(t)e^{-2πift} \mathrm{d}t$$

$$\int_{-\infty}^{+\infty} H(t+1)e^{-2πift} \mathrm{d}t=\int_{-\infty}^{+\infty} H(z)e^{-2πif(z-1)} \mathrm{d}z=V(f)e^{2πif}$$

$$e^{2πif}V(f)-V(f)=-\int_{-\infty}^{+\infty} e^{-2^{t}}e^{-2πift} \mathrm{d}t$$

$$V(f)=\int_{-\infty}^{+\infty} \frac{e^{-2^{t}}e^{-2πift}}{1-e^{2πif}} \mathrm{d}t$$

Now we need to take the inverse Fourier transform

$$H(z)=\int_{-\infty}^{+\infty} V(f) e^{2πifz} \mathrm{d}f=\int_{-\infty}^{+\infty}  e^{2πifz}\int_{-\infty}^{+\infty} \frac{e^{-2^{t}}e^{-2πift}}{1-e^{2πif}} \mathrm{d}t\,\mathrm{d}f $$

The closed form of $H(z)$ in integral expression:
$$H(z)=\int_{-\infty}^{+\infty}  \int_{-\infty}^{+\infty} e^{2πifz} \frac{e^{-2^{t}}e^{-2πift}}{1-e^{2πif}}  \mathrm{d}t\,\mathrm{d}f $$

$$\sum_{k=0}^\infty x^{2^k}=H(\log_2(-\ln x))=\int_{-\infty}^{+\infty}  \int_{-\infty}^{+\infty} e^{2πif\log_2(-\ln x)} \frac{e^{-2^{t}-2πift}}{1-e^{2πif}}  \mathrm{d}t\,\mathrm{d}f$$

Where $0<x<1$

$$\sum_{k=0}^\infty x^{2^k}= \int_{-\infty}^{+\infty} \frac{e^{2πif\log_2(-\ln x)}}{1-e^{2πif}} \int_{-\infty}^{+\infty} e^{-2^{t}-2πift}  \mathrm{d}t\,\mathrm{d}f=\int_{-\infty}^{+\infty}  e^{-2^{t}} \int_{-\infty}^{+\infty}  \frac{e^{2πif(\log_2(-\ln x)-t)}}{1-e^{2πif}}  \mathrm{d}f\,\mathrm{d}t$$

Note: I made the update on 07/29/2016 . Variable change was $x=e^{2^t}$ at Tag (1) and it had problems as @leonbloy 's comment below. Thanks for the comment. Please let me know if you notice something else in definitions.


问:
Motivation Behind c-Convexity in Optimal Transport
I try to get into optimal transport by reading Villani's book "Optimal Transport - old and new" and I came across this definition. Unfortunately, I dont really have an intuition on what motivates them (lines (5.6)-(5.8)). They seem pretty mysterious to me.

答:
The motivation is as follows. For the quadratic cost, i.e. $d(x,y)^2$, there are a bunch of basic optimal transport facts which can be proved using techniques/objects from convex analysis, including: Legendre transform, subdifferential, etc. It would be nice to be able to run those proofs for costs other than $d(x,y)^2$; the solution turns out to be to define a generalization of convexity relative to a given cost $c(x,y)$, such that, in the terminology of $c$-convexity, $d(x,y)^2$-convexity is just regular convexity.

In particular, the $c$-transform is the "right" generalization of the Legendre transform, the $c$-subdifferential is the "right" generalization of the subdifferential, etc.

I can't remember if this is emphasized in Villani's book, but definitely in Santambrogio's book the special, convenient proofs that use convex analysis techniques in the quadratic case are presented, so you can try and compare how the generalization goes through when one switches to a general cost.


问:
Solving equation in one unknown with absolute value
Consider the following equation :
$$\tfrac12e^{-|x|}=e^{-2|x-2|}$$
Now what I have done is:
$$\ln\tfrac12 -|x| = -2 |x-2|$$
But I don't know how to go further because of the absolute.

Body:
$$\ln\frac{1}{2}-|x|=-2|x-2|$$

$$\ln\frac{1}{2}=|x|-2|x-2|$$

Case 1: $x>2, |x|=x,|x-2|=x-2$

$$\ln\frac{1}{2}=-x+4$$

$$x=4+\ln(2)>2$$ so this is a solution

Case 2: $0<x<2, |x|=x,|x-2|=2-x$

$$\ln\frac{1}{2}=3x-4$$

$$0<x=\frac{\ln\frac{1}{2}+4}{3}<2$$ so this is a solution

Case 3: $0>x, |x|=-x,|x-2|=2-x$

$$\ln\frac{1}{2}=x-4$$

$$x=4+\ln\frac{1}{2}>0$$ So there is no solution given this case


问:
How to find the maximum distance between two paths over a time interval (diff eqn) AP Calc Question
Alice and Bob go for a jog in the same direction along a straight path. For $0\leq t \leq20$, Alice’s velocity at time t is given by $A(t)=\frac{6010}{t^2-3t+50.5}$ meters per minute, and Bob’s velocity at time t is given by $B(t)=8.5t^3e^{-0.45t}$ meters per minute. Both of these velocities are always positive. Alice is 12 meters ahead of Bob at time $t=0$, and she remains ahead of Bob for $0\leq t\leq 20$.

What is the maximum distance between Alice and Bob over the time interval $0\leq t \leq 20$?

The answer is $d=1413.23$ when $t=4.58$.

I'm confused on how to get this answer...

I did $\int_0^{20} \frac{6010}{t^2-3t+50.5} - \int_0^{20} 8.5t^3e^{-0.45t}$

$\int_0^{20}\frac{6010}{t^2-3t+50.5} =1232.323$

$\int_0^{20} 8.5t^3e^{-0.45t}=1217.313$

and got $1232.323-1217.313=15.01=d$

I thought that the distance between Alice and Bob is $\int A(t)-\int(B(t)$ so I'm confused where I went wrong.

How can I solve this?

答:
You've calculated the overall difference in the distances they traveled between $t=0$ and $t=20$, not the difference at a specific point in time. What we need is the difference as a function of $t$.

We're given $A(t)=\frac{6010}{t^2-3t+50.5}$ meters per minute, and $B(t)=8.5t^3e^{-0.45t}$. Since we know Alice is ahead of Bob at any given time, and we know that at $t = 0$ Alice is $12$ meters ahead of Bob, we can write our distance function as follows:

$$ d(t) = 12 + \int_0^t A(\tau) - B(\tau) d\tau $$

This is similar to what you wrote, except you integrated from $\tau=0$ to $\tau=20$. Now, the key to this problem is that we want to maximize that distance, so all we have to do is differentiate and set the derivative equal to $0$, which is very easy with the Fundamental Theorem of Calculus:

$$ \frac{d}{dt} d(t) = \frac{d}{dt} \left( 12 + \int_0^t A(\tau) - B(\tau) d\tau \right) = A(t) - B(t)$$
$$ \Rightarrow A(t) - B(t) = 0 $$
$$ \Rightarrow A(t) = B(t) $$
$$ \Rightarrow \frac{6010}{t^2 - 3t + 50.5} = 8.5t^3e^{-0.45t} . $$

We can use a calculator to solve numerically and find that $A(t) = B(t)$ at $t \approx 4.584$ or $t \approx 16.715$. These are the local extrema on the interval $t\in [0,20]$, so plug those values back into $d(t)$, plus check the endpoints at $t = 0$ and $t = 20$, and one of those four values will be your maximum distance.


答:
Hint

The slow, pedantic, pedestrian method:

In a problem like this, before you attempt any math manipulations, immediately create cases to represent each relevant range of $x$.

Here, it is immediate that you have $x < 0$ and $x \geq 0.$

You also have $(x-2) < 0$ and $(x-2) \geq 0.$

So simply compress the 4 possibilities above into 3 distinct cases.  Then, construe each case as a totally separate problem.

Naturally, with experience, you will look for elegant analysis/shortcuts to the slow pedantic pedestrian methods.  However, when starting out, don't be seduced into trying for elegance.

Title: Solving equation in one unknown with absolute value

Body:
Consider the following equation :
$$\tfrac12e^{-|x|}=e^{-2|x-2|}$$
Now what I have done is:
$$\ln\tfrac12 -|x| = -2 |x-2|$$
But I don't know how to go further because of the absolute.


问:
Alternative Quadratic Formula
Well the formula for solving a Quadratic equation is :

$$\text{If }\space ax^2+bx+c=0$$

then

$$x=\dfrac{-b \pm \sqrt{b^2 -4ac} }{2a}$$

But looking at this : [Wolfram Mathworld] (And also in other places)

They give An Alternate Formula:

$$x=\dfrac{2c}{-b \pm \sqrt{b^2 -4ac}}$$

How does one get this?

Also in the first formula (the one we know) , $a \neq 0$ ... but here is it still the case?

Please help, Thanks!

Title: Alternative Quadratic Formula

Body:
\begin{align}
   \dfrac{-b \pm \sqrt{b^2 -4ac} }{2a}
      &= \dfrac{-b \pm \sqrt{b^2 -4ac} }{2a}
         \cdot \frac{-b \mp \sqrt{b^2 -4ac}}{-b \mp \sqrt{b^2 -4ac}} \\
      &= \dfrac{b^2-(b^2-4ac)}{2a(-b \mp \sqrt{b^2 -4ac})} \\
      &= \dfrac{4ac}{2a(-b \mp \sqrt{b^2 -4ac})} \\
      &= \dfrac{2c}{-b \pm \sqrt{b^2 -4ac}} \\
\end{align}


问:
How to prove $f(x+1) = g(x)$ is equivalent to $f(x) = g(x-1)$.
How to prove $f(x+1) = g(x)$ is equivalent to $f(x) = g(x-1)$,
given that both $f(x)$ and $h(x)$ are defend on $\mathbb{R}$.

To me this statement is very intuitive, but I don't know how to justify this rigorously.
I would appreciate any helps and comments.

Body:
$$f(x+1)=g(x)$$

$$y=x-1$$

$$f(y+1)=g(y)$$

$$f(x)=g(x-1)$$


答:
Let $x\in\mathbb{R}$ and define $y=x-1\in\mathbb{R}$. If $f(x+1)=g(x)$ for every $x\in\mathbb{R}$, then, in particular it is true for $y=x-1$, which implies that
$$f(x)=f(x-1+1)=f(y+1)=g(y)=g(x-1).$$

Title: How to prove $f(x+1) = g(x)$ is equivalent to $f(x) = g(x-1)$.

Body:
How to prove $f(x+1) = g(x)$ is equivalent to $f(x) = g(x-1)$,
given that both $f(x)$ and $h(x)$ are defend on $\mathbb{R}$.

To me this statement is very intuitive, but I don't know how to justify this rigorously.
I would appreciate any helps and comments.


答:
We have $\ln\tfrac12 -|x| = -2 |x-2|$. Rearranging, we get $$\ln\tfrac12  = |x|-2 |x-2|$$.

Now we split this equation into cases and start solving.

Case $1$: $x \ge 2$, which means that all expressions in absolute value are positive, so we can remove the absolute value signs to get $$\ln\tfrac12  = x-2 (x-2)$$, which can be easily solved to show that $$x = 4 - ln(\frac12)$$. However, we specified that $x \ge 2$in this case, so we need to make sure this condition is satisfied. Using a calculator, we see that $x \approx 4.69$, so it fits and is thus a solution.

Case $2$: $ 0 \le x \le 2$, which means $x$ is positive, so we may remove the signs from $|x|$, but $x-2$ is negative, so we have to remove the signs from $|x-2|$ and negate it to make it $2-x$. Thus, we have $$\ln\tfrac12  = x-2(2-x)$$. We can easily solve thus to find $$x = {ln(\frac12)+4 \over 3}$$. A quick calculation tells us that $x \approx 1.1$, which is in the domain we specified in this case, and is thus a solution.

Case $3$: $x \le 0$, all absolute values are negative, so we remove all the absolute value signs and negate the expressions inside them. Thus our equation becomes $$\ln\tfrac12  = -x-2(2-x)$$, which we can easily solve to find $x= 4 + ln(\frac12)$. A quick calculation tells us that $x \approx 3.3$, which unfortunately does not satisfy this case requirement of $x \le 0$, so discard this solution.

Thus, in conclusion, our solutions are $$x = 4 - ln(\frac12)$$ & $$x={ln(\frac12)+4 \over 3}$$


问:
In need of help for a transitive property for inequalities!
Need help in determining if $\frac{1}{1+x}$ is larger or smaller than $\frac{1}{x+\sin^2 x}$ for $x \geq 1$.

My reasoning for this question being up is that I don't see too many transitive property for inequalities questions on the site!

Body:
$$\cos^2(x)\geq 0$$

$$\cos^2(x)+\sin^2(x)\geq \sin^2(x)$$

$$1\geq \sin^2(x)$$

$$1+x\geq \sin^2(x)+x$$

$$\frac{1}{1+x} \leq \frac{1}{\sin^2(x)+x}$$


答:
Take the difference and study the resulting function:
\begin{align*}
\frac{1}{1+x} - \frac{1}{x + \sin^{2}(x)} & = \frac{(x+\sin^{2}(x)) - (1+x)}{(1+x)(x+\sin^{2}(x))}\\\\
& = \frac{\sin^{2}(x) - 1}{(1+x)(x+\sin^{2}(x))} \leq 0
\end{align*}

Title: In need of help for a transitive property for inequalities!

Body:
Need help in determining if $\frac{1}{1+x}$ is larger or smaller than $\frac{1}{x+\sin^2 x}$ for $x \geq 1$.

My reasoning for this question being up is that I don't see too many transitive property for inequalities questions on the site!


问:
how to calculate probability that a service exceeds x hours
Can some one help me with the workings for part a and b, i have the answers (0.223 and 0.166), but can't  figure out how it was calculated...

If the time taken to service a truck is exponentially distributed with an average of 5 hours.
(a) Calculate the probability that a service exceeds 7.5 hours;
(b) If one particular truck takes more than 6 hours to service, what is the probability that the service takes at least 10 hours?

Thank you

答:
First, find the CDF of the appropriate exponential distribution.
I show answers using R, where pexp is an exponential CDF and the parameter is the rate $\lambda = 1/\mu$.

(a) $P(X > 7.5) = 1 - P(X \le 7.5) = 1 - F_X(7.5) =  0.2231302.$

```<code>1- pexp(7.5, 1/5)
[1] 0.2231302
</code>```

(b) As stated, you're asking for $$P(X > 10|X  > 6) = P(X > 4) = 1 - P(X\le 4) = 0.449329.$$
Use the no-memory property. Knowing that the truck is not finished by 4 hours, we want the probability of more than 4 additional hours.

```<code>1- pexp(4, 1/5)
[1] 0.449329
</code>```

Either I misunderstand the problem or you have the wrong answer. Your answer is for the probability that it will take between 6 and 10 hours to service the truck
(with no advance knowledge it will take more than 6 hours.)
$$P(6 < X \le 10) =  P(X \le 10) - P(X \le 6)\\ = F_X(10) - F_X(6) = 0.1658589.$$

```<code>diff(pexp(c(6,10), 1/5))
[1] 0.1658589
</code>```

I can't say whether the person who got this answer was confused, or whether something
got lost translating the original problem into English.

Body:
First, find the CDF of the appropriate exponential distribution.
I show answers using R, where pexp is an exponential CDF and the parameter is the rate $\lambda = 1/\mu$.

(a) $P(X > 7.5) = 1 - P(X \le 7.5) = 1 - F_X(7.5) =  0.2231302.$

1- pexp(7.5, 1/5)
[1] 0.2231302

(b) As stated, you're asking for $$P(X > 10|X  > 6) = P(X > 4) = 1 - P(X\le 4) = 0.449329.$$
Use the no-memory property. Knowing that the truck is not finished by 4 hours, we want the probability of more than 4 additional hours.

1- pexp(4, 1/5)
[1] 0.449329

Either I misunderstand the problem or you have the wrong answer. Your answer is for the probability that it will take between 6 and 10 hours to service the truck
(with no advance knowledge it will take more than 6 hours.)
$$P(6 < X \le 10) =  P(X \le 10) - P(X \le 6)\\ = F_X(10) - F_X(6) = 0.1658589.$$

diff(pexp(c(6,10), 1/5))
[1] 0.1658589

I can't say whether the person who got this answer was confused, or whether something
got lost translating the original problem into English.


问:
Prove the monotonicity of the expectation of a binary random variable function
Consider $R$ independent binary random variables $y^1, \ldots, y^R$ over the space $\{-1, +1\}$ such that $\Pr(y^j = 1) = p^j \geq 0.5$ and $\Pr(y^j = -1) = 1 - p^j$, $\forall j = 1,\ldots,R$. Moreover, let $\mathcal{J}$ denote a subset of $\mathcal{Y} = \{1, \ldots, R\}$, i.e., $\mathcal{J} \subseteq \mathcal{Y}$. I would like to prove that:

$f(\mathcal{J} \cup \{j'\}) \geq f(\mathcal{J}) \quad \forall \mathcal{J} \in 2^{\mathcal{Y}}, \forall j' \in \mathcal{Y} \setminus \mathcal{J}$

where the function $f( \cdot )$ is defined as follows:

$f(\mathcal{J}) = E\left(| \sum_{j \in \mathcal{J}} y^j  |\right)$

In other words, I want to verify that the afore-mentioned expectation is monotonically non decreasing in the set $\mathcal{J}$. Thanks.

EDIT: I posed a new question about the same proof in case $f(\mathcal{J})$ is altered so that $f(\mathcal{J}) = E\left(| \sum_{j \in \mathcal{J}} logit(p^j) y^j  |\right)$. The new question can be found here: Where is the fallacy in this coupling argument of two Bernoulli variables?

答:
The result holds. By induction on $R$, it suffices to show that if $X=Y_1+\cdots+Y_n$ for independent $\pm1$ Bernoulli random variables $(Y_k)$ with $P[Y_k=1]=p_k\geqslant\frac12$, and if $Y$ is a $\pm1$ Bernoulli random variable independent of $(Y_k)$ with $P[Y=1]=p\geqslant\frac12$, then
$$
E[|X+Y|]\geqslant E[|X|].
$$
Note that $E[|X+Y|]=pE[|X+1|]+(1-p)E[|X-1|]$ and that, for every integer $x$,
$$
p|x+1|+(1-p)|x-1|-|x|=\mathbf 1_{x=0}+(2p-1)(\mathbf 1_{x\gt0}-\mathbf 1_{x\lt0}).
$$
Since $2p-1\geqslant0$ and $X$ is integer valued, $E[|X+Y|]\geqslant E[|X|]$ as soon as
$$
P[X\gt0]\geqslant P[X\lt0].
$$
The last inequality above is not a priori obvious, fortunately it stems from a coupling argument.

To wit, one can assume that, for each $k$, $Y_k=2\mathbf 1_{U_k\leqslant p_k}-1$ where the sequence $(U_k)$ is i.i.d. uniform on the interval $(0,1)$. Let $Y^0_k=2\mathbf 1_{U_k\leqslant\frac12}-1$, then $Y_k^0\leqslant Y_k$ almost surely and the sequence $(Y^0_k)$ is i.i.d. symmetric Bernoulli. In particular, the distribution of $X^0=Y^0_1+\cdots+Y^0_k$ is symmetric hence $P[X^0\gt0]=P[X^0\lt0]$.

Finally, $X^0\leqslant X$ almost surely hence $[X^0\gt0]\subseteq[X\gt0]$ and $[X\lt0]\subseteq[X^0\lt0]$.
Thus, $P[X\gt0]\geqslant P[X^0\gt0]=P[X^0\lt0]\geqslant P[X\lt0]$, and we are done.


问:
Can the axioms for real closed fields be weakened in this way?
I am told that real closed fields $(F,+,-,*,0,1,<)$ can be axiomatized by the axioms for ordered fields and also an axiom stating that every positive element has a square root and an axiom schema stating that every odd-degree polynomial has a root. I am wondering whether this can be weakened somewhat. Would it suffice to say, instead of the axiom schema for odd-degree polynomials, that for every odd natural number $n$, every element of $F$ has an $n$-th root? And if not, can someone exhibit a model of this weaker theory that is not a real-closed field?

答: Let $R$ be the ordered subfield of $\mathbb{R}$ consisting of all real numbers which are constructible by radicals over $\mathbb{Q}$ (the union of all radical extensions of $\mathbb{Q}$ contained in $\mathbb{R}$). This is an ordered field in which every positive element has a square root and every element has an $n$th root for all odd $n$. But it is not real closed since (by Abel-Ruffini / Galois) there are polynomials of degree $5$ over $\mathbb{Q}$ with real roots which are not constructible by radicals.


问:
Get a result and calculate back to it's Divisor
In the calculation 28 /7 = 4 the result is 4 and the divisor is 7. From the result i want to calculate back into the divisor. In other words, all I have to do in this case is  is 4 + 3 and i get back to the divisor 7.

But that does not work when i do 28 / 2 = 14. In this case I have to do 14 - 12 = 2 to get back to the divisor.

Is there a formula or equation that I can apply that calculates the result back to the divisor every time?

答:
Yes, if you know both the original product and the resulting quotient. If all you're allowed to use is your final result, then no.

Given a calculation of the form $P/D = Q$ (where $P$ is the original product, $D$ is the divisor, and $Q$ is the resulting quotient), you can "calculate the result back to the divisor" by using the formula $D=P/Q$.

Thus, in your first example, we have $P = 28$ and $Q=4$, so the divisor was $D=28/4=7$.

For your second example, we have $P = 28$ and $Q=14$, so the divisor was $D=28/14=2$.


问:
Monty Hall Problem + dice - Why does the probability of whining the car raises after switching doors?
I'm trying to understand the Monty Hall problem. When the contestant switches the door the probability of them winning a car is higher than if they continued with their first choice. Why would the probability change from 1/3 go up to 2/3 if both doors had the same chances of being opened?

Update:
There are many posts about this subject in SO but none of then answer my question. Let me re define my question.

I know that there are many models out there that shows that if you change the doors you have 2/3 of probability (higher than 1/3). After Graham Kemp answer I will update a scenario that I have in my mind.

You choose door1 (1/3) and the host opens the door 2 (the goat). By the current solution the door 3 has  2/3 of probability to be the one with the car. Now let's think a little change. After the host open the door 2 he will give you a coin with numbers 1 and 3 and you should torn it. Now is not a choice, you should stay with the number you get in the coin. In this new scenario based on the previous model the door 3 has 2/3 of success probability than door 1. With clearly makes no sense. In this new scenario why door 3 is not 1/2 of probability? I am pretty sure that a model with 10M attempts will show 50% for each door. Is it correct?

答:
You select a door at random, knowing nothing about the placement of the car, other than that is presumably behind any particular door with equal probability (1/3).

Monty opens one of the other two doors, knowing the location of the car, he always chooses to reveal a goat.

You are asked if you want to switch to the remaining door.

If the car was behind your door, you cannot win if you switch.  It was there with probability 1/3.

If the car was not behind your door, you surely win if you switch.  It wasn't there with probability 2/3.

Therefore the probability that you win if you switch is 1 minus the probability that the car was behind the door which you selected.  IE: 2/3


问:
Supporting hyperplane of a closed ball in $\mathbb{R}^n$
Let $r>0$ and $\overline{B}(0,r)$ be the closed ball of center $0$ and radius $r$ in $\mathbb{R}^n$ with its usual norm.

Let $a$ be such that $\left\|{a}\right\|=r$. Consider the hyperplane $$H=\left\{{x:x\cdot \dfrac{a}{\left\|{a}\right\|}=\left\|{a}\right\|}\right\}.$$

I want to prove that $H$ is a supporting hyperplane of $\overline{B}(0,r)$, in other words I want to prove that

$$\overline{B}(0,r)\subset \left\{{x:x\cdot \dfrac{a}{\left\|{a}\right\|}\le\left\|{a}\right\|}\right\}.$$

But I have been unable to prove this. I tried the following:

Let $x\in\overline{B}(0,r)$. It would be equivalent to proving that $x\cdot a\le \left\|{a}\right\|^2=a\cdot a$, or $(x-a)\cdot a\le 0$. I think something could be done from here but I don't know how.

Any hints? Thank you.

答:
The mapping $f(x) = x \cdot a$ is a linear functional on $\mathbb{R}^n$, hence its kernel is a hyperplane in $\mathbb{R}^n$ and passes through the origin.  This kernel, it helps to think, is the orthogonal complement of $a$.

Let's prove that $H$ is a translate of $\ker f$ by $a$; i.e., that
$$
H = a + \ker f.
$$
Before we go for it, note that $f(a) = ||a||^2$.

Now, let's show that $H$ is contained in $a + \ker f$.  To see this, suppose $y \in H$. Now let
$$
z = y - a.
$$
(Here, it helps to stop and sketch the above.)  It follows that $f(y) = ||a||^2$, and therefore
$$
f(z) = 0,
$$
so $z$ lies in $\ker f$, so $y$ lies in $a + \ker f$.  We have shown that $a + \ker f$ contains $H$.

To show the containment in the other direction, note that for every $z$ in $\ker f$, we have
$$
f(a + z) = ||a||^2,
$$
so $(a + z)$ lies in $H$.

Now, it is fairly straightforward to show that the closed ball $\overline{B}(0, r)$ is contained in the closed half-space $f \leq ||a||^2$ and intersects $H$ at just one point, namely $a$.


问:
critical value set is noncompact
From Sard's theorem, we know that the critical value of smooth "enough" function must be of measure zero. But this does not exclude the case of noncompactness. My question is: is there any smooth function with noncompact critical value set? What about real analytic function? What about real subanalytic function?</p>

答: Let $f(x) = \sin(x)+ x/2$. This real analytic function has unbounded and hence non-compact critical set.

Title: critical value set is noncompact


问:
Why does an exponential function eventually get bigger than a quadratic
I have seen the answer to this question and this one.

My $7$th grade son has this question on his homework:

How do you know an exponential expression will eventually be larger than any quadratic expression?

I can explain to him for any particular example such as $3^x$ vs. $10 x^2$ that he can just try different integer values of $x$ until he finds one, e.g. $x=6$. But, how can a $7$th grader understand that it will always be true, even $1.0001^x$ will eventually by greater than $1000 x^2$? They obviously do not know the Binomial Theorem, derivatives, Taylor series, L'Hopital's rule, Limits, etc,

Note: that is the way the problem is stated, it does not say that the base of the exponential expression has to be greater than $1$. Although for base between $0$ and $1$, it is still true that there exists some $x$ where the exponential is larger than the quadratic, the phrase "eventually" makes it sound like there is some $M$ where it is larger for all $x>M$. So, I don't like the way the question is written.

答:
By shifting and rescaling, you can turn any comparison of a quadratic vs. exponential into a standard one.  Say you are trying to determine which of $f(x) = \alpha(x-\beta)^2$ and $g(x) = ab^x$ grows faster.  By shifting $x\rightarrow x+\beta$ you can turn this into a comparison of $f_{1}(x) = \alpha x^2$ and $g_{1}(x) = a' b^x$.  Then by rescaling $x\rightarrow c x$ where $b^c = 2$, you turn the comparison into $f_2(x) = \alpha' x^2$ and $g_2(x) = a'' 2^x$.

So as long as you can convince your son that $2^x$ grows faster than $a x^2$, all other exponentials must also grow faster than all other quadratics.


问:
Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$
I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:

$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$

I really have no idea why this statement is true. Can someone please explain why this is true and if possible show how to arrive at one given the other?

答:
A combinatorial proof:

Let $S=\{1,2,\dots,(n+1)\},n\ge 2$ and $T=\{(x,y,z)|x.y,z\in S,x< z,y< z\}$.By counting the number of members of $T$ in $2$ different ways I will prove the formula.

$1$st way:

We will at first Choose $z$ form the set $S$.When $z$ is $1$ then there are no choices for $x,y$ so the no. of elements of $T$ with $z=0$ is zero.When $z=2$ the number of choices for $x$ is $1$ and so is for $y$(precisely $x=y=1$).When $z=3$ then $x\in \{1,2\}$ and $y\in \{1,2\}$ so total no. of choices equals $2^2$.In a similar manner when $z=k,(1\le k\le (n+1))$,no. of choices for $x$ equals $(k-1)$ and no. of choices for $y$ is also $(k-1)$. So total no . of elements of T with $z=k$ is $(k-1)^2$.

So we will get the total no. of elements of $T$ by summing $(k-1)^2$ up from $1 $ to $(n+1)$.Hence $$|T|=\sum _{l=1}^{(n+1)}(l-1)^2=\sum_{k=1}^{n}k^2$$

$2$nd way:

Among the elements of $T$ consisting of three numbers from the set $S$, there are elements in $x=y$ and elements in which $x\ne y$.

We can count the no. of elements in which $x=y$ by choosing two distinct nos. from $S$ and assigning $z$ with the lagest no. and $x,y$ with the smallest number. We can choose two distinct numbers from $S$ in $\displaystyle \binom{n+1}{2}$ ways, so the total no. elements having $x=y$ is $\displaystyle \binom{n+1}{2}$.

Now we have to count the number of elements in which $x\ne y$.This means that $x,y$ are dinstict and as they are less than $z$ this means that all the three are distinct. So we can count no. of such elements in $T$ in the following way.At first we will choose three elements from the set $S$ and assign the largest value to $z$ and assign the other two values to $x,y$. Now we can choose three numbers from the set $S$ in $\displaystyle \binom{n+1}{3}$.From each such three element we can get two elements of the set $T$(Assigning the largest to $z$ and then assigning any one of then to $x$ and the other to $y$). So no. of elements of $T$ having $x\ne y$ is $2\displaystyle \binom{n+1}{3}$

So by this method we have $|T|=\displaystyle \binom{n+1}{2}+2\displaystyle \binom{n+1}{3}$

On equating the result obtained from both the methods we have
$$\sum_{k=1}^{n}k^2=\displaystyle \binom{n+1}{2}+2\displaystyle \binom{n+1}{3}=\frac{n(n+1)(2n+1)}{6}$$

Note that this can easily be extended to find the sum of $p$th power of integers.($p\in \mathbb{N})$


问:
Finding the integral of $\frac{1}{x(x^n+a^n)}$
I was working on the problems in Mathematical Methods for Physics and Engineering by Riley,Hobson & Bence.
In Problem 2.34 (d) I'm supposed to find this integral: $$J=\int\frac{dx}{x(x^n+a^n)}.$$
I used partial fractions and arrived at the form
$$J=\frac{1}{a^n}\left[\log x-\int \frac{dx}{x^n+a^n}\right]$$
and now I'm stuck, I don't know how to integrate $1/(x^n+a^n)$.

答:
\begin{align*}
J&=\int\frac{dx}{x(x^n+a^n)}\\
&=\frac1{a^n}\int\left(\frac1{x}-\frac{x^{n-1}}{x^n+a^n}\right)dx\\
&=\frac1{a^n}\ln|x|-\frac1{na^n}\int\frac{nx^{n-1}}{x^n+a^n}dx\\
&=\frac1{a^n}\ln|x|-\frac1{na^n}\ln|x^n+a^n|+C,\\
\end{align*}
where $C$ is the constant of integration.

Alternatively, we have
\begin{align*}
J&=\int\frac{x^{-n-1}dx}{1+a^nx^{-n}}\\
&=-\frac1n\int\frac{(x^{-n})'dx}{1+a^nx^{-n}}\\
&=-\frac1{na^n}\int\frac{(1+a^nx^{-n})'dx}{1+a^nx^{-n}}\\
&=-\frac1{na^n}\ln|1+a^nx^{-n}|+C.\\
\end{align*}


答:
We seek a decomposition of the form
$$\frac{1}{x(x^n + a^n)} = \frac{A}{x} + \frac{Bx^{n-1}}{x^n + a^n} = \frac{(A+B)x^n + Aa^n}{x(x^n+a^n)}.$$  Hence the choice $A = a^{-n}$, $B = -A = -a^{-n}$, yields
$$\frac{1}{x(x^n+a^n)} = \frac{1}{a^n} \left(\frac{1}{x} - \frac{x^{n-1}}{x^n + a^n}\right).$$
The rest is straightforward:

$$\int \frac{dx}{x(x^n+a^n)} = \frac{1}{a^n} \left(\log |x| - \frac{1}{n} \int \frac{nx^{n-1}}{x^n + a^n} \, dx \right) = \frac{1}{a^n} \left( \log |x| -\frac{1}{n} \log |x^n + a^n| \right) + C.$$

Title: Finding the integral of $\frac{1}{x(x^n+a^n)}$

Body:
I was working on the problems in Mathematical Methods for Physics and Engineering by Riley,Hobson & Bence.
In Problem 2.34 (d) I'm supposed to find this integral: $$J=\int\frac{dx}{x(x^n+a^n)}.$$
I used partial fractions and arrived at the form
$$J=\frac{1}{a^n}\left[\log x-\int \frac{dx}{x^n+a^n}\right]$$
and now I'm stuck, I don't know how to integrate $1/(x^n+a^n)$.


问:
Formula for the number of edges of complete $m$-partite graph
The complete $m$-partite graph on $n$ vertex in which each part has either $[n/m]$ ($n/m$ rounded down to an integer) or $\{n/m\}$ ($n/m$ rounded up to an integer) vertices is denoted by $T_{m,n}$. Show that
\begin{equation}
\epsilon(T_{m,n})=\binom{n-k}{2}+(m-1)\binom{k+1}{2},\mathrm{where} \;k=[n/m].
\end{equation}

My efforts:

Use induction. Obviously $\epsilon(T_{2,2})=1$ and the formula is correct. First fix $m$ and vary $n$. Assume the formula holds for $n$. Consider $\epsilon(T_{m,n+1})$. One vertex is added to the partition with [$n/m$] vertices. The new vertex can be connected to any vertex in the other partitions. Thus
\begin{equation}
\begin{split}
\epsilon(T_{m,n+1}) & = \epsilon(T_{m,n})+n-[n/m] \\
 & = \binom{n-[n/m]}{2}+(m-1)\binom{[n/m]+1}{2}+n-[n/m]
\end{split}
\end{equation}
We need to consider two cases: $[(n+1)/m]=[n/m]$ and $[(n+1)/m]=[n/m]+1$.

(1) $[(n+1)/m]=[n/m]$.
\begin{equation}
\begin{split}
\epsilon(T_{m,n+1}) & = \binom{n-[\frac{n+1}{m}]}{2}+(m-1)\binom{[\frac{n+1}{m}]+1}{2}+\binom{n-[\frac{n+1}{m}]}{1} \\
 & = \binom{n+1-[\frac{n+1}{m}]}{2}+(m-1)\binom{[\frac{n+1}{m}]+1}{2}
\end{split}
\end{equation}
The formula is correct.

(2) $[(n+1)/m]=[n/m]+1$. This happens only if $(n+1)/m$ is an integer.
\begin{equation}
\begin{split}
\epsilon(T_{m,n+1}) & = \binom{n+1-[\frac{n+1}{m}]}{2}+\frac{1}{2}(m-1)([\frac{n}{m}]+1)[\frac{n}{m}]+n-[\frac{n}{m}] \\
 & = \binom{n+1-[\frac{n+1}{m}]}{2}+\frac{1}{2}(m-1)([\frac{n+1}{m}])[\frac{n}{m}]+n-[\frac{n}{m}] \\
 & = \binom{n+1-[\frac{n+1}{m}]}{2}+\frac{m-1}{2}([\frac{n+1}{m}])[\frac{n}{m}]+m[\frac{n+1}{m}]-[\frac{n+1}{m}] \\
 & = \binom{n+1-[\frac{n+1}{m}]}{2}+\frac{m-1}{2}([\frac{n+1}{m}])[\frac{n}{m}]+(m-1)[\frac{n+1}{m}] \\
 & = \binom{n+1-[\frac{n+1}{m}]}{2}+\frac{m-1}{2}[\frac{n+1}{m}]([\frac{n}{m}]+2) \\
 & = \binom{n+1-[\frac{n+1}{m}]}{2}+\frac{m-1}{2}[\frac{n+1}{m}]([\frac{n+1}{m}]+1) \\
 & = \binom{n+1-[\frac{n+1}{m}]}{2}+\binom{[\frac{n+1}{m}]+1}{2}
\end{split}
\end{equation}
The formula is again correct.

Then I don't know how to do induction of fixing $n$ and varying $m$. Is there any elegant counting method without using induction?

答:
The idea of this proof is that we can count pairs of vertices in our graph of a certain form. Some of them will be edges, but some of them won't be. When we get a pair that isn't an edge, we will give a bijective map from these "bad" pairs to pairs of vertices that correspond to edges.

Label the parts $X_1, X_2, \ldots, X_m$, and label the vertices in $X_i$ by $x_{i,1}, x_{i,2},\ldots, x_{i,j}$ where either $j=k$ or $j=k+1$. Without loss of generality, assume that $|X_1|=k$, since there must be at least one part of size $k$. The term ${n-k\choose 2}$ counts the number of ways to pick pairs of vertices $\{x_{i,j}, x_{k,l}\}$ from $X_2, X_3, \ldots, X_m$. If $i\neq k$, then $\{x_{i,j}, x_{k,l}\}$ is an edge in the graph. Otherwise, we have $i=k$. We give a map from such pairs of vertices to edges in the graph. Without loss of generality, assume $j<l$. We map the pair $\{x_{i,j}, x_{i,l}\}$ to the pair $\{x_{1,j}, x_{i,l}\}$, which is an edge in our graph. This counts every edge between two vertices not from $X_1$, and every edge of the form $\{x_{1,j}, x_{i,l}\}$ where $i>1$ and $j<l$.

It remains to count the vertices of the form $\{x_{1,j}, x_{i,l}\}$ where $i>1$ and $j\geq l$. First, choose one of the $m-1$ parts other than $X_1$. Let this part be $X_i$. Add a vertex denoted $x_{1,k+1}$ to $X_1$. There are ${k+1\choose 2}$ ways to choose a pair of vertices $\{x_{1,j}, x_{1,l}\}$ from $X_1\cup\{x_{1,k+1}\}$. Without loss of generality, assume $j<l$. If $l<k+1$, then map this pair of vertices to the pair $\{x_{i,j}, x_{1,l}\}$, which is an edge in the graph. Otherwise, if $l=k+1$, we map this pair of vertices to the pair $\{x_{i,j}, x_{1,j}\}$. Hence, we've counted the rest of the edges of the desired form, and these edges are counted by $m{k+1\choose 2}$, completing the proof.


答:
You don't need to separately induct on $m$.  Instead, you can just use a separate base case for each $m$, and then use the induction you have done on $n$ to cover every possible value of $n$.  Assuming you require $n\geq m$, this just means you have to check the case of $T_{m,m}$ for each $m$.  (There's not really any reason to require $n\geq m$, though; there's nothing wrong with parts being empty.  So, really the base case is $T_{m,0}$ for each $m$, which is quite easy to check since you just get $0$ edges.)

That said, you can also just count directly without induction.  Kevin Long's answer shows a clever way to do this to make the answer pop out, but you can also count without any cleverness and then just bash out algebra.   Let $n=km+r$, so we have $r$ parts of size $k+1$ and $m-r$ parts of size $k$.  Each vertex in a part of size $k+1$ is connected to $n-(k+1)$ other vertices, so this gives $r(k+1)(n-(k+1))$ edges.  Each vertex in a part of size $k$ is connected to $n-k$ other verices, giving $(m-r)k(n-k)$ edges.  But this counts each edge twice (once for each vertex), so we have a total of $$\frac{r(k+1)(n-(k+1))+(m-r)k(n-k)}{2}$$ edges in total.  Now it's just a matter of substituting $r=n-km$ and doing a bunch of algebra to verify this is equal to $\binom{n-k}{2}+(m-1)\binom{k+1}{2}$.

Title: Formula for the number of edges of complete $m$-partite graph

Body:
The complete $m$-partite graph on $n$ vertex in which each part has either $[n/m]$ ($n/m$ rounded down to an integer) or $\{n/m\}$ ($n/m$ rounded up to an integer) vertices is denoted by $T_{m,n}$. Show that
\begin{equation}
\epsilon(T_{m,n})=\binom{n-k}{2}+(m-1)\binom{k+1}{2},\mathrm{where} \;k=[n/m].
\end{equation}

My efforts:

Use induction. Obviously $\epsilon(T_{2,2})=1$ and the formula is correct. First fix $m$ and vary $n$. Assume the formula holds for $n$. Consider $\epsilon(T_{m,n+1})$. One vertex is added to the partition with [$n/m$] vertices. The new vertex can be connected to any vertex in the other partitions. Thus
\begin{equation}
\begin{split}
\epsilon(T_{m,n+1}) & = \epsilon(T_{m,n})+n-[n/m] \\
 & = \binom{n-[n/m]}{2}+(m-1)\binom{[n/m]+1}{2}+n-[n/m]
\end{split}
\end{equation}
We need to consider two cases: $[(n+1)/m]=[n/m]$ and $[(n+1)/m]=[n/m]+1$.

(1) $[(n+1)/m]=[n/m]$.
\begin{equation}
\begin{split}
\epsilon(T_{m,n+1}) & = \binom{n-[\frac{n+1}{m}]}{2}+(m-1)\binom{[\frac{n+1}{m}]+1}{2}+\binom{n-[\frac{n+1}{m}]}{1} \\
 & = \binom{n+1-[\frac{n+1}{m}]}{2}+(m-1)\binom{[\frac{n+1}{m}]+1}{2}
\end{split}
\end{equation}
The formula is correct.

(2) $[(n+1)/m]=[n/m]+1$. This happens only if $(n+1)/m$ is an integer.
\begin{equation}
\begin{split}
\epsilon(T_{m,n+1}) & = \binom{n+1-[\frac{n+1}{m}]}{2}+\frac{1}{2}(m-1)([\frac{n}{m}]+1)[\frac{n}{m}]+n-[\frac{n}{m}] \\
 & = \binom{n+1-[\frac{n+1}{m}]}{2}+\frac{1}{2}(m-1)([\frac{n+1}{m}])[\frac{n}{m}]+n-[\frac{n}{m}] \\
 & = \binom{n+1-[\frac{n+1}{m}]}{2}+\frac{m-1}{2}([\frac{n+1}{m}])[\frac{n}{m}]+m[\frac{n+1}{m}]-[\frac{n+1}{m}] \\
 & = \binom{n+1-[\frac{n+1}{m}]}{2}+\frac{m-1}{2}([\frac{n+1}{m}])[\frac{n}{m}]+(m-1)[\frac{n+1}{m}] \\
 & = \binom{n+1-[\frac{n+1}{m}]}{2}+\frac{m-1}{2}[\frac{n+1}{m}]([\frac{n}{m}]+2) \\
 & = \binom{n+1-[\frac{n+1}{m}]}{2}+\frac{m-1}{2}[\frac{n+1}{m}]([\frac{n+1}{m}]+1) \\
 & = \binom{n+1-[\frac{n+1}{m}]}{2}+\binom{[\frac{n+1}{m}]+1}{2}
\end{split}
\end{equation}
The formula is again correct.

Then I don't know how to do induction of fixing $n$ and varying $m$. Is there any elegant counting method without using induction?


问:
How to evaluate partial derivative of $f$ with respect to $x$ as a function of $y$ only?
Hopefully this is just the notation confusing me, but how would I correctly evaluate the following function $g$ below?

$g(x,y) = \frac{\partial f(2y, y)}{\partial x}$ (or equivalently $g(x,y) = f_{x}(2y, y)$), where $f(x,y) = x + y$.

On one hand, $\frac{\partial f(x, y)}{\partial x} = 1$, so we could have $g(x,y) = 1$. On the other hand, $f(2y, y) = 3y$, suggesting $g(x,y) = \frac{\partial (3y)}{\partial x} =0.$

Edit:

To add some context, here is where this came from ($\phi$ below is a function of $x$ and $t$):

Define the mapping $T$ from $C^{\infty}_{0}(\mathbb{R}^2)$ to $\mathbb{R}$ by $T: \phi \mapsto \int_{0}^\infty \phi(t, ct)dt$, where $c$ is a real number. What is $\partial_x T$?

答:
I think the notation is meant to be understood as

$$g(x,y) \enspace = \enspace \frac{\partial f(x,y)}{\partial x} \; \Big|_{x = 2y} $$

So, your answer $g(x,y) = 1$ would be correct. However, I would rather consider the context carefully - hard to say what was really meant here since the whole notation seems a bit off, e.g. the LHS still holds the dependency for $x$, whilst the RHS doesn't. So, I'd probably write it as

$$ g(x = 2y, y ) \enspace = \enspace \frac{\partial f(x,y)}{\partial x} \; \Big|_{x = 2y} $$

$${}$$

EDIT:

With respect to your latest edit, I would understand your map as follows: The $x$-variable of $\phi(t,x)$ is a function of $t$, so one can rewrite it as $\phi \big( t, x(t) \big)$.

Therefore, your derivative can be solved as:

$$ \partial_x T \enspace = \enspace \partial_x T \big( \phi \big) \enspace = \enspace \partial_x T \big( \phi(t,x(t)) \big)$$

$$= \enspace \int_0^{\infty} \partial_x \phi \big( t, x(t) \big) \, dt \enspace = \enspace \int_0^{\infty} \frac{\partial \phi \big( t, x(t) \big)}{\partial x} \, dt$$

So, you would have to derive $\phi(t,ct)$ as if $ct$ was $x$ and then - AFTER deriving it, but still BEFORE integrating it - fill in $x = ct$. However, I am not entirely sure.


答:
\begin{gather*}
Let\ I=\int \frac{dx}{x\left( x^{n} +a^{n}\right)} =\int \frac{dx}{x^{n+1}} \cdotp \frac{1}{1+\frac{a^{n}}{x^{n}}}\\
Let\ 1+\frac{a^{n}}{x^{n}} =t\\
\frac{-n\cdotp a^{n}}{x^{n+1}} dx=dt\\
\frac{dx}{x^{n+1}} =\frac{-dt}{n\cdotp a^{n}}\\
I=\int \frac{-dt}{n\cdotp a^{n}} \cdotp \frac{1}{t} =\frac{-1}{n\cdotp a^{n}}\ln t=\frac{-1}{n\cdotp a^{n}}\ln\left( 1+\frac{a^{n}}{x^{n}}\right)\\
=\frac{-1}{n\cdotp a^{n}}\ln\left(\frac{x^{n} +a^{n}}{x^{n}}\right) =\frac{1}{a^{n}}\left(\ln x-\frac{1}{n}\ln\left( x^{n} +y^{n}\right)\right)\\
\end{gather*}
Hope this helps!