答:
I think you're a little confused about the steps in this problem. Note that, after you've multiplied both sides by the denominator, you need to try and solve the resulting equation, in this case $$7 = x(A+2B) + 2A - 3B.$$ Like terms are the coefficients of identical powers of $x$. Observe that $7 = 0x + 7$. Can you see the similarity now? Had $(A+2B)$ been anything but $0$, you'd have a non-zero $ax$ term on the left-hand side of the equation above. The same logic applies to $(2A-3B)$.

So really you end up with $$A+2B = 0 \\ 2A - 3B = 7$$ which, when solved simultaneously, gives $A= 2$, $B = -1$.

Title: What does it mean to equate the coefficients of like terms when solving for A and B in partial fractions?

Body:
I'm trying to step myself through solving partial fractions in a year 10 book by Cambridge. This is a concept they're introducing early for students who want to challenge themselves and it's pretty light on the explanation.

For example: 7/( x+2 ) ( 2x-3 ) = A/2x-3 + B/x+2. I understand how to work this to the point where I reach 7=x(A+2B)+2A−3B. From there I've read that I need to do something called "equating coefficients. The coefficients near the like terms should be equal, so the following system is obtained: A+2B=0 2A−3B=7.

But I don't understand WHY or how it is valid that we set these parts of the equation to these values. Why not A+2B=7 2A−3B=0 for instance. I've tried looking at YouTube and asking friends, but I can't seem to get my head around it.

I can do it and I can solve for A and B using this method. But I'm really struggling to understand what it is I'm doing at that point in the process.  The phrase that keeps popping up when I look into this is "we can equate the coefficients of like terms".  For example on the wikipedia page on Fraction Decomposition it says "Equating the coefficients of x and the constant (with respect to x) coefficients of both sides of this equation...".  Second example: it says "The coefficients near the like terms should be equal, so the following system is obtained: " on the emathhelp page when I enter the equation 7/( x+2 ) ( 2x-3 ).


问:
Inequality with mean inequality

If $a,b,c > 0$ and $a+b+c = 18 abc$, prove that
$$\sqrt{3 +\frac{1}{a^{2}}}+\sqrt{3 +\frac{1}{b^{2}}}+\sqrt{3 +\frac{1}{c^{2}}}\geq 9$$

I started writing the left member as $\frac{\sqrt{3a^{2}+ 1}}{a}+ \frac{\sqrt{3b^{2}+ 1}}{b} + \frac{\sqrt{3c^{2}+ 1}}{c}$ and I applied AM-QM inequality, but I obtain something with $\sqrt[4]{3}$.

答:
Hints:

Put Let $a=\frac{x}{3\sqrt2},$ $b=\frac{y}{3\sqrt2}$ and $c=\frac{z}{3\sqrt2}.$

And $x=\tan(u)$,$y=\tan(v)$,$z=\tan(w)$ with $u+v+w=\pi$

Then the function $f(u)=\sqrt{3+\frac{18}{\tan(u)^2}}$ is convex on $(0,\pi)$ see here

Remains to apply Jensen's inequality .

ContentLicense: CC BY-SA 4.0

Title: Inequality with mean inequality


问:
Summation of integration terms
If $I = \sum\limits_{k = 1}^{98} {\int\limits_k^{k + 1} {\frac{{k + 1}}{{x\left( {x + 1} \right)}}} dx} $ then

(A) $I>\ln99$

(B) $I<\ln99$

(C) $I<\frac{49}{50}$

(D) $I>\frac{49}{50}$

The official answer is B and D

My approach is as follow

$$I = \sum\limits_{k = 1}^{98} {\int\limits_k^{k + 1} {\frac{{k + 1}}{{x\left( {x + 1} \right)}}} dx} $$

$$\frac{A}{x} + \frac{B}{{x + 1}} = \frac{{A\left( {x + 1} \right) + Bx}}{{x\left( {x + 1} \right)}} = \frac{{x\left( {A + B} \right) + A}}{{x\left( {x + 1} \right)}}$$

$$A = k + 1\text{ and } B =  - \left( {k + 1} \right)$$

$$I = \sum\limits_{k = 1}^{98} {\int\limits_k^{k + 1} {\left( {\frac{{k + 1}}{x} - \frac{{k + 1}}{{x + 1}}} \right)} dx}  \Rightarrow I = \sum\limits_{k = 1}^{98} {\left( {k + 1} \right)\int\limits_k^{k + 1} {\left( {\frac{1}{x} - \frac{1}{{x + 1}}} \right)} dx} $$

$$I = \sum\limits_{k = 1}^{98} {\left( {k + 1} \right)\left. {\ln\frac{x}{{x + 1}}} \right|_k^{k + 1}}  \Rightarrow I = \sum\limits_{k = 1}^{98} {\left( {k + 1} \right)\left( {\ln\frac{{k + 1}}{{k + 2}} - \ln\frac{k}{{k + 1}}} \right)}$$  $$\Rightarrow I = \sum\limits_{k = 1}^{98} {\left( {k + 1} \right)\left( {\ln\frac{{{{\left( {k + 1} \right)}^2}}}{{k\left( {k + 2} \right)}}} \right)} $$

Not able to approach from here

Title: Summation of integration terms

Body: Hint: $$\sum_1^{98} (k+1)\ln\frac{(k+1)^2}{k(k+2)} \\ = \sum_1^{98} (k+1) \left( \ln\frac{k+1}{k} - \ln \frac{k+2}{k+1} \right) \\ = 2(\ln 2-\ln\frac 32) + 3(\ln \frac 32 -\ln \frac 43) +4(\ln \frac 43 -\ln \frac 54) +\dots +99(\ln \frac{99}{98} -\ln \frac{100}{99}) \\ =2\ln 2 + \big( \ln\frac 32 +\ln \frac 43 +\dots +\ln \frac{99}{98} \big) -99\ln\frac{99}{100} \\ =2\ln 2+\ln\frac{99}{2} -99\ln\frac{99}{100}$$


问:
5-adic numbers represented by binary quadratic form
This question comes from Borevich-Shafarevich Number Theory CH.1 Sec.6, from which I am teaching myself about p-adic numbers.

It asks you to find all 5-adic numbers represented by the form $f = 2x^2 + 5y^2$

I have used the Hilbert Symbol to find that $(5,2)_5 = -1$ , and therefore $2x^2 + 5y^2 -z^2$ does not represent $0$, ie I know that the form f does not represent squares in $\mathbb{Q}_5^*$. I have also considered finding an equivalent form and looking at its representations but am not sure this would be much help.

Thank you in advance.

答:
If $\infty\ne n=v(x)\le v(y)$ then $2x^2+5y^2 \in 5^{2n}  2 (\pm 1+5\Bbb{Z}_5)$. Conversely any element of $5^{2n}  2 (\pm 1+5\Bbb{Z}_5)$ is represented by $2x^2$.

If $\infty\ne m=v(y)< v(x)$ then $2x^2+5y^2 \in 5^{2m+1} (\pm 1+5\Bbb{Z}_5)$. Conversely any element of $5^{2m+1} (\pm 1+5\Bbb{Z}_5)$ is represented by $5y^2$.

So the quadratic form represents $$0\cup 5^{2\Bbb{Z}}  (\pm 2+5\Bbb{Z}_5)\cup 5^{2\Bbb{Z}+1} (\pm 1+5\Bbb{Z}_5)$$


问:
Why substitution method does not work for $\int (x-\frac{1}{2x} )^2\, \mathrm dx$?
Why $$\int \ \left(x-\frac{1}{2x} \right)^2 \, \mathrm  dx$$ is easy to integrate once $$\left(x-\frac{1}{2x} \right)^2$$ is expanded, but impossible using substitution method? (tried 5 different subs but of course that is not the proof that there is no suitable substitution) if mathematical results are independent of the logical methods used to derive them, why something so simple works one way but not the other? </p>

答:
The situation you describe is not too dissimilar to the following: imagine standing in front of a wall with a few doors in it. You want to go in. At first you don't notice any doors so you wander up and down in front of the wall, until eventually you realize there was a door right in front of you all along. Then since you believe there's only one room inside, you are surprised that there wasn't a door everywhere else on the wall outside. Does this make clearer why your surprise is unfounded?

Put another way, equivalence of any correct route doesn't show existence of any proposed route.

ContentLicense: CC BY-SA 3.0


问:
modelling smooth square wave function
I would like to model a smooth square wave function (see graph) but be able to control the amount of smoothness from fairly soft as shown to quite hard.

But I want to control the amount of smoothness using a simple function so the more the graph goes to a crest or peak the more compressed it becomes depending on some parameter that I can adjust.

I thought something sine logarithmic as that tends to compress but could not get that to work. I also want to avoid such things as the sgn function or Fourier series.

答:
A generic method for drawing a controllable "smoothed" version of a function that is sharp:

Generate the "sharp" function as a time series $S=\{s_0,s_1,\cdots\}$

Convolve $S$ with a smoothing kernel, e.g. $[{1 \over 4},{1\over 2},{1\over4}]$

If it is not smooth enough, convolve again.

Playing with more sophisticated kernels will give you more freedom and control.

Title: modelling smooth square wave function


问:
primes represented by norm form of cubic with square discriminant
I've asked similar questions before. Currently I would like to have the justification filled in for this: we wish to find the primes integrally represented by the norm form of an abelian totally real cubic field. Which is to say, given a monic irreducible polynomial $x^3 + a x^2 + b x + c$   with discriminant an integer square, I believe that the represented primes are defined entirely by ordinary congruences.

I think this is the easy case.

The more difficult cases of non-square discriminant are discussed in, for instance, Williams

Primes represented by $x^3-21xy^2+35y^3$.

That would be the question, is it really really true that  congruences suffice to determine the represented primes, what sort of arguments are involved, and where might one find full detail? Also, is there a method similar to Gauss reduction that can construct a representation of a given prime? I will play with that for some easy examples, see what might work. Relevant are $x^3 - 3 x + 1$ and $x^3 + x^2 - 2x - 1$ of discriminants $3^4$ and $7^2.$  ADDED: the other very strong property I expect is that, for any prime $q$ that is not represented and does not ramify, the represented number can only be divisible by $q$ when every variable in the norm form is divisible by $q,$  so that the number is divisible by $q^3$

For one of the monic polynomials, take the companion matrix, call it $M,$  after which the norm form is just
$$  f(x,y,z) = \det \left( xI + y M + z M^2 \right)  $$
and see what numbers happen with integer $x,y,z.$  For the first, I get $3$ and all primes $p \equiv \pm 1 \pmod 9$ For the second, I get $7$ and all primes $p \equiv \pm 1 \pmod 7.$

The two examples from the field database represent $1, 8, 55, 62 \pmod {63} $  Taken as polynomials, the second discriminant is $4$ times the first.

The counterexample of Pisco  is cubic, discriminant $163^2$ while the Hilbert class field has degree 12 and is not abelian. All words to look up, I have Cox

答:
Your belief

Given a monic irreducible polynomial $x^3+ax^2+bx+c$ with discriminant an
integer square, I believe that the represented primes (of norm cubic form) are defined
entirely by ordinary congruences.

is unfortunately false in general.

If $K/\mathbb{Q}$ is an abelian cubic field, then your assertion is true for $K$ (i.e. prime represented can be expressed via congruence) iff its Hilbert class field $H$ is abelian over $\mathbb{Q}$. This is true for all cubic fields you cited: $$x^3-3x+1\qquad x^3+x^2-2x-1\qquad x^3-21x-35\qquad x^3-21x-28$$
But false in general, for example, the Hilbert class field (degree $12$) of $x^{3} - x^{2} - 54 x + 169 $ is not abelian over $\mathbb{Q}$.


问:
How can I show this field extension equality?
How can I show this field extension equality   $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$?</p>

答:
A way to do this is computing the degrees:

$$\def\Q{\mathbb{Q}}
[\Q(\sqrt{2}):\Q]=2
$$
because $x^2-2$ is irreducible over $\Q$. Moreover
$$
[\Q(\sqrt{2},\sqrt{3}):\Q(\sqrt{2}]=2
$$
because $x^2-3$ is irreducible over $\Q(\sqrt{2})$ as it is easy to see.

It follows that $[\Q(\sqrt{2},\sqrt{3}):\Q]=4$.

Of course $\Q\subset\Q(\sqrt{2}+\sqrt{3})\subseteq\Q(\sqrt{2},\sqrt{3})$ and so

$$
[\Q(\sqrt{2},\sqrt{3}):\Q(\sqrt{2}+\sqrt{3})]
$$

can be either $1$ or $2$. If it is $2$, then $\sqrt{2}+\sqrt{3}$ would have degree $2$ over $\Q$, hence it would satisfy a monic polynomial

$$
x^2 + ax + b
$$
in $\Q[x]$. If a root is $\sqrt{2}+\sqrt{3}$, the other one is

$$
\frac{b}{\sqrt{2}+\sqrt{3}}=b(\sqrt{3}-\sqrt{2}).
$$

This implies that $\sqrt{3}-\sqrt{2}\in\Q(\sqrt{2}+\sqrt{3})$, so both $\sqrt{2}$ and $\sqrt{3}$ belong to $\Q(\sqrt{2}+\sqrt{3})$, a contradiction.

Therefore

$$
[\Q(\sqrt{2},\sqrt{3}):\Q(\sqrt{2}+\sqrt{3})]=1
$$

and the proof is complete.


问:
Projecting a point from a diameter to the circumference of a circle.
What I want to know is if I have a unit circle centered at the origin with a diameter drawn along the y axis and I know the distance that a point on the diameter has from the center what would be the x coordinates of the points on the circumference with the same y coordinates. further more how would you repeat this process on bigger or smaller circles.</p>

答: Say the circle with radius $r$ has equation $$x^2+y^2 =r^2 $$ and the point you have is $P(0,a)$ ($a$ is the distance of the point from the center) where $-r\le a\le r$. Then you want to solve for the intersection point of $y=a$ with the circle: $$x^2+a^2 =r^2 \implies x=\pm\sqrt{r^2-a^2}$$


问:
How would aggregating over a a large number of transmissions change the length.
In the book: Introduction to coding by Roth, the author makes a claim that the integer effect (which I understood it as being the rounding effect) would phase out as we aggregate over an infinite number of transmissions, while I think  nothing would change (since eventually l would be factored out).
The following shows the author words

答: Let us show an example.  We will let $F=3$, so we have an alphabet of $3$ symbols.  If we want to send one of $16$ messages, we have to send $\lceil \log_3 (16) \rceil=3$ symbols.  This is a waste, we can actually send $3^3=27$ different messages.  $\log_3(16) \approx 2.52$, so we are wasting about a half a symbol or $\frac 16$ of our message.  If we want to send one of $2^{16}=65536$ messages, we need to send $\lceil \log_3 (65536) \rceil=3$ symbols.  We need $11$ symbols and waste $0.91$ of one because the log is just over $10$.  The fraction wasted is $\frac {0.91}{11} \approx \frac 1{12}$ of the message.  As the number of messages increases, the wasted symbols are bounded above by $1$, so the fraction wasted goes down with the number of symbols required.


问:
Approximating continuous function by step functions
Let $f$ be a continuous function on $[0,1]$.  For a given $\epsilon$ we know there are step functions$f_{1}$ and $f_{2}$ such that $f_{1} \leq f \leq f_{2}$ with $0 \leq \int_{0}^{1} (f_{2}(x) - f_{1}(x))  dx \leq \epsilon$.

Now let us assume $g(x)$ is a monotone increasing function.
I wish to prove the following:
For a given $\epsilon$ we know there are step functions$f_{1}$ and $f_{2}$ such that $f_{1} \leq f \leq f_{2}$ with $0 \leq \int_{0}^{1} (f_{2}(x) - f_{1}(x))  dg(x) \leq \epsilon$.

I could prove the first statement. But was unable to mimic the proof for the 2nd part.

Any help would be appreciated. Thanks in advance.

答:
A reasonable strategy is to just take the upper and lower sums approximating $\int_0^1 f\,dg$ and then let $f_2$ and $f_1$ be the step functions corresponding to the max/min on each interval. Unfortunately the existence of the Riemann-Stieltjes integral for $f_2-f_1$ is not so clear in this case. So we need to be careful in how $f_2,f_1$ are constructed.

A common result that will be useful to us is the following: If $h$ is bounded on $[a,b]$, $h$ has only finitely many points of discontinuity on $[a,b]$, and $g$ (which is taken to be monotone increasing, or more generally bounded variation) is continuous at each discontinuity of $h$, then $\int_a^b h\,dg$ exists. This is proven in Rudin and can be found in many other places (for example, Corollary 7.1.24 of the notes here). So all we need to do is choose $f_2$ and $f_1$ such that $g$ is continuous at each discontinuity of $h=f_2-f_1$.

Here is an explicit construction: Since $f$ is continuous on the compact interval $[0,1]$, it is uniformly continuous. Thus given $\epsilon>0$ there is $\delta>0$ such that $|s-t|<\delta$ implies $|f(s)-f(t)|<\epsilon$. Let $\Gamma =\{x_0,x_1,\dotsc, x_n\}$ with $0=x_0< x_1<\dotsm<x_n=1$ be a partition of $[0,1]$ such that $\Delta x_i = x_i-x_{i-1} <\delta$ for each $i$. We further assume that each $x_i$ is not a point of discontinuity for $g$ (this is possible since $g$ is monotone increasing, and thus has at most countably many points of discontinuity). Let $m_i$ and $M_i$ be the minimum and maximum of $f$ on $[x_{i-1}, x_{i}]$ respectively. Define a step function $f_1(x)= m_i$ when $x\in [x_{i-1},x_i)$ and $f_1(x) = m_n$ when $x\in[x_{n-1},x_n]$, and similarly for $f_2$ and $M_i$. Clearly $f_1\leq f\leq f_2$, and $0\leq f_2-f_1\leq \epsilon$ since the maximum and minimum differ by at most $\epsilon$ on each partition subinterval $[x_{i-1},x_i]$ by construction. The possible points of discontinuity of $f_2-f_1$ are the points of the partition $x_i$, but those were chosen such that $g$ was continuous at each of them. Hence $\int_0^1 (f_2-f_1)\,dg$ exists. In fact, for any choice of intermediate points $x_{i-1}<c_i<x_i$, the Riemann sum corresponding to the partition $\Gamma$ has
\begin{align}
R(f_2-f_1,\Gamma,g) &= \sum_{i=1}^n (f_2(c_i)-f_1(c_i))(g(x_i)-g(x_{i-1})) = \sum_{i=1}^n (M_i-m_i)(g(x_i)-g(x_{i-1})) \\
&\leq\epsilon\sum_{i=1}^n(g(x_i)-g(x_{i-1})) =\epsilon (g(1)-g(0)).
\end{align}
It is also easy to see that this estimate remains true for any refinement $\Gamma'\supset\Gamma$. Since
$$\int_0^1 (f_2-f_1)\,dg = \lim_{|\Gamma'|\to 0} R(f_2-f_1,\Gamma',g),$$
we conclude
$$\int_0^1 (f_2-f_1)\,dg\leq \epsilon(g(1)-g(0)).$$
Finally, given any $\epsilon'>0$ we can choose $\epsilon = \epsilon'/(g(1)-g(0))$ in the argument above to get the desired result.


问:
Average and Means
In a group of 10 students, the mean of the lowest 9 scores is 42 while the mean of the highest 9 scores is 47. For the entire group of 10 students, the maximum possible mean exceeds the minimum possible mean by

How I have approached this question till now :-
Let x1,x2,x3,......,x10 be the marks of 10 students such that x1<x2<x3<....<x10.

(x1+x2+x3+....+x9)/9 = 42
==> x1+x2+x3+...+x9 = 378

(x2+x3+x4+...+x10)/9 = 47
==> x2+x3+x4+....+x10 = 423

From the above two equations subtraction :-
x10 - x1 = 45

from the above two equations addition :-

```<code>x1 + 2(x2+x3+....+x9) + x10 = 801
==> x2+x3+...+x9 = [801 - (x10+x1)]/2
</code>```

After this I tried using this relation in the average of 10 students but couldn't work that out towards a solution. Need some help...

答:
I solved the question with exact answer of 4. Below is my solution approach:-

Let the marks of 10 students be x1,x2,x3,...........,x10.

Suppose

x1 < x2 < x3 < x4 <.......< x10

.

As per the question :-

```<code>(x1+x2+x3+....+x9)/9 = 42   Also we can say that x1 <= 42.

==> x1+x2+x3+...+x9 = 378  ----(i)

(x2+x3+x4+...+x10)/9 = 47   Also we can say that x10 >=47

==> x2+x3+x4+....+x10 = 423   ------(ii)
</code>```

Mean of 10 students = (x1+x2+x3+........+x10)/10

```<code>Maximum mean will occur when x1 = 42 :-

[x1 + (x2+x3+x4+.....+x10)]/10 = (42 + 423)/10 = 46.5  From eq.(ii)

Minimum mean will occur when x10 = 47 :-

[(x1+x2+.....x9) + x10]/10 = (378+47)/10 = 42.5   From eq.(i)

Difference = 4
</code>```

Please correct me if I am wrong in my approach.

Title: Average and Means

Body:
I solved the question with exact answer of 4. Below is my solution approach:-

Let the marks of 10 students be x1,x2,x3,...........,x10.

Suppose

x1 < x2 < x3 < x4 <.......< x10

.

As per the question :-

(x1+x2+x3+....+x9)/9 = 42   Also we can say that x1 <= 42.

==> x1+x2+x3+...+x9 = 378  ----(i)

(x2+x3+x4+...+x10)/9 = 47   Also we can say that x10 >=47

==> x2+x3+x4+....+x10 = 423   ------(ii)

Mean of 10 students = (x1+x2+x3+........+x10)/10

Maximum mean will occur when x1 = 42 :-

[x1 + (x2+x3+x4+.....+x10)]/10 = (42 + 423)/10 = 46.5  From eq.(ii)

Minimum mean will occur when x10 = 47 :-

[(x1+x2+.....x9) + x10]/10 = (378+47)/10 = 42.5   From eq.(i)

Difference = 4

Please correct me if I am wrong in my approach.


问:
The elements of $\mathbb{Z}[\sqrt{-5}]/(2,\sqrt{-5}+1)$
I'm really confused with this one...

How can I determine the elements of the module $\mathbb{Z}[\sqrt{-5}]/(2,\sqrt{-5}+1)$? Or its cardinality?

Does $$\mathbb{Z}[\sqrt{-5}]/(2,\sqrt{-5}+1)=\{\bar{0},\bar{1},\overline{0+\sqrt{-5}}\}?$$

答:
Consider the map $\phi: \mathbb{Z}[\sqrt{-5} ] \to \mathbb{Z}_2 $ given by $\phi( a+b\sqrt{-5})= a+b.$ This is a surjective ring homomorphism; the hardest part to check is that is respects multiplication.

$\phi\left( (a+b\sqrt{-5})(c+d\sqrt{-5}) \right) = \phi ( (ac-5bd)+ (ad+bc)\sqrt{5} ) = ac - 5bd + ad+bc$

$\phi( a+ b\sqrt{-5} ) \phi (c+d\sqrt{-5} ) = (a+b)(c+d) = ac + ad+bc+bd$

These are indeed equal in $\mathbb{Z}_2.$ It is clear that $(2,1+\sqrt{-5})\subseteq \ker \phi.$

If $a+b\sqrt{-5} \in \ker \phi $ then $ a=-b \in \mathbb{Z}_2 $ so $a=-b+2k$ for some $k\in \mathbb{Z}.$ Therefore $a+b\sqrt{-5} = 2(k-b) + b(1+\sqrt{-5})\in (2,1+\sqrt{-5})$ so $\ker \phi = (2,1+\sqrt{-5}).$ The first isomorphism theorem gives the result.

As per BenjaLim's deleted answer, we have

$$\begin{eqnarray*} \Bbb{Z}[\sqrt{-5}]/(2,1 + \sqrt{-5}) &\cong& \Bbb{Z}[x]/(x^2 + 5)/(2,1+x)/(x^2 + 5)\\
&\cong& \Bbb{Z}[x]/(2,1+x)\\
&\cong& \Bbb{Z}/2\Bbb{Z}\end{eqnarray*}.$$

OP asked why the last isomorphism holds. It is because $\mathbb{Z}[x]/(1+x) \cong \mathbb{Z}.$ To see this, consider the homomorphism $\varphi: \mathbb{Z}[x] \to \mathbb{Z}$ given by $\varphi( p(x) ) = p(-1).$ It is not hard to verify it is surjective with kernel $(1+x)$ which yields the isomorphism. Then

$$\begin{eqnarray*} \Bbb{Z}[x]/(2,1 + x) &\cong& \Bbb{Z}[x]/(x+1)/(2,1+x)/(x+1)\\
&\cong& \Bbb{Z}/(2)= \Bbb{Z}/2\Bbb{Z}\end{eqnarray*}.$$


问:
Computing $ \int_{-\infty}^{\infty} e^{-ax^4+bx^2} dx$
I was wondering if it was possible to compute $ \int_{-\infty}^{\infty} e^{-ax^4+bx^2} dx$  explicitly.

Completing the square doesn't seem to work, since there is no factor of $x$ in order to use the Gaussian integral.

Any hint would be greatly appreciated.

Thanks!

答:
Your integral is related by a scaling transformation to
$$\int_{-\infty}^{\infty}
\mathrm{e}^{-z(x^4/8-x^2+1)}\mathrm{d}x
= \pi(I_{-1/4}(z)+I_{1/4}(z))\text{,}$$
where $\Re z > 0$ and $I$ denotes a modified Bessel function. To prove this equality,

Write $\mathrm{e}^{zx^2}$ as a Maclaurin series,

Integrate termwise using Euler's gamma function,

Use the duplication formula for the gamma function,

Write the series as a linear combination of two confluent hypergeometric functions, and

Simplify to the modified Bessel functions.


问:
Is any group the group of units of some ring?
I was wondering if there was a way to see any group $G$ as a subgroup of the inverse group of a ring, $G \leq A^\times$?

Thanks!

答:
Consider the group algebra $\mathbb{C}G$ which consist of the formal linear combinations of the form $$\sum_{g \in G}\alpha_g g$$
and where the algebra operations are the obvious ones. Note that we have an obvious inclusion $G \hookrightarrow \mathbb{C}G$.  Then $G \subseteq(\mathbb{C}G)^\times$. Note however that in general the latter inclusion can be strict.


问:
Let $P$ be a polytope of dimension $d$. Then for some $v \notin$ aff($P$), the pyramid $v \ast P$ has dimension $d+1$.

Let $P$ be a polytope of dimension $d$. Then for some $v \notin$ aff($P$), the pyramid $v \ast P$ has dimension $d+1$.

I’m looking for a straightforward argument to prove this statement. All the arguments I’ve come up with so far involve embedding $P$ in $\mathbb{R}^{d+1}$, and analyzing the number of linearly independent vectors. I’m hoping to find a simpler explanation, if possible.

Here are the definitions of the relevant terms in the course I’m doing:

$\cdot$ A polytope is defined as the convex hull of a finite set of points in some $\mathbb{R}^d$. Alternatively, it is a bounded polyhedron - defined as the intersection of finitely many closed halfspaces in $\mathbb{R}^d$.

$\cdot$ Let $P \subset \mathbb{R}^d$ be a (possibly empty) polytope of dimension $< d$. For a point $v \in \mathbb{R}^d \setminus \text{aff}(P)$, we define the pyramid $v \ast P$ with apex $v$ and base $P$ as the convex hull of $P \cup \{v\}$.

答:
When you say a set is of dimension $n$ it means that it can be embedded at least in $\mathbb{R}^n$.
The idea of a dimension is that of this  exercise: you are in $\mathbb{R}^d$; for a point here $v$ not in $\mathbb{R}^d$  we can join $v$ to $P$ by a segment in fact this segment (vector) is the added dimension if you want, this is the generalisation of moving from $\mathbb{R}^2$ to $\mathbb{R}^3$.


问:
Dividing by an almost-surely positive random variable
I am reading Shreve's "Stochastic Calculus for Finance II".  In it, he states (Theorem 1.6.1) that if $Z$ is an almost-surely strictly positive random variable on a probability space $(\Omega, \mathcal F, P)$ with $E(Z) = 1$, then the probability measure defined as $\widetilde P(A) := \int_A Z d P$ satisfies
$$
E(Y) = \widetilde E(Y/Z)
$$
for any random variable Y (here E and $\widetilde E$ are the expected values in the respective measures).  I don't understand the meaning of this expression; $Y/Z$ is not a random variable since there may be $\omega \in \Omega$ upon which $Z$ vanishes.

Is $Y/Z$ a valid expression, or is it shorthand for something more rigorous?

答:
EDIT: Replacing $\tilde \cdot$ (tilde) by $\widetilde \cdot$ (wide tilde); for some reason still does not appear like a (wide) tilde.

To complete Nate's answer, it is important to consider here $Y/Z$ and $\tilde E(Y/Z):=\int_\Omega  {(Y/Z)d\tilde P} $ with respect to the probability measure $\tilde P$. As Nate noted, there is no problem with $Y/Z$ with respect to the probability measure $P$ (since $P(Z=0)=0$). Let's see that there is also no problem with respect to $\tilde P$, which amounts to showing that $\tilde P(Z=0)=0$. Indeed, let $A$ denote the event $\{Z=0\}:=\{ \omega \in \Omega : Z(\omega) = 0\}$. Then,
$$
\tilde P(Z=0) = \tilde P(A) := \int_A {ZdP} = \int_A {Z(\omega )dP(\omega )}.
$$
So, there is no problem here for two (independent) reasons. First, by definition, $Z(\omega) = 0$ for $\omega \in A$, hence
$$
\int_A {Z(\omega )dP(\omega )} = \int_A {0dP(\omega )} = 0.
$$
Second, by assumption, $P(Z=0)=0$, that is $P(A)=0$; hence, $\int_A {ZdP} = 0$ (recall from measure theory that $
\int_E {fd\mu }  = 0$ if $\mu(E)=0$).


问:
Periodic tribonacci-like sequence
How to prove that if $a_n =[(t_{n-3} + 2t_{n-2} + t_{n-1}) a_{1} + (t_{n-3} + t_{n-2} + 2t_{n-1})] \quad (\text{mod}10)$ and $a_{1}, a_{2}, a_{3}$ are consecutive numbers and $t_{1}=0, t_{2}=1$ and $t_{3}=1$ are the first three terms of a tribonacci sequence, then $a_{n} = a_{n+62}$?</p>

答:
The period of $t_n\bmod 5$ is 31. $t_n\bmod 2$ repeats the sequence 0, 0, 1, 1. From this, it is not hard to see that $t_{n+62}=t_n+5$, modulo 10.

The question's formula for $a_n$ contains two parenthesised expressions, each of which remains unchanged (modulo 10) if $n$ is increased by 62 and thus each of the $t$'s is changed by 5. Thus $a_n=a_{n+62}$.

Could the period of $t_n\bmod 5$ be found more easily than by calculating 31 further terms? The tribonacci recurrence is $t_n=t_{n-1}+t_{n-2}+t_{n-3}$. The corresponding characteristic cubic is thus $x^3-x^2-x-1$, and its discriminant is $-44=-4\cdot11$. $5$ is a prime; moreover, modulo $11$, 5 is a square ($5=16=4^2$), so some primitive binary quadratic form of discriminant $-44$ represents 5 [1]. Unfortunately, $h(-44)=3$, that is, there are $3$ equivalence classes of primitive binary quadratic forms of discriminant $-44$. We need to know which one represents 5.

Here follows a more general principle for finding information about the period $R$ of a cubic recurrence modulo a prime $p$, where the recurrence's characteristic cubic's discriminant is $D$, and $h(D)=3$.

If the principal binary quadratic form of discriminant $D$ represents $p$, and $p\nmid D$, then $R\mid p-1$.

If the other primitive binary quadratic form of discriminant $D$ represents $p$, then $R\mid p^2+p+1$. Note that $p^2+p+1=(p^3-1)/(p-1)$.

If $p$ is not a square modulo $D$ (if $D$ is odd) or $D/4$ (if $D$ is even), then $R\mid p^2-1$.

Back to the case in question. The principal binary quadratic form of discriminant $-44$ is $x^2+11y^2$, which does not represent 5. The other one is $g(x, y)=3x^2+2xy+4y^2$, and $g(1, -1)=5$. So the period divides $p^2+p+1$ which here is 31. Which is prime, and the period can't be 1, so it must be 31, as required.

[1] This inference applies only to representation of primes. Like 5, 38 and 82 are congruent to $16=4^2$ modulo 11, but no binary quadratic form of discriminant $-44$ represents 38 or 82. Not even the non-primitive $2x^2+2xy+6y^2$.

Title: Periodic tribonacci-like sequence


问:
How to calculate an integral vector such as $A=\int_{-5}^0 \frac{1}{\left(x^2+2^2\right)^{3/2}} \,dx$ $\{-x,0\}$
Why these integrals have different results ?: $A=\int_{-5}^0 \frac{1}{\left(x^2+2^2\right)^{3/2}} \,dx$ $\{-x,0\}$  and $B=\int_{-5}^0 \frac{-x}{\left(x^2+2^2\right)^{3/2}} \, dx$. Vector $\overset{\to }{u}=\{-x,0\}=-x\overset{\to }{i}+\overset{\to }{0j}$ . In B, I introduced the component ${-x}$ of $\overset{\to }{u}$ into de Integral to then multiply by $\overset{\to }{i}$ , Is it mathematicaly right? or, How I must solve A?

Evaluating in Mathematica these results are: $A=\left\{-\frac{5 x}{4 \sqrt{29}},0\right\}$ and $B=(\frac{1}{2}-\frac{1}{\sqrt{29}})$.  *(it is $B=(\frac{1}{2}-\frac{1}{\sqrt{29}})\overset{\to }{i}$)

EDIT:
in this example, it shows they are solving as I solved B:
image

答:
You can’t take the $x$ inside the integral when multiplying by a vector, as the integral is a scalar and must be evaluated independently. So $A\ne B$.
Consider $A$ as an indefinite integral first.
Substitute $x=2\tan y \implies dx = 2\sec^2 y dy $ in $A$ to get $$\int \frac{2\sec^2y}{8\sec^3 y} dy = \frac 14 \int \cos y dy = \frac 14\sin y + c = \frac{x/2}{4\sqrt{1+(x/2)^2}} + c = f(x)$$

Then, just evaluate $f(x)\big|_{-5}^0=A$ and then your answer would be $\big(- Ax,0\big ).$


问:
Fourier Transform of a function that is zero for $x<0$.
I have this function
$$
f(x)=xH(x)\ln{(x)}e^{-x},
$$
of which I need the Fourier transform, where $H(x)$ is the Heaviside function. Maple tells me the Fourier transform is
$$
{\frac { \left({\omega}+i \right)^2 \left(\ln  {\left(\sqrt{
\omega^2+1}\right)} + i \arctan \left( \omega \right) + \gamma-1 \right) }{ \left( {\omega}^{2}+1
 \right) ^{2}}}
,
$$
where $\gamma$ is Euler's constant.

I have computed it several different ways through Maple and I believe it is the right formula. I tried to do it analytically but I never quite got there. If someone would know the steps to get there, I would really appreciate it.

I need this result because I need to check in which fractional Sobolev spaces $H^s(\mathbb{R})$ does $f$ belong to.

答:
The Fourier transform of a function $f$ is traditionally denoted ${\displaystyle {\hat {f}}}$, by adding a circumflex to the symbol of the function. There are several common conventions for defining the Fourier transform of an integrable function ${\displaystyle f:\mathbb {R} \to \mathbb {C} }$. One of them is:

$$\hat{f}\left(\xi\right):=\int_\mathbb{R}f(x)\exp\left(-2\pi x\xi i\right)\space\text{d}x\tag1$$

for any real number $\xi$ with $i^2=-1$.

Using your function, we can notice that:

$$\hat{f}\left(\xi\right)=\int_0^\infty x\ln\left(x\right)\exp\left(-x\right)\exp\left(-2\pi x\xi i\right)\space\text{d}x\tag2$$

Let's define a second integral:

$$\mathscr{S}_\text{n}\left(\text{s}\right):=\int_0^\infty x\ln\left(x\right)\exp\left(-\text{s}x\right)\exp\left(\text{n}x\right)\space\text{d}x\tag3$$

Using Laplace transform, we can write:

$$\mathscr{S}_\text{n}\left(\text{s}\right)=\mathscr{L}_x\left[x\ln\left(x\right)\exp\left(\text{n}x\right)\right]_{\left(\text{s}\right)}\tag4$$

Using the frequency-domain derivative and frequency shifting property of the Laplace transform, we can write:

$$\mathscr{S}_\text{n}\left(\text{s}\right)=-\frac{\partial}{\partial\text{s}}\left\{\mathscr{L}_x\left[\ln\left(x\right)\right]_{\left(\text{s}-\text{n}\right)}\right\}\tag5$$

Using table of selected Laplace transforms, we can write:

$$\mathscr{S}_\text{n}\left(\text{s}\right)=-\frac{\partial}{\partial\text{s}}\left\{\frac{\ln\left(\text{s}-\text{n}\right)+\gamma}{\text{n}-\text{s}}\right\}=\frac{1-\gamma-\ln\left(\text{s}-\text{n}\right)}{\left(\text{n}-\text{s}\right)^2}\tag6$$

Where $\gamma$ is the Euler–Mascheroni constant.

Going back to $(2)$ we can see that:

$$\hat{f}\left(\xi\right)=\int_0^\infty x\ln\left(x\right)\exp\left(-x\right)\exp\left(-2\pi x\xi i\right)\space\text{d}x=\frac{1-\gamma-\ln\left(1-\left(-2\pi\xi i\right)\right)}{\left(\left(-2\pi\xi i\right)-1\right)^2}=$$
$$\frac{1-\gamma-\ln\left(1+2\pi\xi i\right)}{\left(\left(-2\pi\xi i\right)-1\right)^2}=\frac{1-\gamma-\ln\left(1+2\pi\xi i\right)}{\left(1+2\pi\xi i\right)^2}\tag7$$


问:
If $E = \{ x \in \mathbb{R}: \sin(\frac1{x}) = 1\}$ then $l = 0$ is a limit point of E
If $E = \{ x \in \mathbb{R}: \sin\left(\frac{1}{x}\right) = 1\}$, then $l = 0$ is a limit point of $E$.

I have a proof here but I don't quite understand a few points, I hope someone can explain it a bit : )

To prove that $l = 0$ is a limit point, we need to show for any $\delta > 0 $, there exists $x \in E $ such that $ 0 < |x-0| < \delta $,

$$\sin\left(\frac{1}{x}\right) = 1 \quad \text{implies}  \quad x = \frac{2}{(2k+1)\pi}. $$

If $k = \max \left(1, \left\lceil \frac{1}{\delta} - \frac{1}{2} \right\rceil \right)$ where $\delta > 0 $, then $0 < \left|\frac{2}{(2k+1)\pi} - 0 \right| < \delta $ so $\exists x \in E$ s.t. $0 < |x-0| < \delta$.

So questions are here,

Why max function is used here? and why $1$ is chosen?

Why the ceiling is used for $\left\lceil \frac{1}{\delta} - \frac{1}{2} \right\rceil$?

答:
A simpler way to prove that $0$ is a limit point is to exhibit a sequence $(x_n)$ in $E$ such that $x_n\to0$.

Once this point is clear, I am sure you have an idea about what such a sequence $(x_n)$ could be...


问:
Prove that $\ddot {g} (t) = D^2 f(x+th) (h,h).$

Let $E$ be a Banach space. Let $U \subseteq E$ be open and $f : U \longrightarrow F$ be $C^2.$ Let $x \in U,h \in E$ be such that $B(x, \|h\|) \subseteq U.$ Consider the map $g : (-1,1) \longrightarrow F$ defined by $g(t) = f(x+th),\ t \in (-1,1).$ Then $\ddot {g} (t) = D^2 f(x+t h) (h,h).$

I have proved that $\dot {g} (t) = Df (x + t h) (h).$ In order to find the $\ddot {g} (t)$ I need to compute the limit $$\lim\limits_{s \to 0} \frac {Df (x + th + sh) (h) - Df (x + th) (h)} {s}.$$

How to simplify the above expression? Since $Df$ is Frechet differentiable I know that $$\left \|Df(x + k) - Df (x) - D^2 f(x) (k) \right \| = o(\|h\|).$$

I adopted the following strategy $:$

I know that the above limit exists. Let it be $y \in F.$ But then $$\lim\limits_{s \to 0} \frac {\left \|Df(x + th + sh)(h) - Df(x+th) (h) \right \|} {|s|} = \|y\|.$$ Now $$\begin{align*} \left \|Df(x +th +sh)(h) - Df(x+th)(h) \right \| & = \left \|(Df(x + th + sh) - Df(x + th)) (h) \right \| \\ & = \left \|\left (Df(x + th +sh) - Df(x + th) - D^2f (x+th) (sh) \right )(h) + D^2 f(x + th) (sh) (h) \right \| \\ & \leq \left \|Df(x + th + sh) - Df(x + th) - D^2 f(x + th)(sh) \right \| \|h\|  + \left \|s\ D^2f (x +th) (h)(h) \right \|  \\ & = o(\|sh\|)\|h\| + |s| \left \|D^2f(x + th) (h)(h) \right \| \\ & = o(|s|\|h\|) \|h\| + |s| \left \|D^2f(x + th) (h)(h) \right \| \end{align*}$$ This shows that $$\lim\limits_{s \to 0} \frac {\left \|Df (x + th + sh) (h) - Df (x + th) (h) \right \|} {|s|} \leq \left \|D^2f(x + th) (h) (h) \right \|.$$

How do I proceed now? Any help regarding this will be highly appreciated.

Thanks for your time.

答:
Your problem is you want to show that
\begin{equation*}
\lim_{s \to 0} \left\|\frac{Df(x + th + sh)(h) - Df(x + th)(h) - sD^{2}f(x + th)(h,h)}{s}\right\| = 0.
\end{equation*}

The fact that
\begin{equation*}
\lim_{s \to 0} \left\|\frac{Df(x + th + sh)(h) - Df(x + th)(h)}{s}\right\| = \|D^{2}f(x + th)(h,h)\|
\end{equation*}
is also true, but it's of no particular use here and it's getting you stuck.

This is an easy mistake to make when starting out with Fréchet derivatives.  The 1D habit to write $\frac{f(t + h) - f(t)}{h} \to f'(t)$ becomes a  bit clumsy and it becomes worthwhile to think of $\frac{f(t + h) - f(t) - f'(t)h}{h}$ going to zero instead.

Edit: At the request of the author, I'll show how to compute $\ddot{g}(t)$ using the chain rule.

$\dot{g}(t) = Df(x + th)(h)$.  This is a composition so the chain rule can be applied to find $\ddot{g}(t)$.  The issue is writing this coherently as a composition.

The first operation involved is scalar multiplication and vector addition $t \mapsto x + th$.  Next, application of $Df$: $y \mapsto Df(y)$.  Finally, evaluating the matrix so-obtained (call it $A$) at $h$: $A \mapsto A(h)$.  Let's call this final map $\mathcal{I}_{h} : \mathcal{L}(E,F) \to F$, $\mathcal{I}_{h} : A \mapsto A(h)$.  Putting it all together, we have $\dot{g}(t) = \mathcal{I}_{h}(Df(x + th))$, which is a composition.

By the chain rule, $\ddot{g}(t) = D\mathcal{I}_{h}(Df(x + th)) \circ D^{2}f(x + th)(h)$.  All that is left is computation of $D\mathcal{I}_{h}$.  This is straightforward, though: $\mathcal{I}_{h}$ is linear so $D\mathcal{I}_{h}(A) = \mathcal{I}_{h}$ for all $A \in \mathcal{L}(E,F)$.  Thus, $\ddot{g}(t) = [D^{2}f(x + th)(h)](h)$.


问:
Elementary exposition of the Picard group?
I'm trying t find good references to understand "why" the elliptic curves have a group structure. The canonical answer seems to be that:

Every curve has a group associated to it, the picard group

the elliptic curves' set of solutions in projective $\mathbb Q$ space is in set theoretic bijection with the Picard group.

Hence, we pullback the group structure of the "correct object" (the Picard group) onto the elliptic curve itself, giving the "weird geometric group law" of:

draw a line between $P$ and $Q$, find third point of intersection $R$, reflect $R$ to $R'$. Define $P \cdot Q \equiv R'$.

Can I find an elementary exposition of this somewhere? Silverman expects one to know sophisticated algebraic geometry as far as I can tell. I found picard groups in Hartshorne as well, but it once again seems like quite a lot of effort to get to the idea of a picard group in the book.

Is there an elementary (read: undergrad who knows a first course in varieties/diffgeo/algebra/number theory/topology) source to understand the structure of the Picard group of the elliptic curve, and how it relates to the elementary group law one is taught?

答:
Is this clear to you ?

For an affine elliptic curve defined by $C:y^2=x^3+Ax+B$ then the line passing through $P,Q$ is defined by some equation $ax+by+c=0$ (assuming $P,Q$ are distinct otherwise we are considering the tangent line at $P$)

$ax+by+c$ is a rational function on the curve and its divisor (zeros & poles) on the projective closure (ie. $E=C\cup O$ where $O$ is the point at infinity) is $P+Q+R-3O$,

where $R$ is the 3rd point of the line.

So $P+ Q+R=3O$ in the Picard group $Pic(E)$, choosing $O$ as the neutral element gives $P+Q=-R=R'$ in $Pic(E)/\langle O\rangle$.

Where  $R'=(x_R,-y_R)$ because the divisor of the rational function $x-x_R$ is $R+R'-2O$.


问:
How to solve following problem?
I know it might have something to do with modular exponentiation, but after extensive search and reading, I am completely dumbfounded.

$$\text {Let}\; N = 12 = 2^2 + 2^3$$
Given that $M^2 \equiv 51$ (mod 59), What is $M^{12}$(mod 59)?

答:
An explanation of the accepted answer above for anyone else coming here from WGU.

Given that we know:

$$ [(x\ mod\ n)(y\ mod\ n)]\ mod\ n = [x·y]\ mod\ n $$

and basic exponential power rules:

$$ M^{12} \equiv (M^8)(M^4) \equiv (((M^2)^2)^2)((M^2)^2) $$

we can use $ M^2 $ to find $ M^{12} $:

$$ M^4 \equiv (M^2)^2 \equiv (M^2)(M^2) \equiv [(51\ mod\ 59)(51\ mod\ 59)] \equiv 51^2(mod\ 59) $$

Because $ 51^2 $ is already quite large to compute, we know that via modulo congruence that $ -8 \equiv 51(mod\ 59) $, hence:

$$ M^4 \equiv 51^2(mod\ 59) \equiv (-8)^2(mod\ 59) \equiv 64(mod\ 59) \equiv 5(mod\ 59) $$

Repeat to get us to $ M^8 $ (leaving out the expansion and then later simplification of the exponents):

$$ M^8 \equiv (M^4)^2 \equiv 5^2(mod\ 59) \equiv 25(mod\ 59) $$

Lastly, we can use previously computed values to find $ M^{12} $:

$$ M^{12} \equiv (M^4)(M^8) \equiv [(5\ mod\ 59)(25\ mod\ 59)] \equiv 5(25)(mod\ 59) \equiv 125(mod\ 59) \equiv 7(mod\ 59) $$

Title: How to solve following problem?


问:
Can you find a topology on the set
Can you find a topology on the set of real numbers $\mathbb{R}$ so that all polynomials $p(x)=a_nx^n+...a_1x+a_0$ considered as functions from $\mathbb{R}$ to $\mathbb{R}$ are continuous functions but the cosine function $\cos \left(x\right)$ is not a continuous function from $\mathbb{R}$ to $\mathbb{R}$.

I tried cofinite topology and Zariski Topology but I didn't succesful. Can you help?

答: Take the cofinite topology, in which the closed sets are $\mathbb{R}$ and the finite sets. If $F$ is finite, and $f$ is a polynomial, then $f^{-1}(F)$ is finite. Furthermore, $f^{-1}(\mathbb{R}) = \mathbb{R}$, so $f$ is continous. However, $\cos^{-1}(0)$ is infinite, but not equal to $\mathbb{R}$. Thus $\cos$ is not continuous.

Title: Can you find a topology on the set


问:
About $\mathcal{L}(V,W)$
Let $V,W$ are two vector space and let $S\subseteq V$. Define: $$S^{0}=\{T\in\mathcal{L}(V,W)\mid~T(x)=0, \forall x\in S\}$$ The problem aks me to verify $S^{0}$ is a subspace of $V$ and if $V_1,V_2$ are two subspaces, then $$(V_1+V_2)^{0}=V^0_1\cap V^0_2$$ Please help me to overcome the second part. I did the first one. Thank you!</p>

答:
You can do this by a double inclusion, suppose that $W\in(V_1+V_2)^0$, then for every $x\in V_1$ we have that $W(x)=0$ and similarly for $x\in V_2$, therefore $W\in V_1^0$ and $W\in V_2^0$, so $W$ is in the intersection.

The opposite direction is similar, and I leave it to you. Note that you will have to use the definition of $V_1+V_2$ and the linearity of $W$ here.

Title: About $\mathcal{L}(V,W)$


问:
$\sin(3x)/2x$ to power series?

Write to power series of $$f(x)=\frac{\sin(3x)}{2x}, \quad x\neq 0.$$

I try to get a series for $\sin(3x)$ with $x=0$ and multiplying the series with $2x$.

Is that right?

答:
Just write taylor series for sin(3x) and the divide by 2x
$$\frac{1}{2} \sum_{n=0}^{\infty} (-1)^n \frac{3^{2n+1}x^{2n}}{(2n+1)!}$$

Title: $\sin(3x)/2x$ to power series?


答:
Yes, that's completely fine. The Maclaurin series for $\sin$ is
$$\sin(t)=t-\frac{t^3}{3!}+\frac{t^5}{5!}+...$$
So letting $t=3x$,
$$\frac{1}{2x}\sin(3x)=\frac{1}{2x}\left(3x-\frac{(3x)^3}{3!}+\frac{(3x)^5}{5!}+...\right)$$
You can move the $(2x)^{-1}$ inside:
$$\frac{\sin(3x)}{2x}=\frac{3^1x^1}{2x}-\frac{3^3x^3}{2x\cdot3!}+\frac{3^5x^5}{2x\cdot5!}+...$$
Hopefully you can construct the general term from here?

Title: $\sin(3x)/2x$ to power series?

Body:
Write to power series of $$f(x)=\frac{\sin(3x)}{2x}, \quad x\neq 0.$$

I try to get a series for $\sin(3x)$ with $x=0$ and multiplying the series with $2x$.

Is that right?


问:
The difference between the HCF and LCM of $x$ and $18$ is $120$. Find $x$.

The difference between the HCF and LCM of $x$ and $18$ is $120$. Find $x$.

What I Tried: Let $\gcd(x , 18) = y$ and lcm$(x , 18) = z$ . Also, $z \geq y$ . We have :-
$$\rightarrow 18x = yz$$
And :-
$$\rightarrow z - y = 120$$
$$\rightarrow\frac{18x}{y} - y = 120$$
$$\rightarrow 18x - 120y - 120 = y^2$$
From here, we get that $6 | y^2$ $\rightarrow 6 | y$ .

After doing this, I am stuck. I do get this information but I am not able to use it somehow for $x$ , I could show that $x = 6k$ , but what to do next?

Can anyone help me? Thanks.

答:
Hint:

as $\text{lcm}(x,18)$ is even :$\gcd(x,18)$ is even .That is $\gcd(x,18)=2$ or $\gcd(x,18)=6$  or $\gcd(x,18)=18$

which do you think will work??


问:
Combinatorics w/ binary string
How many string of 16 bit can we build having the following constraints:

There must be exactly 5 one

Between two 1 there must be at least two 0.

Example of valid string:

(1) 1 00 1 00 1 00 1 00 1 000

(3) 0 1 00 1 00 1 00 1 00 1 00

(3) 1 0000 1 00 1 00 1 000 1

I have no clue how to approach problems like these. I know the answer (which is 56) but I do not know the logic behind that.

答:
Because the length of the string is 16 and there are $5$ ones, there must be $11$ zeros.

Now, place the ones on the line, as _ $1$ _ $1$ _ $1$ _ $1$ _ $1$ _, we have 2 gaps at each end, and 4 gaps in the middle of the ones, for a total of 6 gaps. Now denote the number of zeros in each gap $g_i$, with $ i \in \{1,2,3,4,5,6 \}$, then from the given conditions, $g_2,g_3,g_4,g_5$ must be larger than or equal to $2$, and thus $G_i=g_i -2 \ge 0$ for all $i \in \{ 2,3,4,5 \}$

So we have the sum $ g_1 +G_2 +G_3+G_4+G_5 +g_6 = 3$, which means divide $3$ balls in to $6$ groups, each group might have a non-negative number of balls. The formula for this division is $ \binom {8}{3}$, which you may see the explanation in the Star and Bar link mentioned above

Title: Combinatorics w/ binary string


答:
Another way:

WLOG $(x,18)=d>0$  and $\dfrac xp=\dfrac{18}q=d$

$\implies(p,q)=1$ and LCM$(x,18)=pqd$

and $120=pqd-d=d(pq-1)\implies p=\dfrac{\dfrac{120}d+1}q=\dfrac{\dfrac{120}d+1}{\dfrac{18}d}=\dfrac{120+d}{18}$

Clearly, $3|d$ and $2|d\implies d$ must be divisible by LCM$(2,3)=?$

As $d|18$ and $d|120,d$ must divide $(18,120)=6$

So, $d$ has a unique value!

Body:
The difference between the HCF and LCM of $x$ and $18$ is $120$. Find $x$.

What I Tried: Let $\gcd(x , 18) = y$ and lcm$(x , 18) = z$ . Also, $z \geq y$ . We have :-
$$\rightarrow 18x = yz$$
And :-
$$\rightarrow z - y = 120$$
$$\rightarrow\frac{18x}{y} - y = 120$$
$$\rightarrow 18x - 120y - 120 = y^2$$
From here, we get that $6 | y^2$ $\rightarrow 6 | y$ .

After doing this, I am stuck. I do get this information but I am not able to use it somehow for $x$ , I could show that $x = 6k$ , but what to do next?

Can anyone help me? Thanks.


问:
Determine whether origin is a local extrema of a given function
Consider the function $f(x,y)=x^6-2x^2y-x^4y+2y^2$ defined from $\mathbb{R}^2$ to $\mathbb{R}$.

Then $(0,0)$ is a critical point.

Also $f_{xx}$ at origin is zero, $f_{yy}$ at origin is $4$ and $f_{xy}$ at origin is $0.$ The value of the determinant is zero. So it cannot be concluded whether the origin is local minimum or maximum or saddle point.

How can we determine for this particular problem whether the origin is local maximum or local minimum or a saddle point?

Thanks in advance.

答:
$f(x,\lambda x^2)=x^4(2\lambda(\lambda-1) + (1- \lambda)x^2)$ for all $x \in \mathbb R$.

Therefore $$f(x, \frac{1}{2}x) = x^4(-\frac{1}{2} + \epsilon(x)x^2)$$ is negative in a neighborhood of $(0,0)$. While

$$f(x,2x) = x^4 (4 - \delta(x)x^2)$$ is positive in a neighborhood of $(0,0)$.

You have a saddle point.


Title: $\sin(3x)/2x$ to power series?

Body:
Using the Taylor series for the sine:
$$\sin{3x}=\sum_{n=0}^{\infty} (-1)^{n}\frac{(3x)^{2n+1}}{(2n+1)!}$$

We can write the function $f(x)$ as:

$$f(x)=\frac{\sin{3x}}{2x}=\frac{1}{2x}\sum_{n=0}^{\infty} (-1)^{n}\frac{(3x)^{2n+1}}{(2n+1)!} =\frac{1}{2}\sum_{n=0}^{\infty} (-1)^{n}\frac{3^{2n+1}x^{2n}}{(2n+1)!}$$


问:
Irreducible polynomials over $\mathbb{Q}$?
I'm trying to understand how to check if polynomials are irreducible.

For example: Are these polynomials irreducible over $\mathbb{Q}$?
$$f(x)=x^3+4x^2+3\\g(y)=12y^4+5y^3-10y^2+25y+10$$

A rational root $\frac{r}{s}$ for $f(x)=x^3+4x^2+3$ gives $$s|1 \Rightarrow s=\pm 1\\
r|3 \Rightarrow r=\pm 1, 3\\ Possible\ roots:\frac{3}{1},-\frac{3}{1},\frac{1}{1},-\frac{1}{1}$$

None of the possible roots gives $f(x)=0$. Therefor $f(x)$ is irreducible.

A rational root $\frac{r}{s}$ for $g(y)=12y^4+5y^3-10y^2+25y+10$ gives
$$s|12\Rightarrow s=\pm 1,2,3,4,6,12\\ r|10 \Rightarrow r=\pm 1,2,5,10\\
Possible\ roots: \frac{1}{1},-\frac{1}{1},\frac{1}{2},-\frac{1}{2}, etc.$$

None of the possible roots gives $g(y)=0$. Therefore $g(y)$ is irreducible.

Have I misunderstood the concept of irreducible polynomials?

答:
The polynomial $(x^2+1)^2$ has no roots in $\mathbb{Q}$, but it's reducible.

The root criterion only holds for polynomials of degree two or three, because if they can be factored, at least one of the factors must have degree one (hence a root). On the other hand, a polynomial of degree $>3$ can be the product of two nonconstant polynomials of lower degree having no roots and so be reducible.

Your argument is therefore good for the first polynomial, but not for the second one.

The second polynomial is irreducible over $\mathbb{Q}$ by Eisenstein's criterion applied to the prime $5$, which divides all coefficients except the leading one and its square doesn't divide the constant term.


答: $x_{n}:=1/((4n+1)\pi/2)$ gives you a sequence $x_{n} \in E \ \forall n\in\mathbb{N}$ such that $\lim_{n \to \infty}x_n=0$.</p>

Title: If $E = \{ x \in \mathbb{R}: \sin(\frac1{x}) = 1\}$ then $l = 0$ is a limit point of E

Body:
If $E = \{ x \in \mathbb{R}: \sin\left(\frac{1}{x}\right) = 1\}$, then $l = 0$ is a limit point of $E$.

I have a proof here but I don't quite understand a few points, I hope someone can explain it a bit : )

To prove that $l = 0$ is a limit point, we need to show for any $\delta > 0 $, there exists $x \in E $ such that $ 0 < |x-0| < \delta $,

$$\sin\left(\frac{1}{x}\right) = 1 \quad \text{implies}  \quad x = \frac{2}{(2k+1)\pi}. $$

If $k = \max \left(1, \left\lceil \frac{1}{\delta} - \frac{1}{2} \right\rceil \right)$ where $\delta > 0 $, then $0 < \left|\frac{2}{(2k+1)\pi} - 0 \right| < \delta $ so $\exists x \in E$ s.t. $0 < |x-0| < \delta$.

So questions are here,

Why max function is used here? and why $1$ is chosen?

Why the ceiling is used for $\left\lceil \frac{1}{\delta} - \frac{1}{2} \right\rceil$?


答:
Since $y\le 18,y=6,12,18$. Reject $12$ since it is not a factor of $18$.

We get corresponding $z=120+y=126,138$. Reject $138$ since it is not a multiple of $18$.

We are left with $y=6,z=126,x=yz/18=42$.

Title: The difference between the HCF and LCM of $x$ and $18$ is $120$. Find $x$.


问:
Finding coefficients in powers of polynomials $x+2$, $x+1$, $2x+1$.
Question-
Let $n$ be a positive integer.
Select the correct expression for each of the following:

The coefficient of $^$
in $(2+)^ $
is -

The coefficient of $^$
in $(1+)^$
is-

The coefficient of $^$
in $(1+2)^$
is-

I’m having difficulty with these questions- how shall i approach them.
I know that the cardinality of the set is $(2^n)$
But how will this be applied when we have $x^k$ ?

答:
I think you should use Newton's binom formula :
$$ (a+b)^n = \sum\limits_{k=0}^{n} \binom{n}{k} a^k b^{n-k}$$

So in your question :

$(2+x)^n = \sum\limits_{k=0}^{n} \binom{n}{k} x^k2^{n-k}$, so the coefficient in $x^k$ is $\binom{n}{k} 2^{n-k}$

$(1+x)^n = \sum\limits_{k=0}^{n} \binom{n}{k} x^k$, so the coefficient in $x^k$ is $\binom{n}{k}$

$(2x+1)^n = \sum\limits_{k=0}^{n} \binom{n}{k} (2x)^k$, so the coefficient in $x^k$ is $\binom{n}{k} 2^{k}$


答:
One simple approach is to note that the HCF has to be a factor of $18$ and there are only $6$ of those to try.  We also know that the product of the HCF and LCM is $18x$ so we can make a table
$$\begin {array} {r r r r}HCF&LCM&18x&x\\ \hline
1&121&121&NA\\
2&122&244&NA\\
3&123&369&NA\\
6&126&756&42\\
9&129&1161&NA\\
18&138&2484&138\end{array}$$
where the NAs come because $18x$ is odd.  Only $x=42, HCF=6, LCM=126$ works


问:
If $a,b,c$ are three non coplanar vectors, then prove that the vector equation $r=(1-p-q)\vec a+p\vec b +q\vec c$ represents a plane
Let $r=x\vec a + y\vec b+z\vec c$

Comparing with the given equation, we obtain
$$x+y+z=1$$ which is a plane

What I don’t understand is how does this say $r$ is a plane, since r is actually $x \vec a+y\vec b+z\vec c$

答:
A plane $\vec r=(x,y,z)$ that passes the point $\vec a$ and with normal vector $\vec n$ is
$$(\vec r -\vec a)\cdot \vec n =0\tag1$$

Observe that the given vector can be rearranged as

$$\vec r -\vec a  = p(\vec b-\vec a) +q (\vec c-\vec a)$$
which satisfies the plane equation (1) with the choice of $\vec n =(\vec b-\vec a)\times (\vec c-\vec a)$.


问:
Prove divisibility using linear congruences
I need to prove that:

$$10|(53^{53} - 33^{33})$$

I can and should only use linear congruences ($a \equiv b \mod n$) - how can I do this?

Body:
$$53^{53}-33^{33}\equiv 3^{53}- 3^{33}\pmod {10}$$

Now, $$3^4=81\equiv 1\pmod 4\implies 3^{53}- 3^{33}\equiv 3^1-3^1\pmod {10}\equiv0$$ as $33\equiv1\pmod 4, 53\equiv1\pmod 4$

Alternatively, $$3^{53}- 3^{33}\equiv 3^{33}(3^{20}-1)\pmod {10}$$

Now, $3^4\equiv1\pmod 4\implies 3^{20}=(3^4)^5\equiv1^5\pmod {10}\equiv1$


问:
Let $F$ be a field. How do we show that maximal ideals of $F[x]$ are the principal ideals generated by the monic irreducible polynomials?

Let $F$ be a field. How do we show that maximal ideals of $F[x]$ are the principal ideals generated by the monic irreducible polynomials?

In Algebra by Artin, he says this proposition is proven analogously to:

Here, he shows that if $n$ is prime, then $\mathbb Z/(n)$ is a field. Then we use the fact that $R/I$ is a field iff $I$ is maximal, and he concludes that $(n)$ is maximal.

The analogous proof would be that if $f(x)$ is monic irreducible, then $F[x]/(f)$ is a field. The only problem is that he has not proven that $F[x]$ modulo a monic irreducible polynomial is a field.

答:
It is really easy to show that $\mathbb{Z}/p\mathbb{Z}$ is a field when $p$ is prime, because it easily stems from Bézout's identity.

However, Bézout's identity also holds over any Euclidean domain. Let $g(x)\in F[x]$, with $g(x)$ not divisible by $f(x)$. We want to show that $g(x)+(f)$ is invertible in $F[x]/(f)$.

Since $f$ is irreducible, $1$ is a greatest common divisor of $f$ and $g$, hence there are polynomials $u,v\in F[x]$ such that
$$
1=f(x)u(x)+g(x)v(x)
$$
Then, clearly, $v(x)+(f)$ is the inverse of $g(x)+(f)$ in $F[x]/(f)$.

Note that we actually proved that, for every irreducible element $n$ in a Euclidean domain $R$, the quotient ring $R/(n)$ is a field.


问:
Reference Request: Characterization of Contractability
On this page it is claimed that a non-empty topological space $X$, is contractable if and only if given any two $f,g:Y\rightarrow X$ (for any topological space $Y$) $f$ is homotopic to $g$.

I was looking for a reference which states this fact, both so I can quote it in a paper I am working on and to understand the details myself.  Does anyone know where I can find this?

答:
One might find such a thing as an exercise in an algebraic topology book. I'm doubtful that it would be stated as a separate, citable theorem, although I suppose one could comb through the textbook references given on that web page you linked. In lieu of that, here's a proof.

For the "if" direction, use the functions $f,g : X \to X$ where $f$ is the identity and $g$ is a constant.

For the "only if" direction, let $h : X \times [0,1] \to X$ be a homotopy from the identity map to a constant map $c_p : X \to \{p\} \subset X$. It follows that $h_f(x,t)=h(f(x),t)$ defines a homotopy from $f$ to the constant map $d_p : Y \to \{p\} \subset X$, and similarly with $h_g(x,t)=h(g(x),t)$. Since $f$ and $g$ are both homotopic to $d_p$, they are homotopic to each other.


答:
No casework needed.

$\gcd(x,18) + 120 = \text{lcm} (x,18) \implies \gcd(x,18) \equiv -120 \equiv 6 \pmod{18}$

Therefore $\gcd(x,18)=6, x = \gcd(x,18)\cdot \text{lcm}(x,18)/18 = 6 \cdot 126/18=42$.


问:
Find $\int_{1}^{2}\sum_{n=1}^{\infty }\frac{n}{e^{nx}}\, dx$ , why my method doesn't work?
I want to calculate $\int_{1}^{2}\sum_{n=1}^{\infty }\frac{n}{e^{nx}}\, dx$ and I know the answer is $\frac{e}{e^{2}-1}$ but with my method I found something different where is my mistake?

let $\frac{n}{e^{nx}}=n(e^{-x})^{n}$ and $(e^{-x})=y \Rightarrow$ $ny^n=\frac{n}{e^{nx}}\Rightarrow$

$\sum_{m=1}^{\infty }y^n= \frac{1}{1-y}\Rightarrow $, I take the derivative with respect $y\Rightarrow$

$\sum_{m=1}^{\infty }ny^{n-1}= \frac{1}{(1-y)^2}$    $\Rightarrow $ $\sum_{m=1}^{\infty }ny^n= \frac{y}{(1-y)^2}$

$\Rightarrow$ $\int_{1}^{2}\sum_{n=1}^{\infty }\frac{n}{e^{nx}}\, dy =\int_{1}^{2} \frac{y}{(1-y)^2}= \left [  \frac{-1}{1-y}\right ]_{1}^{2}$

and after I did some calculation, my answer was different from $\frac{e}{e^{2}-1}$ can someone explain how to solve this and why this doesn't work?

答: If you do $y=e^{-x}$, with $x$ going from $1$ to $2$, then $y$ goes from $e^{-1}$ to $e^{-2}$. Besides, if $y=e^{-x}$, then $\mathrm dy=-e^{-x}\,\mathrm dx$.


问:
Prove (or disprove) that $\log_2 n^n=O(\log_2 n!)$ and $\log_2 n!=O(\log_2 n^n)$.
I have to prove (or disprove) that $\log_2 n^n=O(\log_2 n!)$ and $\log_2 n!=O(\log_2 n^n)$.

I've tried to solve this by calculating $$\lim_{n \to \infty} \left\lvert \frac{\log_2 n^n}{\log_2 n!} \right\rvert$$ and $$\lim_{n \to \infty} \left\lvert \frac{\log_2 n!}{\log_2 n^n} \right\rvert$$ but I've had difficulty finding these limits.

答:
$n! =1...n \leqslant n.n....n = n^n$ so $log_2 n^n\leqslant (\log_2 n!)$, which gives : $$\log_2 n!=O(\log_2 n^n)$$

As @stupidqnasker remarked, $n!\geqslant n.(n-1)...(n/2)\geqslant (n/2)^{n/2}$ so $\log_2 n! \geqslant n/2 \log_2 n/2 \geqslant n/3 \log_2 n = 1/3 \log_2 n^n$ for $n$ big enough. So we have that $$\log_2 n^n = O(\log_2 n!)$$


问:
Quotient rule for limits of functions
Suppose that, $\lim\limits_{x\to x_0}f(x)$ and $\lim\limits_{x\to x_0}g(x)$ exist and $\lim\limits_{x\to x_0}g(x)\neq0$. Then, prove that $$\lim\limits_{x\to x_0}\frac{f(x)}{g(x)}=\frac{\lim\limits_{x\to x_0}f(x)}{\lim\limits_{x\to x_0}g(x)}$$

答:
Let $K$ and $L$ be the limits of $f$ and $g$ respectively. Let $\epsilon > 0 $ and choose $\delta > 0 $ such that for $|x-x_0| < \delta$ we have $|f(x) - K| < \epsilon$ and $|g(x) - L| < \epsilon.$ We can also conveniently suppose that in this range $|g(x)| \geq  |L|/2 .$ Then

\begin{align} |f(x)/g(x) - K/L| &= \bigg|\frac{f(x)L - Kg(x)}{Lg(x)}\bigg| \\
&= \bigg|\frac{f(x)L - KL + KL - Kg(x)}{Lg(x)}\bigg| \\
&= \bigg|\frac{L(f(x) - K) + K(L - g(x))}{Lg(x)}\bigg| \\
&< \epsilon(L+K)/L|g(x)|\\
&\leq \epsilon2(L+K)/L^2.
\end{align}
Of course if you want all of this to be neatly less than $\epsilon$ then you could change your values of $\delta$ accordingly.

Title: Quotient rule for limits of functions


问:
Formal Definition of Limit and Proofs
I'm having trouble understanding the formal definition of a limit...

Let $f(x)$ be defined on an open interval about $x_0$, except possibly
  at $x_0$ itself. We say that the limit of $f(x)$ as $x$ approaches
  $x_0$ is the number $L$, and write

$\lim\limits_{x \to x_0} f(x) = L$,

if, for every number $\epsilon > 0$, there exists a corresponding
  number $\delta > 0$ such for all $x$,

$0 < |x - x_0| < \delta => |f(x) - L| < \epsilon$.

From Thomas' Calculus 12th Edition, page 58

This definition is then used to prove that $\lim \limits_{x \to x_0} x = x_0$ and $\lim \limits_{x \to x_0} k = k$.

Can anyone give me a little bit of intuition behind the definition and these proofs? I'm really having trouble understanding them.

答:
The definition is quite subtle, but I will explain exactly what

$\lim_{x\to{x_0}} f(x) $

means.

Def. Let $f(x)$ be function defined for all $ x \neq x_0$, letting $L$ be a real number, we have hence:

$\lim_{x\to{x_0}} f(x) = L$

if, for every $\varepsilon > 0$, there is a $\delta > 0$, such that if $0 < |x- x_0|< \delta$, hence $|f(x) - L| < \varepsilon$.

Now, we will make everything clear. What the definition says is: if you pick an $ \varepsilon > 0$ (fixes it!), a positive number in the vertical $y$ axis, then naturally, if you draw a horizontal line in between $L+\varepsilon$ and $L-\varepsilon$, (for a prefixed L obviously), if you touch this line with the $f(x)$, you are going to have a point of intersection, with coordinates $(\delta, \varepsilon)$, however $\delta$ varies, in the sense that you can have each time, smaller intervalls within the intervall $\delta + x_0$ and $ x_0 - \delta$, so you can pick another number $\delta_0$ such that the intervall $(x_0 - \delta_0, x_0 + \delta_0)$ is "inside" the interval $(x_0 - \delta, x_0 + \delta)$.

Now, this $\delta$ exists, and it depends on $\varepsilon$, and it's clear that if x is in a distance to $x_0$ smaller than $\delta$, if we repeat the process of "line association" we've done, then, we'll obtain naturally y's in the y axis closer to $L$ than $\varepsilon$ the distance we have started with. I think that examples are important, but in this case it can diverge, however if we think a bit beyond, we can do the following:

Pick $\varepsilon > 0$

Pick $\delta = \frac{\varepsilon}{k*}$, (which is clearly smaller than $\varepsilon$!), in such a way that when we put in the special property we get the $|x - x_0| < \varepsilon$. This pick a $\delta$ isn't arbitrary, so you have to manipulate the limit in order to find an $\varepsilon$; the thing to keep in mind is to make $|x - x_0|$ appears somewhere.

Then naturally it's clear that the definition we gave will hold.

*k is a number greater equal to 1.

Title: Formal Definition of Limit and Proofs


答:
I think one part of the question has been answered already (i.e. $\log_2 n^n \geq \log_2 (n!)$).

For the second part (i.e. to show that $\log_2 n^n = O(\log_2 n!)$), note that $n! \geq n\cdot (n-1)\cdots (n/2) \geq (n/2)^{n/2}.$ I think you should be able to finish it from here.

Title: Prove (or disprove) that $\log_2 n^n=O(\log_2 n!)$ and $\log_2 n!=O(\log_2 n^n)$.

Body:
I have to prove (or disprove) that $\log_2 n^n=O(\log_2 n!)$ and $\log_2 n!=O(\log_2 n^n)$.

I've tried to solve this by calculating $$\lim_{n \to \infty} \left\lvert \frac{\log_2 n^n}{\log_2 n!} \right\rvert$$ and $$\lim_{n \to \infty} \left\lvert \frac{\log_2 n!}{\log_2 n^n} \right\rvert$$ but I've had difficulty finding these limits.


问:
$\lim_n \frac{1}{n} E(\max_{1\le j\le n} |X_j|) = 0$
If $\{X_n\}$ is a sequence of identically distributed r.v.'s with finite mean, then

$$\lim_n \frac{1}{n} E(\max_{1\le j\le n} |X_j|) = 0$$

The inequality

$$\frac{1}{n}E(\max_{1\le j\le n} |X_j|) \le \frac{1}{n}E(|X_1| + \cdots + |X_n|)=E(|X_1|)$$

suggests that the result has something to do with the "graphs" of the $|X_j|$ overlapping so that $E(|X_j|)=\int_0^\infty |X_j| P(d\omega)$ isn't fully counted $n$ times in the value of $E(\max_{1\le j\le n} |X_j|)$. But how do we make this rigorous? Notice there is no mention of the $X_j$ being independent.

答:
Let $u(x)=P[|X_1|\geqslant x]$ and $M_n=\max\limits_{1\leqslant k\leqslant n}|X_k|$, then $[M_n\geqslant x]=\bigcup\limits_{k=1}^n[|X_k|\geqslant x]$ hence
$$
n^{-1}E[M_n]=\int_0^{+\infty} u_n(x)\mathrm dx,
$$
with
$$
u_n(x)=n^{-1}P[M_n\geqslant x]\leqslant u(x).
$$
Now, $0\leqslant u_n\leqslant n^{-1}$ hence $u_n\to0$ pointwise. Furthermore, $u$ is integrable since
$$
\int_0^{+\infty}u(x)\mathrm dx=E[|X_1|],
$$
hence, by dominated convergence,
$$
\int_0^{+\infty}u_n(x)\mathrm dx\to0.
$$
Thus,
$$
n^{-1}E[M_n]\to0.
$$</p>


问:
If $A$ is simple, does $\|A\|_{\rm op}=\max_{\lambda\in\sigma(A)}|\lambda|$ imply that $A$ is normal?
We know that, in general, the operator norm of a matrix is larger than its spectral radius:  $\|A\|_{\rm op}\ge r(A)\equiv \max_{\lambda\in\sigma(A)}\lvert \lambda\rvert$, where $\sigma(A)$ denotes the spectrum of $A$.

If $A$ is normal, these quantities are clearly the same, as $\|A\|_{\rm op}$ equals the largest singular value of $A$, which equals the largest $|\lambda|$ for $\lambda$ an eigenvalue of $A$.

I can also find examples of $\|A\|_{\rm op}=r(A)$ for $A$ not normal, by building it as direct sum of a non-normal matrix and a matrix with a large eigenvalue. E.g.
$$\begin{pmatrix}2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1\end{pmatrix}.$$
This is not diagonalisable, but satisfies the equality.

Are there such examples with simple matrices? In other words, given a simple $A$ (i.e. a matrix composed of a single Jordan block), can we have $\|A\|_{\rm op}=r(A)$?

答:
Yes, but only when $A$ is $1\times1$. Note that every complex square matrix is unitarily triangulable. Suppose $B=U^\ast AU$ is such a triangularisation. Since all eigenvalues of $B$ are the same, we have $\rho(B)=|b_{jj}|\le\|\mathbf b_j\|_2\le\|B\|_2=\rho(B)$ for every $j$, where $\mathbf b_j$ denotes the $j$-column of $B$. It follows that $|b_{jj}|=\|\mathbf b_j\|_2$ for every $j$, i.e. $B$ is a diagonal matrix. As $B$ is similar to $A$ and the Jordan form of $A$ has only one Jordan block, $B$ and $A$ must be $1\times1$.

More generally, $\|A\|=\rho(A)$ for some submultiplicative matrix norm $\|\cdot\|$ if and only if every eigenvalue of $A$ with maximum modulus is semi-simple. For a proof, please see here.


问:
modelling smooth square wave function
I would like to model a smooth square wave function (see graph) but be able to control the amount of smoothness from fairly soft as shown to quite hard.

But I want to control the amount of smoothness using a simple function so the more the graph goes to a crest or peak the more compressed it becomes depending on some parameter that I can adjust.

I thought something sine logarithmic as that tends to compress but could not get that to work. I also want to avoid such things as the sgn function or Fourier series.

答:
You can obtain a $1$- parameter smooth approximation starting from the approximation for the sign function, i.e:

$$\sin x /\sqrt{\sin^2 x +\varepsilon ^2} \quad |\; \varepsilon \to 0$$

example

Title: modelling smooth square wave function


问:
An experiment consists of rolling a fair die 100 times.
If this experiment is repeated, what is the sampling distribution of the proportion of the 100 tosses that land with one spot in the top face?

the P(getting a 1) = 1/6

how do I find the proportion value if idk how many times of getting a 1? do I understand this question correctly?

Body:
It's a binomial

$$P(X=x/100)=\binom{100}{x}\Big(\frac{1}{6}\Big)^{x}\Big(\frac{5}{6}\Big)^{100-x}$$

$x=0,1,2...,100$


问:
Condition number on perturbations cannot cancel $||x||$ in inequality.
Theorem
Let $A$ be a square matrix, then
$$\lim_{k \to \infty } A_{k} =0 \rightarrow \rho(A)<1$$
Moreover, the geometric series case $\sum_{k=0}^{\infty} A_{k}$ is convergent iff  $\rho(A) < 1$. In such a

$$\sum_{k=0}^{\infty} A_{k} = (I - A)^{-1}.$$
As a result, if $rho(A) < 1$ the matrix $I - A$ is invertible and the following
inequalities hold
$$\frac{1}{1+||A||} \leq ||(I-A)^{-1}|| \leq \frac{1}{1 - ||A||}$$

where $||\cdot||$ is an induced matrix norm such that $||A|| < $1.

From $$(A + \delta A)(x+ \delta x) = b + \delta b$$ it follows that $$(I+A^{-1}\delta A)(x+ \delta x)= x+A^{-1}\delta b.$$
Moreover since $\gamma \mu(A) < 1$ and $||\delta A|| \leq \gamma ||A||$ it turns out that $I-A^{-1}\delta A$ is non singular.
Taking the inverse of such a matrix and passing to the norms we get $$||x+ \delta x || \leq ||(I-A^{-1}\delta A)^{-1}|| (||x|| + \gamma||A^{-1}|| \cdot ||b||)$$.

From theorem then it follows that:
$$||x+ \delta x || \leq  \frac{1}{||(I-A^{-1}\delta A)||}  (||x|| + \gamma||A^{-1}|| \cdot ||b||) $$

Substituting $||b|| \leq ||A|| \cdot ||x||$ then i obtain:

\begin{align*}
    ||x+ \delta x || \leq & \frac{1}{||(I-A^{-1}\delta A)||}  (||x|| + \gamma||A^{-1}|| \cdot ||b||) \\
    \leq & \frac{1}{||(I-A^{-1}\delta A)||}  (||x|| + \gamma||A^{-1}|| \cdot ||b||) \\
    \leq & \frac{1}{||(I-A^{-1}\gamma ||A||)||}  (||x|| + \gamma||A^{-1}|| \cdot ||A|| \cdot ||x||) \\
    \leq & \frac{1}{||(I-\gamma \mu(A)||}  (||x|| + \gamma \mu(A) \cdot ||x||) \\
    \leq & \frac{1}{||(I-\gamma \mu(A)||}  (||x|| + \gamma \mu(A) \cdot ||x||) \\
    \leq & \frac{1}{1-\gamma \mu(A)}  (||x|| + \gamma \mu(A) \cdot ||x||) \\
    \leq & \frac{||x|| + \gamma \mu(A) \cdot ||x||}{1-\gamma \mu(A)}
\end{align*}

But the true inequality is :

$$\frac{||x+ \delta x ||}{||x||} \leq  \frac{1 + \gamma \mu(A) }{1-\gamma \mu(A)}$$

and this $||x||$ cannot avoid it.What I am doing wrong?Any help?

Body:
\begin{align*}
    ||x+ \delta x || \leq & \frac{1}{||(I-A^{-1}\delta A)||}  (||x|| + \gamma||A^{-1}|| \cdot ||b||) \\
    \leq & \frac{1}{||(I-A^{-1}\delta A)||}  (||x|| + \gamma||A^{-1}|| \cdot ||b||) \\
    \leq & \frac{1}{||(I-A^{-1}\gamma ||A||)||}  (||x|| + \gamma||A^{-1}|| \cdot ||A|| \cdot ||x||) \\
    \leq & \frac{1}{||(I-\gamma \mu(A)||}  (||x|| + \gamma \mu(A) \cdot ||x||) \\
    \leq & \frac{1}{||(I-\gamma \mu(A)||}  (||x|| + \gamma \mu(A) \cdot ||x||) \\
    \leq & \frac{1}{1-\gamma \mu(A)}  (||x|| + \gamma \mu(A) \cdot ||x||) \\
    \leq & \frac{||x|| + \gamma \mu(A) \cdot ||x||}{1-\gamma \mu(A)}  \\
     \leq &\frac{||x||(1 + \gamma \mu(A)) }{1-\gamma \mu(A)} \\
   \frac{||x+ \delta x ||}{||x||}\leq &\frac{1 + \gamma \mu(A) }{1-\gamma \mu(A)}
\end{align*}


问:
Realising the non-trivial orientable $S^2$-bundle over $T^2$ as a quotient of $S^2\times T^2$
Using the fact that $\operatorname{Diff}^+(S^2)$ deformation retracts onto $SO(3)$, one can show that over any connected, closed, smooth surface, there are two orientable $S^2$-bundles, the trivial one and a non-trivial one. Moreover, the two bundles can be distinguished by their second Stiefel-Whitney class.

Let $X$ be the total space of the non-trivial orientable $S^2$-bundle over $T^2$. If we pullback $X \to T^2$ by any double covering $T^2 \to T^2$, we obtain the trivial $S^2$-bundle $S^2\times T^2 \to T^2$. It follows that $X$ is double covered by $S^2\times T^2$.

What is the free $\mathbb{Z}_2$ action on $S^2\times T^2$ which has quotient $X$?

It would be enough to realise the total space of an orientable rank two bundle $E \to T^2$ with $w_2(E) \neq 0$ as a $\mathbb{Z}_2$ quotient of $T^2\times\mathbb{R}^2$.

Note that $X$ can be realised as the sphere bundle of $\gamma_1\oplus\gamma_2\oplus(\gamma_1\otimes\gamma_2)$ where $\gamma_i$ is the pullback of the non-trivial line bundle on $S^1$ by projection onto the $i^{\text{th}}$ factor. It is easy to see pulling back by $(z, w) \mapsto (z^2, w)$ (or $(z, w) \to (z, w^2)$), the bundle becomes trivial (as it must). I was hoping these explicit descriptions would help, but so far I have been unsuccessful.

答:
For flat $SO(3)$-bundles over a (connected oriented) surface $\Sigma$ the 2nd SW-class is the obstruction to lifting the associated representation $\rho: \pi_1(\Sigma)\to SO(3)$ to a representation $\tilde\rho: \pi_1(\Sigma)\to SU(2)$.
Now, if $\Sigma=T^2$ with generators of the fundamental group $a, b$, the standard example of a non-liftable representation $\rho: \pi_1(\Sigma)\to SO(3)$ is the one whose image is $Z_2\times Z_2$ (sending the generators $a, b$ to the generators of the direct factors). If one thinks of $S^2$ as the Riemann sphere and $SO(3)$ as acting by linear-fractional transformations, WLOG, the representation $\rho$ is described as
$$
\rho(a): z\mapsto -z, \rho(b): z\mapsto z^{-1}.
$$
Consider, therefore, the product $Y=T^2\times S^2$ and $Z_2\times Z_2$ (with generators $\alpha, \beta$) acting on $Y$ via:
$$
\alpha\cdot (u, v) \times z= (-u, v) \times (-z),
$$
$$
\beta\cdot (u, v) \times z= (u, -v) \times (z^{-1}).
$$
(Here $u, v\in S^1$.) The quotient space of this action is $X$, the total space of  your oriented nontrivial $S^2$-bundle over $T^2$.

One can (in principle) extract from this description the one for a 2-fold covering space
$$
T^2\times S^2\to X.
$$
Namely, first divide $Y$ by the action of the first factor of $Z_2\times Z_2$. Then, one has to trivialize the quotient bundle $W=Y/Z_2\to T^2$.  This can be done by observing that the map $Y\to S^2$, $(u,v)\times z\mapsto z^2$ is $\alpha$-invariant, hence, descends to a map $W\to S^2$ and, therefore, can be used for a trivialization. Then using the resulting (biholomorphic) diffeomorphism $h: W\to T^2\times S^2$ one obtains the (holomorphic) action of the involution $\beta$ on $T^2\times S^2$  by conjugation via $h$. From this, one concludes for instance that there exists a degree 2 holomorphic covering map $T^2\times S^2\to X$. Writing an explicit holomorphic action of $\beta$ on $T^2\times S^2$ from this is doable but feels unpleasant. If you are sufficiently motivated, I am sure you can write down an explicit formula.


问:
How do I know when I can multiply by something on an inequality?
My doubt: How do I know when I can multiply by something on an inequality? and by this I don't mean multiplying by numbers as 3 or -3, I mean multiplying by things like x or $\ln(x)$ or $\cos(x)$, etc. functions in terms of x basically. I know I shouldn't be multiplying in a inequality on both sides like $\frac{x^2+3}{x-1}>0$ by $x-1$ because we don't know what x is and it might be negative while we don't know, hence why the recommendation of not multiplying by x. But came to a limit the other day which was $$\lim_{x\to 0}x\cos(\frac{1}{x})$$ and they did : $-1\leq \cos(\frac{1}{x})\leq 1$ and they proceeded to multiply by x on an inequality? or double inequality? and I was really extranged because this conflicted with my idea of not multiplying by x or whatever function of x on an inequality. Specifically they did $-x\leq x\cos(\frac{1}{x})\leq x$ and then they limit'd the 3 and by the squeeze theorem, blah blah blah, the limit is 0. But I only used this anecdote with the limit to clarify my doubt, the same way I'd like the reason to this, I'd love to know when I can multiply by something and when not on an inequality.

答:
When you have the inequation $f(x)\leqslant g(x)$ for $x\in I$, then you can multiply both side by $h(x)$ (without inversing the $\leqslant$) only for the $x$ such that $h(x)\geqslant0$.

So you must distinguish 2 cases if $h(x)$ is not always positive.

But in your example ($-1\leqslant \cos \frac{1}{x} \leqslant 1$), the two cases lead to the same inequation since $-(-1) = 1$, so if you're not specifically working on inequalities, you can go faster and not distinguish the 2 cases.


答:
As you say, you have to reverse the sense of the inequality when you multiply by something negative.  If you are going to multiply by an expression, you need to determine whether the expression always has the same sign or not.  If it always has the same sign, you can go ahead and reverse the inequality or not as required.  If not, you need to split into cases based on the sign of what you are multiplying by.

Your example is very special because of the symmetry.  You start with $-1 \le \cos \left(\frac 1x \right) \le 1$.  You are guaranteed that $x \neq 0$ because you divide by it already.  If $x \gt 0,$ you can multiply through by it to get $-x \le x\cos \left(\frac 1x \right) \le x$.  If $x \le 0$ you need to reverse the signs of the inequality and you get $x \le x\cos \left(\frac 1x \right) \le -x$.  The only change is the position of the negative sign on $x$ on the ends.

Body: My doubt: How do I know when I can multiply by something on an inequality? and by this I don't mean multiplying by numbers as 3 or -3, I mean multiplying by things like x or $\ln(x)$ or $\cos(x)$, etc. functions in terms of x basically. I know I shouldn't be multiplying in a inequality on both sides like $\frac{x^2+3}{x-1}>0$ by $x-1$ because we don't know what x is and it might be negative while we don't know, hence why the recommendation of not multiplying by x. But came to a limit the other day which was $$\lim_{x\to 0}x\cos(\frac{1}{x})$$ and they did : $-1\leq \cos(\frac{1}{x})\leq 1$ and they proceeded to multiply by x on an inequality? or double inequality? and I was really extranged because this conflicted with my idea of not multiplying by x or whatever function of x on an inequality. Specifically they did $-x\leq x\cos(\frac{1}{x})\leq x$ and then they limit'd the 3 and by the squeeze theorem, blah blah blah, the limit is 0. But I only used this anecdote with the limit to clarify my doubt, the same way I'd like the reason to this, I'd love to know when I can multiply by something and when not on an inequality.


问:
Determine whether origin is a local extrema of a given function
Consider the function $f(x,y)=x^6-2x^2y-x^4y+2y^2$ defined from $\mathbb{R}^2$ to $\mathbb{R}$.

Then $(0,0)$ is a critical point.

Also $f_{xx}$ at origin is zero, $f_{yy}$ at origin is $4$ and $f_{xy}$ at origin is $0.$ The value of the determinant is zero. So it cannot be concluded whether the origin is local minimum or maximum or saddle point.

How can we determine for this particular problem whether the origin is local maximum or local minimum or a saddle point?

Thanks in advance.

答:
$f(x,y)=x^6-2x^2y-x^4y+2y^2$ can be factored as $$(x^2-y)(x^4-2y),$$ which indicates that along a convenient path through $(0,0)$ can be $f(x,y)>0$ and through another path $f(x,y)<0.$

Indeed, $$\begin{align}f(x,x^4)&=(x^2-x^4)(-3x^4)\\&=3x^6(1-x^2)\end{align}$$
which is negative for $x$ near $0,$ while
$$f(x,-x^4)=(x^2+x^4)(3x^4),$$
which is clearly positive for any $x.$

The point $(0,0)$ is a saddle point of the function $f.$


问:
Function of $f : \mathbb{R} → \mathbb{R}$ satisfies $f(x + y) = f(x) + f(y)$ for each $x, y ∈ \mathbb{R}.$

Suppose that $f : \mathbb{R} → \mathbb{R}$ satisfies $f(x + y) = f(x) + f(y)$ for each $x, y ∈ \mathbb{R}.$

$(a).$ Show that $f(nx) = nf(x)$ for all $x ∈ \mathbb{R}$ and $n ∈ \mathbb{Z}.$

$(b).$ Prove that $f(mx) = mf(x)$ for all $x ∈ \mathbb{R}$ and $m ∈ \mathbb{Q}.$

$(c).$ Prove that $f$ is continuous at $0$ if and only if $f$ is continuous on $\mathbb{R}$.

$(d).$ Prove that if $f$ is continuous at $0$, then there is an $m ∈ \mathbb{R}$ such that $f(x) = mx$ for all $x ∈ R$.

$\underline{Attempt}$

When $x=y=0$, we can get $f(0)=0$ and also when $x=-y$, $f(x)=-f(-x)$

$(a).$ For this part I considered  $3$ cases when $n> 0,$ $n<0$ and $n=0$

When $n=0$ it is obviously true.

When $n> 0$, $$f(nx)=f(\underbrace{x+x+x...+x}_{n})=\underbrace{f(x)+f(x)...+f(x)}_{n}=nf(x)$$

When $n<0$, $$f(nx)=f(\underbrace{(-x)+(-x)+(-x)...+(-x)}_{|n|})=\underbrace{f(-x)+f(-x)...+f(-x)}_{|n|}=nf(x)$$

$(b)$. Take $m \in \mathbb{Q}$ such that $m=\dfrac{p}{q}$
$$f(mx)=f \left(\dfrac{px}{q} \right)=f\left(p \left(\dfrac{x}{q}\right)\right)=pf\left(\frac{x}{q}\right)$$ using $y=\dfrac{x}{q}$ substitution $f(yq)=f(x)=qf(y)$ hence
$$f(mx)=f \left(\dfrac{px}{q} \right)=\frac{p}{q}f(x)$$

$(c)$.  If $f$ is continuous at $0$,

$\forall \epsilon>0$ $ \exists \delta>0$  such that $|x-0|<\delta$ and $x \in \mathbb{R}$ imply $|f(x)-f(0)|=|f(x)|< \dfrac{\epsilon}{(1+|p|)}$

let consider  $a$ and $p$ $\in\mathbb{R}$ such that $a=px$, from part $(a)$ and $(b)$

$$|f(x)-f(a)| \leq |f(x)|+|f(a)|=|f(x)|(1+|p|)<\epsilon \text{ whenever }  |x-a|<\delta_1$$

If $f$ is continuous on $\mathbb{R}$ it also continuous at $0$, So we can conclude
$f$ is continuous at $0$ if and only if $f$ is continuous on $\mathbb{R}$.

$(d).$ Define a function $f : \mathbb{R} → \mathbb{R}$ satisfies $f(x + y) = f(x) + f(y)$ for each $x, y ∈ \mathbb{R}$ such that $f(x)=mx$

given that $f$ is continuous at $0$ so from part $(c)$ $f$ is continuous on $\mathbb{R}$ and $m=f(1)$

This is what I tried. If there Is anything wrong correct me and not sure part $(c)$ and $(d)$

答:
To prove part (d) we use sequential criterion of continuity.

Observe that for any rational $x$ there exist $m$ such that $f(x) = mx$.

Consider a sequence $x_n$ which converges to an irrational $x$ and by hypothesis we have $f$ is continous at all real numbers. Consider a subseqence $(p_n)$ of $x_n$ such that it contains all rational terms. Then we have $f(p_n ) = mp_n \to mx = f(x)$. Now if no subseqence with irrational terms exists then we are done else let $q_n$ be subseqence with irrational terms. Given that $f$ is continous at $x$ we must have $f(q_n) \to f(x)$ and by uniqueness of limits we have $f(x) = mx$


问:
Bound on expectation of absolute value in terms of variance
In my book it says that a white noise process $\{Z_t\}$ with mean zero and variance $\sigma^2$ has the following property:

E$|Z_t| \leq \sigma$.

This had me thinking of Jensen's inequality, that $\text{E}(g(X)) \geq g(\text{E}(X))$, for convex functions g.

Since we have that $\text{E}(Z^2_t) = \sigma^2$ and $\text{E}(Z_t) = 0$, we could apply this inequality to to  $\text{E}(Z_t)$ and obtain $\text{E}(|Z_t|) \geq |\text{E}(Z_t)| = 0$, since the absolute value is a convex function.  But we need the upper bound, and the inequality is only for convex functions, so we can't use $\text{E}(Z^2_t) = \sigma^2$ and take square roots on both sides..

答:
As you guessed, this is a consequence of Jensen inequality, but for $X=|Z_t|$ and $g:x\mapsto x^2$. To wit, with these choices, $E(g(X))=E(Z_t^2)$ and $g(E(X))= \left(E|Z_t|\right)^2$ hence Jensen inequality reads $\left(E|Z_t|\right)^2\leqslant E(Z_t^2)=\sigma^2$, QED.

Of course, there are other ways to prove this, such as invoking Cauchy-Schwarz inequality (but for the random variable $|Z_t|$ and not for $Z_t$ itself, since, as you correctly noted, Cauchy-Schwarz for $Z_t$ yields the vacuous inequality $E(Z_t^2)\geqslant0$), or expanding $E((|Z_t|-z)^2)\geqslant0$ for $z=E(|Z_t|)$.


问:
fluid in a vessel
Consider fluid in a cylindrical vessel of radius a which is undergoing solid-body rotation so
that its velocity field is given by $v = (−\sigma y, \sigma x, 0)$. (The z-axis is along the axis of the vessel and vertical.)

(a) Show that to within an additive constant, the pressure distribution is given by
$p=\frac{1}{2} \rho \sigma^{2} (x^{2}+y^{2})-\rho g z$

Hence find the shape of the free surface of the fluid (i.e. the interface between the
rotating fluid and the atmosphere.) And show that the fluid at the center of the vessel
is lower than that at the edges by $\frac{\sigma^{2}a^{2}}{2g}$.

(b) Estimate the magnitude of this difference in heights for a coffee cup of radius 5cm
which is stirred rapidly at a rate of 2 revolutions per second.

(c) If the height of the vessel is equal to its radius a. And if it is filled up to a level 3a/4
before it is set to rotate. What is the maximum angular velocity for which the liquid
stays in the vessel? What is the minimum level of the water at this maximum angular
velocity?

for the(a) question I noticed that you could draw the vessel with the axes in the middle. You can write down $a^{2}=x^{2}+y^{2}$. But now I don't see how you can becom the solution of the pressure. I thought easy to use $p=\rho g h$ but then I'm stuck with how to get a $\sigma$ in the solution.

for the (b) question I can't really describe it.

For the (c) question I thought i needed to calculate the volume of the rotating fluid. The maximum height that the fluid reaches on the edge must be at most equal to the height of the vessel. But because I'm stuck with (a) I can't use the result here.

答:
I can get you started with part (a) and we can take it from there if needed. This problem lends itself naturally to cylindrical coordinates such that "radial" direction is perpendicular to the axis of symmetry, and the $z$-axis does its usual job.

Now let me look at a small piece of fluid somewhere in the vessel. It feels a force from two effects: (i) the fluid rotation, and (ii) gravity. The effect of rotation acts purely radially, and naturally, gravity pulls along $-z$. This lets me write Newton's second law separately for each component: $$m a_r = m \frac{v^2}{r} \hspace{2.54cm} ma_z = -mg$$

In the above, we have $m$ as the (differential) mass of the small piece of fluid, $v$ being its instantaneous velocity, and $a_r$, $a_z$ being the respective acceleration components. In the left equation, note that we may replace $m$ by its equivalent expression $$m = \rho \cdot dA \cdot dr \:,$$ where $dA \cdot dr$ is the differential volume that $m$ occupies. We can also replace $v^2$ by its equivalent expression, $\sigma^2 \left(x^2+y^2\right)$ to write $$ma_r = \rho \cdot dA \cdot dr \cdot \sigma^2 \frac{x^2+y^2}{\sqrt{x^2+y^2}}$$ Divide through by $dA$, and we have $$\frac{ma_r}{dA} = \rho \cdot dr \cdot \sigma^2 \frac{x^2+y^2}{\sqrt{x^2+y^2}} \:.$$ The form on the left is, by definition, the differential "radial" contribution to the pressure (just force over area). On the right, we can substitute $x^2 + y^2 = r^2$ to simplify things greatly: $$dp_{rad} = \rho \cdot \sigma^2 r \: dr$$ Integrate both sides to get $$p_{rad} = \int dp_{rad} = \int \rho \cdot \sigma^2 r \: dr = \frac{1}{2}\rho \sigma^2 r^2 + p_0 \:,$$ where $p_0$ is an arbitrary constant (of the proper units). Restoring the $xy$-variables, we are halfway done: $$p_{rad} = \frac{1}{2}\rho \sigma^2 \left(x^2 + y^2\right) + p_0$$

That takes care of the rotational part, now for the gravitational. Start with the same setup: take a very very small mass $m$ occupying a differential volume $dA \cdot dz$ such that $$m a_z = -mg = - \rho \cdot dA \cdot dz \cdot g \:.$$ Divide through by $dA$ again to write $$dp_{z} = \frac{m a_z}{dA} = - \rho g \: dz \:.$$ Once again, the form on the left is just force over area, a differential contribution to pressure $p_z$. Integrate both sides, and we have $$p_z = \int dp_z = -\rho g \int dz = -\rho g z + \tilde{p}_0 \:,$$ where $\tilde{p}_0$ is another integration constant.

Finally, sum the two pressure values to crank out the answer: $$p = p_{rad} + p_z = \frac{1}{2}\rho \sigma^2 \left(x^2 + y^2\right) - \rho g z + p_0 + \tilde{p}_0$$ Of course, $p_0 + \tilde{p}_0$ is just a constant too, so lets leave it at: $$p = \frac{1}{2}\rho \sigma^2 \left(x^2 + y^2\right) - \rho g z + C$$

... Hope that helps things along. Also - we "should" replace all masses $m$ with the differential notation $dm$, but hopefully the context makes everything clear enough to not litter the page with more $d$'s.

Title: fluid in a vessel


答:
Yes, $A$ must be normal.  Otherwise, given that  $A$ consists of a single Jordan block,  that block is necessarily
nontrivial (i.e. it is of size at least 2) so
there are two nonzero vectors $u$ and $v$ such that
$Au=\lambda u+v$ and $Av=\lambda v$.  Therefore, for every scalar $b$ one has that
$$
  \|A(u+bv)\|^2 = \|\lambda (u+bv) + v\|^2 = $$$$ =
  |\lambda |^2\|u+bv\|^2 + \|v\|^2 + 2 \Re\lambda \langle u+bv, v\rangle  = $$$$ =
  |\lambda |^2\|u+bv\|^2 + \|v\|^2 + 2 \Re\lambda \langle u, v\rangle    + 2 \Re \lambda b \langle v, v\rangle .
  $$

Assuming that $λ≠0$ we may choose $b$ such that
$$
  \|v\|^2 + 2 \Re\lambda \langle u, v\rangle    + 2 \Re\lambda b\langle v, v\rangle >0,
  $$
so we see that
$$
  \|A(u+bv)\| >   |\lambda |\|u+bv\|,
  $$
whence $\|A\|>|\lambda | = r(A)$, contradicting the hypothesis.

On the other hand,  if $λ=0$ we also get a contradiction because    $\|A\|>0 =r(A)$.

Title: If $A$ is simple, does $\|A\|_{\rm op}=\max_{\lambda\in\sigma(A)}|\lambda|$ imply that $A$ is normal?

Body:
We know that, in general, the operator norm of a matrix is larger than its spectral radius:  $\|A\|_{\rm op}\ge r(A)\equiv \max_{\lambda\in\sigma(A)}\lvert \lambda\rvert$, where $\sigma(A)$ denotes the spectrum of $A$.

If $A$ is normal, these quantities are clearly the same, as $\|A\|_{\rm op}$ equals the largest singular value of $A$, which equals the largest $|\lambda|$ for $\lambda$ an eigenvalue of $A$.

I can also find examples of $\|A\|_{\rm op}=r(A)$ for $A$ not normal, by building it as direct sum of a non-normal matrix and a matrix with a large eigenvalue. E.g.
$$\begin{pmatrix}2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1\end{pmatrix}.$$
This is not diagonalisable, but satisfies the equality.

Are there such examples with simple matrices? In other words, given a simple $A$ (i.e. a matrix composed of a single Jordan block), can we have $\|A\|_{\rm op}=r(A)$?


答:
$$ \begin{split}
\left(\frac1 n \sum_{j1}^n \log_2 k\right ) -\log_2 n &= \frac1 n \sum_{j=1}^n (\log_2 k -\log_2 n)\\
&= \frac1 n \sum_{j=1}^n \log_2 \left (\frac k n \right)\\
&\rightarrow \int_0^1 \log_2 x dx=-\frac 1 {\ln 2}
\end{split}
$$

Thus $$\log_2(n!)=\log_2(n^n)-\frac 1 {\ln 2}n+o(n)\sim \log_2(n^n)$$
which proves the result.


答:
This is from the book ' Introduction to Real Analysis ' by S.K. Mapa(7 th edition).

This is a worked out example at the end of chapter 8, Continuity. This is the answer to your question,  only the converse part of (c) is not given . But you can  prove it similarly.  If you still have doubt,  then comment below.

Title: Function of $f : \mathbb{R} → \mathbb{R}$ satisfies $f(x + y) = f(x) + f(y)$ for each $x, y ∈ \mathbb{R}.$

Body:
Suppose that $f : \mathbb{R} → \mathbb{R}$ satisfies $f(x + y) = f(x) + f(y)$ for each $x, y ∈ \mathbb{R}.$

$(a).$ Show that $f(nx) = nf(x)$ for all $x ∈ \mathbb{R}$ and $n ∈ \mathbb{Z}.$

$(b).$ Prove that $f(mx) = mf(x)$ for all $x ∈ \mathbb{R}$ and $m ∈ \mathbb{Q}.$

$(c).$ Prove that $f$ is continuous at $0$ if and only if $f$ is continuous on $\mathbb{R}$.

$(d).$ Prove that if $f$ is continuous at $0$, then there is an $m ∈ \mathbb{R}$ such that $f(x) = mx$ for all $x ∈ R$.

$\underline{Attempt}$

When $x=y=0$, we can get $f(0)=0$ and also when $x=-y$, $f(x)=-f(-x)$

$(a).$ For this part I considered  $3$ cases when $n> 0,$ $n<0$ and $n=0$

When $n=0$ it is obviously true.

When $n> 0$, $$f(nx)=f(\underbrace{x+x+x...+x}_{n})=\underbrace{f(x)+f(x)...+f(x)}_{n}=nf(x)$$

When $n<0$, $$f(nx)=f(\underbrace{(-x)+(-x)+(-x)...+(-x)}_{|n|})=\underbrace{f(-x)+f(-x)...+f(-x)}_{|n|}=nf(x)$$

$(b)$. Take $m \in \mathbb{Q}$ such that $m=\dfrac{p}{q}$
$$f(mx)=f \left(\dfrac{px}{q} \right)=f\left(p \left(\dfrac{x}{q}\right)\right)=pf\left(\frac{x}{q}\right)$$ using $y=\dfrac{x}{q}$ substitution $f(yq)=f(x)=qf(y)$ hence
$$f(mx)=f \left(\dfrac{px}{q} \right)=\frac{p}{q}f(x)$$

$(c)$.  If $f$ is continuous at $0$,

$\forall \epsilon>0$ $ \exists \delta>0$  such that $|x-0|<\delta$ and $x \in \mathbb{R}$ imply $|f(x)-f(0)|=|f(x)|< \dfrac{\epsilon}{(1+|p|)}$

let consider  $a$ and $p$ $\in\mathbb{R}$ such that $a=px$, from part $(a)$ and $(b)$

$$|f(x)-f(a)| \leq |f(x)|+|f(a)|=|f(x)|(1+|p|)<\epsilon \text{ whenever }  |x-a|<\delta_1$$

If $f$ is continuous on $\mathbb{R}$ it also continuous at $0$, So we can conclude
$f$ is continuous at $0$ if and only if $f$ is continuous on $\mathbb{R}$.

$(d).$ Define a function $f : \mathbb{R} → \mathbb{R}$ satisfies $f(x + y) = f(x) + f(y)$ for each $x, y ∈ \mathbb{R}$ such that $f(x)=mx$

given that $f$ is continuous at $0$ so from part $(c)$ $f$ is continuous on $\mathbb{R}$ and $m=f(1)$

This is what I tried. If there Is anything wrong correct me and not sure part $(c)$ and $(d)$


问:
Proving basic facts about sets and proper classes
I have the following statements:

Definition of set. $y$ is a set if and only if there is $z$ such that $y\in z$; otherwise $y$ is a proper class

The comprehension axiom. Given a formula $\psi(x)$ where $x$ occurs free in $\psi$ and a class $y$, we have that $y\in\{x:\psi(x)\}$ if and only if $y$ is a set and $\psi(y)$

Definition of the empty set. $\emptyset$ is defined as $\{x:x\neq x\}$

Definition of the universe. $V$ is defined as $\{x:x=x\}$

At this point I managed to prove that:

no set belongs to $\emptyset$

every set belongs to $V$

$\{x:x\notin x\}$ is a proper class

How can I prove that $\emptyset$ is a set and that $V$ is a proper class? How can I prove that $\{x:x\notin x\}$ is equal to $V$?

Do I need additional statements perhaps?

Thank you.

答:
To prove that $\emptyset$ is a set, use the Empty Set Axiom (if it's on your list).

To prove that $V$ is a proper class, obtain a contradiction if it's a set, using $\{x : x \notin x\}$.

Finally, the Axiom of Regularity should imply that $x \notin x$ for all sets.


问:
Solving modular equations that gives GCD = 1
I have problems with understanding modular equations that gives GCD = 1. For example:

$$3x \equiv 59 \mod 100$$

So I'm getting $GCD(3, 100) = 1$. Now:

$1 = -33*3 + 100$

That's where the first question appears - I always guess those -33 and 1 (here) numbers...is there a way to solve them?

And the second question - the answer to that equation, at least according to the book, is:

{$-33*59 + 100k$} (k - integer) - and why is that so? Where did the 59 came from?

答:
So, $3x=59+100y$ where $y$ is an integer

$\implies 3x=100y+59(100-3\cdot33)$

$\implies 3(x+59\cdot 33)=100(y+59)$

$\implies \frac{3(x+59\cdot 33)}{100}=y+59$ which is an integer

$\implies 100$ divides $3(x+59\cdot 33)$

$\implies 100$ divides $(x+59\cdot 33)$ as $(3,100)=1$

So, $x+59\cdot 33=100\cdot k$ where $k$ is an integer


问:
Alhazen's problem history
I was wondering if anyone has read '' alhazen book of optics'' and has understood his solution of Alhazen's problem.

I know a modern solution of the problem by Dörrie-100 great problems of mathematics, but I'm interested really in ibn alhaytham's solution. I'm just curious about how he went about solving the problem without the modern tools. Also I couldn't find the problem in his book, maybe I didn't look carefully for it, but it would be nice if someone could reference me to it in his book also.

答:
Here is the latin translation of Alhazen's statement, in the edition by Friedrich Risner (Basel 1572):

Visu et visibili a centro speculi sphaerici convexi inaequabiliter
distantibus, punctum reflexionis invenire. [n. 39, p. 150]

That is:

To find the reflection point, for eye and object at different
distances from the centre of a convex spherical mirror.

His solution is shown in figure below. Let $a$ be the eye and $b$ the object whose light is reflected to the eye by a spherical mirror of centre $g$ (the circle is of course the intersection between the sphere and plane $abg$). To find reflection point $d$, construct a segment $mk$ and a point $f$ on it such that
$mf:fk=bg:ga$. Let then $o$ be the midpoint of $mk$ and construct on its perpendicular bisector a point $c$ such that $\angle ock={1\over2}\angle agb$ (I'm assuming $gb>ga$).

Now, construct on line $kc$ a point $p$ such that, if $s$ is the intersection between lines $pf$ and $oc$, then the ratio $sp/pk$ is equal to the ratio between $bg$ and the radius of the sphere. This construction cannot be done by ruler a compass.  However, Alhazen gives a neusis construction for it, as a lemma (n. 38 in the same edition) and also an alternate construction (n. 34) where the neusis is solved by means of a hyperbola.

He also warns the reader that two possible points $p$ can be constructed, in general, but only one of them, at most, is such that $\angle pks>\pi/2$, and this is the one to be considered. If both solutions give $\angle pks\le\pi/2$ then no reflection point exists.

Point $d$ on the circle, such that $\angle bgd=\angle spk$, can be constructed, and Alhazen then proves this is the wanted reflection point, because $\angle pks=\angle bdg=\angle adg$.

Title: Alhazen's problem history


问:
Question about affine coordinate changes
Fulton in his book defines affine coordinate changes:

I'm trying to prove the item (b) of this question:

Let's prove using the induction suggestion.

Suppose $V=V(F_1)$, where $F_1=b_1X_1+\ldots+b_nX_n+b_0$, then $V^T=V(F_1^T)=V(F\circ T)$. I've found this change of coordinates:

$$T'=
        \begin{pmatrix}
        b_1^{-1} & 0 & \ldots & 0 \\
        0        & 0 &    \ldots   & 0  \\
        \vdots & \vdots & \ddots & \vdots \\
        0 & 0 & \ldots & 0 \\
        \end{pmatrix}
$$ and

$$T''=
        \begin{pmatrix}
        -b_1^{-1}b_0 \\
        0 \\
        \vdots  \\
        0 \\
        \end{pmatrix}
$$

So, $F_1^T=X_1$ and $V^T=V(X_1)$, the problem is $T'$ I've found is not invertible, is there anything I can do to fix this?

Thanks

答:
As the OP noticed in his comments, the other answer to this question is incorrect.

Let $F(X) = a_1X_1 + a_2X_2+\dots+a_nX_n + b$.
WLOG, suppose that $a_1\neq 0$.
Then define $T = (T_1,\dots,T_n)$ by
\begin{align*}
T_1(X) &= \frac{1}{a_1}X_1 -\frac{a_2}{a_1}X_2-\cdots-\frac{a_n}{a_1}X_n - \frac{b}{a_1},
\\\\
T_i(X) &= X_i, \text{ if }i\geq 2.
\end{align*}

Then you can check that
$
(F\circ T)(X) = X_1,
$
thus $V^T = V(X_1)$.


问:
Proof of the universal property of the quotient topology.
Would anyone please have a look at my proof for the following statement?

Let $X$ and $Y$ be topological spaces and $p: X \rightarrow Y$ a surjective continuous map. Consider the following property:

(A): For every topological space $Z$ and every continuous function $f: X \rightarrow Z$ such that $f(x)=f(y)$ if $p(x)=p(y),$ there is a unique continuous $g$ such that $f=g \circ p$.

Show that $(\mathrm{A})$ is satisfied if and only if there is a homeomorphism $h: Y \rightarrow X / \sim$ to the quotient space $X / \sim$ with the equivalence relation $x \sim y \Leftrightarrow p(x)=p(y),$ such that $h \circ p$ is the projection $\operatorname{map} X \rightarrow X / \sim$ We say that $(\mathrm{A})$ is the universal property of the quotient topology.

Proof:

[$\Rightarrow$] Let $\Pi: X \to X / \sim$ be the projection map, i.e. $\Pi(x) = [x] := \{z \in X: p(x) = p(z)\}$. For every $x,y \in X$ such that $p(x) = p(y)$, by defintion of $\Pi$, we have $\Pi(x) = \Pi(y)$. Then by the universal property $(A)$, there exists an unique continuous $h: Y \to X / \sim$ such that $\Pi = h \circ p$.

Similarly, if $x, y \in X$ such that $\Pi(x) = \Pi(y)$, then by definition of $\Pi$, we have $p(x) = p(y)$. $\Pi$ is clearly a surjective map, and $p$ is continuous by assumption, so by the universal property, there exists an unique $g: X / \sim \  \to Y$ such that $p = g \circ \Pi$.

$g$ and $h$ are continuous inverses of each other, since we have:
$$p = g \circ \Pi = g \circ (h \circ p) =  (g \circ h) \circ p, \text{ i.e. } g \circ h = \mathbb{I}_Y, \text{ and}$$
$$\Pi = h \circ p = h \circ (g \circ \Pi) =  (h \circ g) \circ \Pi, \text{ i.e. } h \circ g = \mathbb{I}_{X / \sim}.$$

[$\Leftarrow$] Let $Z$ be a topological space and $f: X \to Z$ a continuous function such that $f(x) = f(y)$ if $p(x) = p(y)$. Define $\varphi: X / \sim \ \to Z$ as: $\varphi([x]) = f(x)$. As $\varphi([x]) = \varphi \circ \Pi(x) = f(x)$, and as both $\Pi$ and $f$ are continuous, $\varphi$ is also continuous.

We check that $\varphi$ is well-defined. Assume for a contradiction that it is not. Then there exists $S \in X / \sim$ such that $S$ is mapped by $\varphi$ to both $z_1$ and $z_2$ in $Z$, where $z_1 \neq z_2$. Since $S \in X / \sim$, $S$ is of the form $\{x_1,x_2,... \in X: p(x_1) = p(x_2) = ...\}$. Since $S \mapsto z_1$, this implies $p(x_1) = p(x_2) = ... = z_1$, and, similarly, since $S \mapsto z_2$, this implies $p(x_1) = p(x_2) = ... = z_2$. But this means $p(x_1) \neq p(x_1)$, a contradiction.

Now define $g = \varphi \circ h$. As the composition of continuous functions, $g$ is continuous. $g$ is well-defined since both $\varphi$ and $h$ are well-defined, and for any $x \in X$:
$$g \circ p(x) = \varphi \circ h \circ p(x) = \varphi \circ \Pi(x) = \varphi([x]) = f(x)$$

We show that $g$ is unique: assume $g': Y \to Z$ is a continuous function such that $f = g' \circ p$. For all $y \in Y$, we want to show that $g'(y) = g(y)$. Now since $p$ is surjective, there exists $x \in X$ such that $p(x) = y$. Then we have:
$$g'(y) = g'(p(x)) = g' \circ p(x) = f(x) = g \circ p(x) = g(p(x)) = g(y)$$

答:
Some comments:

In the two parts of the proof of [$⇒$], it is not clear what universal properties you are referring to.

The proof of the fact that $φ$ is well-defined seems to be  unnecessarily twisted. It is enough to directly show that $[x] = [y]$ implies $f(x) = f(y)$.

The claim that $g$ is continuous and the corresponding reasoning is missing in [$⇐$].

The proof of uniqueness of $g$ is unnecessarily complicated given that we have already been right-cancelling surjections at the end of the proof of [$⇒$].