答:
Hint:

If on day $d$ there is a proportion $f$ of fresh water, on day $d+1$ the proportion is $\dfrac{499f+1}{500}$. Hence you have an arithmetico-geometric progression.

Solve the recurrence

$$f_{d+1}=\dfrac{499f_d+1}{500}$$ or $$f_{d+1}-1=\frac{499}{500}(f_d-1)$$

which is a geometric progression.

$$\left\lceil\frac{\dfrac{50}{500}}{\log\dfrac{499}{500}}\right\rceil-1=1150.$$

Title: How long does it take to change $90\%$ of water in an aquarium?

Body:
I am thinking about buying an aquarium and thought about a system that automatically changes water. This led me to think about the following problem:

Given a $500$$\ell$ tank and a system that automatically changes $1$$\ell$ of the water in the tank with fresh water every day, how many days does it take to replace $90\%$ of the water that's in the tank on day $0$? We assume the old water and the fresh water are perfectly mixed every time fresh water is added.

My idea is that on the first day we simply replace $1/500$ of the water. On the second day again  we replace $1/500$ of the water in the tank. However, $1/500$ has already been replaced on the first day and we have to subtract it. So on the second day we replace $\frac{1}{500}-\frac{\frac{1}{500}}{500}=\frac{1}{500}-\frac{1}{500^2}$. Thus, in total we have replaced $\frac{2}{500}-\frac{1}{500^2}$ on the second day. On the third we do the same thing, replace $1/500$ and subtract the water we have already replaced, obtaining $\frac{1}{500}-\frac{\frac{2}{500}-\frac{1}{500^2}}{500}=\frac{1}{500}-\frac{2-\frac{1}{500}}{500^2}$ of fresh water on the third day.

However I can't find the pattern, but I think this is a series, isn't it? So all we would have to do is find an explicit form of the series and find the finite series that equals $450/500$, right?

I think I was posed a similar question in my calculus course once and thought that there's an easy solution to the problem. Is my reasoning above correct? How can I find the solution to my problem?


问:
Irreducibility over $\mathbb{Z}_3$

Suppose $R= \frac{\mathbb{Z}_3[x]}{x^5+2x^4+x^2+2x+1}$, then which of the following is true?

The number of ideals in $R$ is 4.

$R$ has exactly two prime ideals.

R is an integral domain.

R[x] is a PID.

I have tried to check if the polynomial $x^5+2x^4+x^2+2x+1$ is reducible. For that, I have checked for roots and then I assumed that the polynomial is reducible and tried to compare the left-hand side and right-hand side of
$x^5+2x^4+x^2+2x+1=(mx^3+ax^2+bx+d)(nx^2+ex+f)$
It seems that the polynomial is irreducible. Is there any general way of checking irreducibility in Finite fields? I know there is an Algorithm which is a bit lengthy for this type of problem.

答:
There are only three irreducible polynomials of degree $2$ in $\mathbb{Z}_3[x]$:
$$
x^2 + 1,
\quad x^2 + x + 2,
\quad x^2 + 2 x + 2
$$
Try to divide your polynomial by each of these.

ContentLicense: CC BY-SA 4.0

Title: Irreducibility over $\mathbb{Z}_3$


问:
E[1/(Y+1)] of negative binomial distribution
There was an exercise in class asking to solve for the expectation E[1/(Y+1)] when Y is a negative binomial random variable  (number of failures).

How do I solve this? Whatever transformation or substitution I try I just can't seem to derive a constant number that looks like an expectation.

I would appreciate an explanation. Thanks!

答:
$$\mathbb{E}\Bigg[\frac{1}{Y+1}\Bigg]=\sum_{y=0}^{\infty}\frac{1}{y+1}\frac{(r+y-1)!}{y!(r-1)!}\theta^r(1-\theta)^y=$$

$$=\frac{\theta}{(1-\theta)(r-1)}\underbrace{\sum_{(y+1)=0}^{\infty}\frac{[(r-1)+(y+1)-1]!}{(y+1)!(r-2)!}\theta^{r-1}(1-\theta)^{y+1}}_{=1}=\frac{\theta}{(1-\theta)(r-1)}$$

The expression over the brace is 1 because it is the sum of

$$P(Z=r-1)=\binom{(r-1)+z-1}{z}\theta^{r-1}(1-\theta)^{z}$$

over all Z. (it is still a negative binomial counting the number of failures before the $(r-1)$th success


问:
Finding the volume between a cylinder and a hyperboloid
I'm having trouble coming up with the limits needed to do the integration. The equation of the cylinder is $y^2 + z^2 = 1$,
and the equation for the hyperboloid is $z = \sqrt{a^2 + x^2 + y^2}$ with $  0<a<1$.
So far I have z is between a and 1, $\theta$ is between $0$ and $2\pi$, but I can't work out what $r$ is meant to be between.

答:
I think it helps to find the intersection, so we know that $z>0$:
$$z^2=a^2+x^2+y^2$$
$$1-y^2=a^2+x^2+y^2$$
$$1-a^2=2x^2+y^2$$
and the solutions to this equation form an elipse of the form:
$$(\sqrt{2}x)^2+y^2=(1-a^2)$$
so this gives us the upper limits for $x$ and $y$ and $z$ will vary from $a$ to $1$


问:
Solving recurrence relation, $a_n=6a_{n-1} - 5a_{n-2} + 1$
I'm trying to solve this recurrence relation:

$$
a_n = \begin{cases}
    0 & \mbox{for } n = 0 \\
    5 & \mbox{for } n = 1 \\
    6a_{n-1} - 5a_{n-2} + 1 & \mbox{for } n > 1
    \end{cases}
$$

I calculated generator function as:
$$
A = \frac{31x - 24x^2}{1 - 6x + 5x^2} + \frac{x^3}{(1-x)(1-6x+5x^2)} =
\frac{31x - 24x^2}{(x-1)(x-5)} + \frac{x^3}{(1-x)(x-1)(x-5)}
$$

(I'm not sure if that's right)

and its partial fractions decomposition looks like:
$$
A = \left(\frac{-7}{4} \cdot \frac{1}{x-1} - \frac{445}{4} \cdot \frac{1}{x-5}\right) +
\left( \frac{39}{16} \cdot \frac{1}{x-5} + \frac{3}{4} \cdot \frac{1}{(x-1)^2} - \frac{375}{16} \cdot \frac{1}{x-5} \right)
$$
(again - I'm not sure if it's ok)

I'm stuck here... From solutions I know that I should get:
$$
a_n = \frac{-21}{16} - \frac{1}{4}n + \frac{21}{16}5^n
$$

but I have no idea how it's solved... I hope somebody can help me (I spend more than 3h trying to solve this myself...)

答:
Use Wilf's technique from "generatingfunctionology". Define $A(z) = \sum_{n \ge 0} a_n z^n$, and write:
$$
a_{n + 2} = 6 a_{n + 1} - 5 a_n + 1 \qquad a_0 = 0, a_1 = 5
$$
Multiply by $z^n$ and add for $n \ge 0$, which gives:
$$
\frac{A(z) - a_0 - a_1 z}{z^2} = 6 \frac{A(z) - a_0}{z} - 5 A(z) + \frac{1}{1 - z}
$$
Solving for $A(z)$ gives:
$$
A(z) = \frac{5 z - 4 z^2}{1 - 7 z + 11 z^2 - 5 z^3}
     = \frac{21}{16} \cdot \frac{1}{1 - 5 z}
         - \frac{17}{16} \cdot \frac{1}{1 - z}
         - \frac{1}{4} \cdot \frac{1}{(1 - z)^2}
$$
Remember the expansions:
$$
(1 - u)^{-m} = \sum_{k \ge 0} \binom{-m}{k} (-u)^k
             = \sum_{k \ge 0} \binom{k + m - 1}{m - 1} u^k
$$
(the binomial coefficient is just an $m-1$-degree polynomial in $k$) and you are all set:
$$
a_n = \frac{21}{16} \cdot 5^n - \frac{17}{16} - \frac{1}{4} \binom{n + 1}{1}
    = \frac{21 \cdot 5^n - 4 n - 21}{16}
$$

Maxima's help with the algebra is gratefully acknowledged.

ContentLicense: CC BY-SA 3.0


问:
Absolute values in limits
If $f(x)$ is any function, is the limit of $|f(x)|$ at any point equal to the absolute value of the limit of $f(x)$? I can't think of any examples, but I'd like to know if there's a proof for it before using it.

答: Yes, assuming that $\lim_{x\to a}f(x)$ exists. Then, if $\lim_{x\to a}f(x)=b$, we have $\lim_{x\to a}|f(x)|=|b|$, since$$\bigl||f(x)|-|b|\bigr|\leqslant|f(x)-b|$$for every $x$.

Title: Absolute values in limits


问:
1 normal distribution problem
It is known that the diameter of a fig-tree is normally distributed with $\mu=50$ cm, $\sigma=12$ cm.

$(1)$ What is the prob, that a randomly selected fig-tree has diameter $>73cm$?

$(2)$ what proportion of fig-trees have a diameter between 40 and 55 cm?

$(3)$ Find the diameter $d$, such that the $30\%$ of the fig trees exceeds the diameter $d$.

Here is what I tried:

So $1$. $\mathbb P(X>73)= \frac{1}{12\sqrt{2\pi}}\int_{73}^{\infty} e^-{\frac{(x-50)^2}{2\times 144}}$ , right? But I am not able to perform next few steps of this integration to get the final answer, could someone direct me some easy and quick way if exists?

$2.$ $\mathbb P(40<X<55)= \frac{1}{12\sqrt{2\pi}}\int_{40}^{55} e^-{\frac{(x-50)^2}{2\times 144}}$ , right? But I am not able to perform next few steps of this integration to get the final answer, could someone direct me some easy and quick way if exists?

$3.$ I absolutely do not understand the question so far, Thanks for any help.

Thanks.

答: The integral $\int e^{-kx^2}$ can only be calculated numerically. Look up normal tables (or statistics library of your favourite programming language)!

Title: 1 normal distribution problem


问:
How can I find the non-integer smoothness of a continuous and almost everywhere differentiable function?
Take a function such as f($\lambda$) = min($\lambda$, c) for some $\lambda$ in $[0, 1]$ and some constant c in $(0, 1)$.  This function is continuous and almost everywhere differentiable (it is not differentiable at c).

The smoothness of a function ($\alpha$) is a non-integer that depends on a derivative's modulus of continuity (or whether that modulus is of a certain order), which I find hard to determine for an arbitrary function, unlike the question of how many derivatives it has.  If we restrict smoothness classes to integers, the function f above is obviously C0 and not C1.  But in terms of non-integer smoothness, is the function any C$\alpha$ in between?  In particular, if a function is continuous everywhere but not differentiable everywhere, is the modulus of continuity found in terms of the function itself to find what smoothness class it has? (And how can a function's non-integer smoothness class in general be found, besides counting its derivatives?)

Motivation:

This question is important to me because a method given in Holtz et al. (2011) relies on this non-integer notion of smoothness to describe a method to calculate polynomials that converge from above and below to a function, and I would like to apply this method.

REFERENCES:

Holtz, O., Nazarov, F., Peres, Y., "New Coins from Old, Smoothly", Constructive Approximation 33 (2011).

答:
In your specific example, the function is actually Lipschitz continuous. Such functions $f$ are $C^{\alpha}$ for each $0 < \alpha < 1$, and $\|f\|_{C^{\alpha}} \leq K_L$, where $K_L$ is any Lipschitz constant for $f$, for any $0 < \alpha < 1$.

Another example of Lipschitz continuous functions include the restrictions of concave or convex functions to $I=[0,1]$, when they are defined on an open interval containing $I$.

To determine in general a smoothness class, by differentiating the function, we can reduce to the case of a continuous non-$C^1$ function. Then, if the new function is Lipschitz continuous (e.g. differentiable almost everywhere with bounded derivative), the original function is $C^{d + \alpha}$ for each $\alpha < 1$, where $d$ is the number of continuous derivatives of the function.

Otherwise, you need to find a Hölder continuity index. This is harder, but if (for instance) there are finitely many non-$C^1$ points, you can consider the minimum of the Hölder indices in neighborhoods of such points. But it's not exactly the best way to bound the continuity modulus.


问:
Constructing explicit lift of a circle homeomorphism
Studying a book by Luis Barreira in the Universitext Collection -- Dynamical Systems: an Introduction -- I'm told that given $f: S^{1} \to S^{1}$ homeomorphism, it's always possible to construct a lift $F(x) = f(x - \lfloor x \rfloor ) + \lfloor x \rfloor$, where $f(x - \lfloor x \rfloor)$ is taken to be the representative in $[0,1[$. It's easy for me to check that this is indeed a lift, but I can't ensure myself about continuity. It's obviously continuous in $\mathbb{R} \setminus \mathbb{Z}$, so it's left to check that left and right limits coincide in $\mathbb{Z}$. First, $F(k) = f(0) + k$, and $F(k^{+}) = f(0) + k$ as well. Now the author claims that $F(k^{-}) = f(1) + k$, and since $f(1) = f(0)$ we'd be done. I can't, however, verify this assertion: I always conclude that $F(k^{-}) = f(1) + k -1$. Where does the $-1$ goes in his calculations?</p>

答:
I think the function $F(x) = f(x - \lfloor x \rfloor ) + \lfloor x \rfloor$ is not a continuous lift.

Let $S_1$ be $[0, 1]$ with $0, 1$ identified and let $f$ be rotation by $\frac 14$, $f(x) = x + \frac 14 \pmod 1.$  Let $F(x)$ be defined as above. Then

$$\lim_{x \to \frac 34^-}F(x) = 1$$
while
$$\lim_{x \to \frac 34^+}F(x) = 0.$$


问:
Prove that $(x+y)^{p^m}=x^{p^m}+y^{p^m}$
This is an exercise from the book Introduction to algebra by А. И. КОСТРИКИН.

Assume $p$ is a prime, $R$ is a communicative ring with identity element, such that $\forall x\in R, px=0$. Prove that
$$
(x+y)^{p^m}=x^{p^m}+y^{p^m}, \quad m=1,2,\ldots
$$

My attempt was to make induction on $m$, but I failed to find the right way to proceed.

答: Induction helps because if $(x+y)^{p^k}=x^{p^k}+y^{p^k}$ then$$(x+y)^{p^{k+1}}=(x^{p^k}+y^{p^k})^p=(x^{p^k})^p+(y^{p^k})^p,$$provided we've proven the base step $m=1$. @J.W.Tanner hinted checking the base step with$$(x+y)^p-x^p-y^p=\sum_{j=1}^{p-1}\binom{p}{j}x^jy^{p-j}.$$Each binomial coefficient has a factor of $p$ in its numerator $p!$, but not its denominator $j!(p-j)!$, so is divisible by $p$.


答:
Well, if you have a chart of the normal distribution $N(0, 1)$ you should consider that you can use those values to solve your problem, cause in general computing this kind of integrals is complicated.

As you dont have that distribution, you will have to get it "normalized". That means you define $Y = \frac{X-\mu}{\sigma}$ so that $Y$ is just $N(0, 1)$, and then you use that $$P(X > 73) = P\biggr(\frac{X-50}{12}>\frac{73-50}{12}\biggr) = P\biggr(Y>0.12166...\biggr) = 1 - P\biggr(Y \leq 0.12166...\biggr)$$
and as i said, those probabilities are all around the internet and books. For doing the second one you will just apply this. I barely understand the 3) too.

The trick is that knowing probabilities on $N(0, 1)$ we are able to just "translate" our distribution and get normalized for all values of $\mu$ and $\sigma$.

Title: 1 normal distribution problem

Body:
It is known that the diameter of a fig-tree is normally distributed with $\mu=50$ cm, $\sigma=12$ cm.

$(1)$ What is the prob, that a randomly selected fig-tree has diameter $>73cm$?

$(2)$ what proportion of fig-trees have a diameter between 40 and 55 cm?

$(3)$ Find the diameter $d$, such that the $30\%$ of the fig trees exceeds the diameter $d$.

Here is what I tried:

So $1$. $\mathbb P(X>73)= \frac{1}{12\sqrt{2\pi}}\int_{73}^{\infty} e^-{\frac{(x-50)^2}{2\times 144}}$ , right? But I am not able to perform next few steps of this integration to get the final answer, could someone direct me some easy and quick way if exists?

$2.$ $\mathbb P(40<X<55)= \frac{1}{12\sqrt{2\pi}}\int_{40}^{55} e^-{\frac{(x-50)^2}{2\times 144}}$ , right? But I am not able to perform next few steps of this integration to get the final answer, could someone direct me some easy and quick way if exists?

$3.$ I absolutely do not understand the question so far, Thanks for any help.

Thanks.


问:
Prove that $|a|\leq \max\{|b|,|c|\}$ if $b\leq a \leq c$

Prove that $|a| \leq \max\{|b|,|c|\}$ if $b\leq a \leq c$.

I showed that $a\leq c$ and therefore $-c\leq a \leq c$ so that $|a|\leq c$ but then I got stuck.

Is this the right approach?

答: Alternative proof: Consider the function $y=x^2$ where $x\in [b,c]$. By Fermat's theorem (or just property of parabolas) there is no local maximum. Therefore $$a^2 \leqslant \max(b^2,c^2) \implies |a| \leqslant \max(|b|, |c|).$$


问:
Limit of $a_{n+1} = a_{n}(1+1/n^2)$
I read the following exercise :

Let $(a_n)$ be a sequence such as :

$a_1 = 2$, $a_{n+1} = a_n.(1+1/n^2)$

Prove that $\lim_{n \rightarrow \infty} a_n$ exists.

Hint : $\ln : \: (0, \infty) \rightarrow \mathbb{R} \:$ is a continuous, increasing function such as $\ln(1+x) \leq x$ and $\ln(ab) = \ln(a)+\ln(b)$.

It is quite clear that $a_n$ is increasing but then I don't really know how to proceed. Could you please help me ?

Thanks.

答:
hint

By induction, it is easy to see that
$$(\forall n\in \Bbb N) \;\; a_n>0$$
put $$b_n=\ln(a_n)$$

then

$$b_{n+1}=b_n+\ln(1+\frac{1}{n^2})$$
$$\le b_n +\frac{1}{n^2}$$

Title: Limit of $a_{n+1} = a_{n}(1+1/n^2)$


问:
Adjoint test for complex-to-real operator
Given a linear operator $A: \mathbb{C} \rightarrow \mathbb{R}$, and scalars $x \in \mathbb{C}$, $y \in \mathbb{R}$, I think we should have
\begin{equation}
\langle Ax,y\rangle = \langle x,A^*y\rangle,
\end{equation}
where $A^*: \mathbb{R} \rightarrow \mathbb{C}$ is the adjoint of $A$.

However, as a concrete example, take $A$ to be the operator that returns the real part of its argument. In this case, I think an adjoint would be $A^*y = y + 0i$. Inserting into the equation above, we get
\begin{equation}
\langle \Re(x), y\rangle = \langle x, y + 0i\rangle,
\end{equation}
or: $\Re(x)y$ = $xy$, which is not true. Where is my mistake?

答: My mistake (thanks to s.harp for pointing it out in a comment) is that complex-to-real operators, such as the example $Ax = \Re{x}$ that I gave, are $\mathbb{R}$-linear, not $\mathbb{C}$-linear. They satisfy the condition to be linear operators, $A(ax+by) = aA(x) + bA(y)$, only when $a$ and $b$ are real. This apparently implies that we need to perform the inner product differently, so $\langle x, y + 0i \rangle = \Re{x}\cdot y + \Im{x}\cdot0 = \Re{x}\cdot y$, satisfying the desired equality in my question.


答:
As $\ln x$ is increasing, $b_n = \ln a_n$ is an increasing sequence.

Also
$$b_{n+1}= \sum_{k=1}^n \ln(1+1/k^2) \le \sum_{k=1}^n 1/k^2$$

according to the provided hint.

As the series on the RHS is convergent, the sequence $\{b_n\}$ is bounded. Being also increasing, it is convergent. Therefore so is $\{a_n\}$.

Title: Limit of $a_{n+1} = a_{n}(1+1/n^2)$

Body:
I read the following exercise :

Let $(a_n)$ be a sequence such as :

$a_1 = 2$, $a_{n+1} = a_n.(1+1/n^2)$

Prove that $\lim_{n \rightarrow \infty} a_n$ exists.

Hint : $\ln : \: (0, \infty) \rightarrow \mathbb{R} \:$ is a continuous, increasing function such as $\ln(1+x) \leq x$ and $\ln(ab) = \ln(a)+\ln(b)$.

It is quite clear that $a_n$ is increasing but then I don't really know how to proceed. Could you please help me ?

Thanks.


问:
integral transforms: why do roots in frequency domain correspond to eigenvalues in time domain (and how does it help solve differential equations)?
In Wikipedia you can read about integral transforms, esp. the Laplace transform which maps a differential equation in the time domain into a polynomial equation in the complex frequency domain:

roots of the polynomial equations in the complex frequency domain correspond to eigenvalues in the time domain

Can anybody give a simple example and foremost give an intuition why this is so - and why does this help solving the differential equation? I don't see the big picture.

答:
(Linear, homogeneous, constant-coefficient) differential equations are already polynomial equations. That is, let $D$ denote the differential operator $\frac{d}{dx}$. Then linear homogeneous constant-coefficient differential equations are nothing more than equations of the form $p(D) y = 0$, where $y$ is some function of $x$ and $p$ is some polynomial. Writing $p(x) = \prod_{i=1}^n (x - r_i)^{e_i}$ where the roots $r_i$ are distinct, it follows that we can write the equation as

$$(D - r_1)^{e_1} (D - r_2)^{e_2} ... (D - r_n)^{e_n} y = 0.$$

The point here is that $D$ is a linear operator, and it generates a commutative algebra of differential operators. Now, from the above it follows that any solution to $(D - r_i)^{e_i} y = 0$ is a solution to the above, and it turns out that linear combinations of these give all solutions. (This is a general fact about linear operators on finite-dimensional vector spaces; see, for example, Axler's Linear Algebra Done Right.)

In the special case that $e_i = 1$ for all $i$ things are particularly simple: $(D - r_i) y = 0$ just means $Dy = r_i y$, or $y = c_i e^{r_i x}$ for some constant $c_i$. This gives $n$ linearly independent solutions, which is the correct number, hence the general solution has the form

$$y = \sum_{i=1}^n c_i e^{r_i x}.$$

If the multiplicities $e_i$ are higher then we must consider solutions to $(D - r_i)^{e_i} y = 0$. Let $w(x) = e^{-r_i x} y(x)$. Then $D w = e^{-r_i x} (D - r_i) y$, from which it follows that the above holds if and only if $D^{e_i} w = 0$. But this says precisely that $w$ is a polynomial of degree at most $e_i - 1$. Hence

$$y = c_i(x) e^{r_i x}$$

where $c_i(x)$ is a polynomial of degree at most $e_i - 1$. This gives $e_i$ linearly independent solutions, so again taking linear combinations we have found all of them.

None of this requires that you know anything about integral transforms. From the abstract point of view, integral transforms are just a convenient way to diagonalize $D$. The Laplace transform (more or less) takes $D$ to multiplication by $s$, hence turns $p(D)$ into $p(s)$ (the roots of which are the eigenvalues $r_i$ above), but this is just a more concrete way of realizing the abstract structure already evident in the problem: that the set of solutions to $p(D) y = 0$ is naturally acted on by $\mathbb{C}[D]$.


问:
how many empty sets are there?
Would I be correct in saying that in the category of sets, the "class of sets that are isomorphic to the empty set is a proper class"?

In other words, there are LOTS of initial objects in the category of sets, but they're all related to each other via unique isomorphisms?

Similarly, for any category $C$, if $X$ is an object of $C$, then the full subcategory of objects isomorphic to $X$ is a proper class (ie, there are LOTS of them)?

EDIT: Let me elaborate a bit. I'm not entirely sure if I should be considering ZFC or ETCS, etc. I'm asking this question from the point of view of algebraic geometry and its use of category theory. I mean, I know people say that the class of say, all elliptic curves over $\mathbb{C}$ is a proper class, but the class of $\mathbb{C}$-isomorphism classes of elliptic curves is a set. That would seem to imply that every isomorphism class is a proper class, and hence this would seem to say that the isomorphism class (ie, bijection class) of the empty set in the category of sets should also be a proper class, right?

Lana's answer with the extensionality axiom seems to make sense, in that there is only ONE empty set, so that the isomorphism class of the empty set is a singleton set. On the other hand, the isomorphism class of a set with 1 element should be a proper class right?

Generalizing a bit, would it be true that the isomorphism class of any initial (or terminal) object of a category be a singleton set? It seems to me that the isomorphism class of, say $\text{Spec }\mathbb{Z}$ in the category of schemes is a proper class, simply because there are many different presentations of rings isomorphic to $\mathbb{Z}$?

will

答:
Your conclusion is not entirely correct. While it's perfectly fine to have a context where a category has many initial objects which are isomorphic (but not equal), there is a delicate point to the inaccurate statement:

the class of say, all elliptic curves over $\mathbb{C}$ is a proper class, but the class of $\mathbb{C}$-isomorphism classes of elliptic curves is a set. That would seem to imply that every isomorphism class is a proper class

The fact that there is a proper class of different structure is quite trivial to show, because we can always change one element with another and get a proper class of different, but isomorphic, structures that way.

To say that there is only set-many isomorphism classes is also inaccurate, because the isomorphism classes are proper classes, there is no collection of the form $\{A\mid A\text{ is an isomorphism class of ...}\}$. Only sets can be members of other sets. But it does mean something else, it means that there is a set of pairwise non-isomorphic structures that any other structure is isomorphic to one of them. That is, there is a class of representatives which is a set.

Lastly, it does not mean that every equivalence class is a proper class, just that at least one of the equivalence classes are.

The simplest example of this is equinumerousity (in $\sf ZFC$). There is a proper class of sets of every cardinality, except for the class of sets of cardinality zero. There is only one of those.

If your foundational theory is $\sf ZFC$ or some related theory then there exists only one empty set. If your foundational theory is some category based theory which allows many empty sets, but requires them to be isomorphic, then this is a different case altogether.

Title: how many empty sets are there?


答:
By using $\ln (1+x)\le x$ for $x\ge 0$, we have
$$
1+{1\over n^2}\le e^{1\over n^2}
$$
hence $a_n$ is bounded above by another sequence $b_n$ where
$$
b_{n+1}=b_ne^{1\over n^2}\quad,\quad b_1=4
$$
Since $b_n\to b_1e^{\sum_{i=1}^\infty {1\over i^2}}=4e^{\pi^2\over 6}$ then $a_n$ also tends to a limit because it is strictly increasing and bounded from above.

Title: Limit of $a_{n+1} = a_{n}(1+1/n^2)$


问:
Why does the interchanging of the limit and the integral not work here? Is it due to not having uniform convergence?
I have that $ f(x,t) = \frac{2x^2t}{(x^2 + t^2)^2}, t \in [0,1] $ and I am asked to show that
$$ \lim_{x \downarrow 0} \int_0^1 f(x,t) \,dt \neq \int_0^1 \lim_{x \downarrow 0} f(x,t) \,dt $$

I have that the RHS is 1 and the LHS is 0.

I am struggling to answer is why this doesn't work in terms of interchanging the limit and the integral. I think it may be to do with the fact $f(x,t)$ doesn't converge uniformly to $f(x)$ but I don't know how to show this or if it is even the correct reason why the interchanged limit and integral aren't equal.

How would I prove $f_t(x)$ doesn't converge to $f(x)$ uniformly (that is assuming I have correctly identified why it doesn't work)?

答:
Set $x=1/n$. Then you have $\displaystyle\int_0^1\frac{2(1/n)^2t}{(
(1/n)^2 + t^2)^2}dt$ and the fact that the two integrals are not equal proves that the convergence of the integrand sequence is not uniform.

To do it from scratch, note that $f_n(t)=\frac{2(1/n)^2t}{(
(1/n)^2 + t^2)^2}\to 0$ pointwise but with $t_n=1/n,$ we get $\frac{2/n^3}{2/n^4}=n\to \infty$ as $n\to \infty$, so the convergence is not uniform.


问:
Functions that are their own Fourier transform
In Stein's Fourier Analysis, there's the following exercise:

The function $e^{-\pi x^2}$ is its own Fourier transform.   Generate other functions [presumably in the Schwartz space $S(\mathbb{R})$] that, up to a constant multiple, are their own FTs.  What must the constant multiples be?  To decide this, prove that $F^4 = I$ where $F$ is the FT operator.

My problem is that I don't really understand what the question is asking.  Is it asking us to find the class of all functions such that $F(f) = cf$ for constant c and prove the above identity, or is it asking us to just find other examples?  Or, is the $F^4 = I$ identity for any $f \in S(\mathbb{R})$?

Finding every function would mean finding all $f$ such that
$$f(\xi) = \int^{\infty}_{-\infty} f(x)e^{-2\pi i x \xi}dx$$ which seems rather difficult.

I can come up with a specific example: if I can find a function such that $\hat{f}(\xi) = 1$, I can just take $g = f \ast K_{\delta}(x)$ where $K_{\delta}$ is the family of Gaussian functions as approximations of the identity.

答:
Let $f(x) = e^{-\pi x^2}$. Note that multiplication is the differentiation in the transformed space, $$\mathcal{F}(2\pi ixf(x)) = \frac{d}{d\xi}F(f(\xi)) = -2\pi\xi e^{\pi\xi^2}.$$

This can be generalized to give a family of polynomials $\{G_n\}_{n\in\mathbb{N}}$ such that $\mathcal{F}(G_n\cdot f) = i^n (G_n\cdot f)$.

My guess is this was the intention as suggested by "generate" in the prompt.


问:
complete intersection in $\mathbb{P}^{n+2}$ as K3 surface
I am reading D. Huybrechts' lecture "Lectures on K3 surfaces", http://www.math.uni-bonn.de/people/huybrech/K3Global.pdf
He claims that a smooth complete intersection of type $(d_1,\cdots, d_n)$ in $\mathbb P^{n+2}$ is a K3 surface if and only if $\sum {d_i} = n + 3$.

What confuses me is that he then says under the natural assumption that all $d_{i}>1$ there are in fact only three cases (up to permutation):

(a) $n=1, d_1=4$.

(b)$n=2, d_1=2, d_2=3$.

(c)$n=3, d_1=d_2=d_3=2.$

I wonder why we asssme $d_i>1$? What will happen if some $d_i=1$?

Any help would be appreciated. Thanks a lot!

答: If, say, $d_1 = 1$ this complete intersection is equal to a complete intersection in $\mathbb{P}^{n+1}$ of type $(d_2,\dots,d_n)$, so there is no sense in considering it separately.


问:
Is this continuous function injective?
The following question was part of my complex analysis quiz and I was unable to mark 1 option . I am not able to deduce that is it correct on not.

Let  X= {$(a,b)\in \mathbb{R}^2 : a^2+b^2 =1$} and f : X-> $\mathbb{R}
 $ be a continuous function. Then is f injective or not?

I have given ample time to this option but I am unable to find any contradiction or a function satisfying it. So, with no option left I am posting it here.

Please help!

答:
As $X$ is compact and connected, if $f$ is supposed to be continuous and injective, it would be an homeomorphism from $X$ to a compact interval $I \subseteq \mathbb R$ of positive length.

Take $a$ in the interior of $I$. $X \setminus \{f^{-1}(a)\}$ is connected. As the image of a connected set under an homeomorphism is connected, $I \setminus \{a\}$ would be connected. A contradiction.

$f$ can't be injective.

Title: Is this continuous function injective?


答:
Prove by induction that $\ln a_n \le \sum_{k=1}^{n-1}\frac1{k^2}$ for all $n \ge 2$.

For $n=2$ we have
$$\ln a_2 = \ln \frac52 < 1$$
and if we assume that $\ln a_n \le \sum_{k=1}^{n-1}\frac1{k^2}$ for some $n$ then
$$\ln a_{n+1} = \ln a_n + \ln\left(1+\frac1{n^2}\right) \le \sum_{k=1}^{n-1}\frac1{k^2} + \frac1{n^2} = \sum_{k=1}^n \frac1{k^2}$$
which completes the proof.

Hence $$\ln a_{n} \le \sum_{k=1}^\infty \frac1{k^2} = \frac{\pi^2}6 \implies a_n \le e^{\frac{\pi^2}6}$$
so your sequence is bounded.

Title: Limit of $a_{n+1} = a_{n}(1+1/n^2)$


问:
Cardinality of $GL_n(K)$ when $K$ is finite
I don't know how to do the last task of an exercise.

Let $K$ be a field, $G=GL_n(K)$ and $X=K^n\backslash\{0\}$.

First task: Show that $G \times X \to X$, $(A,x)\mapsto Ax$ defines an action of $G$ on $X$. Done.

Second task: find the stabilizer of $\displaystyle x=\left(\begin{eqnarray} 1\\ 0 \\ \vdots \\0 \end{eqnarray}\right)$. Done.

Third task: Show that there is only one orbit. Done.

Last task: Let $|K|<\infty$. Use the last two tasks and induction to find $|G|$.

I know that $|G|=[G:St_G(x)]\cdot|St_G(x)|$ and i know that $[G:St_G(x)]=|\bar{x}|=m^n-1$ but i can't figure out what $|St_G(x)|$ is.

What should i do next?
I appreciate every hint :)

答:
It looks like you should be done. You said you found the stabilizer of $x$. What is it? What is its cardinality?

Here's an alternative approach to figuring out $|G|$.

Clearly, $|GL_1(K)|=|K|-1$.

To get a $2\times 2$ invertible matrix with entries in $K$, we require that the first column be a non-zero vector (there are $|K|^2-1$ such), and once we've chosen a particular vector for the first column, we must choose the second column so that it is not a $K$-scalar multiple of the first column (there are $|K|^2-|K|$ such vectors). Hence, $$|GL_2(K)|=\bigl(|K|^2-1\bigr)\bigl(|K|^2-|K|\bigr).$$ Check out a few more cases, and you'll notice the pattern that $$|GL_n(K)|=\prod_{i=0}^{n-1}\bigl(|K|^n-|K|^i\bigr).$$


问:
can I finish this solution by this method, if I can please tell me how
$m^2+m=n^3+n^2+n$, find solutions in natural numbers. so this was the problem and answer is that it doesn't have solutions in natural numbers.

It was pretty hard to think about this problem, I wanted to analyze the ends
of the numbers, for example, if $m$ ends on 2, then $m^2$ ends on 4, so their sum ends on 6.

First of all, it is easy to prove that n is even. since the left side is even, then right side must be the even as well, so $n$ must be even.  So I analyzed when n is even, at what number sum n^3+n^2+n ends on, for example if $n$ ends on 2 then $n^2$ ends on 4.

Then I analyzed when $n$ ends on 4,6,8. sum ends on 4 or 8. Then I analyzed ending of $m$ and $m^2$ then I got that $m^2+m$ doesn't end on 4 or on 8, it ends on 0 in 3 cases, when m ends on 4,5,9  $m+m^2$ ends on 0. In other cases $m$ doesn't end on 4 or on 8 so I supposed that $m+m^2$ doesn't equal $n+n^2+n^3$, for natural numbers, but unfortunately and I am very sad right now, I forgot to analyze when $n$ ends on 0. when $n$ ends on 0, it sum $n+n^2+n^3$ ends on 0 and unfortunately, I have to analyze 3 cases for $m$, when sum $m+m^2$ ends on 0.

So I really need your opinion about this can I finish this problem after analyzing ending 0 situations, please answer me it will be very helpful for me, I really want to defend this solution.

答:
I am not sure whether your solution can be defended or not, since we could have both $m$ and $n$ to be divisible by $10$, and if you try to remove common factors, the argument won't hold anymore.

One way to go about solving this problem might be to try to find a nice factorization instead. We have:
$$m^2+m-n^2-n=n^3 \implies (m-n)(m+n+1)=n^3$$
Now, writing $m-n=a$ gives $a(a+2n+1)=n^3$. If there exists a common prime factor for $a$ and $a+2n+1$, it would divide $2n+1$, and thus cannot divide $n$, which would be a contradiction. Thus, we must have $a$ and $a+2n+1$ to be relatively prime, and since their product is a cube, they are both perfect cubes. Writing $a=x^3$ gives:
$$x^3(x^2+2n+1)=n^3$$
Since $x^3 \mid n^3$, we can write $n=xk$. This gives:
$$x^3+2xk+1=k^3$$
Again, writing $k=x+y$ gives:
$$x^3+2x^2+2xy+1=x^3+3x^2y+3xy^2+y^3$$
I believe you can continue from here...


问:
Question on positive square root of an operator
Let $T$ is a bounded linear operator on a Hilbert Space $H$. Let there exists a unique positive square root of $T^{*}T$. Let us call it as $P$

Then how can I show that the dimension of the range of $T$ is the same as that of the range of $P$?

For this, I am trying to show that their null spaces are equal.

I started with $x \in KerT$ $\rightarrow$ $Tx=0$ $\rightarrow$ $\langle Tx,Tx\rangle=0$ $\rightarrow$ $\langle T^{*}Tx,x\rangle=0$

From here, I can't imply that $Px=0$.

Help, please

答:
Here is an alternative method of proof.  By the polar decomposition
one  may write
$
  T=UP,
  $
where $U$ is a partial isometry which also satisfies,
$
  U^*T=P.
  $

This implies that $U$ is an isometric isomorphism from  $\text{Ran}(P)$ to $\text{Ran}(T)$  and, in particular,
$$\text{dim}(\text{Ran}(P)) = \text{dim}(\text{Ran}(T)).$$

PS:  Thanks to user "lonza leggiera"
for pointing out my initial misunderstanding of the method based on the fact that $\text{Ker}(T)=\text{Ker}(P)$.   This
implies that
$$
  \text{Ran}(T)\simeq
  H/\text{Ker}(T) =
  H/\text{Ker}(P) \simeq
  \text{Ran}(P),
  $$
thus solving the problem.

However this method only proves that the ranges are algebraically isomorphic and cannot decide, for instance, whether or
not  one
range  is closed in case the other one is known to be closed.


问:
Iterated integral, vertically/horizontally simple
Topic: Iterated integrals:
Let $D \subset \mathbb{R}^2$ bounded by $x+y^2=0, y-x=6$ and let $f: D \rightarrow \mathbb{R}$ be continuous

(1) is D vertically and/or horizontally simple?

(2) Calculate the area of D.
So what I did was the following,
for (1) I have calculated the intersection points by
$$y^2=-x, y=x+6$$
$(x+6)²=-x,  \hspace{0.5cm}  x²+12x+36=-x, \hspace{1cm} x_1 = -4, x_2=-9$

then I have the region $D=\{(x,y) \in \mathbb{R²}: -9 \leq x \leq -4, \hspace{0.5cm}6+x \leq y \leq \sqrt{-x} \}$

Are the boundaries for $y$ correct? So $6+x$ is continuous, and  $\sqrt{-x}$ is continuous on [-9,-4]. So would it make D vertically simple?

and on the other hand, I have

$y²=-x, y-6=x, \hspace{0.5cm} y²+y-6=0, y_{1}=2, y_{2}=-3$

so
$$D=\{(x,y) \in \mathbb{R²}: -3 \leq y \leq 2, \hspace{0.5cm}-y^2 \leq x \leq y-6 \}$$
Are the boundaries for x correct? would from this follow that it is also horizontally simple?

now how can I calculate the area of D?
would it be $\int \int_{D} f(x,y) d(x,y)$?

答:
Please see the diagram. If you integrate wrt $y$ first, you will need to do it twice as for $-4 \leq x \leq 0, $ the area bound is parabola whereas for $-9 \leq x \leq -4,$ the area is bound below by the parabola and above by the plane.

So instead set it up wrt $x$ first. For the second one, your boundary is correct but what you have shown as lower bound of $x$ should be upper bound otherwise you will get negative value. It should be clear from the diagram.

$\displaystyle \int_{-3}^{2} \int_{y-6}^{-y^2} dx \, dy$


答:
A straightforward way to prove it using induction is to assume that it holds for all $m$ such that $1\le m\le n$ and for some given $n$. To show that it holds for $n+1$, we should prove
$$
(x+y)^{p^{n+1}}=x^{^{p^{n+1}}}+y^{^{p^{n+1}}}
$$
proof

Note that
$$
(x+y)^{p^{n+1}}{=
\left[(x+y)^{p^{n}}\right]^p\\
=\left[x^{p^n}+y^{p^n}\right]^p\\
=(x^{p^n})^p+(y^{p^n})^p\\
=x^{p^{n+1}}+y^{p^{n+1}}
}
$$
Done!

Title: Prove that $(x+y)^{p^m}=x^{p^m}+y^{p^m}$

Body:
This is an exercise from the book Introduction to algebra by А. И. КОСТРИКИН.

Assume $p$ is a prime, $R$ is a communicative ring with identity element, such that $\forall x\in R, px=0$. Prove that
$$
(x+y)^{p^m}=x^{p^m}+y^{p^m}, \quad m=1,2,\ldots
$$

My attempt was to make induction on $m$, but I failed to find the right way to proceed.


问:
Idempotents in a Quotient Ring
Let $R=\mathbb{Z}_p[x]/(x^p-x)$.  Show that $R$ has exactly $2^p$ elements satisfying $r^2=r$.

I know that for $f,g\in\mathbb{Z}_p[x]$, we have $f-g\in(x^p-x)$ if and only if $f(a)=g(a)$ for all $a\in\mathbb{Z}_p$.  It follows that $r^2=r\in R$ if and only if $r$ is represented by some $f\in\mathbb{Z}_p[x]$ such that $f(a)^2=f(a)$ for all $a\in\mathbb{Z}_p$.  Thus, it must follow that $f(a)=0$ or $f(a)=1$ for each $a$.

Now, for each of the $2^p$ $p$-tuples $(b_0,\ldots,b_{p-1})$ where each $b_i$ is either $0$ or $1$, does there exist a polynomial $f$ such that $f(a)=b_a$, and is this polynomial uniquely defined in $R$?

答:
As rohit mentions in the comments, we may use Lagrange interpolation to explicitly construct such polynomials satisfying $f(x)=b_x$ for all $x\in{\bf F}_p$ for any vector $(b_0,\cdots,b_{p-1})$ with entries $0$ or $1$.

Explicitly, the method tells us to consider the following:

$$f(x)=\sum_{\substack{\rm all~linear \\ {\rm factors}~(x-u)}}\,b_u\frac{\displaystyle\prod_{\substack{\rm all~linear~factors \\ {\rm except}~(x-u)}}(x-v)}{\displaystyle\prod_{v\ne u}(u-v)}=\sum_{u\in{\bf F}_p}b_u\frac{\displaystyle\frac{x^p-x}{x-u}~~~~~~}{\displaystyle\left.\frac{x^p-x}{x-u}\right|_{x=u}}.$$

To evaluate $f(t)$ for some $t\in{\bf F}_p$ note that $\frac{t^p-t}{t-u}=0$ for all $t\ne u$, so the only summand that has a contribution is with $u=t$, yielding $f(t)=b_t\frac{\rm blah}{\rm blah}=b_t$, exactly as desired. The denominator in the summands can be evaluated exactly symbolically (so $x=u$ can be plugged in directly) as

$$\frac{x^p-x}{x-u}=\frac{(x/u)^p-(x/u)}{(x/u)-1}=\frac{x}{u}\left[\left(\frac{x}{u}\right)^{p-2}+\cdots+\left(\frac{x}{u}\right)+1\right] \\ \implies \left.\frac{x^p-x}{x-u}\right|_{x=u}=p-1=-1.$$

We assume $x\ne0$ as that case is trivial: $(x^p-x)/x=x^{p-1}-1=-1$ at $x=0$.

Uniqueness follows from the fact that these polynomials have different images under a certain homomorphism. In particular, there is a map from ${\bf F}_p[x]$ to the space of set-theoretic maps on our space ${\bf F}_p\to{\bf F}_p$ with kernel generated by $\prod_{t\in{\bf F}_p} (x-t)=x^p-x$. These $f(x)$ polynomials (in the quotient ${\bf F}_p[x]/\ker$) correspond to functions, and since they are distinct functions they are different elements in the domain.

We may also be able to prove the $f(x)$ exist using linear algebra to create a linear system that the coefficients solve; whether or not this is more or less constructive is arguable. If

$$f(x)=a_0+a_1x+\cdots+a_{p-1}x^{p-1}+(x^p-x)$$

then we obtain

$$\begin{pmatrix} 1 & 0 & 0 & \cdots & 0 \\
1 & 1 & 1 & \cdots & 1 \\
1 & 2 & 2^2 & \cdots & 2^{p-1} \\
\vdots & \vdots & \vdots & \ddots & \ddots \\
1 & -1 & 1 & \cdots & (p-1)^{p-1} \end{pmatrix}\begin{pmatrix}a_0 \\ a_1 \\ a_2 \\ \vdots \\ a_{p-1}\end{pmatrix}=\begin{pmatrix}b_0 \\ b_1 \\ b_2 \\ \vdots \\ b_{p-1}\end{pmatrix}. $$

Thus, as long as we can show the Vandermonde determinant is nonzero mod $p$, linear algebra tells us the system has a solution and the $a_i$ may be solved for, so the $f(x)$ exists.

Both of these methods are completely overkill for the problem. Manny's answer is the smarter rather than harder route. But familiarity with linear algebra and interpolation can come in handy other places, so I thought I'd showcase them.

Title: Idempotents in a Quotient Ring


问:
Be $Im(z) \neq 0$. It follow: $Im(z^n) \neq 0, \forall n \in \mathbb{N}$. Why isn't this correct?
Since there is no number $n$ with $a^n = 0$ with $a \neq 0$  this should be correct, but according to my exercise quiz its not. I would be very grateful for a hint, because I haven't found a useful approach so far.

答: Consider $z=i$.  Then Im$(z)\ne0$ but Im$(z^2)=0$.


答:
Take $\ln$ about both sides, we have $\ln(a_{n+1})=\ln(a_n)+\ln(1+\frac{1}{n^2})\leq\ln(a_n)+\frac{1}{n^2}$.

Then $\ln(a_{n+1})-\ln(a_n)\leq\frac{1}{n^2}$.

adding them above $n=1,2,\cdots$
$$\underset{n\to\infty}{\lim\sup}\ln(a_n)-\ln(a_1)\leq\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$$
So,$\underset{n\to\infty}{\lim\sup}\ln(a_n)\leq \ln(a_1)+\frac{\pi^2}{6}$, and $\underset{n\to\infty}{\lim\sup} a_n\leq 2e^{\frac{\pi^2}{6}}$. Therefore, the sequence is bounded. Furthermore, the sequence has a limit point due to increasing.

Title: Limit of $a_{n+1} = a_{n}(1+1/n^2)$


问:
Four-letter words from the word MEDITERRANEAN such that the first letter is R and the fourth letter is E

If the four letter words (need not be meaningful) are to be formed using the letters from the word MEDITERRANEAN such that the first letter is R and the fourth letter is E, then the total number of all such words is :

My attempt:

$4$ different letters, $3E, 2R, 2A, 2N$. i.e. total $8$ varieties, out of $13$ letters.

Since the first and the fourth letter is fixed, so the intermediate two letters are to be filled from $11$ remaining letters. But we can't just do $^{11}C_2$ as all $11$ are not different. Also, $^8C_2$ won't make sense either as that won't include the option of $2N$ or $2A$ etc.

How to go about it?

答:
The two middle places are to be filled from

$$ \{ M,D,T,I,R \} \quad \{ 2\cdot E,2\cdot A,2\cdot N \}$$

We have following two cases :

Different letters : $(^5 P_2) + (^3P_2) + 5\cdot3\cdot2 = 56$

or as pointed by @Intelligenti pauca, simply $8 \cdot 7 = 56$

Same letters : $^3 C_1$

Total $= 59$


问:
How do I find the characteristic polynomial and eigenvalues?
For the following matrix, compute

its characteristic polynomial

its eigenvalues

$$A = \begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 2 & -5 & 4\end{bmatrix}$$

So I think I know to find the characteristic polynomial, I have to compute det(A-$\lambda$I) = 0.
Which gives...

$$A-\lambda I = \begin{bmatrix}-\lambda & 1 & 0 \\ 0 & -\lambda & 1 \\ 2 & -5 & 4-\lambda\end{bmatrix}$$

$$-\lambda\begin{vmatrix}-\lambda & 1 \\ -5 & 4-\lambda\end{vmatrix} - 1 \begin{vmatrix}0 & 1 \\2 & 4-\lambda\end{vmatrix} $$

After calculating the determinant I get, $-\lambda(-4\lambda + \lambda^{2}+5) +2 = 0 $

What do I do next? I'm completely stuck.

答:
Visual inspection tells us that matrix $\rm A$ is a companion matrix and that $1$ is an eigenvalue of $\rm A$. Hence, the characteristic polynomial of $\rm A$ is

$$q (s) := s^3 - 4 s^2 + 5 s - 2 = (s - 1) (s^2 - 3 s + 2) = \color{blue}{(s-1)^2 (s-2)}$$


问:
Proving the limit is 0
$f(x)=e^{-1/x^2}$ if $f(x)\neq0$

Prove that for every $m>0,f(x) / x^m →0$ as x → 0.

How should I prove this? I tried to use L'Hospital's Rule, but for $e^{-1/x^2}$, it'll become quite complicated to write its derivatives after m times. I tried to use taylor expansion of $e^x$ as well. But I'm stuck again on how to calculate this limit.

答:
It is not hard to prove by induction that $f$ is a $C^\infty$ function and each $f^{(n)}$ can be written as $P_n\left(\frac1x\right)e^{-1/x^2}$ for some polynomial function $P_n(x)$; furthermore, $f^{(n)}(0)=0$.

So, the Taylor polynomial $Q_m(x)$ of order $m$ of $f$ at $0$ is the null polynomial and, since$$\lim_{x\to0}\frac{f(x)-Q_m(x)}{x^m}=0$$(this is a general property of the Taylor polynomials), we have $\lim_{\to0}\frac{f(x)}{x^m}=0$.

You can also do it using that fact that$$\lim_{x\to0}\frac{e^{-1/x^2}}{x^m}=\lim_{x\to\pm\infty}\frac{e^{-x^2}}{\left(\frac1x\right)^m}=\lim_{x\to\pm\infty}\frac{x^m}{e^{x^2}}$$and it is easy to see that this limit is $0$.

Title: Proving the limit is 0


问:
Example of non-degenerate random variable with odd moments = 0
Think of one example of a random variable which is non-degenerate for which all the odd moments are identically zero. Give the probability mass function of the random variable and state a quantity it could represent.

At first I thought of rolling a die since it's non-degenerate, but I don't believe its odd moments are 0. Is anyone able to come up with an example and briefly explain it?

答:
1. Continuous Random Variable

A standard gaussian, $X\sim N(0;1)$ works.

$$\mathbb{E}[X^{2n+1}]=0$$

$\forall n \in \mathbb{N}$

The proof is quite easy expanding its MGF in Taylor series and derivating

It can represent the measurement error when measuring the length of the following stick

2. Discrete Random Variable

$Y$ is a random variable taking the values $Y=\pm1$ with probability $\mathbb{P}[Y=-1]=\mathbb{P}[Y=1]=\frac{1}{2}$

$$\mathbb{E}[Y^{2n+1}]=\frac{1}{2}[(-1)^{2n+1}+1^{2n+1}]=0$$

$\forall n \in \mathbb{N}$

$Y$ represents the following function

$$Y=2X-1$$

Where $X\sim B\Big(\frac{1}{2}\Big)$, a Bernoulli rv with parameter 0.5

It can represent the random gain when playing "toss a fair coin game"  winning $\$1$ if H and loosing $\$1$  if T


问:
Finding a solution to $u_{tt} - c^2 u_{xx} = 0$ with Neumann boundary conditions using the method of separation of variables?
I am a second-year undergraduate taking an introductory course on partial differential equations. I am trying to apply the method of separation of variables to find a solution to the wave equation
$$
u_{tt} - c^2 u_{xx} = 0, \qquad 0 < x < l, \quad t > 0,
$$
with initial conditions $u(x, 0) = 0$, $u_t(x, 0) = 1$, and Neumann boundary conditions $u_x(0, t) = 0 = u_x(l , t)$.

I start by assuming the solution is of the form $u(x, t) = X(x) T(t)$, and substitute it into the wave equation, and separate the variables to get two eigenvalue problems:
$$
X''(x) + \lambda X(x) = 0, \qquad T''(t) + c^2 \lambda T(t) = 0.
$$
The given boundary conditions imply that $X'(0) = 0 = X'(l)$, so when $\lambda < 0$ there is no non-trivial solution, so I assume $\lambda \geq 0$.

When $\lambda > 0$, we have $X(x) = A \cos(\sqrt{\lambda}x) + B\sin(\sqrt{\lambda}x)$, and the boundary conditions imply $B = 0$ and $\lambda = n^2\pi^2 / l$. So we get the eigenfunction $X_n(x) = A_n \cos(n \pi x / l)$ for the eigenvalue $\lambda_n = n^2 \pi^2 / l$. For this eigenvalue, we solve the other EVP to get $T_n(t) = A_n' \cos(n \pi c t / l) + B_n'(n \pi c t / l)$.

When $\lambda = 0$, we have $X(x) = A + Bx$ and the boundary conditions imply $B = 0$, so we get the eigenfunction $X_0(x) = A_0$ for the eigenvalue $\lambda_0 = 0$. For this eigenvalue, we solve the other EVP to get $T_0(t) = A_0' + B_0't$.

Hence, $u_n(x, t) = X_n(x) T_n(t)$ is a solution to the wave equation with the given boundary conditions for every $n \geq 0$, and so is $u(x, t) = \sum_{n = 0}^\infty u_n(x, t)$ (provided the initial data is "nice" enough, though I'm not fully sure what those conditions are).
Combining the constants together, we get
$$
u(x, t) = a_0 + b_0 t + \sum_{n = 1}^\infty \cos(n \pi x / l) \bigl( a_n \cos(n \pi c t / l) + b_n \sin(n \pi c t / l) \bigr).
$$
Now, applying the initial condition $u(x, 0) = 0$, we get $0 = a_0 + \sum_{n = 1}^\infty a_n \cos(n \pi x / l)$, which imply $a_n = 0$ for all $n \geq 0$. Next, applying the initial condition $u_t(x, 0) = 1$, we get $1 = b_0 + \sum_{n = 1}^\infty (b_n n \pi c / l) \cos(n \pi x / l)$. This is a Fourier cosine series, so the formulae for the coefficients are:
$$
b_0 = \frac{1}{l} \int_0^l 1\, dx = 1, \qquad \frac{b_n n \pi c}{l} = \frac{2}{l} \int_0^l 1 \cdot \cos(n \pi x / l)\, dx = 0 \implies b_n = 0\ \ \forall\ n \geq 1.
$$

Thus, we get the solution $u(x, t) = t$.

Can someone please verify for me whether I have solved this problem correctly? I have only followed the outline given in my lecture notes for using the method of separation of variables, but the final answer puzzles me.

答:
You applied the method correctly, and it produced a solution to the problem, as you noticed by just substituting $u=t$ and seeing that it obviously satisfies all the requirements. (a) Why your TA thought it might be complicated--I think you would have to ask him or her--possibly your TA made a mistake about that; we all do. (b) How to understand the solution--there are at least two ways. One way is to think of some application of the problem and see whether the solution fits your expectation of the physics. For example, the pde can describe vibration of a string but the boundary conditions say the the ends of the string at $x=0,\ell$ are not pinned down, This is a little bit nonphysical but since the string starts with uniform velocity $u_t(x,0) = 1$ upward, and no curvature since $u(x,0) = 0$ implies that $u_{xx}(x,0) = 0$. So without curvature there is no restoring force initially. Maybe that makes it plausible that there is nothing to restrain the string and so it keeps moving upward at speed 1. That argument is ok, but not great in my view.

Another way to understand the solution, more mathematically, is to ask whether there could be any other solution besides the one you found. This is asking whether the problem has a unique solution. It does, but this is not usually addressed in an introductory course. A proof of uniqueness can be based on conservation of energy: let $$E(t) = \int_0^\ell (\tfrac{1}{2}u_t^2+\tfrac{1}{2}c^2u_x^2)dx.$$ Here is a way-too-short outline of the argument: It is not too hard to prove that $E'(t) = 0$. Taking $u$ to represent the difference of two suposed solutions, one finds that the energy of $u$ is initially and therefore always $0$, and deduce that $u$ is identically zero, thus the two solutions are the same. I.E. the solution you found is the only one there is.

A last comment is that once you know this, you can sometimes avoid a long computation. With hindsight you might think that $u_t(x,0) = 1$ suggests $u=t$ at least initially, and then realize by subtituting that you have already found the answer! Of course you don't think this way when you are first learning the separation method.


问:
Derive the general formula for the Taylor series of $\frac{1}{1-x}$

The Maclaurin series for $\frac{1}{1-x}$ is simply the geometric series. However, this series converges if and only if $\lvert x \rvert < 1$, despite the fact that $f(x)$ is defined for all $x\neq 1$. In this problem, you will explore other representations of $f(x)$ by Taylor series.

(a) Give a general formula for the Taylor series of $f(x) = \frac{1}{1-x} $ at $a \neq 1$

For (a), I create this formula by degree-$n$ Taylor polynomials and the remainder. But I'm not sure whether it's a general formula.

(b) For what values of $x$ does this Taylor series converge? Express your answer as an interval, in terms of $a$.

For (b), I tried to prove the remainder $\to 0$,  as $n\to\infty$. My formula for the remainder is

$$ (n+1)! \int_{a}^{x} \frac{(x-t)^n}{(1-t)^{n+2}} \, \mathrm{d}t.$$

But I don't know how to show the value of $x$ when this remainder is approaching 0 as $n\to\infty$.

答:
Since

$$ f^{(n)}(x)=\frac{n!}{(1-x)^{n+1}}, $$

the Taylor series for $f(x)$ about $x=a$ (with $a\neq1$) is given by

$$ \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n = \sum_{n=0}^{\infty}\frac{(x-a)^n}{(1-a)^{n+1}}. $$

This is nothing but a geometric series with the common ratio $r=\frac{x-a}{1-a}$, and so, it converges precisely when $\left| r \right| < 1$. So the interval of convergence is

$$ (a-\left|1-a\right|, a+\left|1-a\right|). $$


问:
Find out functions of the form $g(x,y) = \int f(x,t) f(y,t) \lambda(dt)$
I am interested in the following question.

Given a symmetric function $g: \mathbb R^n \times \mathbb R^n \rightarrow \mathbb R$ or $\mathbb R_{+}^{n}\times \mathbb R_{+}^{n} \rightarrow 0$.  I am interested in finding out whether $g$ can be written as the following form:

$$g(x,y) = \int f(x,t) f(y,t) \lambda(dt),$$
where $\lambda$ is some measure but not necessarily the standard Lebesgue measure.

For example, $g(x,y) = \min\{|x|,|y|\}$ can be written as the above form for $f(x,t) = \mathbb I(0<t<|x|)$. $g(x,y) = \frac{1}{|x|+|y|}$ can also be written as above for $f(x,t) = e^{-|x|t}\mathbb I(t>0)$.

I am wondering if there is any necessary or sufficient condition to describe the set of functions which satisfies the above assumption. One necessary condition is $g(x,y)$ needs to be positive symmetric definite.

Thanks a lot!!

答:
My two cents: more a (very long) comment than an answer, but there is something I know about classes of functions that can be represented by the following integral
$$\newcommand{\ffunc}{\odot} \newcommand{\sfunc}{\circledcirc}
g(x,y) = \int f(x,t) \color{red}{f(t,y)} \lambda(\mathrm{d}t),\label{1}\tag{1}
$$
similar to the one in the question, coinciding with it if $f$ is supposed symmetric.

A large class of integral operators of type \eqref{1}, when $\lambda$ is the Lebesgue measure, were introduced by Vito Volterra around 1910 (see references [1], [2] and [3], chapter IV, pp. 99-137) and studied by him and coauthors (notably Joseph Peres) under the name of "compositions" or "functional compositions". Volterra consider only the case $n=1$ and defines two kind of compositions:
$$
\begin{align}
g(x,y) &= \int\limits_x^y f(x,t) h(t,y)\, \mathrm{d}t\triangleq f\ffunc h (x,y)\quad   x<y,\;\; x,y \in {\Bbb R},\label{c1}\tag{C1}\\
g(x,y) &= \int\limits_a^b f(x,t) h(t,y)\, \mathrm{d} t\triangleq f\sfunc h (x,y)\quad   a<b,\;\; a,b \in \overline{\Bbb R}.\label{c2}\tag{C2}\\
\end{align}
$$
Volterra and coworkers call "functional composition of the first kind" the integral operator \ref{c1} and the Functional composition of the second kind integral operator \ref{c2} (I use the symbols $\ffunc$ and $\sfunc$ since Volterra's notation is not easily manageable). Volterra's study focuses on first kind functional compositions, to the point that he's able to solve many classes of nonlinear integral equations. The following one is our main interest here, since it is similar to the one in the question and reduces to \eqref{1} when $n= 2$:
$$
g(x,y) = \underbrace{f \ffunc f \ffunc \cdots \ffunc f}_{n} \triangleq f^{\ffunc n}(x,y)\quad n\in\Bbb N_+ \label{2}\tag{2}
$$

Volterra's solution of \eqref{2} (see [3], chapter IV, §11, pp. 118-119).

Definition 1 (see [1], Lecture 1, p. 11, [2] §1, p. 182, [3] chapter IV, §1, p. 100) Two functions $f, g\in L^1(\Omega)$ are called permutable of the first kind if
$$
f \ffunc g = g \ffunc f \quad
$$
i.e. if their functional composition is commutative. An analogous definition was given by Volterra also for functional compositions of the second kind

Definition 2 (see [3] chapter IV, §7, p.110). A function $f\in L^1(\Omega)$ is said to be a function of order $\alpha\in \Bbb R_+$ if it can be written as
$$
f(x,y)=\frac{(y-x)^{\alpha-1}}{\Gamma(\alpha)}\varphi(x,y)
$$
where $\varphi\in C^0(\Omega)$ is bounded and such that $\varphi (x,x)\neq 0$ for all $x\in \Omega$.

Theorem. Let $g\in L^1(\Omega)$ a function of order $\beta$: if there exists a function $\theta\in L^1 (\Omega)\cup C^1(\overline\Omega)$ of order ${\beta}/{n}$ permutable with $g$ then we can solve integral equation \eqref{2} i.e. there exists precisely a function $f$ of order ${\beta}/{n}$ such that
$$
f(x,y)\triangleq g^{\ffunc\frac{1}{n}}(x,y)\label{3}\tag{3}
$$

Note that, as stated above, this result is proved by Volterra only for functional compositions of the first kind: he does not proves it for composition of the second kind, that would be more interesting from the question point of view as it would be possibly generalizable to the case $n>1$.

Notes

The right side term of \eqref{3} is constructed in a fairly explicit way by Volterra: however if the order of $g$ is 1, an even more explicit solution is given in [2] §13, pp. 188-190.

In the exposition above I tried to translate Volterra's definitions in a more modern language since his notation is a bit outdated: it is worth remember that,
while he does not state this explicitly, he work mostly in the space $C^0(\Omega)$ where $\Omega\subseteq \Bbb R^2$ equipped by a structure of Banach space by the uniform norm, but in this case he assumes integrability of the functions involved in his developments, allowing them to have some integrable singularity (see the definition of the order of a function given below) so I stated them in $L^1(\Omega)$. It would also have been possible to consider $L^1_\text{loc}(\Omega)$ since Volterra considers only compact sets. However, this would have lead to a useless generalization in this context.

The condition $\theta \in  L^1 (\Omega)\cup C^1(\overline\Omega)$ should be introduced in order to deal with functions $g$ of general order: Volterra uses this hypothesis ([3] chapter IV, §11, p. 119) in order to transform a linear Volterra equation of the first kind in a readily solvable linear Volterra equation of the second kind, and thus explicitly construct $f$.

References

[1] Volterra, Vito, Theory of permutable functions, Princeton: Princeton University Press, pp. 66 (1915).

[2] Volterra, Vito, "Functions of composition", Rice Institute Pamphlet - Rice University Studies, 7, no. 4, pp. 181-254 (1920), JFM 48.1258.01.

[3] Volterra, Vito, Theory of functionals and of integral and integro-differential equations. Dover edition with a preface by Griffith C. Evans, a biography of Vito Volterra and a bibliography of his published works by Sir Edmund Whittaker. Unabridged republication of the first English translation, New York: Dover Publications, Inc. pp. 39+XVI+226 (1959), MR0100765, ZBL0086.10402.


答:
Let $m \in \mathbb{R}^+$, now we study the limit of $\dfrac{f(x)}{x^m}$ for $x \in \mathcal{U}(0)$.

$$\dfrac{e^{-1/x^2}}{x^m}=e^{\log\bigg(\dfrac{e^{-1/x^2}}{x^m}\bigg)}=e^{\log(e^{-1/x^2})-\log(x^m)}=e^{-1/x^2-m\log(x)}\to e^{-\infty}\to 0.$$
Note that $-1/x^2-m\log(x)=-\dfrac{1+mx^2\log x}{x^2}$, where $mx^2\log x= \dfrac{m\log x}{\dfrac{1}{x^2}} \longrightarrow _{D.H.}0$, so $-\dfrac{1+mx^2\log x}{x^2}\longrightarrow_{x \to 0} -1/x^2 \longrightarrow_{x \to 0}-\infty$.

Title: Proving the limit is 0

Body:
$f(x)=e^{-1/x^2}$ if $f(x)\neq0$

Prove that for every $m>0,f(x) / x^m →0$ as x → 0.

How should I prove this? I tried to use L'Hospital's Rule, but for $e^{-1/x^2}$, it'll become quite complicated to write its derivatives after m times. I tried to use taylor expansion of $e^x$ as well. But I'm stuck again on how to calculate this limit.


问:
Find a lattice with exactly three congruence relations
A lattice $L$ is a partially ordered set such that any two elements $a$ and $b$ have a least upper bound $a\lor b$ and a greatest lower bound $a\land b$.  A congruence relation on $L$ is an equivalence relation $\sim$ on $L$ that is compatible with the lattice operation in the sense that $a_1\sim b_1$ and $a_2\sim b_2$ implies $a_1\lor a_2\sim b_1\lor b_2$ and similarly for $\land$.

In any lattice there are always two trivial congruence relations, the congruence relation where each element is its own equivalence class (block), and at the other extreme the congruence relation with a single block.

What's an example of a lattice with exactly three congruence relations?

This is Ex 3.40 in Graetzer (Lattice Theory: Foundation).  So it is asking to find a lattice with exactly one non-trivial congruence relation.  Any example I can think of always has at least two non-trivial congruence relations.

For example, for the linear poset with three elements $a\prec b\prec c$, one congruence relation has $a,b$ in one block and $c$ alone, and another has $a$ alone and $b,c$ together in one block.  Another example, the standard diamond lattice with four elements (consisting of all subsets of a two element set ordered by inclusion) also has two non-trivial congruence relations.

答:
Proof: The elements of the
diamond right under the top must be either separated or in a single class,
because diamond has only trivial congruences.

If they are in a single class, then $d\sim b$ implies $a\sim 0$ and $b\sim e$
implies $c\sim 0$. Thus, $a\sim c$ and $0\sim a\sim a\vee c=1$ and this clearly implies that the congruence is trivial.

Similarly, it is easy to see that $a,b,c$ must be separated, otherwise some pair of elements in the diamond is not separated and we have a trivial congruence again.

So the only possible pairs we can relate are $a\sim d$, $b\sim 0$ $e\sim c$.
It is easy to see that any of them forces the other two, so we are left with the single equivalence that is a candidate for a congruence, namely
$\{\{0,b\},\{a,d\},\{c,e\},\{f\},\{1\}\}$.

The remaining tedious checking that this is indeed a congruence is omitted.


答:
$Im(z^n) \ne (Im(z))^n$ so there is no reason that $a\ne 0\implies a^n\ne 0$ should mean that $(a + bi)^n \ne d + 0i$.  It does mean that $(a+bi)^n \ne 0 $ and that means that $Re((a+b)^n)$ and $Im((a+b)^n)$ can't both be equal to $0$ but that doesn't mean one can and the other doesn't.

For a quiz the must obvious example should but $i$ where $Im(i) = 1$ and $i^2 = -1$ and $Im(-1) = -1$.

In general $(a+bi )^n = a^n + na^{n-1}bi - {n\choose 2}a^{n-1}b^2 + ......... + nab^{n-1}i^{n-1} + b^ni^n$.  And the if you add up that long equation there is no reason the imabinare part (which happens to be the sum of the odd terms) should be zero.

Example   $(1 + \sqrt 3i)^3  = 1^3 + 3*1^2\sqrt 3i + 3*1*(\sqrt 3)^2(i)^2 + (\sqrt 3)^3i^3 =$

$1 + 3\sqrt 3i - 9  - 3\sqrt 3i= (1-9) + (3\sqrt 3-3\sqrt 3)i = -8 +0i = -8$.

The imaginary part is a combination of powers of the real and imaginary part that combine to $0$.  And the $(Im z)^3i^3$ wasn't zero but only combined in part to get the Imaginary part of $0$.

Title: Be $Im(z) \neq 0$. It follow: $Im(z^n) \neq 0, \forall n \in \mathbb{N}$. Why isn't this correct?

Body: Since there is no number $n$ with $a^n = 0$ with $a \neq 0$  this should be correct, but according to my exercise quiz its not. I would be very grateful for a hint, because I haven't found a useful approach so far.


答:
With $t:=\dfrac1{x^2}$ we have

$$\lim_{x\to 0}\frac{e^{-1/x^2}}{x^m}=\lim_{t\to\infty}\frac{t^{m/2}}{e^{t}},$$ which is easily tractable by L'Hospital (or just knowing that the exponential grows faster).

Title: Proving the limit is 0


答:
$\ln(e^{-1/x^2}/x^m)
=\ln(e^{-1/x^2})-\ln(x^m)
=-\frac 1{x^2}-m\ln(x)
=\underbrace{-\frac 1{x^2}}_{\to-\infty}\bigg(1+ \underbrace{mx^2\ln(x)}_{\to 0}\bigg)\to-\infty$

Indeed you can use this result: $\ \lim\limits_{u\to 0}\ u\ln(u)=0\ $ there are plenty of proofs on this site.

Therefore $\quad mx^2\ln(x)=\frac m2x^2\ln(x^2)\to 0$ applied for $u=x^2$

And you can conclude that $f(x)=o(x^m)$ for all $m$.

Title: Proving the limit is 0


问:
The number of words that can be made by permuting the letters of _MATHEMATICS_ is
The number of words that can be made by permuting the letters of MATHEMATICS is

$1) 5040$

$2) 11!$

$3) 8!$

$4) 4989600$

First of all I do not understand the statement of the problem, I would like if some one tell me with an example.

答:
To understand the question we assume a simpler word say BALL and see how many permutations of the word exist.

BALL can be permuted as ABLL, ALBL, ALLB, BALL, BLAL, BLLA, LABL, LALB, LBAL, LBLA, LLAB and LLBA. A total of $12$ possible permutations.

The permutations of a word is given by $$\frac{\text{(Total number of alphabets)}!}{\text{(Repetitions of A)}!\text{(Repetitions of B)}!\text{(Repetitions of C)}!...\text{(Repetitions of Z)}!}$$

The number of words that can be made by permuting the letters of MATHEMATICS is

$$\frac{11!}{2!2!2!}=4989600$$


问:
Is there a simple reason for why the two stationary points in a cubic polynomial has its second derivative equal in magnitude but opposite in sign.
Given a polynomial $f(x) = x^3 + ax^2 + bx + c$, then

$$f'(x) = 3x^2 + 2ax + b$$
$$f''(x) = 6x + 2a$$

The stationary points in the curve will have $x = \frac{-a \pm \sqrt{a^2 - 3b}}{6}$, but then the two points have their second derivative equal in magnitude but opposite in sign since

$$f''(\frac{-a + \sqrt{a^2 - 3b}}{3}) = \sqrt{a^2 - 3b} = -f''(\frac{-a - \sqrt{a^2 - 3b}}{3})$$

I remembered that I once had an intuition why this is true but I forgot now. Could somebody tell me why? Thanks in advance

答: If you translate the graph so that the single point of inflection is the origin then the resulting function is odd.</p>


问:
Is the trace of a product of positive definite matrices bigger than the sum of the product of their corresponding diagonal elements?
Let $\mathbf{S},\mathbf{\Theta}\in\mathbb{R}^{p\times p}$ by symmetric, positive definite matrices. Denote their corresponding $(i,j)$-th component by $S_{ij},\Theta_{ij}$. Is it true that
\begin{align*}
\text{Trace}(\mathbf{S\Theta})\geq\sum_{i=1}^{p}S_{ii}\Theta_{ii}.
\end{align*}

答:
This is not true.
Take $$A= \begin{bmatrix} 2& 1 \\ 1 & 2 \end{bmatrix}$$ and $$B= \begin{bmatrix} 2& -1 \\ -1 & 2 \end{bmatrix}$$


问:
Finding standard form of parabola equation
I have a little problem with figuring it out how find standard form of a parabola equation from the given values. I tried googling and watching video in youtube but I don't understand how to actually go about it.

Given:

Directrix is $ x = -2$ ; focus is $(2,0)$

Axis along the $x$ axis; vertex at the origin; passing through $(2,5)$

Vertex at $(-2,-2)$; focus at $(-2,2)$

I don't know how to formulate this, some hints would be appreciated.

Edit. I read this but I don't understand.

答:
Note that it is important to know the definition of a parabola, what the curve signifies besides the geometry. A parabola is a curve (equation) where any point is equidistant from a fixed point (focus) and a fixed line (directrix). Have a look at the image below to get familiar with the terms, first.

Your question seems to describe 3 different parabolas.

Parabola 1. Directrix $\equiv x=-2$ , focus $\equiv (2,0)$

We go with the definition. Let $(h,k)$ be an arbitrary point on the parabola. Then, it must be equidistant from the directrix, $x+2=0$ and focus, $(2,0)$ i.e., it satisfies
\begin{align}|h+2|=\sqrt{(h-2)^2+k^2}&\Rightarrow (h+2)^2=(h-2)^2+k^2\\
&\Rightarrow k^2=(h+2)^2-(h-2)^2=8h\\
&\Rightarrow k^2=8h.\end{align}
Since, $(h,k)$ was an arbitrary point on the curve, we can say that the equation of the parabola is nothing but
$$y^2=8x.$$

Parabola 2. Axis $\equiv y=0$ ($x$ axis) , vertex $\equiv (0,0)$ , $(2,5)$ is a point lying on the curve

The ''axis of symmetry'' or simply, the axis of a parabola is the line perpendicular to the directrix and passing through the focus of the parabola. The vertex has been defined as the point where the axis meets the curve itself.

Any parabola with $y=0$ as its axis of symmetry and vertex $(0,0)$ is of the form
$$y^2=4px\mathrm{~where~}(p,0)\mathrm{~is~the~focus}.$$

(Why is this?)

The focus must be a point lying on the $x$ axis and hence, of the form $(p,0)$. Since, the vertex is a point on the parabola, it must be equidistant from the directrix and the focus. The axis is perpendicular to the directrix, passing through the focus and vertex. Suppose the axis meets the directrix at point $D$ and let $F$ be the focus. Then, the vertex, $(0,0)$ is the midpoint of the line segment, $FD$ and $D$ lies on the axis. We get,
$$D\equiv (-p,0).$$
We use this to get the equation of the parabola, as we did in part 1:
\begin{align}|x+p|=\sqrt{(x-p)^2+y^2}&\Rightarrow (x+p)^2=(x-p)^2+y^2\\
&\Rightarrow y^2=(x+p)^2-(x-p)^2=4px\\
&\Rightarrow y^2=4px.\end{align}

We know that $(2,5)$ also lies on the parabola. Therefore, it satisfies the equation $y^2=4px$ i.e.,
$$5^2=4p\cdot2\Rightarrow p=25/8.$$
Equation of the parabola becomes
$$y^2=\frac{25}{2}x\Leftrightarrow 2y^2=25x.$$

Parabola 3. Vertex $\equiv (-2,-2)$, focus $\equiv (-2,2)$

The focus and vertex of a parabola lie on the axis and, the vertex is the midpoint of the line segment joining the focus and the point where the axis meets the directrix (let us call this point $D$). The directrix will be a line perpendicular to the axis and passing through $D$.

First we find $D\equiv(h,k)$. Then,
$$\frac{h-2}{2}=-2\Rightarrow h=-2\mathrm{~and~}\frac{k+2}{2}=-2\Rightarrow k=-6.$$
$\therefore D\equiv(-2,-6).$

Next, we observe that the line $x=-2$ passes through $(-2,-2)$ and $(-2,2)$. Since, any two points in $\mathbb{R}^2$ have exactly one straight line passing through both of them, we can conclude that the equation of the axis is
$$x+2=0.$$

Any line perpendicular to some line, $ax+by+c=0$ is of the form $bx-ay+k=0.$

The directrix is of the form
$$y-k=0$$
and it passes through the point $D\equiv(-2,-6).$ Hence, $k=-6$ and the directrix is $y+6=0$. Using these, we get that the equation of the parabola is
\begin{align}|y+6|=\sqrt{(x+2)^2+(y-2)^2}&\Rightarrow (y+6)^2=(x+2)^2+(y-2)^2\\
&\Rightarrow (x+2)^2=(y+6)^2-(y-2)^2=8(2y+4)\\
&\Rightarrow (x+2)^2=16(y+2).\end{align}

Note. Once you get familiar with parabolas, you'll learn some standard parabolas and their properties. You can then solve for parabola 3 using @Jan-Magnus Økland's comment under your question.


问:
Half order derivative of $ {1 \over 1-x }$
I'm new to this "fractional derivative" concept and try, using wikipedia, to solve a problem with the half-derivative of the zeta at zero, in this instance with the help of the zeta's Laurent-expansion.

Part of this fiddling is now to find the half-derivative  $$ {d^{1/2}\over dx^{1/2}}{1 \over 1-x}$$

First I would like to understand, whether there is a short/closed form for this ot whether I have to express the fraction as a power series first and then to differentiate termwise.

Next I would like to know the value at $x=0$.

答:
Use what you know about whole number derivatives. Inductively, you can prove $$\frac{d^n}{dx^n}\frac1{1-x}=\frac{n!}{(1-x)^{n+1}}$$ Now express $n!$ using the $\Gamma$ function ($\Gamma(n+1)$), and you can extend the definition to non-integral $n$: $$\frac{d^{1/2}}{dx^{1/2}}\frac1{1-x}=\frac{\Gamma(3/2)}{(1-x)^{3/2}}$$ At $x=0$, this is just $\Gamma(3/2)$.

To confirm that this method works, observe that you can also inductively prove $$\begin{align}\frac{d^n}{dx^n}\frac1{(1-x)^{3/2}}&=\frac{\frac{(2n+1)!}{4^n\cdot n!}}{(1-x)^{n+3/2}}\\&=\frac{\frac{\Gamma(2n+2)}{4^n\Gamma(n+1)}}{(1-x)^{n+3/2}}\end{align}$$ and extend to nonintegral $n$, so that $$\begin{align}\frac{d^{1/2}}{dx^{1/2}}\frac{d^{1/2}}{dx^{1/2}}\frac1{1-x}&=\frac{d^{1/2}}{dx^{1/2}}\frac{\Gamma(3/2)}{(1-x)^{3/2}}\\&=\frac{\Gamma(3/2)\frac{\Gamma(3)}{2\Gamma(3/2)}}{(1-x)^{2}}\\&=\frac{\frac{2!}{2}}{(1-x)^{2}}\\&=\frac{1}{(1-x)^2}\\&=\frac{d}{dx}\frac{1}{1-x}\end{align}$$ and all is as it should be.


问:
Finding the conditional probabilities of a latent dirichlet allocation model
Let's say I'm defining a LDA as the following:

For each doc $m$:

Sample topic probabilities $\theta_m \sim Dirichlet(\alpha)$

For each word $n$:

Sample a topic $z_{mn} \sim Multinomial(\theta_m)$

Sample a word $w_{mn} \sim Multinomial(\beta)$

where $\alpha, \beta$ are fixed hyperparameters.

Now, to do this, I'm going to use an EM algorithm. For the 'E' step, given $\alpha$ and $\beta$, I infer $z_{mn}$ for all $n$ and $\theta_m$ for all $m$ given $w, \alpha, \beta$ using Gibbs sampling. Equivalently, I'm trying to find $p(z, \theta | w, \alpha, \beta) = \frac{p(z, \theta, w | \alpha, \beta)}{p(w|\alpha, \beta)}$ using Gibbs sampling.

At a high level, I understand what Gibbs Sampling is and what this LDA model does. We assigns topics to each word and document (assuming all others are correct). We do repeat this process in a chain until we maximize the probability of each $m$ and $n$ belonging to a particular category.

However, I'm having trouble representing the conditional probabilities of the Gibbs Sampler for this model. Where do I even start and what am I looking for when asked to find the conditional probabilities of this sampler?

答: If you are finding it tricky getting started, I suggest you have a look at this paper by Griffiths and Steyvers here. It was well-known in the ML community for using Gibbs sampling as an alternative to variational inference for topic models. If you are looking for verbal exposition, have a look at the slides and video at around the 40:00 min mark of Lecture 14 on MCMC of this series here. The instructor briefly walks through Gibbs sampling algorithmically and in context of the paper above, and also briefly provides intuition on one of the equations there.</p>


问:
What does circle symbol between two functions means?
I have two function $f$ and $r$, such that: $y = f(x)$, $r(y) = x$. I read that it means that: $f \circ r = I_{U}$, where $x \in U, y \in U$, where $U$ is a finite set. What does $f \circ r = I_{U}$ mean? What does this circle mean and how could i read all this writing $f \circ r = I_{U}$ ?

答:
The $\circ$ symbol corresponds to the function composition. For any two functions $f:X\to Y$ and $g:Y\to Z$, the new function "$g$ composed with $f$" (denoted by $g\circ f$) is a new function defined as $(g\circ f)(x)=g(f(x))$.

In your case, you have X=Y=Z=U and $g=r$. Then, $I_U=r\circ f$ seems to be the identity function on $U$.


问:
Finding a polynomial with integer coefficients that has $\sqrt[100]{\sqrt{3} + \sqrt{2}} + \sqrt[100]{\sqrt{3} - \sqrt{2}}$ as a root
I'm looking for an (ideally) elementary way to find a polynomial with integer coefficients that has $$\sqrt[100]{\sqrt{3} + \sqrt{2}} + \sqrt[100]{\sqrt{3} - \sqrt{2}}$$ as a root.

What I have so far is that we can write $\sqrt 3 - \sqrt 2$ as $1/(\sqrt 3 + \sqrt 2)$ and then we can quite easily check that both $\sqrt 3 + \sqrt 2$ and $\sqrt 3 - \sqrt 2$ solve $$x^4-10x^2+1=0.$$

So then we have that $\sqrt[100]{\sqrt{3} + \sqrt{2}}$ and $\sqrt[100]{\sqrt{3} - \sqrt{2}}$ both solve $$x^{400}-10x^{200}+1=0$$

but I am not sure how to find an integer-coefficient polynomial with the sum of those two values as a root. It sort of looks like Vieta formulas could somehow be relevant, but I may be missing something obvious here.

答:
Hint

Denote

$$\begin{cases}
a &= \sqrt[100]{\sqrt 3 + \sqrt 2}\\
b &= \sqrt[100]{\sqrt 3 - \sqrt 2}
\end{cases}$$

and notice that $ab=1$. We have $a^{100} + \frac{1}{a^{100}} = 2\sqrt 3$.

Now you can write

$$a^{100} + \frac{1}{a^{100}} = \left(a + \frac{1}{a}\right)^{100} - (p(a)+p(1/a))$$ where $p(x)$ is a polynomial of degree at most equal to $99$ with integer coefficients. You can repeat this process by induction to finally write $a^{100} + \frac{1}{a^{100}} = q(a+1/a)$ where $q$ is a polynomial of degree $100$ with integer coefficients.

At the end you get

$$q(a+b) =2 \sqrt 3$$ which implies that $a+b$ can't be a rational number as $\sqrt 3$ isn't. And that $a+b$ is a root of the polynomial with integer coefficients $$q^2(x) - 12=0$$ as desired.


问:
There are 3 sequences $a_n $, $b_n$, $c_n$. Prove that limits in $\infty$ are equal and that 2 of them are convergent
There are 3 sequences $a_n, $, $b_n$, $c_n$ with $n \in \mathbb{N} \setminus \{0\}$ given by recursions:

$a_{n+1} = \frac{a_n+b_n+c_n}{3}$

$b_{n+1} = \sqrt[3]{a_n*b_n*c_n}$

$c_{n+1} = \frac{3}{\frac{1}{a_n} + \frac{1}{b_n}+\frac{1}{c_n}}$

Elements $a_1, b_1$ and $c_1$ are all positive numbers. Prove that at least $2$ of those $3$ sequences are convergent and that: $$\lim_{n \to +\infty} a_n = \lim_{n \to +\infty} b_n = \lim_{n \to +\infty} c_n $$

The first thing that I see is that those seqences are made of subsequent elements of functions known as arithmetic, geometric and harmonic means. Because of inequality of means, I know that:
$$a_{n+1} = \frac{a_n+b_n+c_n}{3} \geq b_{n+1} = \sqrt[3]{a_n*b_n*c_n} \geq c_{n+1} = \frac{3}{\frac{1}{a_n} + \frac{1}{b_n}+\frac{1}{c_n}}$$

Then, I know for sure that:

every subsequent element of $a_n$ must be smaller than the former one because $b_n$ and $c_n$ are smaller than $a_n$ for every n

every subsequent element of $c_n$ must be greater than the former one because $a_n$ and $b_n$ are bigger than $c_n$ for every n

Additionally, I know that the limit of each one of those means is reached when all 3 elements in the mean are equal. In my case all 3 elements are the same for all 3 means so, the limit must be equal as well.

But here comes my problem - how to write it all down, to present my thinking in a correct way? Is there any scheme that I that could apply in my case (I only know how to formally find limits and prove convergence in cases of $\lim_{n \to +\infty}$ for sequences of $n$ elements)?

答: If you put $a_n \gt$ in front of your inequalities and $\gt c_n$ afterward you have a proof of convergence for $a$ and $c$ as the sequences are monotone and bounded.


问:
Inverse of a selfpower matrix
Suppose that $A$ is a matrix and we have $ A ^ 2 = A $ and also $ \gamma \neq 1 .$   Prove that $ I - \gamma A $ is invertible and find its inverse.

答:
Hint:

Solve $(I-\gamma A)(I+\beta A)=I$ for $\beta$.

Another method when $|\gamma| $ is small enough:

$\dfrac 1{1-\gamma A}=1+\gamma A+\gamma^2 A^2+\gamma^3 A^3+\cdots$

$=1+A(\gamma+\gamma^2+\gamma^3+\cdots)=1+A\beta$.

Title: Inverse of a selfpower matrix


答:
No. If it is true, by a continuity argument, the inequality should still be true when $S$ and $\Theta$ are positive semidefinite, but it is not:
$$
S=\pmatrix{1&-1\\ -1&1},
\ \Theta=\pmatrix{1&1\\ 1&1}
\ \Rightarrow\ \operatorname{tr}(S\Theta)=0<2=\sum_iS_{ii}\Theta_{ii}.
$$

Title: Is the trace of a product of positive definite matrices bigger than the sum of the product of their corresponding diagonal elements?

Body:
Let $\mathbf{S},\mathbf{\Theta}\in\mathbb{R}^{p\times p}$ by symmetric, positive definite matrices. Denote their corresponding $(i,j)$-th component by $S_{ij},\Theta_{ij}$. Is it true that
\begin{align*}
\text{Trace}(\mathbf{S\Theta})\geq\sum_{i=1}^{p}S_{ii}\Theta_{ii}.
\end{align*}


问:
What is the relationship between cartesian vectors and tuples?
As far as I have understood the cartesian product, $\mathbb{R}^2 = \{(x,y) \ \vert \ x \in \mathbb{R} \ \land \ y \in \mathbb{R} \}$, contains ordered pairs, represented as tuples.

But think I have also seen $\vec{v} \in \mathbb{R}^2$.

Firstly, is that correct?

Secondly, does that mean that the elements of $\mathbb{R}^2$ can be represented as vectors as well?

In general, what is the relationship between cartesian vectors and tuples?

答:
Tuples (in general points in $R^n$) are points in space.  Vectors have a length and direction.  A one to one correspondence between vectors and points in space can be made by fixing the tail of a vector to the origin and the head of the vector at that point in space.

Note that vectors themselves are free floating  - allowing addition, etc.


答:
we have
 $5^2 \equiv 1\ (mod\ 3) \Rightarrow 5^{2(3n+3)} \equiv 1\ (mod\ 3) \Rightarrow 5^{6n+6} \equiv 1\ (mod\ 3) $

and we have
$2^2 \equiv 1\ (mod\ 3) \Rightarrow 2^{2(n+1)} \equiv 1\ (mod\ 3) \Rightarrow 2^{2n+2} \equiv 1\ (mod\ 3) $
 Thus
 $$5^{6n+6}-2^{2n+2} \equiv 0\ (mod\ 3)$$

Title: Why $3\mid(5^{6n+6}-2^{2n+2})$ for all natural $n$?

Body: How we can easily show that $3\mid(5^{6n+6}-2^{2n+2})$ for all natural $n$. These conditions continue $3\mid(5^{6n+5}-2^{2n+3})$ and $3\mid(5^{6n+4}-2^{2n+4})$ and $3\mid(5^{6n+3}-2^{2n+1})$ and $3\mid(5^{6n+2}-2^{2n+2})$ and $3\mid(5^{6n+1}-2^{2n+1})$.


问:
How can this $T(n) = T(n-1)+T(n-2)+3n+1$ non homogenous recurrence relation be solved
How are can the above recurrence relation be solved?

I've reached here:
$(x^{2}-x-1)(x-3)^2(x-1)$

And then here:

$$a_n = l_1 \cdot (x_1)^n+l_2 \cdot (x_2)^n+l_3 \cdot (x_3)^n+l_4\cdot n \cdot (x_3)^n+l_5\cdot (x_4)^n$$

And we are given that these: $T(0) = 2, T(1) = 3$

So, for $n = 0$ and $n = 1$ we get two equations but we need 3 more, yet we don't have any more constants.

答:
Use "generatingfunctionology" techniques. Define $G(z) = \sum_{n \ge 0} T(n) z^n$, write your recurrence as:
$$
T(n + 2) = T(n + 1) + T(n) + 3 n + 3
$$
Multiply by $z^n$, add for $n \ge 0$. Remember that:
$$
\sum_{n \ge 0} (n + 1) z^n = \frac{1}{(1 - z)^2}
$$
and you get:
$$
\frac{(G(z) - T(0) - T(1) z}{z^2}
  = \frac{G(z) - T(0)}{z} + G(z) + 3 \cdot \frac{1}{(1 - z)^2}
$$
Solve for $G(z)$, expand in partial fractions:
$$
G(z) = \frac{2 - 3 z + z^2 + z^3}{1 - 3 z + 2 z^2 + z^3 - z^4}
     = \frac{4 + 2 z}{1 - z - z^2} - \frac{1}{1 - z} - \frac{1}{(1 - z)^2}
$$
We have the last two terms, the first one is for Fibonacci numbers:
$$
T(n) = 4 F_{n + 1} + 2 F_n - 1 - (n + 1) = 4 F_{n + 1} + 2 F_n - n - 2
$$</p>


问:
Dividing 12 students into classes of 4
In how many ways can you divide 12 students in 3 distinct classes, if a maximum number of students in each class is 4?

If I pick 4 students out of 12 for the first class in C(12,4) ways, then C(8,4) and C(4,4), do I need to divide my result with 3! ? Would dividing by 3! mean that I make no difference between the classes and think of them as three indistinct groups?

答:
The answer depends on whether the classes are labeled.  If we only care about which students are in the same class, then dividing by $3!$ is appropriate.  However, if the classes are labeled, you should not divide by $3!$ since it matters which class which student takes.  Since the classes in your problem are distinct, you should not divide by $3!$ in this case.  The correct answer is
$$\binom{12}{4}\binom{8}{4}\binom{4}{4}$$

Title: Dividing 12 students into classes of 4


问:
Showing $[H(x + t) − H(x − t)]\,\delta'(t) = −2\delta(x)\delta(t)$, for Dirac delta $\delta$ and Heaviside $H$
How to show that
$$[H(x + t) − H(x − t)]\,\delta'(t) = −2\delta(x)\delta(t)$$
where $\delta$ represents the Dirac delta function and $H$ represents the Heaviside function?

I have tried testing the convolution on both sides with a test function $\phi(x,t)$ but it doesn’t seem to work.

答:
Define the 2D distributions
$$ u[f]~:=~ \int_{\mathbb{R}^2} \! \mathrm{d}x \mathrm{d}t f(x,t) [H(x+t)-H(x-t)],\tag{1}$$
and
$$ v[f]~:=~ \int_{\mathbb{R}^2} \! \mathrm{d}x \mathrm{d}t f(x,t) \delta(t)~:=~\int_{\mathbb{R}^2} \! \mathrm{d}x ~ f(x,0),\tag{2}$$
for a test function $f$.

The product distribution $$w~=~uv~=0\tag{3}$$ is well-defined.
In fact $w$ is the zero distribution. Proof of eq. (3):
$$\begin{align} u[f]~:=~&vw[f]\cr
~=~& \int_{\mathbb{R}^2} \! \mathrm{d}x \mathrm{d}t f(x,t) [H(x+t)-H(x-t)]\delta(t)\cr
~=~&0. \end{align}\tag{4}$$

OP's title relation is a consequence of that the derivative
$$0~=~\partial_tw~=~(\partial_tu) v + u(\partial_tv) \tag{5}$$
vanish as well, and that
$$(\partial_tu) v~=~2\delta^2. \tag{6}$$
Proof of eq. (6):
$$\begin{align}(\partial_tu) v[f]~=~&\int_{\mathbb{R}^2} \! \mathrm{d}x \mathrm{d}t f(x,t) \partial_t[H(x+t)-H(x-t)]\delta(t)\cr
~=~&\int_{\mathbb{R}^2} \! \mathrm{d}x \mathrm{d}t f(x,t) [\delta(x+t)+\delta(x-t)]\delta(t)\cr
~=~&2f(0,0)~=~2\int_{\mathbb{R}^2} \! \mathrm{d}x \mathrm{d}t f(x,t)\delta^2(x,t)\cr
~=:~& 2\delta^2[f].\end{align} \tag{7}$$


问:
Easy way to see that $(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (x^2 + 5x + 12)(x^2 + 5x - 2)$?
As the title suggests, is there an easy way to see that$$(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (x^2 + 5x + 12)(x^2 + 5x - 2)$$that doesn't require expanding in full? Is there a trick?

Body: For $a=x^2+5x+4$ we get to factor $a(a+2)-48=a^2+2a-48=(a+8)(a-6)$.


Title: Easy way to see that $(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (x^2 + 5x + 12)(x^2 + 5x - 2)$?

Body: As the title suggests, is there an easy way to see that$$(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (x^2 + 5x + 12)(x^2 + 5x - 2)$$that doesn't require expanding in full? Is there a trick?

Body: $$(x^2 + 5x + 5-1)(x^2 + 5x + 5+1) - 48 = (x^2 + 5x + 5)^2-7^2$$


问:
Expected Value : Poker
You are dealt $4$ cards face down on a table. You know that $2$ are red and $2$ are black. Let’s say you are given $100 to bet on this game and you are able to bet after each card is dealt and before the next one is on whether the outcome will be red or black. If you play this game optimally, what’s your expected value?

Thinking along the lines that at first there is 1/2 probability of red or black. If first is red then second round has probability of picking black 2/3. If second round is black then 1/2 probability that 3 rd round has black or red. Final round will have probability 1. I was thinking that this was optimal play.

Edit - actually I think I messed up round 3 probability.

答:
With an initial bankroll of $\$100$, the simple strategy to go all-in when the outcome is certain, i.e. when there is only one color left, else bet nothing, will yield the maximal expected value of $\$267$, with a standard deviation of $\$94.3$.

In fact, there are in infinite number of strategies that will yield the same optimal expectation (but different variance).

While we cannot increase the expectation further, we can find the strategy of these that has the lowest variance. In fact, there is a unique strategy that will yield a guaranteed final bank-roll of $\$267$ (zero variance). This strategy is to always bet on guaranteed outcomes (as before), except for the second card where you bet 1/3 of your bankroll on the color not seen on the first card. This kills off the variance and ensured a non-stochastic final bankroll.

Figure 1. The state is the number of colored cards remaining (black, red). The green function $E(black, red; x)$ is the expected final bankroll if starting with $x$ in that state.

(0, 0)
Consider starting with $x$ when there are no cards left (0, 0). There is nothing do bet on, so the final bankroll will be unchanged at $x$. So
$$E(b,r,x) = E(0, 0, x) = x.$$

(1, 0)
Consider starting with $x$ when there is one black cards left (1, 0). Outcome is certain so we go all-in, so the final bankroll will be $2x$. So
$$E(1, 0,x) = 2x.$$

(0, 1)
Same as (1,0) due to symmetry. So
$$E(0,1,x) = 2x.$$

(2, 0)
Consider starting with $x$ when there are two black cards left (2, 0). Next outcome is certain so we go all-in twice in a row, so $E(2,0,x) = 4x$. Note that this is the same as going all on the first card, realize $2x$ and start a new game at (1,0). So
$$E(2,0,x) = 1 \cdot E(1,0, 2x) = 2 (2x) = 4x.$$

(0, 2)
Same as (2,0) due to symmetry. So
$$E(0,2,x) = 4x.$$

(1, 1)
Consider starting with $x$ when there is one black and one red card left (1, 1). Next outcome is uncertain with 50/50 and the value at the next state is the same for either outcome (symmetry), so the value is invariant to what we bet. Note that while the expected value doesn't depend on what we bet, the variance is increasing by betting, so we prefer to bet zero. So
$$E(1,1,x) = E(1,0, x) = 2x.$$

To convince yourself, assume you bet $0 \leq w\leq x$ on red; if you win you have $x+w$ to invest at state $(1,0)$; if you lose you have $x-w$ to invest at state $(0,1)$. If you bet on red the expected value is $E(1,1,x)_r = E(1,1,x)| \text{"bet } w \text{ on red"} = \tfrac{1}{2} E(1,0, x+w) + \tfrac{1}{2} E(0,1, x-w) = 2x$. Due to symmetry, $E(1,1,x)_b = E(1,1,x)| \text{"bet } w \text{ on black"} = 2x$. Note that both values are equal, and invariant to the betting size. So we are indifferent to whether we bet or not.

Since we have the liberty to chose the bet arbitrarily without affecting the expected value, let us see if we can decrease the variance instead. The 2nd moment is given by $M_2(1,1,x)| \text{bet } w \text{ on red} = \tfrac{1}{2} E(1,0, x+w)^2 + \tfrac{1}{2} E(0,1, x-w)^2 = 4(x^2 + w^2)$, and so variance is $V(1,1,x)| \text{"bet } w \text{ on red"} = 4w^2.$ So we set $w=0$ to have minimum variance (remember expectation is unaffected). We do the same for betting black, byt due to symmetry, we see that we would also bet zero on black (betting zero on red = betting zero on black).

(2, 1)
Consider starting with $x$ when there are two black and one red card left. If you bet on red then $E(2,1,x)| \text{"bet } w \text{ on red"} = \tfrac{1}{3}E(2,0, x+w) + \tfrac{2}{3}E(1,1, x-w) = \tfrac{8}{3}x.$ If you bet on black then $E(2,1,x)| \text{"bet } w \text{ on black"} = \tfrac{1}{3}E(2,0, x-w) + \tfrac{2}{3}E(1,1, x+w) = \tfrac{8}{3}x.$ Value is invariant of what card you bet on and betting size.

Since we have the liberty to chose the bet arbitrarily without affecting the expected value, let us see if we can decrease the variance instead.
If you bet on red then $V(2,1,x)| \text{"bet } w \text{ on red"} = \tfrac{1}{3}E(2,0, x+w)^2 + \tfrac{2}{3}E(2,0, x-w)^2 - \left(\tfrac{8}{3}x \right)$ has minimum variance of $\tfrac{8}{9}x^2$ at $w=x$. If you bet on black then $V(2,1,x)| \text{"bet } w \text{ on black"} = \tfrac{1}{3}E(2,0, x-w)^2 + \tfrac{2}{3}E(2,0, x+w)^2 - \left(\tfrac{8}{3}x \right)$ has zero minimum variance at $w=x/3$. So we pick betting on black since expected value is the same for both bets, but we can kill off the variance by betting on black.

$$E(2,1,x) = \tfrac{8}{3}x.$$

(2, 2)
Due to similar reasoning as (1,1), we bet nothing
$$E(2,2,x) = \tfrac{8}{3}x = 2.67x$$

In general we have

\begin{align}
E(b, r, x) &= \max \left(\max_w E(b,r,x)_{\text{red}}, \;\max_w E(b,r,x)_{\text{black}} \right)
\end{align}
where
\begin{align}
 E(b,r,x)_{\text{red}} &= \tfrac{r}{b+r}E(b, r-1, x+w) + \tfrac{b}{b+r}E(b-1, r, x-w), \\
 E(b,r,x)_{\text{black}} &= \tfrac{r}{b+r}E(b, r-1, x-w) + \tfrac{b}{b+r}E(b-1, r, x+w).
\end{align}
and boundary conditions
\begin{align}
E(0,0) &= x \\
E(r,b) &= 0 \text{ if } r<0 \text{ or } b<0
\end{align}

```<code>from scipy.optimize import minimize

def pr_black(b, r):
    return b / (b + r)

def pr_red(b, r):
    return r / (b + r)

def E_bet_black(w, x, b, r):
    w = w[0]
    return pr_black(b,r) * E(b-1, r, x + w) + pr_red(b,r) * E(b, r-1, x - w)

def E_bet_red(w, x, b, r):
    w = w[0]
    return pr_black(b,r) * E(b-1, r, x - w) + pr_red(b,r) * E(b, r-1, x + w)

cache = {}
def E(b, r, x):
    if (b, r) in cache:
        return cache[(b,r)]*x # Assume linear value mapping.
    elif r==0 and b==0:
        return x
    elif r<0 or b<0:
        return 0.0
    else:
        bnds = [(0, x)]
        res_black = minimize(lambda w,x,b,r: -E_bet_black(w,x,r,b), x0=x/2, args=(x,b,r), bounds=bnds)
        res_red = minimize(lambda w,x,b,r: -E_bet_red(w,x,r,b), x0=x/2, args=(x,b,r), bounds=bnds)
        val_black, w_black = -res_black.fun, res_black.x
        val_red, w_red = -res_black.fun, res_black.x
        color,w,value = ('black', w_black, val_black) if val_black > val_red else ('red', w_red, val_red)
        cache[(b, r)] = value/x
        return value

b, r = (2, 2)
x = 100

print(    'E(b,r,x) = E({},{},{}) = {}'.format(b,r,x, round(E(b, r, x), 2))    )
</code>```
```<code>E(b,r,x) = E(2,2,100) = 266.67
</code>```

Title: Expected Value : Poker

Body:
With an initial bankroll of $\$100$, the simple strategy to go all-in when the outcome is certain, i.e. when there is only one color left, else bet nothing, will yield the maximal expected value of $\$267$, with a standard deviation of $\$94.3$.

In fact, there are in infinite number of strategies that will yield the same optimal expectation (but different variance).

While we cannot increase the expectation further, we can find the strategy of these that has the lowest variance. In fact, there is a unique strategy that will yield a guaranteed final bank-roll of $\$267$ (zero variance). This strategy is to always bet on guaranteed outcomes (as before), except for the second card where you bet 1/3 of your bankroll on the color not seen on the first card. This kills off the variance and ensured a non-stochastic final bankroll.

Figure 1. The state is the number of colored cards remaining (black, red). The green function $E(black, red; x)$ is the expected final bankroll if starting with $x$ in that state.

(0, 0)
Consider starting with $x$ when there are no cards left (0, 0). There is nothing do bet on, so the final bankroll will be unchanged at $x$. So
$$E(b,r,x) = E(0, 0, x) = x.$$

(1, 0)
Consider starting with $x$ when there is one black cards left (1, 0). Outcome is certain so we go all-in, so the final bankroll will be $2x$. So
$$E(1, 0,x) = 2x.$$

(0, 1)
Same as (1,0) due to symmetry. So
$$E(0,1,x) = 2x.$$

(2, 0)
Consider starting with $x$ when there are two black cards left (2, 0). Next outcome is certain so we go all-in twice in a row, so $E(2,0,x) = 4x$. Note that this is the same as going all on the first card, realize $2x$ and start a new game at (1,0). So
$$E(2,0,x) = 1 \cdot E(1,0, 2x) = 2 (2x) = 4x.$$

(0, 2)
Same as (2,0) due to symmetry. So
$$E(0,2,x) = 4x.$$

(1, 1)
Consider starting with $x$ when there is one black and one red card left (1, 1). Next outcome is uncertain with 50/50 and the value at the next state is the same for either outcome (symmetry), so the value is invariant to what we bet. Note that while the expected value doesn't depend on what we bet, the variance is increasing by betting, so we prefer to bet zero. So
$$E(1,1,x) = E(1,0, x) = 2x.$$

To convince yourself, assume you bet $0 \leq w\leq x$ on red; if you win you have $x+w$ to invest at state $(1,0)$; if you lose you have $x-w$ to invest at state $(0,1)$. If you bet on red the expected value is $E(1,1,x)_r = E(1,1,x)| \text{"bet } w \text{ on red"} = \tfrac{1}{2} E(1,0, x+w) + \tfrac{1}{2} E(0,1, x-w) = 2x$. Due to symmetry, $E(1,1,x)_b = E(1,1,x)| \text{"bet } w \text{ on black"} = 2x$. Note that both values are equal, and invariant to the betting size. So we are indifferent to whether we bet or not.

Since we have the liberty to chose the bet arbitrarily without affecting the expected value, let us see if we can decrease the variance instead. The 2nd moment is given by $M_2(1,1,x)| \text{bet } w \text{ on red} = \tfrac{1}{2} E(1,0, x+w)^2 + \tfrac{1}{2} E(0,1, x-w)^2 = 4(x^2 + w^2)$, and so variance is $V(1,1,x)| \text{"bet } w \text{ on red"} = 4w^2.$ So we set $w=0$ to have minimum variance (remember expectation is unaffected). We do the same for betting black, byt due to symmetry, we see that we would also bet zero on black (betting zero on red = betting zero on black).

(2, 1)
Consider starting with $x$ when there are two black and one red card left. If you bet on red then $E(2,1,x)| \text{"bet } w \text{ on red"} = \tfrac{1}{3}E(2,0, x+w) + \tfrac{2}{3}E(1,1, x-w) = \tfrac{8}{3}x.$ If you bet on black then $E(2,1,x)| \text{"bet } w \text{ on black"} = \tfrac{1}{3}E(2,0, x-w) + \tfrac{2}{3}E(1,1, x+w) = \tfrac{8}{3}x.$ Value is invariant of what card you bet on and betting size.

Since we have the liberty to chose the bet arbitrarily without affecting the expected value, let us see if we can decrease the variance instead.
If you bet on red then $V(2,1,x)| \text{"bet } w \text{ on red"} = \tfrac{1}{3}E(2,0, x+w)^2 + \tfrac{2}{3}E(2,0, x-w)^2 - \left(\tfrac{8}{3}x \right)$ has minimum variance of $\tfrac{8}{9}x^2$ at $w=x$. If you bet on black then $V(2,1,x)| \text{"bet } w \text{ on black"} = \tfrac{1}{3}E(2,0, x-w)^2 + \tfrac{2}{3}E(2,0, x+w)^2 - \left(\tfrac{8}{3}x \right)$ has zero minimum variance at $w=x/3$. So we pick betting on black since expected value is the same for both bets, but we can kill off the variance by betting on black.

$$E(2,1,x) = \tfrac{8}{3}x.$$

(2, 2)
Due to similar reasoning as (1,1), we bet nothing
$$E(2,2,x) = \tfrac{8}{3}x = 2.67x$$

In general we have

\begin{align}
E(b, r, x) &= \max \left(\max_w E(b,r,x)_{\text{red}}, \;\max_w E(b,r,x)_{\text{black}} \right)
\end{align}
where
\begin{align}
 E(b,r,x)_{\text{red}} &= \tfrac{r}{b+r}E(b, r-1, x+w) + \tfrac{b}{b+r}E(b-1, r, x-w), \\
 E(b,r,x)_{\text{black}} &= \tfrac{r}{b+r}E(b, r-1, x-w) + \tfrac{b}{b+r}E(b-1, r, x+w).
\end{align}
and boundary conditions
\begin{align}
E(0,0) &= x \\
E(r,b) &= 0 \text{ if } r<0 \text{ or } b<0
\end{align}

from scipy.optimize import minimize

def pr_black(b, r):
    return b / (b + r)

def pr_red(b, r):
    return r / (b + r)

def E_bet_black(w, x, b, r):
    w = w[0]
    return pr_black(b,r) * E(b-1, r, x + w) + pr_red(b,r) * E(b, r-1, x - w)

def E_bet_red(w, x, b, r):
    w = w[0]
    return pr_black(b,r) * E(b-1, r, x - w) + pr_red(b,r) * E(b, r-1, x + w)

cache = {}
def E(b, r, x):
    if (b, r) in cache:
        return cache[(b,r)]*x # Assume linear value mapping.
    elif r==0 and b==0:
        return x
    elif r<0 or b<0:
        return 0.0
    else:
        bnds = [(0, x)]
        res_black = minimize(lambda w,x,b,r: -E_bet_black(w,x,r,b), x0=x/2, args=(x,b,r), bounds=bnds)
        res_red = minimize(lambda w,x,b,r: -E_bet_red(w,x,r,b), x0=x/2, args=(x,b,r), bounds=bnds)
        val_black, w_black = -res_black.fun, res_black.x
        val_red, w_red = -res_black.fun, res_black.x
        color,w,value = ('black', w_black, val_black) if val_black > val_red else ('red', w_red, val_red)
        cache[(b, r)] = value/x
        return value

b, r = (2, 2)
x = 100

print(    'E(b,r,x) = E({},{},{}) = {}'.format(b,r,x, round(E(b, r, x), 2))    )

E(b,r,x) = E(2,2,100) = 266.67


答:
If $x^2 + 5x+12 = 0$ then
$$(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (-8)(-6)-48 = 0.$$

If $x^2 + 5x-2 = 0$ then
$$(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = 6\cdot 8-48 = 0.$$

Therefore LHS and RHS have the same zeroes. Since RHS has no double roots and both sides are polynomials of degree $4$, we conclude LHS = RHS.


问:
Tangent plane to a surface parallel to a plane
I can't solve the following question :
Write the coefficient D of tangent plane $ x+By+Cz=D $ to the surface defined by equation $ 2x^2+3y^2+z^2=13/6 $
parallel to the plane $ x+y+z=1 $. Assume that the intersection of the plain and the surface happens in the positive octant.

So I calculate the gradient of the surface's equation, let's write  $ \nabla (f) = 4x +6y + 2z $.

It needs to be a multiple of plane equation $ x+y+z=1 $ so we get the following equation system:

$ 4x= \lambda ; 6y = \lambda  ; $ and $2z= \lambda $

I solve this equation and I get $(\frac{\lambda}{4};\frac{\lambda}{6};\frac{\lambda}{2}) $ but then what I am supposed to do ?

I read that we need to plug it into the equation but why ?

thanks for any help.

as helper told me, answer is 1.99

答:
HINT: As the point is on your curve,

$2(\frac{\lambda}{4})^2 +3 (\frac{\lambda}{6})^2 + (\frac{\lambda}{2})^2 = \frac{13}{6}$

Find $\lambda$.

The equation of the tangent plane will be given by

$(x - \frac{\lambda}{4}) + (y - \frac{\lambda}{6}) + (z - \frac{\lambda}{2}) = 0$

You can resolve it to find the value of $D$.


问:
Using $p$-adic representation of a number in $\mathbb{Z}^{d}$
Let $L,K >1$ and $d \ge 1$ be all integers and consider:
$$\Lambda_{K} = \{0,1,2,...,L^{K}-1\}^{d} \subset \mathbb{Z}^{d}$$
This is an hypercube with cardinality $|\Lambda_{K}| = L^{dK}$. My professor said that any $x \in \Lambda_{k}$ can be written as:
$$x = \sum_{k=1}^{K}\alpha_{k}L^{k-1}$$
where $\alpha_{k} \in \{0,...,L-1\}^{d}$. I don't understand this representation. How can one construct such a representation and what does it mean?

答:
An element of $\Lambda_K$ is a $d$-tuple $(u_1,\dots,u_d)$ of nonnegative integers $<L^K$.

The $i$th coordinate of $\alpha_k$ will be the $k$th digit (from the right) of $u_i$ when represented in number system of base $L$.

These can be determined by using division and remainders, just as in the familiar case when $L=10$.