问:
Boolean algebra's filter as a partially order's filter
This is a basic question I'm trying to figure out: why the Boolean's filter definition corresponds to the order-theoretic definition of filter ?

Here follows the relevant definitions.

Definition 1 (Filter in a partially ordered-set)
A nonempty subset $F$ of a partially ordered-set $(P, \le)$ is a filter on $P$ if the
following conditions hold:

$F$ is downward directed: for every $x, y \in F, \exists z \in F$
s.t. $z \le x$ and $z \le y$.

$F$ is upward-closed: for every $x \in F$ and $y \in P$, if $x \le y$ then $y \in F$.

Definition 2 (Filter in a Boolean algebra)
A nonempty subset $F$ of a Boolean algebra $(\mathbb{B}, \vee, \wedge, \bot, \top, \neg)$ is a
filter if it is

downward directed: $\forall x, y \in F, x \wedge y \in F$;

upward-closed: $\forall x \in F, y \in \mathbb{B}, x \vee y \in F$.

Boolean algebra seen as a lattice can be presented with a relation order $\le$ with the following equivalence:
$$x \le y \iff x \vee y = y \iff x \wedge y = x.$$

My question is, under a Boolean algebra, why the two definitions of filter aforementioned are equivalent ?

It's pretty clear to me that the property of upward closure is equivalent as $x \vee y = \sup\{x, y\}$. But I have doubt about the first property (downward directed), especially the direction (Definition 1 $\implies$ Definition 2). For the reverse direction, we can take $z = x \wedge y$.

Draft:

$z \le x \textrm{ and } z \le y \iff z = z \wedge x \textrm{ and } z = z \wedge y$, and by substitution we have $z = z \land (x \land y)$. How can I prove that $z = x \land y$ and thus by hypothesis $x \land y \in F$ ?

答:
Say we're working with a definition 1 filter.

Let's let $x,y \in F$. We want to show $x \land y \in F$.

By definition 1, we know there's some $z \in F$ with $z \leq x$ and $z \leq y$.
Then by the definition of $\land$, we must have $z \leq x \land y$ too.
Lastly, by upwards closure, we see $x \land y \in F$.

I hope this helps ^_^


问:
Finding density function of max(X,Y), min(X,Y) where X and Y are independent exponential random vars with parameters $\lambda, \mu$
Let X and Y be independent exponential random variables with parameters λ > 0 and µ > 0, respectively. Let $U = max(X,Y)$ and $V = min(X,Y)$. Find the density function $f_{(U,V)}(u,v)$.

I've done this problem but I'm not confident in my answer.

I got $f_{(U,V)}(u,v) = λe^{(-λv- \mu u)}\mu + \mu e^{(-\mu v - λu)}λ - λe^{(-λv - \mu u)}\mu$

Could someone verify my answer or walk me through how they would solve it. I would really appreciate it! Thanks!

答:
I would proceed by a limit argument. Let $a < b$ and let $h$ be such that $a + h < b$. Then
\begin{align}
P(V \in (a, a + h), U \in (b, b + h)) &= P(X \in (a, a + h), Y \in (b, b + h)) + P(Y \in (a, a + h), X \in (b, b + h)) \\
&= P(X \in (a, a + h))P(Y \in (b, b + h)) + P(Y \in (a, a + h))P(X \in (b, b + h))
\end{align}
Thus by dividing both sides by $h^2$ and letting $h \to 0^{+}$ we get
\begin{align}
f_{V, U}(a, b) = f_{X}(a)f_{Y}(b) + f_{Y}(a)f_{X}(b).
\end{align}
This argument assumed $f_{V, U}$ would be continuous at $a, b$, but you can arrive at the same result without this assumption by computing $F_{V, U}(a, b)$ and then using
$$f_{V, U}(a, b) = \frac{\partial}{\partial a \partial b}F_{V, U}(a, b).$$

ContentLicense: CC BY-SA 4.0


问:
Using iterated integrals compute the area of the set $D \subset \mathbb{R}^2$, when $D$ is bounded by $y=\frac12 x^2$ and $y=2x.$

Using iterated integrals compute the area of the set $D \subset \mathbb{R}^2$, when $D$ is bounded by $y=\frac12 x^2$ and $y=2x.$

So I need to compute $\iint_D 1 \ dy \ dx$ right?

It's easy to see that the intersection of the parabola and the line is at $x=4$ and $y=8$.

Thus should I integrate $$\iint_D 1 \ dy \ dx = \int_{0}^{4}\int_{0}^{2x}1 \ dy \ dx = 16?$$

I'm quite confused by the term iterated integral? I assume they mean just a double integral?

答:
You need to find area bound between two given curves so your lower bound should be $y = \frac{1}{2}x^2$. The way you have set up, you will get the area of the triangle bound by line $\,y = 2x$ and $x$-axis for $0 \leq x \leq 4$.

$\displaystyle \iint_D 1 \ dy \ dx = \int_{0}^{4}\int_{x^2/2}^{2x}1 \ dy \ dx$

Here is a sketch


问:
Logarithmic equation. Need to know if i am teaching right
Two of my friends is studying for a test. They asked me about a simple question. But they told me that i was wrong on a question. I could be wrong. But i need you guys to make sure that they learn the right stuff. So if i was right. I then can tell them how to do the equations

The question:
Assume that $\log x = 3$ and $\log y = 4$
Calculate the following equation $\log x^4 + 2\log y - \log(xy)$

I got it to be

$$\log \left( \frac{x^4  y^2}{x y} \right)$$

Then they just change the $x$ and $y$ to the assumed value.
So am i right or am i wrong

答: You're right. For logarithms in general it holds that $log(a)+log(b)=log(ab)$ and $log(a)-log(b)=log(a/b)$. Furthermore it holds that $alog(b)=log(b^a)$. Hence, your answer is correct.</p>

ContentLicense: CC BY-SA 3.0


问:
What are ways to compute polynomials that converge from above and below to a continuous and bounded function on $[0,1]$?

Main Question

Suppose $f:[0,1]\to [0,1]$ is continuous and belongs to a large class of functions (for example, the $k$-th derivative, $k\ge 0$, is continuous, Lipschitz continuous, concave, strictly increasing, bounded variation, and/or in the Zygmund class, or $f$ is real analytic).

Then, compute the Bernstein coefficients of a sequence of polynomials ($g_n$) of degree 2, 4, 8, ..., $2^i$, ... that converge to $f$ from below and satisfy: $(g_{2n}-g_{n})$ is a polynomial with non-negative Bernstein coefficients once it's rewritten to a polynomial in Bernstein form of degree exactly $2n$. Assume $0\lt f(\lambda)\lt 1$ or $f$ is polynomially bounded.

The convergence rate must be $O(1/n^{r/2})$ if the class has only functions with Lipschitz-continuous $(r-1)$-th derivative.  The method may not introduce transcendental or trigonometric functions (as with Chebyshev interpolants).

See "Strategies", below, for different ways to answer this question.

In this question, a polynomial $P(x)$ is written in Bernstein form of degree $n$ if it is written as— $$P(x)=\sum_{k=0}^n a_k {n \choose k} x^k (1-x)^{n-k},$$ where $a_0, ..., a_n$ are the polynomial's Bernstein coefficients.

The degree-$n$ Bernstein polynomial of an arbitrary function $f(x)$ has Bernstein coefficients $a_k = f(k/n)$.  In general, this Bernstein polynomial differs from $f$ even if $f$ is a polynomial.

Background

I asked this question in order to solve the so-called Bernoulli factory problem, described next. We're given a coin that shows heads with an unknown probability, $\lambda$. The goal is to use that coin (and possibly also a fair coin) to build a "new" coin that shows heads with a probability that depends on $\lambda$, call it $f(\lambda)$. This is the Bernoulli factory problem, and it can be solved only if $f$ is continuous (Keane and O'Brien 1994).

However, since I asked this question I have found a Bernoulli factory algorithm that I believe is general enough to cover all the cases that this question would help solve.

Since this question may be of broader interest, though, I leave this question open.  See also my other open questions about the Bernoulli factory problem.

Polynomials that approach a factory function

An algorithm simulates a factory function $f(\lambda)$ via two sequences of polynomials that converge from above and below to that function.  To use the algorithm, however, the polynomial sequences must meet certain requirements; among them, the sequences must be of Bernstein-form polynomials that converge from above and below to a factory function.  Specifically:

For $f(\lambda)$ there must be a sequence of polynomials ($g_n$) in Bernstein form of degree 1, 2, 3, ... that converge to $f$ from below and satisfy: $(g_{n+1}-g_{n})$ is a polynomial with non-negative Bernstein coefficients once it's rewritten to a polynomial in Bernstein form of degree exactly $n+1$ (see end notes; Nacu and Peres 2005; Holtz et al. 2011).  For $f(\lambda)=1-f(\lambda)$ there must likewise be a sequence of this kind.

A Matter of Efficiency

However, ordinary Bernstein polynomials converge to a function at the rate $\Omega(1/n)$ in general, a result known since Voronovskaya (1932) and a rate that will lead to an infinite expected number of coin flips in general.  (See also my supplemental notes.)

But Lorentz (1966) showed that if the function is positive and has a continuous $k$-th derivative, there are polynomials with nonnegative Bernstein coefficients that converge at the rate $O(1/n^{k/2})$ (and thus can enable a finite expected number of coin flips if the function is "smooth" enough).

Thus, people have developed alternatives, including linear combinations and iterated Boolean sums of Bernstein polynomials, to improve the convergence rate. These include Micchelli (1973), Guan (2009), Güntürk and Li (2021a, 2021b), the "Lorentz operator" in Holtz et al. (2011), Draganov (2014), and Tachev (2022).

These alternative polynomials usually include results where the error bound is the desired $O(1/n^{k/2})$, but nearly all those results (e.g., Theorem 4.4 in Micchelli; Theorem 5 in Güntürk and Li) have hidden constants with no upper bounds given, making them unimplementable (that is, it can't be known beforehand whether a given polynomial will come close to the target function within a user-specified error tolerance). (See end notes.)

A Conjecture on Polynomial Approximation

The following is a conjecture that could help reduce this problem to the problem of finding explicit error bounds when approximating a function by polynomials.

Let $f(\lambda):[0,1]\to(0,1)$ have $r\ge 1$ continuous derivatives, let $M$ be the maximum of the absolute value of $f$ and its derivatives up to the $r$-th derivative, and denote the Bernstein polynomial of degree $n$ of a function $g$ as $B_n(g)$. Let $W_{2^0}(\lambda), W_{2^1}(\lambda), ..., W_{2^i}(\lambda),...$ be a sequence of functions on [0, 1] that converge uniformly to $f$.

For each integer $n\ge 1$ that's a power of 2, suppose that there is $D>0$ such that— $$|f(\lambda)-B_n(W_n(\lambda))| \le DM/n^{r/2},$$ whenever $0\le \lambda\le 1$.  Then there is $C_0\ge D$ such that for every $C\ge C_0$, the polynomials $(g_n)$ in Bernstein form of degree 2, 4, 8, ..., $2^i$, ..., defined as $g_n=B_n(W_n(\lambda) - CM/n^{r/2})$, converge from below to $f$ and satisfy: $(g_{2n}-g_{n})$ is a polynomial with non-negative Bernstein coefficients once it's rewritten to a polynomial in Bernstein form of degree exactly $2n$. (See end notes.)

Equivalently (see also Nacu and Peres 2005), there is $C_1>0$ such that the inequality $(PB)$ (see below) holds true for each integer $n\ge 1$ that's a power of 2 (see "Strategies", below).

My goal is to see not just whether this conjecture is true, but also which value of $C_0$ (or $C_1$) suffices for the conjecture, especially for any combination of the special cases mentioned at the end of "Main Question", above.

Strategies

The following are some strategies for answering these questions:

For iterated Boolean sums of Bernstein polynomials ($U_{n,k}$ in Micchelli 1973; see also Güntürk and Li), find an explicit bound, with no hidden constants, on the approximation error for functions with continuous $r$-th derivative, or verify my proof of those error bounds.

For linear combinations of Bernstein polynomials (Butzer 1953, Tachev 2022), verify my proof of those error bounds in my Proposition B10.

For the "Lorentz operator", find an explicit bound, with no hidden constants, on the approximation error for the operator $Q_{n,r}$ and for the polynomials $(f_n)$ and $(g_n)$ formed with it, and find the hidden constants $\theta_\alpha$, $s$, and $D$ as well as those in Lemmas 15, 17 to 22, and 24 in the paper.  Or verify my proof of the order-2 operator's error bounds in my Proposition B10A.

Find other polynomial operators meeting the requirements of the main question (see "Main Question", above) and having explicit error bounds, with no hidden constants, especially operators that preserve polynomials of a higher degree than linear functions.

Find a sequence of functions $(W_n(f))$ and an explicit and tight upper bound on $C_1>0$ such that, for each integer $n\ge 1$ that's a power of 2— $$\left|\left(\sum_{i=0}^k W_n\left(\frac{i}{n}\right)\sigma_{n,k,i}\right)-W_{2n}\left(\frac{k}{2n}\right)\right|=|\mathbb{E}[W_n(X_k/n)] - W_{2n}(\mathbb{E}[X_k/n])|\le \frac{C_1 M}{n^{r/2}},\tag{PB}$$ whenever $0\le k\le 2n$, where $M = \max(L, \max|f^{(0)}|, ...,\max|f^{(r-1)}|)$, $L$ is $\max|f^{(r)}|$ or the Lipschitz constant of $f^{(r-1)}$, $X_k$ is a hypergeometric($2n$, $k$, $n$) random variable, and $\sigma_{n,k,i} = {n\choose i}{n\choose {k-i}}/{2n \choose k}=\mathbb{P}(X_k=i)$ is the probability that $X_k$ equals $i$. (See end notes as well as "Proofs for Polynomial-Building Schemes.)

References

Łatuszyński, K., Kosmidis, I., Papaspiliopoulos, O., Roberts, G.O., "Simulating events of unknown probabilities via reverse time martingales", arXiv:0907.4018v2 [stat.CO], 2009/2011.

Keane, M. S., and O'Brien, G. L., "A Bernoulli factory", ACM Transactions on Modeling and Computer Simulation 4(2), 1994.

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

Nacu, Şerban, and Yuval Peres. "Fast simulation of new coins from old", The Annals of Applied Probability 15, no. 1A (2005): 93-115.

Micchelli, C. (1973). The saturation class and iterates of the Bernstein polynomials. Journal of Approximation Theory, 8(1), 1-18.

Guan, Zhong. "Iterated Bernstein polynomial approximations." arXiv preprint arXiv:0909.0684 (2009).

Güntürk, C. Sinan, and Weilin Li. "Approximation with one-bit polynomials in Bernstein form" arXiv preprint arXiv:2112.09183 (2021).

C.S. Güntürk, W. Li, "Approximation of functions with one-bit neural networks", arXiv:2112.09181 [cs.LG], 2021.

Draganov, Borislav R. "On simultaneous approximation by iterated Boolean sums of Bernstein operators." Results in Mathematics 66, no. 1 (2014): 21-41.

Farouki, R.T., and Rajan, V.T., "Algorithms for polynomials in Bernstein form", Computer Aided Geometric Design 5(1), 1988.

Tachev, Gancho. "Linear combinations of two Bernstein polynomials", Mathematical Foundations of Computing, 2022.

Lee, Sang Kyu, Jae Ho Chang, and Hyoung-Moon Kim. "Further sharpening of Jensen's inequality." Statistics 55, no. 5 (2021): 1154-1168.

Butzer, P.L., "Linear combinations of Bernstein polynomials", Canadian Journal of Mathematics 15 (1953).

Note 5: This condition is also known as a "consistency requirement"; it ensures that not only the polynomials "increase" to $f(\lambda)$, but also their Bernstein coefficients do as well.  This condition is equivalent in practice to the following statement (Nacu & Peres 2005). For every integer $n\ge 1$ that's a power of 2, $a(2n, k)\ge\mathbb{E}[a(n, X_{n,k})]= \left(\sum_{i=0}^k a(n,i) {n\choose i}{n\choose {k-i}}/{2n\choose k}\right)$, where $a(n,k)$ is the degree-$n$ polynomial's $k$-th Bernstein coefficient, where $0\le k\le 2n$ is an integer, and where $X_{n,k}$ is a hypergeometric($2n$, $k$, $n$) random variable.  A hypergeometric($2n$, $k$, $n$) random variable is the number of "good" balls out of $n$ balls taken uniformly at random, all at once, from a bag containing $2n$ balls, $k$ of which are "good".  See also my MathOverflow question on finding bounds for hypergeometric variables.

Note 6: If $W_n(0)=f(0)$ and $W_n(1)=f(1)$ for every $n$, then the inequality $(PB)$ is automatically true when $k=0$ and $k=2n$, so that the statement has to be checked only for $0\lt k\lt 2n$.  If, in addition, $W_n$ is symmetric about 1/2, so that $W_n(\lambda)=W_n(1-\lambda)$ whenever $0\le \lambda\le 1$, then the statement has to be checked only for $0\lt k\le n$ (since the values $\sigma_{n,k,i} = {n\choose i}{n\choose {k-i}}/{2n \choose k}$ are symmetric in that they satisfy $\sigma_{n,k,i}=\sigma_{n,k,k-i}$).
This question is a problem of finding the Jensen gap of $W_n$ for certain kinds of hypergeometric random variables (see Note 5).  Lee et al. (2021) deal with a problem very similar to this one and find results that take advantage of $f$'s (here, $W_n$'s) smoothness, but unfortunately assume the variable is supported on an open interval, rather than a closed one (namely $[0,1]$) as in this question.

答:
While the following does not fully answer my question, I make the following notes.

Polynomial Schemes

This section relates to finding polynomials for certain functions.

But since they involve the ordinary Bernstein polynomials, they converge at a rate no faster than $O(1/n)$ in general, rather than the desired $O(1/n^{k/2})$ for functions with $k$ continuous derivatives. Other users are encouraged to add other answers to my question.

The following inequality from Nacu and Peres 2005 is useful: $$|\mathbb{E}[f(X/n)]-f(k/(2n))| \le \kappa_f(n), \tag{1}$$ where—

$\kappa_f(n)$ is a function that depends only on $f$ and $n$ and makes the inequality hold true for all $f$ belonging to a given class of functions (such as Lipschitz continuous or twice-differentiable functions), and

$X$ is a hypergeometric($2n$, $k$, $n$) random variable.

Notably, if a function $f$ is such that (1) holds true, then in general, for all $n$ that are powers of 2— $$a(k, n) = f(k/n)-\delta(n) \text{ and } b(k,n) = f(k/n)+\delta(n),$$ where $n$ is the polynomial's degree and $\delta(n)$ is a solution to the following functional equation: $$\delta(n) = \delta(n*2) + \kappa_f(n), $$or equivalently, the linear recurrence $\delta(n) = \delta(n+1) + \kappa_f(2^n)$.

For example:

(Nacu and Peres 2005, Proposition 10) If $f$ is Lipschitz continuous with Lipschitz constant $C$ —

$\kappa_f(n) = C/\sqrt{2*n}$ , so

$\delta(n) = (1+\sqrt{2})C/\sqrt{n}$ , and

(Nacu and Peres 2005, Proposition 10) If $f$ is twice differentiable and $M$ is not less than the highest value of $|f′′(x)|$ for any $x$ in [0, 1]—

$\kappa_f(n) = M/(4*n)$, so

$\delta(n) = M/(2*n)$

Due to Jensen's inequality (see, for example, Nacu and Peres 2005, Lemma 8):

If $f$ is concave in $[0, 1]$, then $a(k,n) = f(k/n)$.

If $f$ is convex in $[0, 1]$, then $b(k, n) = f(k/n)$.

I have also found results to derive polynomial-building schemes for Hölder continuous functions and other functions with an arbitrary modulus of continuity.  For example, if $f$ is $\alpha$-Hölder continuous with Hölder constant $M$—

$\kappa_f(n) = M(1/(7n))^{\alpha/2}$, so

$\delta(n) = \frac{M(2/7)^{\alpha/2}}{(2^{\alpha/2}-1)n^{\alpha/2}}$.

Other schemes of this kind are not mentioned here because they can be impractical when $f$ is not Lipschitz continuous.  Indeed, Nacu and Peres showed that a finite number of coin flips is possible only if $f$ is Lipschitz continuous.

For proofs and additional notes on polynomial-building schemes, see my supplemental notes.

Other Polynomials

Now for other polynomials in Bernstein form, not necessarily Bernstein polynomials.

It can be seen that the following inequality is useful, based on (10) and (11) of Nacu & Peres.  The goal is to find a simple function $\kappa_f(n)$ such that— $$\max_{0\le k\le 2n}|\mathbb{E}[W_n(X/n)]-W_{2n}(k/(2n))| \le \kappa_f(n),$$ or equivalently— $$\max_{0\le k\le 2n}\left|\left(\sum_{i=0}^k \left(W_n\left(\frac{i}{n}\right)\right) {n\choose i}{n\choose {k-i}}/{2n \choose k}\right)-W_{2n}\left(\frac{k}{2n}\right)\right|\le \kappa_f(n),$$ for every integer $n\ge 1$ that's a power of 2, where—

$W_n$ is a function whose Bernstein polynomial of degree $n$ approximates $f$ with the desired error bound $O(1/n^{k/2})$ (note that $W_n$ need not equal $f$),

$\kappa_f(n)$ is a function that depends only on $f$ and $n$ and makes the inequality hold true for all $f$ belonging to a given class of functions, and

$X$ is a hypergeometric($2n$, $k$, $n$) random variable.

Then, for certain choices of $\kappa_f(n)$, condition 3 will be met if the following series converges: $$\sum_{m\ge\log_2(n)}\kappa_f(2^m).$$  See Theorem 1 of "Proofs for Polynomial Building Schemes" for details.

As an example of $W_n$, there are error bounds for iterated Bernstein polynomials, which can be seen as a Bernstein polynomial of another function $W_n$ that approximates $f$ appropriately (see Güntürk and Li 2021, sec. 3.3).  Unfortunately, those error bounds can't be used as is to bound $\kappa_f(n)$.

References

Nacu, Şerban, and Yuval Peres. "Fast simulation of new coins from old", The Annals of Applied Probability 15, no. 1A (2005): 93-115.

Flajolet, P., Pelletier, M., Soria, M., "On Buffon machines and numbers", arXiv:0906.5560 [math.PR], 2010.

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

C.S. Güntürk, W. Li, "Approximation of functions with one-bit neural networks", arXiv:2112.09181 [cs.LG], 2021.


问:
A convergent series for the Trigamma function $\psi_1(n) =\sum_{k=n}^{\infty} \frac1{k^2} $
I just came up with
the following
convergent series
for the Trigamma function
defined by

$\psi_1(n)
=\sum_{k=n}^{\infty} \frac1{k^2}
$.

\begin{align*}
\psi_1(n)
&=\lim_{m \to \infty} \sum_{j=1}^m \frac{(j-1)!}{j\prod_{i=0}^{j-1}(n+i)}\\
&=\frac1{n}+\frac{1}{2n(n+1)}+\frac{2}{3n(n+1)(n+2)}+\ldots\\
\end{align*}

This contrasts with
the usual asymptotic series
for $\psi_1(n)$
which is asymptotic,
does not converge,
and involves the
Bernoulli numbers.

I'm sure this isn't new,
but I could not find it here.

So,
my questions are,
as is often the case,

(1) Is this new?

(2) Is there a reasonably simple proof?
(Mine is moderately messy.)

I'll post my proof
in a few days
if anyone is interested.

Thanks

Body:
Observe that $\prod_{i=0}^{j-1}(n+i)=\Gamma(n+j)/\Gamma(n)$, so that, as claimed,
\begin{align*}
\sum_{j=1}^\infty\frac{(j-1)!}{j\prod_{i=0}^{j-1}(n+i)}
&=\sum_{j=1}^\infty\frac1j\mathrm{B}(n,j)\\
&=\sum_{j=1}^\infty\frac1j\int_0^1 t^{n-1}(1-t)^{j-1}\,dt\\
&=\int_0^1 t^{n-1}\frac{-\log t}{1-t}\,dt\\
&=\int_0^1 t^{n-1}(-\log t)\sum_{j=0}^\infty t^j\,dt\\
&=\sum_{k=n}^\infty\int_0^1 t^{k-1}(-\log t)\,dt=\sum_{k=n}^\infty\frac1{k^2}.
\end{align*}


问:
Friedberg Linear Algebra problems
In Friedberg's Linear Algebra 4th edition, I have trouble understanding 2 related questions:

Section 2.1 problem 34:
"Let $V$ and $W$ be vector spaces over a common field, and let $\beta$ be a basis for $V$. Then for any function $f: \beta \to W$ there exists exactly one linear transformation $T: V \to W$ such that $T(x) = f(x)$ for all $x \in \beta$.

I don't understand what the restriction of $T$ to the basis space $\beta$ does at all, and how does one prove that there is "exactly one" transformation? Directly related to this,

Section 2.7 problem 18: "Let $V$ be a nonzero vector space over a field $F$ and let $S$ be a basis for $V$. Let $\Phi: V^* \to \mathcal{L}(S,F)$ be the mapping defined by $\Phi(f)=f_S$, the restriction of $f$ to $S$. Prove that $\Phi$ is an isomorphism. Hint: Apply Exercise 34 of Section 2.1. (the above problem)

I don't see how the two problems are related except for the fact that there is a linear transformation and there is a linear transformation restricted to a subset containing only basis vectors.

答:
For the first exercise, this is simply the fact that a linear transformation is completely determined by its effect on a basis.  One proves that there is exactly one by using linearity.

The second problem is a direct application of that fact.  It basically says that dual space is isomorphic to the space of linear functions on $V$ into $\mathcal F$.  These are also known as linear functionals.   Just let $W=\mathcal F$, the one dimensional vector space over $\mathcal F$, in the first part.

It might have been better if Freidberg had written $\mathcal L(\color{blue}{V},\mathcal F)$, in the second question, I think.  Since a basis is not a vector space.

Title: Friedberg Linear Algebra problems


问:
General form of parabola using equation of Tangent at Vertex and Axis
I had read in my notes that the equation of parabola can be given by

(Equation of axis)$^2$ = (Length of Latus rectum)*(Equation of tangent at vertex)

(I don't know the systematic proof. Is there something I am missing in the equation?)

Now take look at this very basic equation of a parabola

$ y^2=4ax $

Here the equation of axis of parabola is $(y=0)$ and that of tangent at vertex is $(x=0)$

I can also write the equation of axis as $(ny=0)$ and tangent at vertex as $(mx=0)$

(where m and n are constants)

And hence using the first equation I can write the equation of parabola as

$(ny)^2 = 4a(mx)$

which gives me a completely different parabola.

I don't know where I have gone wrong. Please guide me.

答:
Here, it is not equation of axis, it is perpendicular distance of any random point on parabola say (h,k) from axis and also perpendicular distance from vertex and not equation of vertex
So for general equation
Perpendicular distance from axis of point (h,k) is k and from vertex, it's h
So equation will be:

k²=4ah

Replacing h with x and k with y, we get equation of parabola
y²=4ax


问:
Determine the final stage of a game
Peter and Chris are playing a game. At the beginning, each of them has some balls, where each ball is red, blue, or green. There is also a ball pool which contains more than enough balls of each colour for Peter and Chris to draw.

Peter and Chris take turns to move, and Peter moves first. After each turn, if Peter has exactly one ball and Chris has none, the game will end immediately.

Each move consists of the following steps:

If the player has less than two balls, he draws red balls from the pool until he has two.

The player chooses two of his balls and throws them into the pool.

If the two chosen balls are of the same colour, the opponent player gets from the pool one ball of the same colour. Otherwise the opponent player gets from the pool one ball of colour different from the colours of both chosen balls.

For example, if Peter chooses two red balls, Chris will get a red ball from the pool. If Peter chooses a red ball and a blue ball instead, Chris will get a green ball from the pool.

Determine the colour of Peter's final ball, given $R_1, G_1, B_1, R_2, G_2, B_2$.

$R_1, G_1, B_1$ represent the number of red, green and blue balls, respectively, that Peter has at the beginning of the game.
$R_2, G_2, B_2$ represent the number of red, green and blue balls, respectively, that Chris has at the beginning of the game.

The answer claims that if $$(G_1 + B_2 - B_1 - G_2) \equiv 0 (\mod 3)$$, then the final ball is in red, if the same expression $\equiv 1 (\mod 3)$ then the final ball is green. Otherwise, it will be in blue.

I cannot wrap my head around it for a while, can anyone explain this to me please? Thanks in advance.

答:
Let $g_1(n)$ be the number of green balls that Peter has after $n$ moves have been played, $g_2(n)$ the number of green balls that Chris has after $n$ moves have been played, $b_1(n)$ the number of blue balls that Peter has after $n$ moves have been played, and $b_2(n)$ the number of blue balls that Chris has after $n$ moves have been played. By examining the six possibilities you can show that

$$g_1(n+1)+b_2(n+1)-b_1(n+1)-g_2(n+1)\\
=g_1(n)+b_2(n)-b_1(n)-g_2(n)\tag{1}$$

no matter what move is made. For example, suppose that $n$ is even, so that it is Peter’s turn to move, and Peter chooses a blue and a red ball. Chris gets a green ball, so

$$b_1(n+1)=b_1(n)-1$$

and

$$g_2(n+1)=g_2(n)+1\,,$$

and $(1)$ holds. This implies that

$$g_1(n)+b_2(n)-b_1(n)-g_2(n)\\
=g_1(0)+b_2(0)-b_1(0)-g_2(0)\tag{2}$$

Say that the game ends after $m$ moves. At that point Chris has no balls, so $b_2(m)=g_2(m)=0$, and $(2)$ tells us that

$$g_1(m)-b_1(m)=g_1(0)+b_2(0)-b_1(0)-g_2(0)\,.$$

If you now use the fact that Peter has exactly one ball at this point, you should be able to finish the argument from here.

Title: Determine the final stage of a game


问:
What is the power set of the set {{∅}}?
The power set, as I was informed, includes the empty set and the set being discussed itself.

I am quite unsure with the answer, specifically with the braces.

It could be {∅, {∅}}

or it could be {∅, {{∅}}}

If someone could help me clear up my confusion, I would appreciate it.

答:
$
\newcommand{\set}[1]{\left\{ #1 \right\}}
\newcommand{\PP}{\mathcal{P}}
$
You might find it simpler to define $A := \set \varnothing$. This makes it perfectly clear you're dealing with a singleton (namely, one whose sole element is a set containing an empty set), and helps remove confusion about braces.

Then you want to find $\PP(\set A)$. Obviously, this is given by

$$\PP(\set A) = \big\{ \varnothing, \{A\} \big\}$$

After all, $\set A$ has cardinality one, and these are the only subsets of it. But then, recalling what $A$ is,

$$\PP \left(\big\{\{\varnothing\} \big\} \right) = \Big\{ \varnothing, \big\{ \{ \varnothing \} \big\} \Big\}$$

Title: What is the power set of the set {{∅}}?


答:
But they aren’t given the values of $x$ and $y$: they’re given the values of $\lg x$ and $\lg y$. Specifically, $\lg x=3$ and $\lg y=4$, so

$$\lg x^4+2\lg y-\lg(xy)=4\lg x+2\lg y-(\lg x+\lg y)=12+8-7=13\;.$$

It’s perfectly true that

$$\lg x^4+2\lg y-\lg(xy)=\lg\frac{x^4y^2}{xy}\;,$$

but this does not really help them to solve the problem.

Title: Logarithmic equation. Need to know if i am teaching right

Body:
Two of my friends is studying for a test. They asked me about a simple question. But they told me that i was wrong on a question. I could be wrong. But i need you guys to make sure that they learn the right stuff. So if i was right. I then can tell them how to do the equations

The question:
Assume that $\log x = 3$ and $\log y = 4$
Calculate the following equation $\log x^4 + 2\log y - \log(xy)$

I got it to be

$$\log \left( \frac{x^4  y^2}{x y} \right)$$

Then they just change the $x$ and $y$ to the assumed value.
So am i right or am i wrong


问:
if $x+y+z=0$ then $(\zeta(x) +\zeta(y) + \zeta(z))^2= \wp(x) + \wp(y) +\wp(z)$?
I found that if $x+y+z=0$, one can prove that $(\zeta(x) +\zeta(y) + \zeta(z))^2= \wp(x) + \wp(y) +\wp(z)$ using Liouville's theorem (here $\wp$ is the Weierstrass elliptic function)

How one can do this?

答:
$L=w_1\Bbb{Z}+w_2\Bbb{Z}$.

$\wp_L$ is the unique $L$-periodic meromorphic function whose poles are double at $L$ and $\wp_L(z)=z^{-2}+O(z)$.

$\zeta_L'=-\wp_L$ and $\zeta_L(z)=z^{-1}+O(z)$. It is the unique meromorphic function whose poles are simple at $L$, $\zeta_L(z)=z^{-1}+O(z)$ and $\zeta_L(z+w_j)=\zeta_L(z)+c_j$.

$$f_{L,y}(z)=(\zeta_L(-z-y) +\zeta_L(y) + \zeta_L(z))^2$$ is meromorphic doubly periodic with double poles at $L$ and $L-y$ where $f_{L,y}(z)=z^{-2}+O(1),f_{L,y}(z-y)=z^{-2}+O(1)$, thus $$g_{L,y}(z)=f_{L,y}(z)-\wp_L(z)-\wp_L(y-z)$$ is entire and $L$-periodic, it is constant.

Since $\zeta_L'=-\wp_L'$ and $\wp_L$ is even then $\zeta_L(z)=z^{-1}+O(z^2)$ and $$g_{L,y}(z)= (\wp_L(y) z+z^{-1}+O(z^2))^2-z^{-2}-\wp_L(-y)+O(z)=\wp_L(y)+O(z)$$
and hence $$g_{L,y}(z)=\wp_L(y)$$


问:
Find $\lim_{\alpha \to \infty} \int^{\pi/4}_{0} \alpha e^{x} (\tan x)^{\alpha} dx$
Can you help me show that
$\lim_{\alpha \to \infty} \int^{\pi/4}_{0} \alpha e^{x} (\tan x)^{\alpha} dx=\frac{e^{\pi/4}}{2}$
?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
This is a Laplace Method example:

The main contribution to the integral comes from $\ds{x \lesssim \pi/4}$.

Under the change $\ds{x \mapsto \pi/4 - x}$, the main contribution to the integral comes from $\ds{x \gtrsim 0}$.

Namely,
\begin{align}
&\bbox[5px,#ffd]{\lim_{\alpha \to \infty}\int_{0}^{\pi/4}
\alpha\expo{x}\tan^{a}\pars{x}\,\dd x} =
\lim_{\alpha \to \infty}\bracks{a\int_{0}^{\pi/4}
\expo{\pi/4 - x}\tan^{a}\pars{{\pi \over 4} - x}\,\dd x}
\\[5mm] = &\
\expo{\pi/4}\lim_{\alpha \to \infty}\bracks{a\int_{0}^{\pi/4}
\expo{-x}\exp\pars{a\ln\pars{\tan\pars{{\pi \over 4} - x}}}
\,\dd x}
\\[5mm] = &\
\expo{\pi/4}\lim_{\alpha \to \infty}\bracks{a\int_{0}^{\infty}
\expo{-0}\expo{-2ax}\,\dd x} =
\expo{\pi/4}\lim_{\alpha \to \infty}\pars{a\,{1 \over 2a}}
\\[5mm] = &\
\bbx{{1 \over 2}\,\expo{\pi/4}} \approx 1.0966 \\ &
\end{align}


问:
Why do we use different equations in these two questions?

Suppose you have 5 balls. How many ways can you put them in 3 boxes, when no box can stay empty?

Suppose you have 5 letters. How many ways can you put them in 3 boxes?

My book has different answers for these questions but to me both questions seem the same.

1: 150

2: 243

Now the math about balls was done using Stirling numbers of second kind maybe. The solution looks like it but as we are in high school we are not told about its origin. And the question about letters was solved like $3^5$.

答: For the second, each letter can go in three boxes.  By the multiplication principle, there are $3^5=243$ ways that can happen.  For the first, you have the condition that no box is empty.  That condition means the answers must be different.  Now we subtract the cases where all the balls are in two boxes.  There are $3$ ways to choose the two boxes that are occupied, then $2^5=32$ ways to arrange the balls, so $243-3\cdot 32=147$.  We have removed the cases where all the balls are in one box twice, so we need to add them back in once.  $147+3=150$.


问:
Find polynomials $r(x)$ and $s(x)$ such that $r(x)(x^2+x+1)+s(x)(x^3-x-1)=1$.
I realize this has something to do with the Euclidean algorithm over $\mathbb{Q}[x]$, and I know how to show that the $\gcd(x^2+x+1, x^3-x-1)=1$, but I have no idea where to even begin looking for polynomials satisfying the Euclidean algorithm.

Is it largely just trial and error? Or is there some methodic process that I can go about solving for these polynomials...

答:
Another way. You know from inspection that the degree of $r(x)$ has to be exactly one higher than that of $s(x)$.

By substituting $x= \omega$ and $x = \overline \omega$ in turn (the conjugate complex roots of one), the term with $x^2 + x+1$ vanishes, and you immediately get:

$s(\omega) = - \frac 1{\omega} = \omega +1$

and $s(\overline \omega) = \overline \omega + 1$

which is consistent with a linear polynomial $s(x) = x+1$

We now need to test if we can get a consistent quadratic solution for $r(x)$ corresponding to this, then the problem is solved.

By inspection, the lead term for $r(x) = - x^2$ and the constant term is $2$. All we need is the middle term, let its coefficient be $b$, i.e. $r(x) = -x^2 +bx+2$.

Expanding, we get:

$-x^4 +x^3(b-1)+x^2(b+1)+x(b+2)+2 + x^4 +x^3-x^2-2x-1 =1$

which immediately allows us to conclude that $b=0$ is a consistent solution.

This gives $r(x) = -x^2 +2,\ s(x)=x+1$.


问:
Is there a characterization of which posets arise as the transitive closure of a set?
I'm self-studying Kunen's "new" Set Theory and I just got to the transitive closure of a set.
This is (somewhat abusively) defined as the sets below $x$ in the transitive closure of $\in$.
The example given (shown below) is a lattice, and this got me wondering if $\text{trcl}(\{x\})$ is always a lattice with top element $x$ and bottom element $\emptyset$.

I have convinced myself that this probably isn't the case1, but it did get me wondering if there is a characterization of which posets can arise as $\text{trcl}(x)$. I suspect that is too complicated a question, so I would be happy for partial results. I suspect there are lots of posets and lattices which we know can be embedded into $\text{trcl}(x)$ for some $x$, for instance.

Here is the lattice

Say $x \cap y = \emptyset$, and $z \neq \{x,y\}$. Then if $\{x,y\}$ and $\{x,y,z\}$ are both in the transitive closure of some set, $x$ and $y$ might not have a well defined join

Thanks in advance ^_^

答:
Belatedly expanding on my comment above:

We can show by a variation of the proof of Mostowski's collapse lemma the following:

If $P$ is a well-founded partial order then there is some set $X$ such that the membership relation restricted to elements of $X$ is isomorphic to $P$.

In particular, this means that every well-founded partial order embeds into $trcl(\{x\})$ for some $x$.

Suppose $P$ is well-founded. Mostowski would have us consider the map $\rho$ defined recursively by $\rho(p)=\{\rho(q): q<_Pp\}$. This recursive definition makes sense because $P$ is well-founded. However, it doesn't behave well in this context since $P$ may not be extensional: $\rho$ will not in general be injective.

Instead we have to fold in some additional "labelling." Fix some bijection $b:P\rightarrow A$. We define the map $\rho^b$ recursively by $$\rho^b(p)=\{\rho^b(q): q<_Pp\}\cup\{b(p)\}.$$ The map $\rho^b$ is clearly injective, and we have $$p<_Pq\leftrightarrow \rho^b(p)\in \rho^b(q).$$ So we just need to check that we don't have any unwanted elementhood-nesses. Of course we might if we pick the labelling function $b$ poorly. So we have to go back and be more careful about our $b$: we want a set $A$ which is "totally unrelated" to $P$ so that no annoyances will crop up. One choice which will work is to take an $A$ whose elements are each cardinals greater than $\vert P\vert$; the point then is that the elements of $A$ are so big that we can't "accidentally" build one of them via the recursive construction of $\rho^b$.


问:
Switching the Derivative From One Function to the Other Inside an Integral
Let $\Omega \subset \mathbb{R^n}$ be open and bounded by $\partial \Omega$ sufficiently smooth. Let $u \in C^k(\bar{\Omega})$ and $\phi \in C_0^\infty (\Omega)$ for a positive integer k. Show that:
$$
\int_\Omega uD^{\alpha}\phi dx = (-1)^{|a|}\int_{\Omega}\phi D^\alpha udx
$$
for any multi-index $\alpha$ with $|\alpha| <k$.

To me, this problem looks like an application of one of the fundamental theorems of calculus (stoke's, divergence, etc.). However, I might be wrong. I tried to work out the LHS but couldn't get to the RHS.

答: It's integration by parts, if you assume that all derivatives of $\phi$ go to zero at the boundary.


问:
A question about the definition of$\sigma|\approx p$ for a sequence $\sigma$ and a proposition $p$
I am reading an article which talks about temporal semantics.

For a sequence $$\sigma=s_{0},s_{1},\ldots$$ and a proposition
$p$ the article defines a few things such as $$\sigma|\approx W_{1}\vee W_{2}\text{ iff }\sigma|\approx W_{1}\text{ or }\sigma|\approx W_{2}$$ and
$$\sigma|\approx\sim W\text{ iff }\sigma|\not\approx W$$

However, there is one definition that I don't understand $$\sigma|\approx p\text{ iff }s_{0}|\approx p$$

What is the reasoning behind looking only at the first element in
the sequence ?

答: Roughly: What does it take for $p$ to be true now (according to a model $\sigma$, where $\sigma$ represents a world-history)? $p$ needs to be true with respect to the world-in-the-model which represents the present, i.e. the world indexed $s_0$ in the possible from-now-on history $s_0, s_1, s_2, s_3, \ldots$. Thus $\sigma$ makes $p$ currently true iff $p \in s_0$ (or $p$ is true at $s_0$, i.e. $p$ is true at at the element of $\sigma$ which represents the present).


问:
Let $ABCD$ be a convex quadrilateral with $M \in AD, N \in BC$
Let $ABCD$ be a convex quadrilateral and $M \in AD, N \in BC$ such that $\dfrac{AB}{CD}=\dfrac{AM}{MD}=\dfrac{BN}{NC}$ and $BA \cap CD=P$. Prove that the angle bisector of $\angle BPC$ and $MN$ are parallel.

I wanted to prove this geometrically, yet I don't know how to tackle this problem. Any help would be appreciated.

答:
Let $F$ be so that $ABCF$ is a parallelogram and $CE$ the angle bisector of $\angle FCD$:

It follows that
$$DE:FE = CD:FC = CD:AB = DM:AM$$
so $EM\parallel FA$, and
$$EM : FA = MD:AD = MD:(MD + AM) = CN:(CN + NB)$$

That shows $EM$ is parallel and equal to $CN$. It follows that $EMNC$ is a parallelogram, or $MN\parallel CE$ and we are done.


问:
Calculate the double integral: $\int \int_R \frac{1}{\sqrt{x}} \ dx \ dy$
Calculate the double integral:

$$\int \int_R \frac{1}{\sqrt{x}} \ dx \ dy$$

With  the region R  : $ y^2 \leq 8x$ and $y \leq 2x $ and  $ y+4x \leq 24$
So, the problem that I have is about the graph of the region. What I've tried is to write $ y^2 \leq 8x$ as $ y \leq \sqrt{8x}$ . I used  $ y \leq \sqrt{8x}$ and $y \leq 2x $ to try to graph the region .
I' ve attached an image with the graph that I drew . I would like to know if this is the correct graph and also the bounds of $x$.Thank you!

答:
To confirm your hand sketch, you can use one of the online tools. Some of them are free and are very easy to use. Here is the sketch of your region. You need to find the shaded area. Now if you integrate wrt $y$ first, you have $3$ sub-regions to integrate over but if you integrate over $x$ first, you have only two. So I would suggest setting up your integral as below.

As you can find, the intersection point of parabola $y^2 = 8x$ and line $y = 2x$ is $(2,4)$.

The intersection point of parabola $y^2 = 8x$ and line $y + 4x = 24$ is $(\frac{9}{2},6)$.

Limits of integral -

$\frac{y}{2} \leq x \leq 6 - \frac{y}{4} \,$ for $0 \leq y \leq 4$

$\frac{y^2}{8} \leq x \leq 6 - \frac{y}{4} \,$ for $4 \leq y \leq 6$

Order of Your integral o- first over $dx$ and then $dy$.


问:
Roll a die 10 times. What is the probability that the number of ones and sixes equals four?
Suppose a die is tossed 10 times. Let $X$ be the number of ones and let $Y$ be the number of sixes.

How would you find the probability that $P\left(X+Y=4\right)$?

My attempt so far:

Applying the Binomial Distribution, I can find the probability that $X$ or $Y$ equals $4$, but I do not know how to put these pieces together to find $P\left(X+Y=4\right)$.

答: You have to consider all the cases, $(X=4,Y=0); (X=3,Y=1); (X=2,Y=2); (X=1,Y=3); (X=0,Y=4).$  There are two pairs that are equal.  Alternately, you can group the sides of the die into the groups of interest:  (1 or 6) with probability $\frac 13$ and something else with probability $\frac 23$ and do the binomial distribution for $4$ of the first case and $6$ of the second.


问:
Normalizer of cyclic subgroup of $S_5$

In $S_5$, consider $\langle \sigma \rangle = \langle (12)(345) \rangle$. Find the normalizer of $\langle g \rangle$ in $S_5$.

Since $\langle \sigma \rangle$ is cyclic, it is abelian, so its normalizer equales its centralizer, and the center of $S_5$ is a subgroup of the centralizer. I can therefore state that
$$Z(S_5) \leq C_{S_5} (\langle \sigma \rangle) = N_{S_5} (\langle \sigma \rangle) \leq S_5$$
and that
$$\langle \sigma \rangle \leq N_{S_5} (\langle \sigma \rangle).$$
By Lagrange's theorem, I have that $6$ divides the order of the normalizer and that the normalizer divides $120$, the order of $S_5$. So the order is some positive integer multiple of $6$. The only elements that are obviously in it are the identity (the center of $S_5$), $(12)(345)$ itself, and its inverse. Is there a way to find the normalizer other than brute force?

答:
A couple comments first:

Since ⟨σ⟩ is cyclic, it is abelian, so its normalizer equales its centralizer

This is false. For instance consider a nontrivial cyclic subgroup of a dihedral group; the reflections normalize the cyclic subgroup but are not in its centralizer.

the center of S5 is a subgroup of the centralizer

This is a good observation in general, but useless here since $Z(S_5)$ is trivial.

Now let's let $H=\langle\sigma\rangle$ and $N=N_{S_5}(H)$ and $C=C_{S_5}(H)$.

The cyclic group $H\cong\mathbb{Z}_6$ has two generators, $\sigma$ and $\sigma^{-1}$. Therefore, any $n\in\ N$ either swaps these or fixes them. That is, we have a homomorphism $N\to\mathbb{Z}_2$. The kernel is $C$ (why?), so if we can find any element $g\in N\setminus C$ we will know that $N=C\sqcup gC$ by Lagrange's theorem (why?).

I will let you exhibit a $g\in S_5$ for which $g\sigma g^{-1}=\sigma^{-1}$, in which case $g\in N\setminus C$. You will need to know how conjugation-by-$g$ affects a permutation $\sigma$'s cycle notation; are you familiar with that? You will also need to determine $C$. Show any $\tau\in C$ must preserve the sets $\{1,2\}$ and $\{3,4,5\}$ and do a little more work to show the centralizer is simply $C=\langle(12),(345)\rangle=\langle (12)(345)\rangle$.

Title: Normalizer of cyclic subgroup of $S_5$


问:
Existence or construction of kernel of order $m$
I'm looking for a function $K$ satisfying $\int K(y)dy=1$, $\int y^jK(y)dy=0$ for $1\leq j<m$, and $\int|y|^mK(y)dy<\infty$, and finally $\int|y|K^2(y)dy<\infty$.

In the statistical literature such a function is known as a kernel of order $m$. Finding a second order kernel is easy: just take any symmetric probability density; popular choices are Gaussian, $\mathrm{Uni}([-\frac12,\frac12])$, Epanechnikov $K(x)=\frac34(1-x^2)\mathbf1_{|x|\leq1}$, Biweight $K(x)=\frac{15}{16}(1-x^2)^2\mathbf1_{|x|\leq1}$ or Triweight $K(x)=\frac{35}{32}(1-x^2)^3\mathbf1_{|x|\leq1}$.

An idea I had was Gram-Schmidt orthogonalization: suppose we use $K(x)=\frac1{\pi\sqrt{1-x^2}}$, then the Chebyshev polynomial of degree $m$, $T_m$, satisfies $\langle T_m,x^j\rangle_K=0$ for $0\leq j<m$, meaning that all conditions are satisfied, expect for the first one: if we use $\tilde K(x)=T_m(x)K(x)$, $$\int \tilde K(y)dy=\langle T_m,1\rangle_K=0\neq1,$$ and I do not see how to fix this problem. EDIT: the condition $\int|y|K^2(y)dy<\infty$ is neither satisfied, but this can be fixed, for example, by orthogonalizing w.r.t. the Gaussian density, yielding the Hermite polynomials. Then the above applies: we still have $\int \tilde K(y)dy\neq1$.

At different sources online, it is stated that ``higher-order kernels are obtained by multiplying a second-order kernel by an $(2m-1)$-order polynomial in $x^2$''. However, I do not see why this yields a function satisfying the needed conditions.

Any help or reference is much appriciated. I'm looking either for a construction or a specific example, if such a closed form expression is valid for every $m$.

EDIT: This question came from the context of density estimation in nonparametric estimation, where we use kernel estimators. If we let $K$ be a p.d.f. of a random variable with mean zero and unit variance, then it is true that the mean integrated square error of the kernel estimator for an optimal bandwith of order $n^{-1/5}$ is of order $n^{-4/5}$. Then it is stated that using a parametric model, we could achieve an error of order $n^{-1}$. However, using a kernel of order $m$, which I'm asking about in this question, we can take bandwith of order $n^{-1/(2m+1)}$, leading to a mean integrated square error of order $n^{-(2m)/(2m+1)}$. Therefore, by taking a kernel of sufficiently high order, the parametric rate can be approximated to the desired accuracy. See $\S5.2$ in these lecture notes based on the book Asymptotic Statistics by van der Vaart (they are English, ignore first 2 pages) for the details. Here it is stated that we can just take such a kernel, but it is not explained why such a kernel should exist.

答:
Let $K(x)$ be a probability density function and take the ansatz $\tilde{K}(x)=p(x)K(x)$ for our higher order kernel. Notation wise, let $p(x)=\sum_i x^{i}p_i$, and let  $M_i$ be the ith moment of $K$, thus $M_i=\int x^{i} K(x)dx$.

$0=\int x^l \tilde{K}(x)dx$ becomes the linear constraint $\sum_i p_i M_{i+l}=0$. Similarly we have the constraint $\sum_i p_iM_i=1$. So we want to show that the matrix $\left(\begin{array} {ccc} M_0 & \dots & M_{m+1}\\ \vdots & \vdots & \vdots\\ M_{l} &\dots & M_{l+m+1}\\ \vdots & \vdots & \vdots\\ M_{m+1} &\dots& M_{m+1+m+1} \end{array}\right)$ is invertible.

A matrix of this form is called a Hankel matrix, and is often invertible. For example, if we take $K(x)=1_{x\in [0,4]}\sqrt{-1+4/x}/2\pi$, then $M_i$ is a Catalan number, and it is well known that the matrix always has determinant 1 (see the wikipedia article on Catalan numbers for example). Moreover, the finiteness conditions will be satisfied because $K$ has finite support.


问:
Prove that $\sqrt{3} + \sqrt{7}$ is a primitive element of $\mathbb{Q}(\sqrt{3},\sqrt{7})$
I want to find a primitive element of $\mathbb{Q}(\sqrt{3},\sqrt{7})$ over $\mathbb{Q}$. I try with $\sqrt{3} + \sqrt{7}$.

Let $X = \sqrt{3} + \sqrt{7}$ we get

$$ X^4 - 20X^2 + 16 = 0 $$

We can prove that $[\mathbb{Q}(\sqrt{3},\sqrt{7}) : \mathbb{Q}] = 4$ since $\sqrt{3}$ and $\sqrt{7}$ are the roots of $X^4 - 10X^2 + 21$ and it is irreducible over $\mathbb{Q}$ by the Eisenstein criterion with $p = 3$.

Let $P(X) = X^4 - 20X^2 + 16$. Since $\sqrt{3} + \sqrt{7}$ is a root of $P$ then $[\mathbb{Q}(\sqrt{3}+\sqrt{7}) : \mathbb{Q}] \leq 4$.
If we prove that $P$ is irreducible over $\mathbb{Q}$ then the degree is equal to 4. But the Eisenstein criterion does not seem to work...

So, I try another thing. Let $\alpha = \sqrt{3} + \sqrt{7}$, we get $\alpha^2  = 10 + 2 \sqrt{21}$ and $\mathbb{Q} \subset \mathbb{Q}(10 + 2 \sqrt{21}) \subset \mathbb{Q}(\sqrt{3} + \sqrt{7})$. Now

$$ [\mathbb{Q}(10 + 2 \sqrt{21}) : \mathbb{Q}] = 2 \Rightarrow 2 \mid [ \mathbb{Q}(\sqrt{3} + \sqrt{7}) : \mathbb{Q} ] $$

The degree must be 1, 2 or 4.

But  $\mathbb{Q}(10 + 2 \sqrt{21}) \subset \mathbb{Q}(\sqrt{3} + \sqrt{7})$ and $ [ \mathbb{Q}(\sqrt{3} + \sqrt{7}) : \mathbb{Q}(10 + 2 \sqrt{21}) ] > 1$ then the degree must be 2 or 4. I don't know how to conclude with the formula

$$ [ \mathbb{Q}(\sqrt{3} + \sqrt{7}) : \mathbb{Q} ] = [\mathbb{Q}(\sqrt{3} + \sqrt{7})  :\mathbb{Q}(10 + 2 \sqrt{21}) ] [\mathbb{Q}(10 + 2 \sqrt{21}) : \mathbb{Q}]  $$

Thank you. I think I did a mistake somewhere.

答:
I don't know what is your background in Galois theory, but essentially, this theory was conceived to replace tedious (and most of the time blind) polynomial calculations by natural conceptual manipulations of groups. Here we deal with quadratic and biquadratic fields. General quadratic fields are of the form $\mathbf Q(\sqrt a)$, with $a \in \mathbf Q^*, \notin \mathbf {Q^*}^2$, and it is straightforward to show that to quadratic fields $\mathbf Q(\sqrt a),\mathbf Q(\sqrt b)$ are equal iff $ab \in \mathbf {Q^*}^2$. It follows that, if $ab \notin \mathbf {Q^*}^2$, then the extension  $K=\mathbf Q(\sqrt a,\sqrt b)$, which is the compositum of two distinct quadratic fields, is actually biquadratic, i.e. galois with group $G\cong C_2 \times C_2$. Obviously $G$ contains exactly three subgroups of order two, isomorphic to $C_2$. Then the first fundamental theorem of Galois theory  gives us all the subfields of $K$, which are $\mathbf Q, K$ and  three quadratic subfields, $\mathbf Q(\sqrt a),\mathbf Q(\sqrt b),\mathbf Q(\sqrt ab)$ exactly. It follows that the subfield $\mathbf Q(\sqrt a +\sqrt b)$ must coincide with $K$.

Of course we can apply this to $a=3, b=7$, but the galois approach gives much more. For instance it answers one recurrent question in this thread, which is the determination of all the subextensions of $\mathbf Q(\sqrt p_1),...,(\sqrt p_n)$, where the $p_i$ are $n$ distinct primes, see e.g. https://math.stackexchange.com/a/3444776/300700.


问:
Quick Conditional Probability Question
Been going through lecture notes (ones I made myself, so who knows how accurate they are - or rather - they most certainly are not) and can't seem to understand this one example

Example: Two horses - A and B. A wins with probability 0.5, B wins with probability 0.3. What is the probability that B wins?

The calculation that follows:
$$P(B|A^C)=\frac{P(B\cap A^C)}{P(A^C)}=\frac{0.3}{0.5}=0.6 $$

Now, I have no trouble with the theory, just understanding what is actually meant by the question. My questions:

1) Should the question really be "What is the probability that B wins, given that A doesn't?"

2) Is $P(B)=P(B\cap A^C)$ since $P(A\cap B)=\emptyset$ (or is that true?)

Perhaps 1) and 2) are obviously true and I just forgot to add that sentence in my notes. But I want to be sure (and be able to sleep at night :) )

Thanks for any help.

答: The answers to both 1) and 2) seem to be 'yes'.  Certainly the calculation makes sense in that context.</p>

Title: Quick Conditional Probability Question


问:
Lyapunov Function for system with $f(x,y) = x' = \sin y$
I have the system

$x'=\sin y = f(x,y)$

$y'=-2x-3y = g(x,y)$

I am given the Lyapunov function $V(x,y)=15x^2+6xy+3y^2$.

Obviously, $V(0,0)=0$. Also, $V(x,y)=6x^2+9x^2+2 \cdot 3x y + y^2 + 2y^2 = 6x^2+2y^2+(3x+y)^2>0$,
$ \forall (x,y) \in R^2 \setminus \{(0,0)\}$.

From Maple I know that the origin is asymptotically stable.
But I can't show that

$\dot V(x,y) = \displaystyle \frac{\partial V}{\partial x} f(x,y) + \displaystyle \frac{\partial V}{\partial y} g(x,y)=(30x+6y)\sin y + (6x+6y)(-2x-3y) = ... <0$.

I would appreciate any help. Thanks.

答:
A Lyapunov function can be found by linearizing the original system at the origin. The linearized system is
$$\tag{1}
\left\{\begin{array}{lll}
\dot x &=& y\\
\dot y&=&-2x-3y
\end{array}\right.
$$
or
$$
\left(\begin{array}{c}
\dot x\\ \dot y
\end{array}\right)=A
\left(\begin{array}{c}
x\\y
\end{array}\right),
\quad\mbox{where }
A=\left(\begin{array}{rr}
0 & 1\\ -2& -3
\end{array}\right).
$$
A Lyapunov function for (1) is also a Lyapunov function for the original system (why this is so, is described, for example, in this answer). The quadratic form
$$
V(x,y)=\left(\begin{array}{cc}x&y\end{array}\right)
P
\left(\begin{array}{r}x\\y
\end{array}\right)
$$
is a Lyapunov function for (1) if and only if $P>0$ and the matrix $Q=A^TP+PA$ is negative definite. We have
$$
P=\left(\begin{array}{cc}
15 & 3\\ 3& 3
\end{array}\right),\quad
Q=A^TP+PA=\left(\begin{array}{rr}
-12 & 0\\ 0& -12
\end{array}\right),
$$
thus, $V(x,y)$ is a Lyapunov function for the original system.


问:
Calculation of $ \int_0^{2\pi} \frac{\sqrt{a^2+2a\cos\theta+1}}{(a\cos\theta+1)^2} d\theta $
This integral formula came from the calculation in the spherical coordinate. Here, $a$ is a real number greater than 1.
$$ I = \int_0^{2\pi} \frac{\sqrt{a^2+2a\cos\theta+1}}{(a\cos\theta+1)^2}d\theta $$
Since $a^2+2a\cos\theta+1=\mathrm{Re}(a+e^{i\theta})(a+e^{-i\theta})$ and $a\cos\theta+1=\mathrm{Re}(ae^{i\theta}+1)$, the integration becomes
$$ I = \mathrm{Re}\int_0^{2\pi} \frac{\sqrt{(a+e^{i\theta})(a+e^{-i\theta})}}{(ae^{i\theta}+1)^2}d\theta. $$
Substitution $z=e^{i\theta}$ yields
$$ I = \mathrm{Re}\oint_C \frac{\sqrt{(a+z)(a+1/z)}}{iz(az+1)^2}dz, $$
where $C$ is an unit circle centered at the origin.

At this moment, I tried to decompose the integrand into partial fractions and to use Cauchy's integral formula, but the numerator of integrand, $\sqrt{(a+z)(a+1/z)}$, is not continuous on $C$ because of the multivaluedness of complex square root. How can I approach to the next step?

答:
As mentioned by Felix Marin in comments, the integral diverges for $a\ge1$. However we shall verify ComplexYetTrivial's claim that $I=\frac{4E(a^2)}{1-a^2}$ if $a\in[0,1)$. Elliptic integrals follow Mathematica/mpmath conventions in this answer.

Using Byrd and Friedman 289.09:
$$I=\int_0^{2\pi}\frac{\sqrt{a^2+2a\cos\theta+1}}{(a\cos\theta+1)^2}\,d\theta=2\int_0^\pi\frac{a^2+1+2a\cos\theta}{(a\cos\theta+1)^2}\frac1{\sqrt{a^2+1+2a\cos\theta}}\,d\theta\newcommand{sn}{\operatorname{sn}}$$
$$=\frac4{a+1}\int_0^{K(m=4a/(a+1)^2)}\frac{a^2+1+2a(1-2\sn^2u)}{(a(1-2\sn^2u)+1)^2}\,du$$
$$=\frac4{(a+1)^2}\int_0^{K(m)}\left(\frac{a-1}{(1-\frac{2a}{a+1}\sn^2u)^2}+\frac2{1-\frac{2a}{a+1}\sn^2u}\right)\,du$$
Now using B&F 336.01 and .02:
$$\int_0^\varphi\frac1{1-n\sn^2u}\,du=\Pi(n,\varphi,m)$$
$$\int_0^\varphi\frac1{(1-n\sn^2u)^2}\,du=\frac1{2(n-1)(m-n)}\left(nE(\varphi,m)+(m-n)F(\varphi,m)+(2mn+2n-n^2-3m)\Pi(n,\varphi,m)-\frac{n^2\sn u\operatorname{cn}u\operatorname{dn}u}{1-n\sn^2u}\right)$$
we get after much simplification (note that $\varphi=\pi/2$ here)
$$I=2\left(\frac{K(m)}{a+1}-\frac{E(m)}{a-1}\right)$$
which is in turn simplified by a Gauss transformation to
$$I=\frac{4E(a^2)}{1-a^2}$$


问:
Volume of Revolution...
I'm studying for a test and am stuck on one volume problem.

The figure is rotated about the x-axis and is bounded by
$$
y=x^2, y=\frac{(3-x)}{2},y=0.
$$

I thought the solution would be:
$$
\pi\int_{0}^{1}(\frac{3-x}{2})^{2}-(x^{2})^{2}dx
$$
However, the key is indicating that this is wrong. Any insights as to where I may have gone wrong would be appreciated.

Additional question: Is the area integrated between the two curves and the x-axis then rotated about the x-axis? Or is the area integrated between the two curves and the y axis then rotated about the x-axis? Which area am I "shading?" Unsure if that made sense but I believe that is where I'm fundamentally misunderstanding the problem

答:
hints

$f(x)= x^2.$

$g(x) = (3-x)/2.$

where does:

$f(x)$ hit the $x$-axis?

$g(x)$ hit the $x$-axis?

$f(x)$ hit $g(x)$?

given a height $f(x)$, from $a$ to $b$, what is the formula for the volume of revolution generated by $f(x)$?

Addendum

Per OP's comment/question:

The stated problem actually involves a region that has one height for $x \leq 1$ and a different height for $x \geq 1.$  This is how you should be interpreting the instructions:

The figure is rotated about the x-axis and is bounded by
$$
y=x^2, y=\frac{(3-x)}{2},y=0.
$$

This means that the volume of revolution is completely solid, with no (inner gap).  This means that the cross-sectional area is $\pi\times R^2$, where $R$ is either equal to $f(x)$ or $g(x)$, depending on whether $x \leq 1$ or $x \geq 1$.

Contrast that with a different problem:

Let $s(x) = 2.$

Let $t(x) = 1.$

Consider the region bounded by $x=0$, $x=3$, $s(x)$ and $t(x)$.

If you were considering the volume of revolution for this region rotated about the $x$-axis, you would realize that the 3 dimensional solid formed has a hole in it, like a hollow cylinder.

Here, the formula for the cross-sectional area would be $\pi \times (R^2 - r^2)$, where $R$ is the outer radius of $s(x)$ and $r$ is the inner radius of $t(x)$.

Title: Volume of Revolution...


问:
Can a Pythagorean Triple ever have TWO identical exponents > 1?
For example, the triple

$(12, 16, 20)$

has these exponents:

$(12^1, 4^2, 20^1)$ or $(12^1, 2^4, 20^1)$

denoted by

$(1,2,1)$ or $(1,4,1)$

Here are more triples (ignore simple (1,1,1) cases):

I ran millions of triple tests looking for two identical exponents > 1, but found no matches.

Is there an existing proof or can someone prove it?

Thanks!

答:
EDIT: The original answer implicitly made the assumption that the sides of the triangle were relatively prime, which is not part of the original post. I've edited to explain why this assumption is valid.

Your question is equivalent to seeking an integer solution to $a^x+b^y=c^z$ in positive integers $a,b,c,x,y,z$, where $x,y,z$ are even and either $x=y\ge 4$ or $y=z\ge 4$. Note that since any $2k$th power is also a square, we can assume WLOG that the third exponent in either case is $2$.

First, let's see that we can reduce to the case where $a,b,c$ are relatively prime. Suppose that our bases have a common divisor $d>1$, and write them as $da,db,dc$ where $a,b,c$ are relatively prime. Then if $x=y$, we have

$$d^x(a^x+b^x)=d^2c^2\implies a^x+b^x=\left(\frac{c}{d^{x/2-1}}\right)^2$$

(Recall that $x$ is even and at least $4$, so $x/2-1$ is a positive integer.) So then $a^x+b^x$, an integer, is equal to the square of the rational number $c/(d^{x/2-1})$. But since the square of a rational number is an integer only when the original rational number is, the tuple $(a,b,c/(d^{x/2-1}))$ is another solution, but this time without any common factors.

A very similar analysis shows that in the case $y=z$, we can reduce to $(a/(d^{z/2-1}),b,c)$ and obtain a solution with relatively prime $a,b,c$. (This sort of reduction does not work if all three exponents are different; if it did, Oscar Lanzi's answer could be reduced to yield a solution with relatively prime $a,b,c$ and falsify Beal's conjecture.)

Having made this reduction, we can find several results in the literature tackling sums of powers of relatively prime positive integers.

The case $x=y$ has been proven impossible by Darmon and Merel (1995); see this citation in Wikipedia.

The case $y=z$ can be reduced to $z=2p$ for some prime $p$, since an $n$th power is an $m$th power if $m$ divides $n$. (Hence, we can ignore $k=8,12,16,18,20,24,28,30,\ldots$.)

$z=4$ follows from a relatively elementary proof by infinite descent, and dates to Fermat (see here for a proof).

$z=6,10$ follow from a result of Bennett, Chen, Dahmen and Yazdani (from Wikipedia's list of partial results on the Beal conjecture.)

As far as I am aware, $z\ge 14$ remains open (and is likely very difficult); one should at least note that the Fermat-Catalan conjecture implies there are at most finitely many solutions, and the ten known solutions to the Fermat-Calatan equation do not have even exponents, so at least no examples are known (very likely none exist).


问:
How do you represent the following sentences in first-order logic?
Represent the following sentences in first-order logic, using a consistent vocabulary(which you must define):

a) Some students took French in spring 2001.

b) Every student who takes French passes it.

c) Only one student took Greek in spring 2001.

d) The best score in Greek is always higher than the best score in French.

e) Every person who buys a policy is smart.

f) No person buys an expensive policy.

g) There is an agent who sells policies only to people who are not insured.

答:
Let me address the first three sentences (since they form a maximal coherent collection of sentences).

We have a use for the following five predicates:

$\mathsf{Student}(x)$: $x$ is a student;

$\mathsf{French}(y), \mathsf{Greek}(y)$: $y$ is the course in French, resp. Greek;

$\mathsf{Take}(x,y)$: $x$ (student) takes $y$ (course);

$\mathsf{Pass}(x,y)$: $x$ (student) passes $y$ (course);

$\mathsf{TakeInSpring2001}(x,y)$: $x$ (student) takes $y$ (course) in Spring 2001.

Now we can formulate the sentences as follows:

a) $\exists x\exists y: \mathsf{Student}(x) \land \mathsf{French}(y) \land \mathsf{TakeInSpring2001}(x,y)$

b) $\forall x\forall y: (\mathsf{Student}(x) \land \mathsf{French}(y) \land \mathsf{Take}(x,y)) \implies \mathsf{Pass}(x,y)$

c) $\exists x\exists y\forall z: \mathsf{Student}(x) \land \mathsf{Greek}(y) \land \mathsf{TakeInSpring2001}(x,y) \land ((\mathsf{Student}(z) \land \mathsf{TakeInSpring2001}(z,y))\implies x = z)$

The other five shouldn't be hard once one understands the above three. A different approach would be to use more specific predicates (e.g. $\mathsf{FrenchInSpring2001}(x)$ as suggested in the comments), but that would defeat the "consistent vocabulary" requirement of the question IMO.


问:
Filter and ultrafilter properties in Boolean algebra
In a Boolean algebra $(\mathbb{B}, \vee, \wedge, 0, 1, \neg)$, a ultrafilter $U \subseteq \mathbb{B}$ satisfies

If $a \wedge b \in U \iff a \in U$ and $b \in U$.

If $a \vee b \in U \iff a \in U$ or $b \in U$.

The first one is actually true for all filters ($\Rightarrow$ upward closure, $\Leftarrow U$ downward directed). For the second one, $\Leftarrow$ can be proven by upward closure, but how about the $\Rightarrow$ direction which seems to be true only for ultrafilter ?

NOTE: The only definition of ultrafilter I have at this stage is: an ultrafilter $U$ is a filter satisfying $\forall x \not\in U, \neg x \in U$ (i.e. $\forall x \in \mathbb{B}, x \in U \leftrightarrow \neg x \not\in U$).

答:
Suppose that $a\lor b\in U$ but $b\notin U$. Then $\neg b\in U$, so $(a\lor b)\land\neg b\in U$. And

$$(a\lor b)\land\neg b=a\land\neg b\,,$$

so $a\in U$ by upward closure.


答:
alternative approach

$E$ is the event of a 1 or a 6

$p(E) = 1/3$.

Let $p = 1/3.$

Let $q = 1 - p$.

General formula for exactly $k$ successes in $n$ tries, where

$n \in \mathbb{Z^+}$ and $k \in \{0,1,\cdots, n\}$ is

$$\binom{n}{k}p^kq^{(n-k)}$$

Here, $n=10$ and $k=4.$

Title: Roll a die 10 times. What is the probability that the number of ones and sixes equals four?

Body:
Suppose a die is tossed 10 times. Let $X$ be the number of ones and let $Y$ be the number of sixes.

How would you find the probability that $P\left(X+Y=4\right)$?

My attempt so far:

Applying the Binomial Distribution, I can find the probability that $X$ or $Y$ equals $4$, but I do not know how to put these pieces together to find $P\left(X+Y=4\right)$.


问:
How to evaluate an integral in a multiple connect region
a while ago I was faced by this problem:

Let $D$ be a multiple connected region interior to $C$ and exterior to
$C_{1}$ and $C_{2}$ (See figure). We know that $Q_{x} - P_{y} = 0$.

What is the value of $\oint_{C} F. dr$?

If $\oint_{C_{1}} P dx + Q dy = 2$ and $\oint_{C_{2}} P dx + Q dy = 3$, what is the value of $\oint_{C} P dx + Q dy$?

This is the region:

My attempt:

Well, if $Q_{x} - P_{y} = 0$, then $F$ is conservative, so, $\oint_{C} F. dr = 0$.

For the second part, I don't know how to proceed, I was thinking in sum both of the $C_{i}$ integrals.

Can someone help me with this?
Thanks in advance!

答:
If ${\bf F} = \langle P,Q\rangle$, Green's theorem  (which remains true for multiply connected regions) gives
$$ \int_{\partial D} {\bf F}\cdot d{\bf r} = \int_{\partial D}P\,dx+Q\,dy =\iint_D (Q_x-P_y)\,dx\,dy = 0.$$
With the given orientations, we have $\partial D = C+C_1+C_2$ (keep in mind the inner boundary orientations need to clockwise rather than counterclockwise, which we have in this case). Thus
$$ \int_{\partial D} {\bf F}\cdot d{\bf r} = \oint_{C} {\bf F}\cdot d{\bf r}+\oint_{C_1} {\bf F}\cdot d{\bf r}+\oint_{C_2} {\bf F}\cdot d{\bf r}.$$
It follows that
$$\oint_{C} {\bf F}\cdot d{\bf r} = -\oint_{C_1} {\bf F}\cdot d{\bf r}-\oint_{C_2} {\bf F}\cdot d{\bf r}.$$
So there isn't a numerical answer to the first problem, since it depends on the value of the integrals along the inner boundaries. I imagine you thought ${\bf F}$ was conservative because $Q_x = P_y$ in $D$, but that implication only works in a simply-connected region (and $D$ is not). You would be able to conclude the integral around $C$ is $0$ if $Q_x = P_y$ everywhere inside $C$, but I don't think this is what they meant (since in this case, the situation of $2$ would be impossible; the integrals around the inner boundaries could not be nonzero numbers).

Anyway, for 2, the above formula implies
$$\oint_C P\,dx+Q\,dy = -2-3 = -5.$$


问:
Find all positive integer solutions of the equation
Find all positive integer solutions of the equation $1650_x = 1131_3$$_y$ where $x$ and $3y$ are the bases of the numbers on the two sides.

I have this example in my book but have never come across an equation like this and I can't seem to find anything about these online.

Am I correct to by starting with

$1650_x = 1*x^3 + 6*x^2 +5*x^1 + 0*x^0 = x^3+6x^2+5x$

$1131_3$$_y = (3y)^3 + (3y)^2 + 9y + 1$

If so, where do I go from here?  If not, than I am lost.  Any help would be appreciated.  Thanks.

Body: Hint: What are the residues of $$1650_x = x^3+6x^2+5x = x(x+1)(x+5)$$ and $$1131_{3y} = (3y)^3+(3y)^2+3(3y)+1 = 3(9y^3+3y^2+3y)+1$$ modulo $3$?


问:
Why $C^2\geq n$ for any real $n\geq 0$?
I am trying to prove the following statement: if $C=⌊n^{1/2}⌋+1$ then $C^2>n$ for any real $n\geq 0$.
It seems very obvious, but I don't understand how to prove it. Hope you can help.

P.S. I believe that next statement will also be correct: $C=⌊n^{1/k}⌋+1$ then $C^k>n$ for any real $n\geq 0$ and k>0. If it so, then it is better to prove the generalized statement.

P.P.S. What happens if k is less than 0?

答:
Counterexample: Let $n=2$, then $C=2$, then we have $C^2=n^2$.

If what we wanted to show is

$$C^k > n$$

Guide:

Step 1: Show that $\lfloor x \rfloor > x-1$.

Step 2: Show that $f(x)=x^k$ is an increasing function for nonnegative $x$.

Title: Why $C^2\geq n$ for any real $n\geq 0$?


问:
Show that limit of a sequence goes to 0.
Assume that $\lim_{x\to\infty} f’(x) = 0$. Let $a_n = f(n +1) - f(n)$. Show that $\ lim_{n\to\infty} a_n = 0$.

I can’t seem to think how to even begin with this. Can someone help me out?

答: $\lim_{n\to\infty}f(n+1)-f(n)$ can be shown as $\lim_{n\to\infty}\frac{1}{n}[f'(x)+f'(x+1/n)+f'(x+2/n)+...+f'(x+1)]$ which is $\lim_{n\to\infty}\frac{1}{n}(n \times0)=1\times0=0.$


问:
For which values of $p,q$ does the integral $\int_0^1 x^p (\ln\frac{1}{x})^qdx$ converge?
For which values of $p,q$ does the integral $\int_0^1 x^p (\ln\frac{1}{x})^qdx$ converge?

I use the substitution $t=1/x$ to obtain this better looking integral: $\int_1^\infty \frac{(\ln t)^q}{t^{p+2}}$. Integration by parts gives me a recurence formula for the integral, but $p,q$ are not necessarily integers so I don't think that's the right approach.

答:
I understand this is a bit late.

But we can substitute $x=e^{-t}$ to make the integral nice looking: $$I=\int_{0}^{\infty}{e^{-(p+1)t}t^q}dt$$
We can split this into $I_1=\int_{0}^{1}{e^{-(p+1)t}t^q}dt$ and $I_2=\int_{1}^{\infty}{e^{-(p+1)t}t^q}dt$.

$I_1$ converges for $q>-1$ (LCT with $t^q$).

Now for $q>-1$, we consider $I$. The only point of concern is $\infty$.

$p < -1: e^{-(p+1)t}t^q\rightarrow \infty$ as $x \rightarrow \infty \implies I$ diverges

$p = -1: I=\int_{0}^{\infty}t^qdt$ diverges for $q>-1$

$p > -1: \int_{0}^{\infty}{e^{-(p+1)t}t^q}dt=\frac{1}{(p+1)^q}\Gamma(q+1)$ converges.

Thus $I$ converges for $p>-1, q>-1$.


答:
Calculate

$$
\lim_{n\to\infty} a_n = \lim_ {n\to\infty} \frac{f(n + 1) - f(n)}{n + 1 - n} = \lim_{n\to\infty} f'(n + \xi) = \lim_{x\to\infty} f'(x) = 0
$$

where in the first step we use the definition of $a_n$, in the second step we use the mean value theorem with $\xi \in (0, 1)$ and in the third step we notice that $x = n + \xi$ tends to infinity when $n$ tends to infinity.

Title: Show that limit of a sequence goes to 0.

Body:
Assume that $\lim_{x\to\infty} f’(x) = 0$. Let $a_n = f(n +1) - f(n)$. Show that $\ lim_{n\to\infty} a_n = 0$.

I can’t seem to think how to even begin with this. Can someone help me out?


问:
Friedberg Linear Algebra problems
In Friedberg's Linear Algebra 4th edition, I have trouble understanding 2 related questions:

Section 2.1 problem 34:
"Let $V$ and $W$ be vector spaces over a common field, and let $\beta$ be a basis for $V$. Then for any function $f: \beta \to W$ there exists exactly one linear transformation $T: V \to W$ such that $T(x) = f(x)$ for all $x \in \beta$.

I don't understand what the restriction of $T$ to the basis space $\beta$ does at all, and how does one prove that there is "exactly one" transformation? Directly related to this,

Section 2.7 problem 18: "Let $V$ be a nonzero vector space over a field $F$ and let $S$ be a basis for $V$. Let $\Phi: V^* \to \mathcal{L}(S,F)$ be the mapping defined by $\Phi(f)=f_S$, the restriction of $f$ to $S$. Prove that $\Phi$ is an isomorphism. Hint: Apply Exercise 34 of Section 2.1. (the above problem)

I don't see how the two problems are related except for the fact that there is a linear transformation and there is a linear transformation restricted to a subset containing only basis vectors.

答:
First, there is a typo in the statement of the second problem: it should be $\mathcal F(S,F)$ instead of $\mathcal L(S,F)$. Recall that, following the notation of the book, $\mathcal F(S,F)$ is the set of all functions from $S$ to $F$; naturally $\mathcal F(S,F)$ is a vector space over $F$ with the addition and the scalar multiplication defined pointwise.

Now, once observed this, if you put $W = F$ and $\beta=S$ in the first problem, this one says that:

If $V$ is a vector space over a field $F$ and $S$ is a basis for $V$, then for any function $f : S \to F$ there is a unique linear map $\tilde{\vphantom{}f} : V \to F$ such that $\tilde{\vphantom{}f}|_S = f$, in other words, any function in $\mathcal F(S,F)$ can be extended to a function in $V^*$ in a unique way.

Thus, if $\Phi : V^* \to \mathcal F(S,F)$ is the function given by $\Phi(g) =  g|_S$, the statement written above means precisely that $\Phi$ is a bijection (the existence part means that $\Phi$ is a surjection and the uniqueness part means that $\Phi$ is an injection). Therefore, it only rest to prove that $\Phi$ is linear.

Title: Friedberg Linear Algebra problems


问:
Showing minimum between x and y is a primitive recursive function with initial functions
for $f(x)=min(x,y)$ with this formula:
\begin{equation}
min(x,y) =
\begin{cases}
x & \text{  x<y} \\
 y & \text{ y<x} \
\end{cases}
\end{equation}
how we can show it's primitive recursive with initial functions?

答:
For showing $f(x)$ function is primitive recursive with any formula we must generate $f(x)$ function with initial functions or functions that can built with initial functions. in this problem, we can do it by the following steps.

Extended subtraction is primitive recursive

$
x ∸  y =
\begin{cases}
x-y & \text{  x ≥ y} \\
 0 & \text{ x < y} \
\end{cases}
$

Negation function is primitive recursive

$
\alpha(x) =
\begin{cases}
0 & \text{  x = y} \\
 1 & \text{ x ≠ y} \
\end{cases}
$

So by combining these two we can have

$
\alpha(x ∸  y) =
\begin{cases}
1 & \text{  x ≥ y} \\
 0 & \text{ x < y} \
\end{cases}
$

And multiplication is primitive recursive and we can generate this

$y.\alpha(x ∸ y) =
\begin{cases}
y & \text{  x ≥ y} \\
 0 & \text{ x < y} \
\end{cases}$

And the final step is this

$min(x,y) = y.\alpha(x ∸ y) + x.\alpha(y ∸ x)$

So min(x,y) is premitive recursive.


问:
Parabolic, Hyperbolic, Elliptic
The terms "parabolic," "hyperbolic" and "elliptic" are used to classify certain differential equations. The terms "hyperbolic" and "elliptic" are also used to describe certain geometries. Is there a connection between these usages, and, if so, what is it?</p>

答:
I have just added an entry on my blog to explain this.

Let's consider the space $\mathbb{M}^4$, that is, the manifold $\mathbb{R}^4$ with coordinates $(x,y,z,t)$ but with the pseudo-Riemannian metric given in each tangent space by
$$
ds^2=dx^2+dy^2+dz^2-dt^2.
$$
This metric is left invariant by the group $O(3,1)$ when acting on $\mathbb{M}^4$. It also leaves invariant the cone
$$
t^2-x^2-y^2-z^2=0
$$
so is logical that we pay attention to it.

Now, following TRTR from Penrose page 423, we can consider the family of hyperplanes inside $\mathbb{M}^4$ given by
$$
z+t+\lambda(t-z)=2
$$
intersecting our cone in different 2-dimensional submanifolds $P_{\lambda}$ of $\mathbb{M}^4$.

Here we have a picture from that book, where we consider only $(x,z,t)$ for purposes of drawability.

If we analyse the case $\lambda=0$, we observe that we obtain the submanifold $P_0$, $E$ in the picture with a shape of a parabola, given by the embedding

\begin{align*}
  \phi \colon \mathbb{R}^2 &\longrightarrow \mathbb{M}^4\\
  (u_1,u_2) &\mapsto (u_1, u_2, 1-\frac{u_1^2+u_2^2}{4},1+\frac{u_1^2+u_2^2}{4})
\end{align*}

We can look for the (possibly pseudo) metric inherited by $P_0$. Since
$$
d\phi=\begin{pmatrix}
1 & 0 \\
0 & 1\\
-\frac{u_1}{2} &-\frac{u_2}{2}\\
\frac{u_1}{2} &\frac{u_2}{2}
\end{pmatrix}
$$
we conclude that $P_0$ consists of $\mathbb{R}^2$ with the euclidean metric:
$$
g=\begin{pmatrix}
1 & 0 \\
0 & 1\\
\end{pmatrix}
$$

That is, although we see it like a parabola, intrinsically is only the Euclidean plane, and that is the reason the latter is sometimes called parabolic geometry.

But, what happens for others $\lambda$? Let's see. We have a far more complicated parametrization of our embedded manifold:
$$
\phi(u_1,u_2)=(u_1,u_2, \frac{-1+{ \lambda }+\sqrt{-(1+{ \lambda })^{2}\left(-1+{\lambda}\left(u_1^{2}+u_2^{2}\right)\right)}}{2{ \lambda }},
$$
$$
\frac{2+{ \lambda }+\sqrt{-(1+{\lambda })^{2}\left(-1+{\lambda}\left(u_1^{2}+u_2^{2}\right)\right)+\frac{1-\sqrt{-(1+{\lambda})^{2}\left(-1+{\lambda}\left(u_1^{2}+u_2^{2}\right)\right)}}{{ \lambda }}}}{2(1+{ \lambda })})
$$
(computations made with Mathematica).

If we compute the (possibly pseudo) metric in this chart we obtain the inherited metric in $\mathbb{R}^2$

$$
g_{\lambda}=\begin{pmatrix}
\frac{1-{ \lambda } u_2^{2}}{1-{ \lambda }\left(u_1^{2}+u_2^{2}\right)} & \frac{\lambda u_1 u_2}{1-{ \lambda }\left(u_1^{2}+u_2^{2}\right)} \\
\frac{\lambda u_1 u_2}{1-{ \lambda }\left(u_1^{2}+u_2^{2}\right)} & \frac{1-{ \lambda } u_1^{2}}{1-{ \lambda }\left(u_1^{2}+u_2^{2}\right)}\\
\end{pmatrix}
$$

Next step is to show that for $\lambda>0$ we have an isometry into the usual sphere, and that for $\lambda<0$ we have an isometry into the hyperbolic space. This would justify the terms elliptic and hyperbolic for those geometries, as you can see at the picture above ($S$ and $H$ respectively). In order to keep it simple we can work with $\lambda=1$ and $\lambda=-1$.

For $\lambda=1$, we have
$$
g=\begin{pmatrix}
\frac{1- u_2^{2}}{1-\left(u_1^{2}+u_2^{2}\right)} & \frac{ u_1 u_2}{1-\left(u_1^{2}+u_2^{2}\right)} \\
\frac{u_1 u_2}{1-\left(u_1^{2}+u_2^{2}\right)} & \frac{1- u_1^{2}}{1-\left(u_1^{2}+u_2^{2}\right)}\\
\end{pmatrix}
$$

But if we look at a typical chart of the usual sphere in $\mathbb{E}^3$
$$
\psi: (x,y)\longmapsto (x,y,\sqrt{1-x^2-y^2})
$$
and compute the metric we obtain just
$$
g=\begin{pmatrix}
\frac{1- y^{2}}{1-\left(x^{2}+y^{2}\right)} & \frac{ xy}{1-\left(x^{2}+y^{2}\right)} \\
\frac{xy}{1-\left(x^{2}+y^{2}\right)} & \frac{1- x^{2}}{1-\left(x^{2}+y^{2}\right)}\\
\end{pmatrix}
$$
so they are the same.

And for $\lambda=-1$ we have
$$
g=\begin{pmatrix}
\frac{1+ u_2^{2}}{1+u_1^{2}+u_2^{2}} & \frac{ -u_1 u_2}{1+u_1^{2}+u_2^{2}} \\
\frac{-u_1 u_2}{1+u_1^{2}+u_2^{2}} & \frac{1+ u_1^{2}}{1+u_1^{2}+u_2^{2}}\\
\end{pmatrix}
$$

But if we look at the typical model for hyperbolic plane that consists of the pseudosphere embedded in $\mathbb{M}^3$:
$$
x^2+y^2-z^2=-1
$$
with a chart given by
$$
\psi: (x,y)\longmapsto (x,y,\sqrt{1+x^2+y^2})
$$
we obtain the inherited metric in $\mathbb{R}^2$ (keep an eye: inherited from the Minkowski metric, not the Euclidean one):
$$
g=\begin{pmatrix}
\frac{1+ y^{2}}{1+x^{2}+y^{2}} & \frac{ -xy}{1+x^{2}+y^{2}} \\
\frac{-xy}{1+x^{2}+y^{2}} & \frac{1+ x^{2}}{1+x^{2}+y^{2}}\\
\end{pmatrix}
$$
so we are done.

Title: Parabolic, Hyperbolic, Elliptic


问:
Help in applying the boundaries of an elliptic integral
I need urgent help in determining the final result of the definite integral:

$$\int_0^{2\pi}\left (b\cos(x) + \sqrt{a^2 - b^2\sin^2(x)}\right)\, \mathrm{d}x.$$
I know, by using an online integral calculator, the indefinite integral gives:

$$|a| E\left(x\left|\frac{b^2}{a^2}\right.\right) + b \sin(x) + C$$

But, since it is an elliptical integral (of the second kind with parameter $m=k^2$), which is out of the scope of the mathematics I have studied and been familiar with, I cannot literally do anything about it.

Could someone, acquainted with elliptic integrals, apply the boundaries of the integral for me, please?

Please, I need this so badly. Any help would be appreciated.

答:
Note that
$$\int_0^{2\pi}\sqrt{a^2-b^2\sin^2x}\,dx=4|a|\int_0^{\pi/2}\sqrt{1-(b/a)^2\sin^2x}\,dx=4|a|E(k=b/a)$$
and this is the complete elliptic integral. The other term in the original integral evaluates to zero, so the final answer is
$$4|a|E(b/a)$$


问:
How to determine the residue of an (arithmetic) Dirichlet series
Consider a multiplicative function $f(n)$ that we write in the form $f(n) = h(n)n^a$ for a certain $a>0$ and $h(n)$ a multiplicative function such that $h(n) \asymp 1$ (more precisely, we can take $h(p^k) = (1-p^{-2})^3$ for all $k>0$, even if in fact I am interested in this function with two different values of $h(p)$ and $h(p^2)$). I denote the Dirichlet series $D_f(s)$ associated to $f$.

My question is: what is known about the poles and residues of $D_f$?

I can write what I believe to be true, but maybe extra hypotheses are needed on the function $h(n)$. At least $D_f$ converges for $\Re(s)>a+1$ by the bound on $f(n)$. I think it has a pole at $s=a+1$, and meromorphic continuation up to $\Re(s)>a$. I am particularly interested in the residue of $D_f(s)$ at $s=a$. I guess it is
$$R := \prod_p \left(1 - \frac 1p \right) \left( 1 + \frac{h(p)}{p} + \frac{h(p^2)}{p^2} + \cdots \right)$$

I would like to know if this is true, even up to adding some hypothesis, and why it is so. I thought of it as follows: we know the asymptotics
$$\sum_{n<x} f(n) = \sum_{n<x} h(n)n^a \sim R \frac{x^{a+1}}{a+1}$$
and we know that the Dirichlet series is related by Mellin transform to this partial sum, more precisely
$$D_f(s) = s \int_1^\infty x^{-s-1} \left(\sum_{n<x} f(n) \right) dx \sim \frac{Rs}{a+1} \int_1^\infty x^{a - s} dx \sim \frac{Rs}{(a+1)(a + 1 - s)} $$

Here the residue at $s=a+1$ is obviously $R$, but of course I cannot substitute that roughly the equivalent.

答:
With $h(n)=\prod_{p| n} (1-p^{-2})^3$, $H(s)=\sum_n h(n)n^{-s}= \prod_p (1+(1-p^{-2})^3\frac{p^{-s}}{1-p^{-s}})$, $G(s)=\frac{H(s)}{\zeta(s)}=\sum_n g(n)n^{-s}$ you get that
$$G(s)= \prod_p (1-p^{-s})(1+(1-p^{-2})^3\frac{p^{-s}}{1-p^{-s}})=\prod_p (1-p^{-2s}+(1-p^{-s})((1-p^{-2})^3\frac{p^{-s}}{1-p^{-s}}-p^{-s})) $$
from which $g(p)=0,g(p^k)=O(1)$ so that $\sum_{n\le x} |g(n)| = O(x^{1/2+\epsilon}),\sum_{n> x} g(n)/n = O(x^{-1/2+\epsilon})$ and

$$\sum_{n\le x}h(n)= \sum_{d\le x}g(d) \lfloor x/d\rfloor= \sum_{d\le x}x\frac{g(d)}{d}+\sum_{d\le x} O(g(d))=x (G(1)+O(x^{-1/2+\epsilon}))+O(x^{1/2+\epsilon})$$

Next $\log (1+(1-u)^3 \frac{v}{1-v})$ is analytic for $|u|<2r,|v|<2r$ (we don't really care of $r$ just that it is $\in (0,1/2)$) so that for $|u|\le r,|v|\le r$, $\log (1+(1-u)^3 \frac{v}{1-v})=\sum_{m,l} c_{m,l} u^l v^m=\sum_m c_{m,0} v^m+O(u)$ from some coefficients $c_{m,l}\in \Bbb{Q}$ and for $|p^{-s}|\le r$ $$\log (1+(1-p^{-2})^3\frac{p^{-s}}{1-p^{-s}})=\sum_{m\le M} c_{m,0}p^{-sm} +O(p^{-sM})+O(p^{-2})$$

Thus for $\Re(s) > 1/M$ $$\log H(s)-\sum_{p\le r^{-M}}\log (1+(1-p^{-2})^3\frac{p^{-s}}{1-p^{-s}})= \sum_{m\le M}c_{m,0}\sum_{j\le M} \frac{\mu(j)}{j} \log \zeta(smj))+\sum_p O(p^{-sM})$$ $$=\sum_{d\le M} \log \zeta(sd)\sum_{m| d} c_{m,0} \frac{\mu(d/m)}{d/m}+\sum_p O(p^{-sM})$$

Since the $\sum_p O(p^{-sM})$ term is analytic for $\Re(s)>1/M$ you got the analytic continuation of $H(s)$ to $\Re(s)>1/M$ with some pole/zero/branch point at $1/d$ and $\rho/d$ whenever $\sum_{m| d} c_{m,0} \frac{\mu(d/m)}{d/m}$ is non-zero and/or not an integer, plus some trivial zeros coming from the Euler factors.

It is supposedly obvious that $\sum_{m| d} c_{m,0} \frac{\mu(d/m)}{d/m}$ won't be zero for all $d$ large enough so that $H(s)$ has infinitely many poles/zeros/branch points accumulating near $\Re(s)=0$ (a natural boundary with no analytic continuation beyond).


问:
Roll a die 10 times. What is the probability that the number of ones and sixes equals four?
Suppose a die is tossed 10 times. Let $X$ be the number of ones and let $Y$ be the number of sixes.

How would you find the probability that $P\left(X+Y=4\right)$?

My attempt so far:

Applying the Binomial Distribution, I can find the probability that $X$ or $Y$ equals $4$, but I do not know how to put these pieces together to find $P\left(X+Y=4\right)$.

答:
You can calculate the probability also in an elementary way as follows:

You are interested in sequences of length $10$ consisting of the numbers $1,2,3,4,5,6$:

Number of all possible such sequences: $\color{blue}{6^{10}}$

Now for the number of sequences with the sum of occurrences of $1$'s and $6$'s equals $4$:

number of ways to choose 4 places out of $10$ for $1$'s and $6$'s: $\color{blue}{\binom{10}4}$

number of possible fillings of these $4$ places with $1$'s and $6$'s: $\color{blue}{2^4}$

number of ways to fill the remaining $6$ places with the other $4$ numbers: $\color{blue}{4^6}$

All together:

$$P(X+Y=4)= \color{blue}{\frac{\binom{10}4\cdot 2^4\cdot 4^6}{6^{10}}}$$

Or more general for $k=0,\ldots ,10$:

$$P(X+Y=k)=\frac{\binom{10}k\cdot 2^k\cdot 4^{10-k}}{6^{10}}=\binom{10}k\left(\frac 13\right)^{k}\left(\frac 23\right)^{10-k}$$
which corresponds to the other answers using Bernoulli directly.


问:
Inequality solving as a function of other variable
Suppose I need to show that the following holds for any $x \in D$,
$$ Ay^2(x) + By(x) + Cx + E \geq 0, $$
but the expression becomes quite complicated when I insert $y(x)$. Is the following attempt mathematically correct, if so possibly why? (It seems to me intuitively correct.)

Attempt: Pretend (i) the left-hand side as as function of $y$, and (ii) $x$ as a parameter. Then solve for $y$ in which inequality holds (the condition will be as a function of $x$ at the end, e.g., $M_1(x) \leq y(x) \leq M_2(x)$, where $M$'s are some functions.) Then, I will prove the conditions with $M$ for any $x \in D.$

What do you think?

答:
Lets try to see what happens with this, and WLOG we may set $A \to 1, B \to 2B$.  Then essentially you are setting:
$$(y + B)^2 \geqslant B^2-Cx - E$$

Now if $RHS \leqslant 0$, this holds trivially for any $y(x)$, so $Cx \geqslant  B^2-E$ is part of the solution set, and this is easy to get knowing the sign of $C$.

Else, we have $z(x) = \sqrt{B^2-Cx-E} > 0$, and $(y+z+B)(y-z+B)\geqslant 0$ so you do get "boundaries" as $ y \in \Bbb R \setminus (B-z, B+z)$.

While we know $y, z$, inverting those functions and finding the remaining solution set for $x$ doesn't seem particularly easier than the original equation...


问:
The domain of the functions f(x)=√log1/2(2x-x^2) is?
Given that $$f(x)=\sqrt{\log(2x-x^2)}$$ then find the domain of $f(x)$

My solution:

$\log(2x-x^2)\geq0$

$\implies 2x-x^2\geq1 \implies x\leq1$

Also,

$2x-x^2\gt0$

$\implies x\gt0$ ,$\ x\lt2$

On finding the intersection,

$x\in(0,1]$

I can't figure out where I went wrong here, The given solution is ${\{1\}}$

答:
Notice that:

$$2x - x^2 \ge 1 \iff x^2-2x+1 \le 0 \iff (x-1)^2\le 0 \iff x=1$$

Besides, the second inequality $2x - x^2 > 0$ is implied by the first.


问:
Quick Conditional Probability Question
Been going through lecture notes (ones I made myself, so who knows how accurate they are - or rather - they most certainly are not) and can't seem to understand this one example

Example: Two horses - A and B. A wins with probability 0.5, B wins with probability 0.3. What is the probability that B wins?

The calculation that follows:
$$P(B|A^C)=\frac{P(B\cap A^C)}{P(A^C)}=\frac{0.3}{0.5}=0.6 $$

Now, I have no trouble with the theory, just understanding what is actually meant by the question. My questions:

1) Should the question really be "What is the probability that B wins, given that A doesn't?"

2) Is $P(B)=P(B\cap A^C)$ since $P(A\cap B)=\emptyset$ (or is that true?)

Perhaps 1) and 2) are obviously true and I just forgot to add that sentence in my notes. But I want to be sure (and be able to sleep at night :) )

Thanks for any help.

答: Both observations are correct and "obvious". But what's obvious at the time (or to an expert) isn't always obvious when you're revising. Let this be a lesson to you about high quality note-taking. (Even if it makes me a massive hypocrite)</p>

Title: Quick Conditional Probability Question


问:
How to solve $ \lim_{R \rightarrow \infty} \int_{-R}^{R} \frac{x}{1+x^3}$
$$ \lim_{R \rightarrow \infty} \int_{-R}^{R} \frac{x}{1+x^3} \ dx$$

It is allowed to use Taylor series if that helps in any way.

答:
Using sum of cubes we have $x^3+1 = (x+1)(x^2-x+1)$. This can be used to find the partial fraction decomposition
$$\frac{x}{x^3+1} = \frac{x+1}{3(x^2-x+1)}-\frac{1}{3(x+1)}.$$
Now, the first term on the right has finite positive integral on the interval $(-1,0)$ (it is not too hard to check that it is nonnegative and bounded). The second term, however, has divergent integral on this interval. Indeed, with $u=x+1$, we have
$$\int_{-1}^0 \frac{dx}{x+1} = \int_0^1\frac{du}{u},$$
and it is a standard exercise that the latter is infinite (compute proper integrals on $(\epsilon, 1)$ as $\epsilon\to 0^+$ to see this).

Since our original function is the difference of something with finite positive integral and divergent integral on $(-1,0)$, it has divergent integral there as well. Thus if $R>1$ the integral inside your limit diverges.


问:
Confusing functional equation problem (application of Cauchy equation when $ f \left( \frac 1 x \right) = \frac { f ( x ) } { x ^ 2 } $)
I came across the following problem concerning functional equations.

Let $ f : \mathbb R \to \mathbb R $ be an additive function such that $ f \left( \frac 1 x \right) = \frac { f ( x ) } { x ^ 2 } $. Prove that $ f ( x ) = c x $, $ c \in \mathbb R $.

The following solution was given.

Since $ f $ is an additive function we have $ f ( x - y ) = f ( x ) - f ( y ) $ for all $ x , y \in \mathbb R $. Let $ x \ne 0 , 1 $. Then
$$ \frac 1 { x - 1 } - \frac 1 x = \frac 1 { x ( x - 1 ) } \text , $$
and we get
$$ f \left( \frac 1 { x - 1 } \right) - f \left( \frac 1 x \right) = f \left( \frac 1 { x ( x - 1 ) } \right) \text . $$
Hence
$$ \frac { f ( x - 1 ) } { ( x - 1 ) ^ 2 } - \frac { f ( x ) } { x ^ 2 } = \frac { f \big( x ( x - 1 ) \big) } { x ^ 2 ( x - 1 ) ^ 2 } \text , $$
which is equivalent to the identity
$$ x ^ 2 f ( x - 1 ) - ( x - 1 ) ^ 2 f ( x ) = f \left( x ^ 2 - x \right) \text . $$
Thus
$$ x ^ 2 \big( f ( x ) - f ( 1 ) \big) - ( x - 1 ) ^ 2 f ( x ) = f \left( x ^ 2 \right) - f ( x ) \text , $$
and therefore
$$ f \left( x ^ 2 \right) = 2 x f ( x ) - x ^ 2 f ( 1 ) \text . \tag 1 \label 1 $$
Now, replace $ x $ by $ x + x ^ { - 1 } $ in this identity to obtain
$$ f \left( x ^ 2 + x ^ { - 2 } + 2 \right) = 2 \left( x + x ^ { - 1 } \right) \left( x + x ^ { - 1 } \right) - \left( x ^ 2 + x ^ { - 2 } + 2 \right) f ( 1 ) \text . $$
Thus
$$ f \left( x ^ 2 \right) + x ^ { - 4 } f \left( x ^ 2 \right) + f ( 2 ) = \\
2 x f ( x ) + 4 x ^ { - 1 } f ( x ) + 2 x ^ { - 3 } f ( x ) - \left( x ^ 2 + x ^ { - 2 } + 2 \right) f ( 1 ) \text , $$
and using \eqref{1} again we get
$$ f ( x ) = \left( \frac { f ( 2 ) + 2 f ( 1 ) } 4 \right) x $$
for $ x \ne 0 , 1 $. But $ f ( 2 ) = 2 f ( 1 ) $ and therefore $ f ( x ) = f ( 1 ) x $, which holds also for $ x = 0 , 1 $.

I am actually not able to understand the following:

Since $ f $ is additive, why did the author choose to consider $ f ( x - y ) = f ( x ) - f ( y ) $ rather than $ f ( x + y ) = f ( x ) + f ( y ) $?

How did the author conclude from $ \frac 1 { x - 1 } - \frac 1 x = \frac 1 { x ( x - 1 ) } $ that $ f \left( \frac 1 { x - 1 } \right) - f \left( \frac 1 x \right) = f \left( \frac 1 { x ( x - 1 ) } \right) $?

I spent a considerable amount of time to understand the aforementioned, but failed. A detailed and easiest possible explanation would really be helpful.

答:
Second bullet first.  In
$$  f(u - v) = f(u) - f(v)  $$
make the substitution \begin{align*}
u &\mapsto \frac{1}{x-1}  \\
v &\mapsto \frac{1}{x}  \text{.}
\end{align*}
Then
$$  f\left(\frac{1}{x-1} - \frac{1}{x}\right) = f\left(\frac{1}{x-1}\right) - f\left(\frac{1}{x}\right)  \text{.}  $$
Of course, the argument to $f$ on the left-hand side is $\displaystyle \frac{1}{x(x-1)}$.

First bullet:  (1) depends on a great deal of cancellation among the terms of the previous equation.  This cancellation requires subtraction; addition does not exhibit the necessary cancellation.

Finding the useful expression usually takes some experimentation.  Pick a few simple expressions and see that happens when you replace $x$ with those expressions.  You have a mysterious operator in front of you start studying how it distorts the inputs.

Notice that the functional equation is equivalent to
$$  x f(x) = \frac{1}{x} f(1/x)  \text{.}  $$

If additivity is going to be of any use on the left, you need something like $f(a-x)$ to cancel against the $x f(x)$.
\begin{align*}
(a-x) f(a-x) &= \frac{1}{a-x} f(1/(a-x))  \\
(a-x)^2 (f(a) - f(x)) &= f(1/(a-x))
\end{align*}
This doesn't work because we do not get $-x f(x)$, the minus sign from the two instances of $-x$, always cancel.  So try more things: raise the degree, try rational functions.  Look at logs and exponentials -- try things.


问:
Can non-irreducible Markov chains converge?
I know that Markov chains that are irreducible and aperiodic are guaranteed to converge and have an invariant distribution, but can a non-irreducible one do too? If so, what would be an example?

Also, is there a different between an invariant/stationary distribution and a convergence for MC?

答:
Concerning your first question:

Just take an irreducible aperiodic chain and add a transient state from which you can only "flow" into the states of the irreducible chain (which then form a recurrent, aperiodic communication class). The so obtained chain is reducible.

Concerning your second question:

Yes, there is a difference. There are Markov chains with several stationary distributions where the initial distribution decides where the chain converges to. And there are Markov chains with only one stationary distribution but where convergence depends heavily on the initial distribution (for the last satement see for example here p. 31).


问:
Show that $f’(0)$ exists and equals 1.
Let $f : \mathbb{R} \to \mathbb{R}$ be continous. Assume that $f’(x)$ exists for all $x \neq 0$ and $ \lim_{x\to\ 0} f'(x) = 1$. Show that $f’(0)$ exists and $f’(0) = 1$

My attempt:
$$1 = \lim_{x\to0} \lim_{h\to0}\frac{f(x + h) - f(x)}{h} = \lim_{h\to0}\frac{f(0 + h) - f(0)}{h} = f’(0)$$

I don’t think that the limit interchange that I have done is correct. Can someone help me out with how to do this.

答:
I think the post Martin R linked says something similar, but this is a standard application of the MVT: Fix $h>0$ and consider $\frac{f(h) - f(0)}{h}$, then by the mean value theorem you can find a point $a \in (0,h)$ such that $\frac{f(h) - f(0)}{h} = f'(a)$. Now take $h \to 0$. What happens to $a$? Keep in mind that $a$ depends on $h$.

Also, interchanging limits is not a good idea unless you're appealing to a specific theorem/result that lets you do this. In general even "easy" limits can't be changed.

Title: Show that $f’(0)$ exists and equals 1.


问:
Finding $\int \frac {e^x \sqrt{e^x - 1}}{e^x+3}\,dx$ by substituting $t^2=e^x-1$
Let's consider the following integral:

$$\int \frac {e^x \sqrt{e^x - 1}}{e^x+3}\,dx$$

And let's take the substitution $t^2=e^x-1$. Then, we get $2tdt=e^xdx$ and so:

$$\int \frac {e^x \sqrt{e^x - 1}}{e^x+3}\,dx = 2\int \frac {t \sqrt{t^2}}{t^2+4}\,dt$$

So, can we take $\sqrt{t^2}=t$ there? It seems to me that we could do that only if we say at the beginning, when we took the substitution, that $t$ is non-negative. However, I've never seen that anybody has  written anything like that in any solution of antiderivative problems. So, if we want to be formal, should we write that we take non-negative $t$?

答:
It makes a difference on how we define our substituition.By taking $e^x-1=t^2$ we could have $t=\pm\sqrt{e^x-1}$.

To avoid this problem I would take $\sqrt{e^x-1}=t$ which would automatically mean $t\ge 0$. Hence we have to evaluate $$2\int \frac{t^2dt}{t^2+4}$$ which is easy to do by hand...


问:
Darboux chart compatible with Lagrangian submanifold
Let $(M,\omega)$ be a symplectic manifold of dimension $2n$ and let $L\subset M$ be a Lagrangian submanifold, that is $\dim L=n$ and $$\omega_p(u,v) = 0$$ for all $u,v\in T_pL$ holds for every point $p\in L.$ We wish to show that for each $p\in L,$ there exist local coordinates $x_1,\dots,x_n,y_1,\dots,y_n$ for a neighborhood $U\subset M$ of $p$ so that $$\omega = \sum_{i=1}^ndy_i\wedge dx_i$$ and $$L\cap U = \{x_1=\cdots=x_n=0\}.$$ The first condition is that the coordinates are a Darboux chart, which we know exists from Darboux's theorem. I am familiar with the proof of this theorem using linear algebra and Moser's trick, but am unsure how to adapt it so that the second condition holds. A different approach described here uses the Weinstein Lagrangian neighborhood theorem to take $M=T^*L$ without loss, in which case the usual coordinates for the cotangent bundle meet our conditions. However, I am unfamiliar with a lot of what is being used in the proofs I have found of the Lagrangian neighborhood theorem and want to find a more direct approach. Does anyone have any suggestions? Thanks!

答:
Pick any local Darboux coordinates. We can suppose (after a bit of renaming) that $L$ is given in these coordinates by equations $x_i=g_i(p_1,\dots,p_n)$ for some functions $g_i$. $L$ being Lagrangian is equivalent to $\sum_i dp_i \wedge dg_i=0$. So just set $x'_i=x_i-g_i$ and you have your Darboux coordinates $x'_i,p_i$.

(As I wrote in the comment above, Weinstein LN theorem is more difficult and requires a version of the Moser's trick.)


问:
General Solution of Diophantine equation
Having the equation:

$$35x+91y = 21$$

I need to find its general solution.

I know gcf $(35,91) = 7$, so I can solve $35x+917 = 7$ to find $x = -5, y = 2$. Hence a solution to $35x+91y = 21$ is $x = -15, y = 2$.

From here, however, how do I move on to finding the set of general solutions? Any help would be very much appreciated!

Cheers

答:
Hint: If $35x + 91y = 21$ and $35x^* + 91y^* = 21$ for for some $(x,y)$ and $(x^*, y^*)$, we can subtract the two equalities and get $5(x-x^*) + 13(y-y^*) = 0$. What does this tell us about the relation between any two solutions?

Now, $5$ and $13$ share no common factor and we're dealing with integers, $13$ must divide $(x-x^*)$. In other words, $x = x^* + 13k$ for some integer $k$ and substituting it into the equality yields $y = y^* - 5k$. Thus, once you have one solution $(x^*,y^*)$, all of them can be expressed as $(x^*+13k, y^*-5k)$.


问:
Prove that $f(x,y)$ is continuous on $\mathbb{R^2}$
Prove that $$f(x,y) =\begin{cases}  e^{\frac{-1}{|x-y|}} &\text{for } x\neq y\\
          0  &\text{ for  } x =  y\end{cases}$$ is continuous on $\mathbb{R^2}$. How much $proof$ has to go into showing that when $x \neq y$ $f(x,y)$ is continuous. It seems clearly true for since we have the constraint that $x\neq y$. For the $f(x,y) =0$ case where can I take this?

答:
Hint

$f(x,y)=g(x-y)$ where

$$g(x)=e^{-\frac{1}{\vert x \vert}}$$ for $x \neq 0$ and $g(0)=0$.

$g$ is continuous as $\lim\limits_{x \to 0} g(x)=0$. Finally $f$ is continuous being the composition of two continuous maps.