问:
Determine the smallest positive integer n, which has exactly 144 distinct divisors and there are 10 consecutive integers among these divisors.
We know that after the prime factorization of that number when the powers of those prime factors +1 when multiplied with each other would result in 144.
144 = (a+1)(b+1). . .
We also know that there are 10 consecutive numbers among these divisors so the number must be divisible by 2,3,5,7.

I don't know what to do now , what to do ?

答: What you say is true, but you should apply it.  Factor $144$.  You know that there are at least $4$ distinct prime factors.  There are not many patterns of exponents of primes to choose from, then not many ways to choose the primes so they include $2,3,5,7$.  Make a list of the patterns of exponents, then the smallest numbers that satisfy each pattern.  Sort them and list the divisors of each, starting from the smallest.  I suspect you will not need to search far.  What is the factorization of $LCM(1,2,3\ldots 10)?$  What is the minimum you need to multiply it by to get $144$ distinct factors?  That gives an upper bound to the answer.


问:
How many students in the school?

If they put the rows of 13, there are 8 students left; if they put the rows of 15, there are 3 students left and if they put the rows of 17, there are 9 students left. How many students are there given that the total students are < 5000?

The following congruences are: $$x\equiv 8 \pmod {13} \\ x\equiv 3 \pmod {15} \\ x\equiv 9 \pmod {17}$$

I'm still new to this, how can I apply Chinese Remainder Theorem to these congruences to find the total students?

ContentLicense: CC BY-SA 4.0

Title: How many students in the school?

Body:
Let $N$ be the number of students. Let $M = 13\times 15 \times 17 = 3315$

$\qquad\ M_1 = 15 \times 17 = \color{#c00}{255},\,\ \ y_1 = 255^{-1} \equiv \ \ 8^{-1}\equiv \ \ \color{#c00}{5}\ \pmod{13} $

$\qquad\ M_2 = 13 \times 17 = \color{#0a0}{221},\,\ \ y_2 = 221^{-1} \equiv 11^{-1} \equiv \color{#0a0}{11}\pmod{15} $

$\qquad\ M_3 = 13 \times 15 = \color{#90f}{195},\,\ \ y_3 = 195^{-1} \equiv\ \ 8^{-1}\equiv \color{#90f}{15} \pmod{17} $

Thus $\ N \equiv 8 \times \color{#c00}{255 \times 5}\, +\, 3 \times \color{#0a0}{221 \times 11}\, +\, 9 \times \color{#90f}{195 \times 15}\,\equiv\, 723\pmod{3315} $

So the number of students $\, N =  723 + 3315 \times k $

With $\,N < 5000,\ N = 4038\,$ with $\,k = 1$

You can read more here (Vietnamese wiki page on CRT)


问:
Show that $\lim_{n\to \infty}\int_{-\pi}^{\pi}f(x)g(nx)\,dx=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)\,dx\int_{-\pi}^{\pi}g(x)\,dx$
Let $f,g\in PC(2\pi)$ continuous. Show that $$\lim_{n\to \infty}\int_{-\pi}^{\pi}f(x)g(nx)\,dx=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)\,dx\int_{-\pi}^{\pi}g(x)\,dx$$

Where if $f\in PC(2\pi)$ then $\int_{-\pi}^{\pi}f(x)\,dx=\int_c^{c+2\pi}f(x)\,dx \quad\forall c\in\mathbb{R}.$

I've been struggling with this proof. Any suggestions would be great!

答:
Here's a sketch:

Let $\mu_g = \int_{-\pi}^{\pi} g(x) \,dx$.  Fix $\epsilon > 0$.  Since $f$ is continuous, it's also uniformly continuous, so let $n$ be big enough so that $|f(x) - f(y)| < \epsilon$ whenever $|x-y| < 2\pi/n$.  Now we can write
\begin{align*}
\int_{-\pi}^{\pi} f(x) g(nx) \,dx \ &= \ \frac1n \int_{-n\pi}^{n\pi} f(y/n) g(y) \,dy \ = \ \sum_{j=-n}^{n-2} \frac1n \int_{j\pi}^{(j+2)\pi} f(y/n) g(y) \,dy \\
&= \ \sum_{j=-n}^{n-2} \frac1n \int_{j\pi}^{(j+2)\pi} \Big(  f(j\pi/n) \pm \epsilon \Big) g(y) \,dy \ = \ \mu_g \cdot \sum_{j=-n}^{n-2} \frac1n  \Big( f(j\pi/n) \pm \epsilon \Big).
\end{align*}

Now the summation can just be recognized as a Riemann sum which converges to $\frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \,dx$ as $n \to \infty$.


问:
How to compute $ P.V. \int_{-\infty}^{\infty} \frac{e^{iwx}}{|w|} \,dw $?
I came across the integral:
$$
\mbox{P.V.}\int_{-\infty}^{\infty} {\mathrm{e}^{\mathrm{i}\omega x} \over
\left\lvert\omega\right\rvert}
\,\mathrm{d}\omega
$$
in my research, and can’t figure it out after several hours of trial and error.

I didn’t know how to use Contour integration techniques on it because the pole is an absolute value, so I tried generating a power series for the exponential kernel, but ended up with a log function that blew up.

May you please explain how I can evaluate such an integral $?$. Wolfram has the answer, but I care more about the thought process than anything.

Many thanks in advance $!$.

答:
1. The integral in question does not exist in the Cauchy principal value sense

$$ \lim_{\substack{\epsilon \to 0^+ \\ R \to \infty}} \int_{\epsilon < |x| < R} \frac{e^{i\omega x}}{\left| x \right|} \, \mathrm{d}x $$

precisely because of the logarithmic divergence mentioned in @Felix Marin's answer.

2. If we interpret the integral as the Fourier transform of a suitable distribution (a.k.a. generalized function), however, then it can be shown that

$$ \mathcal{F}\left[\operatorname{pf}\frac{1}{\left|x\right|}\right](\xi) = -2(\gamma + \log\left|\xi\right|). \tag{*} $$

Here, $\operatorname{pf}\frac{1}{\left|x\right|}$ is the distributional derivative of $\operatorname{sgn}(x)\log\left|x\right|$ and $\mathcal{F}$ is the Fourier transform of tempered distributions which extends

$$ \hat{f}(\xi) = \mathcal{F}\left[f\right](\xi) = \int_{-\infty}^{\infty} f(x)e^{i\omega x} \, \mathrm{d}x $$

for Schwarz functions $f \in \mathcal{S}(\mathbb{R})$. (More precisely, for a tempered distribution $T$ over $\mathbb{R}$, its Fourier transform $\hat{T}$ is defined via the relation $\langle \hat{T}, \varphi \rangle = \langle T, \hat{\varphi} \rangle$ for all $\varphi \in \mathcal{S}(\mathbb{R})$.)

We first note that

$$ \operatorname{pf}\frac{1}{\left|x\right|} = \bigl[ \left|x\right| (\log\left|x\right|-1) \bigr]'' $$

in distributional sense. Since $\left|x\right|(\log\left|x\right|-1)$ is a continuous function with polynomial growth, it follows from the structure theorem that $\operatorname{pf}\frac{1}{\left|x\right|}$ is a tempered distribution, hence its Fourier transform is well-defiend.

We first demonstrate a less rigorous argument. Using the idea of regularization, we may heuristically argue that

\begin{align*}
\mathcal{F}\left[\operatorname{pf}\frac{1}{\left|x\right|}\right](\xi)
&= \lim_{\epsilon \to 0^+} \left< \operatorname{pf}\frac{1}{\left|x\right|}, e^{-\epsilon x^2 + i\xi x} \right> \\
&= - \lim_{\epsilon \to 0^+} \left< \operatorname{sgn}(x)\log\left|x\right|, \bigl( e^{-\epsilon x^2 + i\xi x} \bigr)' \right> \\
&= - \lim_{\epsilon \to 0^+} \int_{-\infty}^{\infty} \operatorname{sgn}(x)\log\left|x\right| (-2\epsilon x + i\xi) e^{-\epsilon x^2 + i\xi x} \, \mathrm{d}x.
\end{align*}

Then

\begin{align*}
&\int_{-\infty}^{\infty} \operatorname{sgn}(x)\log\left|x\right| (-2\epsilon x + i\xi) e^{-\epsilon x^2 + i\xi x} \, \mathrm{d}x \\
&= 2 \int_{0}^{\infty} \log x (-2\epsilon x \cos(\xi x) - \xi \sin(\xi x)) e^{-\epsilon x^2} \, \mathrm{d}x \\
&= 2 \int_{0}^{1} \log x \bigl( e^{-\epsilon x^2} \cos(\xi x) - 1 \bigr)' \, \mathrm{d}x + 2 \int_{1}^{\infty} \log x \bigl( e^{-\epsilon x^2} \cos(\xi x) \bigr)' \, \mathrm{d}x \\
&= 2 \int_{0}^{1} \frac{1 - e^{-\epsilon x^2} \cos(\xi x)}{x} \, \mathrm{d}x - 2 \int_{0}^{1} \frac{e^{-\epsilon x^2} \cos(\xi x)}{x} \, \mathrm{d}x.
\end{align*}

Letting $\epsilon \to 0^+$ gives

\begin{align*}
\mathcal{F}\left[\operatorname{pf}\frac{1}{\left|x\right|}\right](\xi)
&= - 2 \int_{0}^{1} \frac{1 - \cos(\xi x)}{x} \, \mathrm{d}x + 2 \int_{0}^{1} \frac{\cos(\xi x)}{x} \, \mathrm{d}x \\
&= -2(\gamma + \log\left|\xi\right|)
\end{align*}

as desired.

Here is a more rigorous proof. Let $\varphi \in \mathcal{S}(\mathbb{R})$. Then by the definition and the properties of the pairing $\langle \cdot, \cdot \rangle$,

\begin{align*}
\left< \mathcal{F}\left[\operatorname{pf}\frac{1}{\left|x\right|}\right], \varphi \right>
&= \left< \operatorname{pf}\frac{1}{\left|x\right|}, \hat{\varphi} \right>
= - \left< \operatorname{sgn}(x) \log\left|x\right|, \hat{\varphi}' \right> \\
&= - \int_{-\infty}^{\infty} \hat{\varphi}'(x) \operatorname{sgn}(x) \log\left|x\right| \, \mathrm{d}x,
\end{align*}

where the last integral is the usual Lebesgue integral. Then

\begin{align*}
&= - \int_{0}^{\infty} (\hat{\varphi}'(x) - \hat{\varphi}'(-x)) \log x \, \mathrm{d}x \\
&= - \int_{0}^{1} (\hat{\varphi}(x) + \hat{\varphi}(-x) - 2\hat{\varphi}(0))' \log x \, \mathrm{d}x \\
&\qquad  - \int_{1}^{\infty} (\hat{\varphi}(x) + \hat{\varphi}(-x))' \log x \, \mathrm{d}x \\
&= \int_{0}^{\infty} \frac{\hat{\varphi}(x) + \hat{\varphi}(-x) - 2\hat{\varphi}(0) \mathbf{1}_{[0,1]}(x)}{x} \, \mathrm{d}x \\
&= \lim_{\substack{\epsilon \to 0^+ \\ R\to\infty}} \int_{\epsilon}^{R} \frac{\hat{\varphi}(x) + \hat{\varphi}(-x) - 2\hat{\varphi}(0) \mathbf{1}_{[0,1]}(x)}{x} \, \mathrm{d}x. \tag{1}
\end{align*}

Assuming $0 < \epsilon < 1 < R$ without losing the generality, the last integral can be computed by the Fubini's theorem:

\begin{align*}
&\int_{\epsilon}^{R} \frac{\hat{\varphi}(x) + \hat{\varphi}(-x) - 2\hat{\varphi}(0) \mathbf{1}_{[0,1]}(x)}{x} \, \mathrm{d}x \\
&= \int_{-\infty}^{\infty} \varphi(\xi) \biggl( \int_{\epsilon}^{R} \frac{e^{ix\xi} + e^{-ix\xi} - 2 \cdot \mathbf{1}_{[0,1]}(x)}{x} \, \mathrm{d}x \biggr) \, \mathrm{d}\xi \\
&= 2 \int_{-\infty}^{\infty} \varphi(\xi) \left( \operatorname{Ci}(R\left|\xi\right|) - \operatorname{Ci}(\epsilon\left|\xi\right|) - \log(1/\epsilon) \right) \, \mathrm{d}\xi,
\end{align*}

where $\operatorname{Ci}(\cdot)$ is the cosine integral. Then by using the asymptotic formulas for $\operatorname{Ci}(\cdot)$,

$$ \operatorname{Ci}(R\left|\xi\right|) = (\log(R\left|\xi\right|)) \mathbf{1}_{\{\left|\xi\right|< 1/R\}} + \mathcal{O}(1) $$
$$ - \operatorname{Ci}(\epsilon\left|\xi\right|) - \log(1/\epsilon) = -(\log\left|\xi\right|) \mathbf{1}_{\{\left|\xi\right| < 1/\epsilon\}} + (\log \epsilon) \mathbf{1}_{\{\left|\xi\right| > 1/\epsilon\}} + \mathcal{O}(1) $$

uniformly in $\epsilon$, $R$, and $\xi$. Since $\varphi$ is rapidly decreasing, it is not hard to show that

\begin{align*}
&2 \int_{-\infty}^{\infty} \varphi(\xi) \left( \operatorname{Ci}(R\left|\xi\right|) - \operatorname{Ci}(\epsilon\left|\xi\right|) - \log(1/\epsilon) \right) \, \mathrm{d}\xi \\
&= 2 \int_{-\infty}^{\infty} \varphi(\xi) \left( \operatorname{Ci}(R\left|\xi\right|) - \operatorname{Ci}(\epsilon\left|\xi\right|) - \log(1/\epsilon) \right)\mathbf{1}_{\{\epsilon < \left|\xi\right|<R\}} \, \mathrm{d}\xi + o(1)
\end{align*}

uniformly as $\epsilon \to 0^+$ and $R\to \infty$, and then by the dominated convergence theorem, the limit in $\text{(1)}$ is computed as

\begin{align*}
\text{(1)}
&= 2 \int_{-\infty}^{\infty} \varphi(\xi) \lim_{\substack{\epsilon \to 0^+ \\ R\to\infty}} \left( \operatorname{Ci}(R\left|\xi\right|) - \operatorname{Ci}(\epsilon\left|\xi\right|) - \log(1/\epsilon) \right)\mathbf{1}_{\{\epsilon < \left|\xi\right|<R\}} \, \mathrm{d}\xi \\
&= -2 \int_{-\infty}^{\infty} \varphi(\xi) (\gamma + \log\left|\xi\right|) \, \mathrm{d}\xi.
\end{align*}

By the definition of the Fourier transform, this implies $\text{(*)}$ as desired.


问:
Show the inequality $(n+1)^z-n^z \geq n^{z-1}(z-1)$ for $n, z \in \mathbb{R}_{>0}$
How can I show that the inequality $(n+1)^z-n^z \geq n^{z-1}(z-1)$ for $n, z \in \mathbb{R}_{>0}$ always holds? Since $z$ is not a natural number we can not use the Binomial Theorem, otherwise this would be trivial.

答:
In this case, you would just use the Bernoulli's inequality:

$$\left(1+\frac 1n\right)^z - 1\geq 1+\frac zn - 1 = \frac zn > \frac{z-1}{n}$$

EDIT:

The above works when $z\leq 1$ when $z\in\mathbb{R}.$ When $z<1$, RHS is negative so the inequality is trivial.


问:
Characteristic function of a random variable with absolute value in its density function
If I want to find the characteristic function of $X$ with density function $f(x)=\frac {1}{2}e^{-\lvert x\rvert}$, is it true that I will get two characteristic functions depending on the value of $x$?

Body:
no, you won't get two different cf's:

\begin{align*}
E[e^{itX}]
&= \frac{1}{2}\int_{-\infty}^{\infty} e^{itx} e^{-|x|} dx\\
&= \frac{1}{2}\int_{-\infty}^{0} e^{itx} e^{x} dx +  \frac{1}{2}\int_0^{\infty} e^{itx} e^{-x} dx\\
&= \frac{1}{2}\int_{-\infty}^{0} e^{x(it+1)}dx +  \frac{1}{2}\int_0^{\infty} e^{x(it-1)} dx\\
&=\frac{1}{2} \left (\frac{1}{it+1} - \frac{1}{it-1} \right )\\
&= \frac{1}{2} \left(\frac{-2}{-t^2-1} \right )\\
&= \frac{1}{t^2+1}
\end{align*}


问:
Averaging a list of variances
I have a list of means and variances. I want to average them both. For mean I simply put:
$$\mu = \frac{\mu_1+\mu_2+\cdots+\mu_n}{n}$$

However, for the variance, I can't figure out to average $n$ variances. For only two I found this answer here

$$\sigma^2 = \frac{1}{2} \left(\sigma^2_1 + \sigma^2_2\right) + \left(\frac{\mu_1-\mu_2}{2}\right)^2$$

But how about $n$ variances?
thank you

答:
I will surmise that what you mean is the variance of the mean of $n$ random variables, and each has $k$ possible values.

$$\text{It's } \frac{\sigma_1^2 + \cdots + \sigma_n^2} n + \frac{(\mu_1-\overline \mu)^2 + \cdots +(\mu_n-\overline \mu)^2} n \text{ where } \overline\mu = \frac{\mu_1 + \cdots + \mu_n} n.$$

Let $x_{i,1},\ldots ,x_{i,k}$ be the possible values of the $i$ random variable for $i=1,\ldots,n.$ Then
\begin{align} \require{cancel}
& \mu_i = \frac{x_{i,1} + \cdots + x_{i,k}} k \\[10pt]
& \sigma_i^2 = \frac{(x_{i,1} - \mu_i)^2 + \cdots + (x_{i,k} - \mu_i)^2} k \\[10pt]
& \sigma_i^2 + (\mu_i-\overline \mu)^2 = \frac 1 k \sum_{j=1}^k \big( (x_{i,j} - \mu_i)^2 + (\mu_i - \overline\mu)^2 \big) \\[10pt]
= {} & \frac 1 k \sum_{j=1}^k \big( (x_{i,j} - \mu_i)^2 - 2(x_{i,j} - \mu_i) (\mu_i - \overline\mu) + (\mu_i - \overline\mu)^2 \big) \tag 1 \\
& \textbf{Why? See below.}
\end{align}
This last equality is true because
\begin{align}
& \sum_{j=1}^k \big( (x_{i,j} - \mu_i) (\mu_i - \overline\mu) \big) \\[8pt]
= {} & (\mu_i-\overline \mu)\sum_{j=1}^k (x_{i,j} - \mu_i)\\
& \text{since $\mu_i-\overline\mu$ does not change as $j$ goes from $1$ to $k$} \\[10pt]
= {} & (\mu_i-\overline\mu)\cdot0.
\end{align}

Then line $(1)$ above becomes
\begin{align}
& \frac 1 k \sum_{j=1}^k \big( (x_{i,j} - \mu_i)^2 - 2(x_{i,j} - \mu_i) (\mu_i - \overline\mu) + (\mu_i - \overline\mu)^2 \big) \\[8pt]
= {} & \frac 1 k \sum_{j=1}^k \big( (x_{i,j} - \mu_i) - (\mu_i - \overline\mu) \big)^2 = \frac 1 k \sum_{j=1}^k (x_{i,j} - \overline \mu)^2.
\end{align}

Title: Averaging a list of variances


问:
If $a^2+a b+b^2=40$ and $a^2-\sqrt{a b}+b=5$, then find $a^2+\sqrt{a b}+b$
I was given this problem to solve with elementary methods (High School level).

Knowing that
$$\begin{align}
a^2+a b+b^2 &=40 \\
a^2-\sqrt{a b}+b &=\phantom{0}5
\end{align}$$
find
$$a^2+\sqrt{a b}+b$$

I tried to look for $\sqrt{a b}$, since the requested quantity is

$$a^2+\sqrt{a b}+b=(a^2-\sqrt{a b}+b)+2\sqrt{ab}=5 + 2\sqrt{ab}$$
So I set
$$x=a^2;\;y=\sqrt{ab};\;b=z$$
and the system became
$$x+y^2+z^2=40;\;x-y-z=5$$
subtracting the two equations I got
$$z^2-z+y^2+y-35=0$$
which has one real solution when the discriminant is zero.

That is

$1-4(y^2+y-35)=0$

and then

$y=\frac{1}{2} \left(-1\pm\sqrt{142}\right)$

and finally
$$a^2+\sqrt{a b}+b=4\pm\sqrt{142}$$

I know that there are other solutions because I've found them with Wolfram Mathematica, but I couldn't find them with elementary methods.

Any help will be appreciated.

答:
One way to do this is to graph out these equations of lines.
Find the values of $a,b$ and plug into the desired output equation:

Equation of a line:
see link here.

Look at the possible values and

plug answers into the final equation:


问:
Inductive proof - Inequality
I am trying to do mathematical induction to prove an inequality. However, I got stuck in the middle of the proof. I have started my proof from the inductive step $3^{k+1}=3^k*3 ≥3k*3$. I am stuck in here. The below picture is about what I did. I also tried in a separate way. Picture number $2$ is the alternate way that I tried to prove my equation. The main question is attached to image $1$.
Using mathematical induction, prove:

$〖3n≤3〗^n$ for all natural numbers $n$

Can anyone please help me out with that?

Here is the snip of the problem(First attempt)

Second attempt

Title: Inductive proof - Inequality

Body:
\begin{gather*}
3^{n}  >3n\\
3^{n} +3 >3( n+1)\\
Now,\ 3^{n+1}  >3^{n} +3,\ \\
because\ 2\times 3^{n}  >3\ for\ all\ n.\\
Hence,\ 3^{n+1}  >3( n+1) ,\ hence\ proved.\\
\\
\end{gather*}


问:
Riemann Integration and Squeeze Theorem
Let $[a,b]\subseteq \mathbb{R}$ be a non-degenerate closed bounded interval, and let$f,g,h:[a,b]\to\mathbb{R}$ be functions. Suppose that $f$ and $h$ are integrable, and that $\int_a^bf(x)dx=\int_a^bh(x)dx$. Prove that if $f(x)\leq g(x) \leq h(x)$ for all $x\in[a,b]$, then $g$ is integrable and $\int_a^bg(x)dx=\int_a^bf(x)dx$.

I am not allowed to use squeeze theorem. Only use the definition of Riemann integral.

答:
Write $U(P, g)$ and $L(P, g)$ to denote the lower and upper sums for function $g$ over specified partition $P$ of $[a,b]$. Pick a partition $P_1$ and $P_2$ such that
$$ U(P_1, h)-\int_a^b h(x)dx<\epsilon$$
and
$$ \int_a^b f(x)dx-L(P_2, f)<\epsilon$$
This is possible because both $f$ and $h$ are Riemann integrable. Let $P = P_1\cup P_2$ (common refinement of $P_1$ and $P_2$). Observe that $f(x)\le g(x)\le h(x)$ imply
$$ U(P, g)-L(P, g)\le U(P, h) - L(P, f)\le U(P_1, h)-L(P_2, f) < 2\epsilon $$
This shows that $g$ is Riemann integrable. Can you proceed from here?</p>

ContentLicense: CC BY-SA 3.0

Title: Riemann Integration and Squeeze Theorem


问:
Let R be the relation defined on the set of bijective functions from the set (1 to n) to the set (1 to n) by $fRg$ whenever $f = g^{-1}$
I am not too sure as to what the relation is but I think $R = \{(1, 2), (2, 3), (3, 4), ..., (n - 1, n)\} $.

Any guidance would be appreciated.

答: For example, if $f,g : \{1,\dots,n\} \to \{1,\dots,n\}$ are defined by $$f(i) = \begin{cases} i+1 & \textrm{if } i < n \\ 1 & \textrm{if } i=n \end{cases} \qquad g(i) = \begin{cases} n & \textrm{if } i = 1 \\ i-1 & \textrm{if } i>1 \end{cases}$$ it is easy to see that $f$ and $g$ are inverses to each other, and then the pair $(f,g)$ is an element of $R$.


问:
Expected value of process involving the square of the sum of uniform random variables

Suppose that you are given two uniform random variables in the interval [0,1], call them $X_1, Y_1$. If we have that $(X_1 + Y_1)^2 > 1$, then the process is over. Otherwise, continue until $\sum_{i = 1}(X_i + Y_i)^2 > 1$ for the first time. What is the expected number steps in this process?

I tried looking at the distribution of the sum of two uniform random variables but the pdf is piecewise and that doesn't seem like a good approach. I also considered finding the expected value as $\sum_{a \geq 0} \mathbb{P}(N > a)$ where $N$ is desired variable, but again I can't find an expression for this probability.

答:
HINT

Let $X,Y \sim \mathcal{U}(0,1)$, and since $X,Y > 0$ we have
$$
(X+Y)^2>1 \iff X+Y > 1
$$
Now we have
$$
\begin{split}
\mathbb{P}[X+Y<1]
 &= \int_0^1\mathbb{P}[X+Y<1,Y=y] f_Y(y)dy \\
 &= \int_0^1\mathbb{P}[X<1-y] f_Y(y)dy \\
 &= \int_0^1 F_X(1-y) f_Y(y)dy \\
\end{split}
$$

Can you complete this?

HINT 2

If we can assume $X_i, Y_i$ are independent of $X_k,Y_k$ for $i \ne k$, define $(X_k+Y_k)^2 = N_k$ and you are asking if $n$ is the minimum such that $\sum_{i=1}^n N_i > 1$, then what is $\mathbb{E}[n]$.

By Wald's Theorem,
$$
\sum_{i=1}^n N_i = \mathbb{E}[n] \mathbb{E}[N_i],
$$
so $\mathbb{E}[n] \ge \lceil 1/\mathbb{E}[N_i] \rceil$.

To find the distribution of $N_i$ you can use a trick like in the above first hint, i.e.
$$
F_N(x) = \mathbb{P}[N_i < x] = \mathbb{P}\left[X_i + Y_i < \sqrt{x}\right],
$$
which you can now condition on $Y_i$.

To find the expectation, you don't need the entire distribution:
$$
\mathbb{E}[N_k]
 = \mathbb{E}\left[(X_k+Y_k)^2\right]
 = \int_0^1 \int_0^1(x+y)^2dA = \frac76.
$$


问:
How to solve this cubic equation (where I must take the square root of a negative)?
I am trying to solve the following set of equations for $L'$ as described in this publication excerpt:

So basically, to re-write that, I have the following series of equations:

$a=X/2$

$b=-(75X^2 + 147Y^2)/80$

$c = -9X^3/16$

$P = 2a^3−9ab+ 27c$

$Q = P^2 -4(a^2-3b)^3$

$w_1=−0.5 +\frac{i\sqrt{3}}{2}$

$w_2=−0.5 -\frac{i\sqrt{3}}{2}$

Which lead to the following three possible solutions:

$L'_1 = -\frac{1}{3} * [a+\sqrt[3]{0.5*(P+\sqrt{Q})} +\sqrt[3]{0.5*(P-\sqrt{Q})}]$

$L'_1 = -\frac{1}{3} * [a+w_2\sqrt[3]{0.5*(P+\sqrt{Q})} +w_1\sqrt[3]{0.5*(P-\sqrt{Q})}]$

$L'_1 = -\frac{1}{3} * [a+w_1\sqrt[3]{0.5*(P+\sqrt{Q})} +w_2\sqrt[3]{0.5*(P-\sqrt{Q})}]$

I am now trying to solve $L'$, but I'm not actually sure how to do it.

My problem is for any input of $X$ or $Y$, $P$ I believe can end up positive or negative, but $Q$ is always a negative.

So let's say hypothetically $P = 2$ and $Q = -3$, then how do you solve:

$\sqrt[3]{0.5*(P-\sqrt{Q})} = \sqrt[3]{1-0.5*i\sqrt{3}}$

What would be the next step?

Basically, I don't know how to solve these equations if $Q$ is negative or if $0.5*(P-\sqrt{Q})$ is negative. I don't know much about complex math. I've used Euler's formula in other situations but I'm not sure here.

Any help? Thanks.

答:
For this equation, I should strongly recommend to use the trigonometric method for three real roots since
$$\Delta=6914880 \left(175 X^4 Y^2+230 X^2 Y^4+147 Y^6\right) >0 \quad \forall X,Y$$
$$p=-\frac{49}{240} \left(5 X^2+9 Y^2\right)\qquad\qquad  q=\frac{49}{4320} X \left(27 Y^2-35 X^2\right)$$ which gives for the roots
$$L'_{k}=-\frac X 6+\frac{7 \sqrt{5 X^2+9 Y^2}}{6 \sqrt{5}}\times $$ $$\cos \left(\frac{2 \pi  k}{3}-\frac{1}{3} \cos ^{-1}\left(-\frac{\sqrt{5} X
   \left(27 Y^2-35 X^2\right)}{7 \left(5 X^2+9 Y^2\right)^{3/2}}\right)\right)$$ with $k=0,1,2$.


问:
multivariable limit question
Is this an acceptable solution?
$$\lim_{(x,y)\rightarrow(0,0)}\frac{\sin(2(x^2+y^2))}{x^2+y^2}$$
$$t=x^2+y^2$$
So $t\rightarrow0$. Now I change the limit to:
$$\lim_{t\rightarrow0}\frac{\sin(2t)}{t}=2$$
This solution makes sense to me, but I am still having doubts.</p>

答:
Or, indeed, the technique used for evaluating $$\lim_{x\to 0}\frac{\sin ax}{x}.$$
Any time $u\to 0$ as $x\to 0$, it is easy to check with the official limit definition that
$$\lim_{x\to 0} f(u(x)) = \lim_{u\to 0} f(u)\,.$$
The same applies when $x$ is a vector.</p>

Title: multivariable limit question


问:
Is there a method to determine the integration limits using spherical and cylindrical analytically?
I know it's best to draw a picture and figure it out intuitively. But I was wondering if there was a more formal method for determining bounds of integration on triple integrals in spherical and cylindrical coordinates. For example,say I wanted to evaluate the following integral in spherical coordinates

$$\int_{0}^{1}dx\int_{0}^{\sqrt{1-x^2}}dy\int_{0}^{\sqrt{1-x^2-y^2}}(x^2+y^2+z^2)^2dz$$

How could I figure out the bounds on the triple integral in spherical coordinates WITHOUT drawing a picture?

答:
Ultimately, you want to work out a system of inequalities without drawing anything. That may be difficult to accomplish even in one dimension.

Following your example, you want to work out

$$\left\{
    \begin{array}{ll}
      0 \leq x \leq 1\\
      0 \leq y \leq \sqrt{1-x^2}\\
      0 \leq z \leq \sqrt{1-x^2-y^2}\\
     \end{array}
    \right.
 \: = \:
\left\{
    \begin{array}{ll}
      0 \leq \rho \cos{\theta} \cos{\varphi} \leq 1\\
      0 \leq \rho \sin{\theta} \cos{\varphi} \leq \sqrt{1 -\rho^2 \cos^2{\theta} \cos^2{\phi}}\\
      0 \leq \rho \sin{\varphi} \leq \sqrt{1-\rho^2 \cos^2{\phi}}\\
     \end{array}
    \right.$$

into

$$\left\{
    \begin{array}{ll}
      0 \leq \rho \leq 1\\
      0 \leq \theta \leq 90°\\
      0 \leq \varphi \leq 90°\\
     \end{array}
    \right.$$

knowing that

$$\left\{
    \begin{array}{ll}
      0 \leq \rho < \infty\\
      -180° \leq \theta \leq +180°\\
      0 \leq \varphi \leq 90°\\
     \end{array}
    \right.$$

unambiguously maps

$$\left\{
    \begin{array}{ll}
      -\infty < x < \infty\\
      -\infty < y < \infty\\
      -\infty < z < \infty\\
     \end{array}
    \right.
 \: = \:
\left\{
    \begin{array}{ll}
      -\infty < \rho \cos{\theta} \cos{\varphi} < \infty\\
      -\infty < \rho \sin{\theta} \cos{\varphi} < \infty\\
      -\infty < \rho \sin{\varphi} < \infty\\
     \end{array}
    \right.$$.

If you want to be even more rigorous, you may treat the spherical coordinate system as a mere change of variables. In that case you would have to work out the unambiguous mapping in the first place.

I don't think it's worth a try, but if you do, remember that there are more inequalities from sine and cosine definitions.


问:
Find bijection $\phi:[n]\times S_{n-1}\to S_n$ and conclude that $\vert S_n\vert = n!$
$[n] = \{1,...,n\}$

I am struggling with finding the bijection.

$\underline{\text{Assume } \mathbf{n=3:}}$

Than we have $A:=[n]\times S_{n-1} = \{(1,\pi_1), (2,\pi_1),(3,\pi_1),(1,\pi_2), (2,\pi_2),(3,\pi_2)\}$,

with $\pi_1 = id$ and $\pi_2 = (12)$

$S_3 = \{(123), (12)(3), (13)(2), (23)(1), (132), id\}$

So we are looking for some $\phi:A\to S_3$. The mapping for $a\in A \mapsto \sigma\in S_3$ is not obvious to me.

Has someone a hint?

The conclusion should be easy if the bijection is found.

Edit:

Due to your hints and further research I found

$\phi:[n]\times S_{n-1}\to S_n, (i,\pi_j)\mapsto \pi_j\circ (in)$, $i\in[n], j\in [n-1]$

My current status on $\phi^{-1}$ is:

$\phi^{-1}:S_n\to [n]\times S_{n-1}, \pi \mapsto (\pi^{-1}(n), ???)$

I am still struggling with $\phi^{-1}$. Does someone has a hint?

答:
A permutation $\pi$ of $S_n$, which is a bijective map from $[n]$ to $[n]$, is completely characterized by specifying $\pi(i)$ for all $i\in[n]$. The idea is that we can break this into two parts, (1) the specification of $\pi(n)$, and (2) the specification of $\pi(i)$ for $i\in[n-1]$. If $\pi(n)=n$ then $\pi$ restricted to $[n-1]$ is just an element of $S_{n-1}$. On the other hand if $\pi(n)=j\ne n$, then $\pi(i)\ne j$ for any $i\in[n-1]$ but $\pi(k)=n$ for some $k\in[n-1]$. So $\pi$ restricted to $[n-1]$ is a bijective map from $[n-1]$ to $[n]\setminus\{j\}$, which is not so different from a bijective map from $[n-1]$ to $[n-1]$, that is, from an element of $S_{n-1}$. (Both are bijective maps between sets of size $n-1$.)

To realize a concrete bijection between these two kinds of bijective map, we send the map $\pi$, restricted to $[n-1]$, to an element of $S_{n-1}$ by composing $\pi$ with the swap $(jn)$. Note that composition with $(jn)$ is an involution: composing $\pi$ with $(jn)$ twice in succession gives $\pi$.

So for $\pi\in S_n$ use the map $\pi\mapsto(j,\pi\circ(jn))$, where $j=\pi(n)$ and $\pi\circ(jn)$ means the permutation $\pi$ is followed by the swap $(jn)$. The map $\pi\circ(jn)$ is here understood, by restricting to $[n-1]$, as an element of $S_{n-1}$, which is possible since it maps $n$ to itself. The definition of this map applies even when $j=n$ since then $(jn)$ is the identity permutation.

For the inverse, just use the fact that composition with $(jn)$ is an involution. Let $\rho\in S_{n-1}$. Then use the map $(j,\rho)\mapsto\rho\circ(jn)$. In this definition $\rho$ is understood as an element of $S_n$ that sends $n$ to itself.


问:
How should I interpret the gradient of a tangent in the complex plane?
I have seen a visual demonstration of $e^{i\pi}+1=0$ that goes like this:

Consider a particle that after $t$ seconds has the position $(\Re(e^{it}),\Im(e^{it}))$ in the complex plane. Since $\frac{d}{dt}(e^{it})=ie^{it}$, the velocity of the particle must be at right angles to its displacement at all times.

If your velocity is always at right angles to your displacement, then the path you take must be circular. (I tried proving this by solving the differential equation $\frac{dy}{dx}=-\frac{x}{y}$.)

Given that your starting position is $(1,0)$, the radius of the circle is $1$. This means that the magnitude of your displacement is $1$, and since $|e^{it}|=1$, $|ie^{it}|=1$, i.e. you are moving with unit speed counterclockwise around the circle.

After $\pi$ seconds, you must have travelled $\pi$ units counterclockwise around the circle, meaning that your position is $(-1,0)$.

Hence, $e^{i \pi}+1=0$.

I think I mostly understand this explanation, but what bothers me is geometrically understanding the statement
$$
\frac{d}{dt}(e^{it})=ie^{it} \, .
$$
If I draw a tangent to an arbitrary point $(\Re(e^{it}),\Im(e^{it}))$, then the gradient of the tangent doesn't seem like it would be $ie^{it}$, if that even makes sense. Since both axes are being used to describe the particle's position, I am unsure about the geometric significance significance of this tangent, if any. In the 'complex world', is there an analogue to drawing a tangent as you would when plotting a real-valued function?

答:
Drawing a tangent to an arbitrary point does not make sense because you can have infinite lines passing through it. Thinking of $(ℜ(^{}),ℑ(^{}))$ as a position vector makes it easier: When you differentiate this vector with respect to time, you will get the velocity vector. And since, $$\frac{}{}(^{})=.^{},$$
it is conclusive that the velocity vector should be perpendicular to the position vector.*
Remember that we differentiate with respect to time, but we have two position axes. In a normal position-time graph, the derivative of the position function (w.r.t time) gives the velocity function, which is the same as saying the slope of the tangent of the position curve gives the velocity curve. The slope here is defined as: $$\frac{\delta x}{\delta t}$$

You can see easily how this slope equals the velocity through the units. Since we have two position axes, it would not make sense to say that the slope of the tangent to the position vector gives us velocity- distance divided by distance does not give the units of velocity. The velocity vector here, is just another vector which happens to be the time derivative of the position vector. Considered this way, the tangent vector is just the derivative. The fact that it is tangential to the original vector tells us that for any value of $t$, the path traced by the particle has to be circular- which is a significant result, at least for proving the equation you have mentioned.

If you think of tangent vectors to curves as complex numbers you can recast conformality in terms of complex numbers. See:1. Conformal mapping is extremely important in complex analysis, as well as in many areas of physics and engineering. 2

*Both the position co-ordinates are multiplied by $i$, which means the final vector can be written as $i.(original\_vector)$. Multiplication by $i$ gives a 90° counterclockwise rotation about $0$.


问:
Proof of Hopf Lax formula
Consider the Hopf-Lax semigroup given by
$$
Q_tf(x) = \inf_{y \in \mathbb{R}^d}  \:\{f(y) +\frac{1}{2t}|x-y|^2\}
$$
for $f \in C_b(\mathbb{R}^d)$.

Prove that

$$
\frac{d}{dt} Q_tf(x) + \frac{1}{2} |\nabla Q_t f(x)|^2 = 0
$$

So far I only showed that $Q_t f \to f$ as $t\to 0^+$ but I have problems with computing the derivatives because of the infimum. I also thought of dealing with the infimum by thinking of the minimizers being in the direction of the steepest descent (given by the opposite of the gradient) but could not make it work. Any help?

答:
You've guessed it right but more careful treatment is needed. Furthermore, within my limited knowledge, in general, your statement is not rigorous as we can not be sure of the differentiablity of $Q^tf$.

So I decided to reformulate it as follows.

Main theorem

Theorem

if $f$ is a Lipschitz continuous function, then for almost every $(t,x) \in \mathbb{R}_+ \times \mathbb{R}^n$ (wrt to Lesbeque measure), we have:
$$ \frac{d}{dt}Q^tf(x)+ \dfrac{1}{2} \Vert \nabla f(x) \Vert^2 = 0$$

Primilinaries

Now, I present some ingredients for the demonstration.

Theorem 1

$g_t(x):= \frac{1}{2t}\Vert x \Vert^2 -Q^tf(x)$ is a convex function$\square$

Theorem 2

For a real convex function $g$ defined on $\mathbb{R}^n$, if $g$ is differentiable at $x$ then we have the following inequality for all $y$:

$$ g(y)-g(x) \ge \langle \nabla g(x), y-x) \quad \square$$

Theorem 3 (Rademacher theorem- Almost everywhere differentiability)

If $f$ is a locally Lipschitz continuous function from  $\mathbb{R^n} \rightarrow \mathbb{R^m}$, then $f$ is differentiable almost everywhere.

Answer

We'll start by answering partially your question.

Theorem 4

For any $s>0$ and $x$ such that $Q^sf$ is differentiable at $x$, then we have:
$$\dfrac{d}{ds^+} Q^sf(x) =-\dfrac{1}{2} \Vert \nabla Q^sf(x) \Vert^2$$

Demonstration

For simplicity, we let $h:= Q^tf$

For any $t>0$, by choosing $y:= x-t\nabla h(x)$, we imply that
$$Q^th(x) -h(x) \le h(x-t\nabla h(x)) -h(x) +\dfrac{t}{2} \Vert \nabla h(x) \Vert^2 $$
Thus by letting $t$ converge to $0^+$, we have:
$$ \dfrac{d}{dt^+} Q^tf(x)|_{t=0} \le -\dfrac{1}{2} \Vert \nabla Q^tf(x) \Vert^2$$

Secondly, because $h$ is differentiable at $x$, $g^t$ is also differentiable at $x$, so we have:
$$ h(y) -h(x) \underbrace{=}_{\text{definition of} g^t}  g^t(x)-g^t(y) -\dfrac{1}{2s}\left( \Vert x \Vert^2 - \Vert y \Vert^2 \right) \ge \langle \nabla g^t(x), x-y \rangle-\dfrac{1}{2s}\left( \Vert x \Vert^2 - \Vert y \Vert^2 \right)   $$

Thus,
$$ h(y) +\frac{1}{2t}\Vert x-y \Vert^2 \ge h(x)+u(x)+ \left[v(y) +\frac{1}{2t}\Vert x-y \Vert^2\right] $$

where $u(x) :=   \langle \nabla g^t(x), x \rangle-\frac{1}{2s} \Vert x \Vert^2 $ and $v(y)= - \langle \nabla g^t(x),y\rangle +\frac{1}{2s} \Vert y \Vert^2$
So,
$$ Q^th(x)-h(x) \ge  Q^tv(x) -v(x) $$
(Note that v(x)=u(x))
The function $v$ is clearly continously differentiable, and we can reconfirm that:
$$\dfrac{d}{dt}Q^t(v)|_{t=0^+} = -\frac{1}{2}\Vert \nabla v (x) \Vert^2 = -\frac{1}{2}\Vert \nabla h (x) \Vert^2 $$
Done.

So now what is left the regularity of $Q^tf$ and indeed, we see that $Q^tf$ is lipschitz continuous by the following theorem :

Theorem 5 (Regularity of $Q^tf$ )

The function $F(t,x) = Q^t(x) $ is $\alpha$ lipschitz continous on $\mathbb{R}_{+} \times \mathbb{R^n}$.

Hence it is differentiable almost everywhere.

Demonstration
For all points $x,y$ and $t>0$,

Let $\tilde{y}$ be the minimizer of $Q^tf(y)$, and $z:= x+\tilde{y}-y$

We see that:

$$ Q^t(x)-Q^t(y) \le f(z)+\dfrac{1}{2t}\Vert z-x\Vert^2 - f(\tilde{y})-\dfrac{1}{2t}\Vert \tilde{y}-y\Vert^2 = f(z)-f(\tilde{y}) \le \alpha \Vert x - y \Vert $$
From then we can imply that:
$$|Q^t(x)-Q^t(y)| \le \alpha \Vert x - y \Vert $$
Then we see that:

$Q^tf(x) \le f(x)$ , and

$Q^tf(x) = f(x) + \inf\{ f(y)-f(x)+\frac{1}{2t} |x-y|^2 \} \ge f(x)+\inf \{ -\alpha |y-x|+\frac{1}{2t} |x-y|^2 \} = f(x)-t \dfrac{\alpha^2}{2}$

Thus $F$ is Lipschitz $\square$

Remark 6 Note that $Q^sQ^t =Q^{t+s}$

From theorem 4 and theorem 5, we have the conclusion in your main theorem which is stated in the previous section.

Comments

The main idea is to bound $Q^tf$ by a reasonable bound. While the upper bound is pretty straightforward if we know how the minimizers look like, the treatment of the lower bound is a bit trickier.

In fact, the idea for the lower comes from another mathematical object called subdifferential. To be honest, I have written a short introduction to that object for you, however as it turned out to be not that important, I decided to leave that part aside.

Professor Evans's PDE book also provides an excellent short chapter on this, which may be different than mine (though I learned a lot from it). It's a great reference if you are serious about PDEs.

Disclaimer I'm not an expert in PDEs

Title: Proof of Hopf Lax formula


问:
How can a sum be limited up and down by an Integral?

I have this sum $\sum_{n=1}^{k^2-1} \sqrt {n}$.
I saw that this can be written as
$$
\int_0^{k^2-1} \sqrt {x} \,dx \leq \sum_{n=1}^{k^2-1}  {\sqrt {n}} \leq \int_1^{k^2}  \sqrt {x} \,dx
$$

Can someone explain why is this true?

答: For $n\le x \le n+1$, $\sqrt{n}\le \sqrt{x} \le \sqrt{n+1}$, so $\sqrt{n}\le \int_n^{n+1} \sqrt{x}dx$, while $\int_{n-1}^n \sqrt{x}dx\le \sqrt{n}$.  Take the sums over $n$ to get the desired result.


问:
Total variation of a Fourier series
Let $f(x) = f(x+2\pi)$ be a bounded real function given by the Fourier series of the form

$$
f(x) = \sum_{k=1}^N a_k \sin(kx + \phi_k).
$$
What is the total variation $V(f)$ of this function over one period? In this case, one should be able to use that $V(f) = \int |f'(x)|dx$ and that
$$
f'(x) = \sum_{k=1}^{N} k a_k \cos(kx + \phi_k),$$
but how?

If instead the function is given by an infinite Fourier series, then what are the conditions on the $a_k$ terms for the total variation to be finite?

答:
A periodic function is of bounded variation if and only if it is antiderivative of a finite signed measure on $[0,2\pi)$ (or, better, on the circle $\mathbb T$) with total mass $0$. Therefore, $\sum_{n\in \mathbb Z} c_n e^{inx}$ is the Fourier series of a function of bounded variation if and only if $\sum_{n\in\mathbb Z} in c_n e^{inx}$ is the Fourier series of a finite signed measure. Let $b_n=i n c_n$ to simplify notation. The following result can be found, for example, in An Introduction to Harmonic Analysis by Katznelson.

Theorem (Herglotz). $\sum_{n\in\mathbb Z} b_n e^{inx}$ is the Fourier series of a positive measure if and only if the sequence $(b_n)$ is positive definite. The latter means that
$$\sum_{n,m}b_{n-m}z_n\overline{z_m}\ge 0\quad \text{ for all sequences }\ z_n\in \mathbb C \tag1 $$ where only finitely many $z_n$ are nonzero.

Here "positive" and "positive definite" are understood in nonstrict sense.

Hence, $\sum_{n\in \mathbb Z} c_n e^{inx}$ is the Fourier series of a function of bounded variation if and only if the sequence $(nc_n)$ is the difference of two positive definite sequences. I don't think this is a practical condition, but then, neither is (1).

Title: Total variation of a Fourier series


问:
Nullhomotopic loops in $R^3$ minus two disjoint circles.
Suppose we have two disjoint circles in $\mathbb{R}^3$: $C_1=\{(x,y,1)\in\mathbb{R}^3:x^2+y^2=1\}$ and $C_2=\{(x,y,2)\in\mathbb{R}^3:x^2+y^2=1\}$. Define $X=\mathbb{R}^3-C_1$, $Y=\mathbb{R}^3-C_2$, $Z=\mathbb{R}^3-C_1-C_2$.

I am trying to find a loop that is not nullhomotopic in $Z$, but is nullhomotopic in $X$ and $Y$. I have no idea how to do that... Any hint please?

答:
You can interpret this problem by way of another: how can you hang a painting with two pins and a string so that if either pin is removed the painting falls? This might seem completely out of left field but there is a direct analogy here. For one, replace these pins with rubber bands (perhaps being held up on the wall by tape), so that wrapping the string around the pin is replaced by inserting it into the loop of the rubber band. These rubber bands will serve as our two disjoint circles. Furthermore, the painting will contain a path connecting the two endpoints of the string, and it can just deformation retract onto this, so the painting and the string serve as the loop.

Let's consider first the initial configuration of the painting hanging. It's not falling because somehow the string is looped into the rubber bands in such a way that it cannot be removed without cutting. The string cannot cut through the rubber band itself, in the same way that a loop going through the center of a circle is not null homotopic. The only way for the painting to fall is for the string to be able to unhook from the inside of the rubber bands, thus allowing the string and painting to be completely separate from the rubber bands. This is essentially saying that the corresponding loop is null homotopic!

Furthermore, going from $Z$ to $X$ or $Y$ is analogous to removing one of the rubber bands from the system. Adding in $C_1$ or $C_2$ allows our loop to pass through that area freely, just as removing a rubber band allows our string to fall through the area where it used to be.

I hope this convinces you that this problem is the same as the painting one I'm describing. For me, I find this much easier to visualize and understand. For one, you can pretty easily get some string and pins and go for it! Unfortunately, I don't think I have the ability to paint a mental picture of such a construction, nor can I draw one. So I'll leave you with this fun little video which shows a solution to this painting hanging problem and does a good job of motivating how you can arrive at such a solution.


问:
What is an $f(x,c)$ that gets very big very fast as $x$ moves away from $c$ in $[0,1]$?
I want a function such that, say if $c = .3$, the graph of $f(x,.3)$ looks like this:

That is actually a graph of $f(x,c) = -\frac{(x - c)^2}{(1-x)x}$

But this function increases too slowly and is too flat around the minimum, $x=c$.
What is a function that is steeper around the minimum and much steeper as you move away from it? That is, one that moves quickly to 100, 1000, etc.

答: Would something like $f(x,c) = e^{\tfrac{\alpha(x-c)^2}{x(1-x)}}$ for some big number $\alpha > 0$ satisfy your conditions?


问:
How many times do you get "k" face of a die after n rolls?
If I ask how many times will I get a five (face not sum) in two rolls the answer is 11 of the 36 combinations will contain a five.

What is the formula to calculate the 11 rolls which include a five? For two rolls you can list all 36 combinations and count them? But if it is 6, 7, or 8 rolls that is quite a bit of combinations to list.

This is somewhat similar to the following post:
Probability of getting $5$ once in rolling a dice two times if...

One formula used was where you take the complement, ex 1 - (5/6)^n where n is the number of rolls.

But if the question is how many times will you roll a 5 in 8 rolls, what formula gives you that value?  What formula gives you just the numerator?  Thank you

答: The basic formula you need is the binomial distribution.  The probability that you will get exactly  $k$ fives in $n$ rolls is given by $\binom{n}{k}p^k(1-p)^{n-k}$, where $p=\frac{1}{6}$ is the probability of getting a five on one roll.


问:
Recursion relation for Euler numbers
I am trying to solve the following:

The Euler numbers $E_n$ are defined by the power series expansion

$$\frac{1}{\cos z}=\sum_{n=0}^\infty \frac{E_n}{n!}z^n\text{ for }|z|<\pi/2$$

(a)  Show that $E_n=0$ when $n$  is an odd integer.

(b) Establish a recursion relation for the sequence.

Through research it seems that part a seems well established but I couldn't find a proof.
Part (b) I found a formula, but couldn't find how to establish it. Any help appreciated.

答:
For part (a): since $\cos(z)$ is an even function (that is, $\cos(z) = \cos(-z)$), then it follows that $\dfrac{1}{\cos(z)}$ is also even (why?).  You should be able to convince yourself that the Taylor series of an even function $f_e(z)$ can't have any odd terms (try differentiating both sides of the identity $f_e(-z) = f_e(z)$ at $z=0$ repeatedly using the chain rule).

For part (b): You know that $\cos(z)\times\dfrac{1}{\cos(z)} = 1$ identically; expand both terms on the LHS in terms of their Taylor series, multiply them out using the usual method for multiplying series, and then set the left and right sides of the equation to be equal as Taylor series — that is to say, look at the terms in $z^n$ on both sides (note that on the RHS, that term is $1$ if $n=0$ and $0$ otherwise).  You should find that this gives you a recursive formula for the Euler numbers.

Title: Recursion relation for Euler numbers


问:
First and second derivative of a function log*(n) ("log star").
I would like to ask what are the derivative values (first and second) of a function "log star": $f(n) = \log^*(n)$?

I want to calculate some limit and use the De'l Hospital property, so that's why I need the derivative of "log star":
$$\lim_{n \to \infty}\frac{\log_{2}^*(n)}{\log_{2}(n)}$$

More about this function: https://en.wikipedia.org/wiki/Iterated_logarithm

答: Hint: take $n=2^m$ in your limit (and make sure you understand why you can do this!). Then (using $\lg(x)$ for $\log_2(x)$, which is a common convention) $\lg^*(n)=\lg^*(m)+1$, whereas $\lg(n)=m$. So your limit is $\lim\limits_{m\to\infty}\dfrac{\lg^*(m)+1}{m}$. But now we can take $m=2^r$, similarly, and get $\lim\limits_{r\to\infty}\dfrac{\lg^*(r)+2}{2^r}$. And since $\lg^*(n)\lt n$ for all $n\gt 1$ (this should be easy to prove) this gives you the limit you're after.


问:
Find the next term of $\frac14,1,\frac94,2+\frac2{\sqrt{3}}, \frac{5}{2} + \frac{5}{2 \sqrt{2}}, 3 + 3 \sqrt{\frac{1}{10} (5 + \sqrt{5})}$
You are given the sequence:
$$\frac{1}{4}, 1, \frac{9}{4}, 2 + \frac{2}{\sqrt{3}}, \frac{5}{2} + \frac{5}{2 \sqrt{2}}, 3 + 3 \sqrt{\frac{1}{10} (5 + \sqrt{5})}$$

What is the next number? Explain

Hint: The next number is rational.

This is my own puzzle

答:
I'm not sure, if this is the intended one. But from $9/4$ onwards we can write:
$$\small{\begin{align}& \frac{9}{4}, 2 + \frac{2}{\sqrt{3}}, \frac{5}{2} + \frac{5}{2 \sqrt{2}}, 3 + 3 \sqrt{\frac{1}{10} (5 + \sqrt{5})}\\ \Rightarrow&\frac{3}{2}\left(1+\frac{1}{2\left(1\right)}\right),\frac{4}{2}\left(1+\frac{1}{2\left(\frac{\sqrt{3}}{2}\right)}\right),\frac{5}{2}\left(1+\frac{1}{2\left(\frac{1}{\sqrt{2}}\right)}\right),\frac{6}{2}\left(1+\frac{1}{2\left(\sqrt{\frac{5-\sqrt{5}}{8}}\right)}\right) \\ \Rightarrow& \frac{3}{2}\left(1+\frac{1}{2}\csc\left(\frac{\pi}{2}\right)\right),\frac{4}{2}\left(1+\frac{1}{2}\csc\left(\frac{\pi}{3}\right)\right),\frac{5}{2}\left(1+\frac{1}{2}\csc\left(\frac{\pi}{4}\right)\right),\frac{6}{2}\left(1+\frac{1}{2}\csc\left(\frac{\pi}{5}\right)\right)\end{align}}$$

If this is the pattern, then the next term will be:
$$\frac{7}{2}\left(1+\frac{1}{2}\csc\left(\frac{\pi}{6}\right)\right)=7$$

Which is also rational. I'll update on this later.


问:
How to integrate $\int_{0}^{1}(y^2+y)\sqrt{1+(2y+1)^2}dy$
I'm trying to solve this integral and I can't figure out how to do it using the basic methods. Can someone help out?</p>

答:
$$I=\int(y^2+y)\sqrt{1+(2y+1)^2}\,dy$$
Let
$$(2y+1)=\sinh(t) \implies y=\frac{1}{2} (\sinh (t)-1)\implies dy=\frac{1}{2}\cosh (t)$$ which make
$$I=\frac{1}{8} \int\left(\sinh ^2(t)-1\right) \cosh ^2(t)\,dt$$

Now, use the double angle formulae to face very simple things.


问:
If $a^2+a b+b^2=40$ and $a^2-\sqrt{a b}+b=5$, then find $a^2+\sqrt{a b}+b$
I was given this problem to solve with elementary methods (High School level).

Knowing that
$$\begin{align}
a^2+a b+b^2 &=40 \\
a^2-\sqrt{a b}+b &=\phantom{0}5
\end{align}$$
find
$$a^2+\sqrt{a b}+b$$

I tried to look for $\sqrt{a b}$, since the requested quantity is

$$a^2+\sqrt{a b}+b=(a^2-\sqrt{a b}+b)+2\sqrt{ab}=5 + 2\sqrt{ab}$$
So I set
$$x=a^2;\;y=\sqrt{ab};\;b=z$$
and the system became
$$x+y^2+z^2=40;\;x-y-z=5$$
subtracting the two equations I got
$$z^2-z+y^2+y-35=0$$
which has one real solution when the discriminant is zero.

That is

$1-4(y^2+y-35)=0$

and then

$y=\frac{1}{2} \left(-1\pm\sqrt{142}\right)$

and finally
$$a^2+\sqrt{a b}+b=4\pm\sqrt{142}$$

I know that there are other solutions because I've found them with Wolfram Mathematica, but I couldn't find them with elementary methods.

Any help will be appreciated.

答:
We could try to "brute force" it at the high school level. Start with the second equation
$$a^2 - \sqrt{ab} + b = 5$$
$$a^2  + b -5 = \sqrt{ab}$$
$$a^4 + b^2 + 25 + 2a^2 b -10a^2 -10b = ab$$
$$b^2 + ( 2a^2 -10 -a) b + (a^4 -10a^2 +25) =0$$
$$b = \frac{-(2a^2 -10 -a) \pm \sqrt{( 2a^2 -10 -a)^2 - 4(a^4 -10a^2 +25))}}{2}$$
$$b = \frac{-(2a^2 -10 -a) \pm \sqrt{a (-4 a^2 + a + 20)}}{2}$$
Now let's just call $b=x \pm \sqrt{y}.$
$$b^2 = x^2 \pm 2x\sqrt{y} + y$$
$$ab = ax \pm a\sqrt{y}$$
Plug this into the second equation:
$$a^2 +ab + b^2 =40$$
$$a^2 + ax \pm a\sqrt{y} + x^2 \pm 2x\sqrt{y} + y =40 $$
$$(\pm a \pm 2x)\sqrt{y}=40 -a^2 - ax - x^2 -y$$
$$(a^2 \pm 2xa + 4x^2)y=(40 -a^2 - ax - x^2 -y)^2$$
Now plug in for $x, y$ and solve for $a$. What you end up with is:
$$-225 + 950 a - 290 a^2 - (255 a^3)/2 - (53 a^4)/4 - (45 a^5)/2 +
 14 a^6 + 2 a^7 - a^8 =0$$ This is not tractable, and it is likely the problem has a typo.

If you type in Mathematica

```<code>Solve[{a^2 + a*b + b^2 == 40, \[Sqrt](a*b) + 5 == a^2 + b}, {a, b}]
</code>```

It will spit out 3 gross looking expressions possible answers. If you then evaluate these for $a^2 + b + \sqrt{ab}$

```<code>a^2 + b + Sqrt[a b] /. %OUTPUT NUMBER
</code>```

Then the resulting expression doesn't simplify nicely. (Which is not a guarantee; Mathematica's simplify function is far from perfect).

Body:
We could try to "brute force" it at the high school level. Start with the second equation
$$a^2 - \sqrt{ab} + b = 5$$
$$a^2  + b -5 = \sqrt{ab}$$
$$a^4 + b^2 + 25 + 2a^2 b -10a^2 -10b = ab$$
$$b^2 + ( 2a^2 -10 -a) b + (a^4 -10a^2 +25) =0$$
$$b = \frac{-(2a^2 -10 -a) \pm \sqrt{( 2a^2 -10 -a)^2 - 4(a^4 -10a^2 +25))}}{2}$$
$$b = \frac{-(2a^2 -10 -a) \pm \sqrt{a (-4 a^2 + a + 20)}}{2}$$
Now let's just call $b=x \pm \sqrt{y}.$
$$b^2 = x^2 \pm 2x\sqrt{y} + y$$
$$ab = ax \pm a\sqrt{y}$$
Plug this into the second equation:
$$a^2 +ab + b^2 =40$$
$$a^2 + ax \pm a\sqrt{y} + x^2 \pm 2x\sqrt{y} + y =40 $$
$$(\pm a \pm 2x)\sqrt{y}=40 -a^2 - ax - x^2 -y$$
$$(a^2 \pm 2xa + 4x^2)y=(40 -a^2 - ax - x^2 -y)^2$$
Now plug in for $x, y$ and solve for $a$. What you end up with is:
$$-225 + 950 a - 290 a^2 - (255 a^3)/2 - (53 a^4)/4 - (45 a^5)/2 +
 14 a^6 + 2 a^7 - a^8 =0$$ This is not tractable, and it is likely the problem has a typo.

If you type in Mathematica

Solve[{a^2 + a*b + b^2 == 40, \[Sqrt](a*b) + 5 == a^2 + b}, {a, b}]

It will spit out 3 gross looking expressions possible answers. If you then evaluate these for $a^2 + b + \sqrt{ab}$

a^2 + b + Sqrt[a b] /. %OUTPUT NUMBER

Then the resulting expression doesn't simplify nicely. (Which is not a guarantee; Mathematica's simplify function is far from perfect).


问:
Inductive proof - Inequality
I am trying to do mathematical induction to prove an inequality. However, I got stuck in the middle of the proof. I have started my proof from the inductive step $3^{k+1}=3^k*3 ≥3k*3$. I am stuck in here. The below picture is about what I did. I also tried in a separate way. Picture number $2$ is the alternate way that I tried to prove my equation. The main question is attached to image $1$.
Using mathematical induction, prove:

$〖3n≤3〗^n$ for all natural numbers $n$

Can anyone please help me out with that?

Here is the snip of the problem(First attempt)

Second attempt

答:
Since you are told to prove by induction, I will proceed by induction on $n$

For the base case we know  for $n=1$, $3^1 = 3$. Thus assume the assertion holds for $n=k$ with $1 \leq k  \leq n$. Consider $n=k+1$,

We have $3^{k+1} = 3^k.3 \geq 3k.3 = 3k+3k+3k >  3k+3 = 3(k+1)$  establishing $n = k+1$.

Title: Inductive proof - Inequality


问:
Calculus II, Curve length question.
Find the length of the curve
$x= \int_0^y\sqrt{\sec ^4(3 t)-1}dt, \quad 0\le y\le 9$

A bit stumped, without the 'y' in the upper limit it'd make a lot more sense to me.

Advice or solutions with explanation would be very appreciated.

答:
$$\frac{dx}{dy} = \sqrt{\sec^4{3 y}-1}$$

Arc length is then

$$\begin{align}\int_0^9 dy \sqrt{1+\left ( \frac{dx}{dy} \right )^2}  &= \int_0^9 dy\, \sec^2{3 y} \\ &= \frac13 \tan{27} \end{align}$$

Title: Calculus II, Curve length question.


问:
Complex equation gives unexpected roots
The following question requires you to find the number of distinct matrices in S:

Let ω ≠ 1 be a complex cube root of unity and S be the set of all non singular matrices of the form \begin{bmatrix}
1&a&b \\
\omega&1&c \\
\omega^2&\omega&1\
\end{bmatrix} where each of a, b, and c is either ω or ω2.

From solving the matrix, we get $ac\omega^2-(a+c)\omega+1\neq0$. Now, both $ac\neq1$ and $(a+c)\neq1$ have to be simultaneously true, so I substituted $a= \frac{1}{c}$ in the second inequality and solved it; logically, this should yield the values a and c shouldn't become.

But I got $\omega$ and $\omega^2$ as the solutions.

I know there are better ways you solve this, but where did I mess up here?

答:
The condition for invertibility is that $\omega^2ac-(a+c)\omega+1\ne0$. I.e. $(1-\omega a)(1-\omega c)\ne 0$, i.e. $a\ne\omega^2\land c\ne\omega^2$.

Which means $2$ matrices.

Title: Complex equation gives unexpected roots


问:
Books to improve mathematical problem solving abilty
I don't want books like "How to solve it". Instead I am looking for books that actually have problems in them. And preferably solutions too. Basically books with problems that help you develop mathematical problem solving ability.

For my background, I have graduated from high school and know high school math fairly well. But lot of the problems at high school level are reiterative and can be done by memorizing stuff. I am looking for problems that challenge you and at the same time make you a better problem solver.

答: Turns out that there is a very good book that does what you want: "How to Prove It" by Daniel Velleman. It will not be as challenging as olympiad problems, but is definitely a good place to get started.</p>


问:
Choose a random number between $0$ and $1$ and record its value. Keep doing it until the sum of the numbers exceeds $1$. How many tries do we need?
Choose a random number between $0$ and $1$ and record its value. Do this again and add the second number to the first number. Keep doing this until the sum of the numbers exceeds $1$. What's the expected value of the number of random numbers needed to accomplish this?</p>

答:
The derivations involving the multidimensional integral for the probability of finding a sum smaller than 1, can be simplified by restoring permutation symmetry in the integration before evaluating it. The probability that the sum of $n$ uniformly distributed random numbers between 0 and 1 does not exceed 1 is given by:

$$P(n) = \int_{0\leq s_1\leq s_2\leq\cdots \leq s_n\leq 1} ds_1 ds_2\cdots ds_n $$

The integration variables here are the partial sums of the first $n$ random variables, which are therefore ordered as indicated. But because these are dummy variables, we can permute these variables in any way we like. This means that integrating over the variables without imposing the ordering will yield $n!$ times the value of the original integral. We thus have:

$$P_n = \frac{1}{n!} \int_0^1\int_0^1\cdots\int_0^1 ds_1 ds_2\cdots ds_n = \frac{1}{n!}$$

The expectation value then follows e.g. using the derivation given in this answer by Dinesh.


问:
Proposition 6.6 in Hartshorne II: Bijectivity
Proposition 6.6 in Hartshorne II seeks to show that $\operatorname{Cl} (X \times \mathbb A^1) \simeq \operatorname{Cl} X$ for $X$ Noetherian, integral, separated, and regular in codimension 1.

A point of codimension 1 in $X \times \mathbb A^1$ is called type I if its image $y$ under the projection to $X$ is codimension 1. We can see such an $x$ is the generic point of the fiber of $y$.

A point of codimension 1 in $X \times \mathbb A^1$ is called type II if its image is the generic point of $X$.

I understand on a technical level, thanks to other questions on this site, the first two paragraphs of the proof. However, when Hartshorne goes to show the map

$$\pi^* : \operatorname{Cl} X \to \operatorname{Cl} (X \times \mathbb A^1)$$

given by $\sum n_i Y_i \mapsto \sum n_i \pi^{-1} Y_i$, where $\pi$ is the projection to $X$, I do not understand the following statements:

Injectivity: If $D = \operatorname{div} f \in \operatorname{div} X$, then $\pi^*D$ involves only prime divisors of type I. Why?

Injectivity: The above forces $f \in K$, the function field of $X$. Why? He gives the reasoning that otherwise $f = g/h$, with $g, h \in K[t]$, which I understand, but then claims that if both are not in $K$, the divisor of $f$ will involve some type II divisor. Why is this true?

Surjectivity: Why does it suffice to check that prime type II divisors are linearly equivalent to sums of prime type I divisors? Do we already know we are surjective on type I divisors?

Surjectivity: Why does localizing a type II prime divisor at the generic point give us a prime divisor in the spectrum of $K[t]$?

Assuming the above statement, we see this divisor corresponds to a principal prime $P = (f)$ in $K[t]$. I see that just fine. Hartshorne then claims the divisor of $f$ is $Z$ plus possibly some purely type I divisors. Why?

I get the sense I am not thinking about divisors in general correctly, and, more specifically, I am not truly understanding the definition and implementation of type I and II divisors.

答:
Writing $D=\sum n_iD_i$ for $D_i$ prime, we have $\pi^*D=\sum n_i\pi^*D_i$. But $\pi^*D_i=\pi^{-1}(D_i)$, and $\pi(\pi^{-1}(D_i))=D_i$, hence the image of $D$ does not contain the generic point of $X$, and all the $\pi^{-1}D_i$ are divisors of type I.

If $f=g/h$ in lowest terms for $g,h\in K[t]$ so that at least one is in $K[t]\setminus K$, then the valuation of $f$ at a maximal ideal containing a zero of whichever $g,h$ is in $K[t]\setminus K$ is not zero. Such a maximal ideal is a point in the fiber of $X\times\Bbb A^1$ over the generic point of $X$, so the divisor of $f$ contains a type II divisor.

Yes, we already know we're surjective on type I divisors: if $D\subset X\times\Bbb A^1$ is a prime divisor of type 1, then $\pi(D)\subset X$ is a prime divisor which maps to $\pi^{-1}(\pi(D))=D$.

Your statement is not super clear, but the point is that if $D$ is a prime divisor of type II, then it must intersect the fiber of $X\times\Bbb A^1\to X$ over the generic point of $X$ by definition of a type II divisor. As this fiber is exactly $\operatorname{Spec} K[t]$ and the divisor cannot contain the generic point of $\operatorname{Spec} K[t]$ (this is the generic point of $X\times\Bbb A^1$), we get a collection of closed points in $\operatorname{Spec} K[t]$. If we had multiple points in this, we'd have multiple irreducible components of our divisor, so there's only one point, and thus we get a prime divisor on $\operatorname{Spec} K[t]$. (Alternatively, any prime divisor on a scheme $X$ has a unique point with $\dim \mathcal{O}_{X,p}=1$ and no points with $\dim = 0$.)

Take $X=\operatorname{Spec} k[x]$ and $f=xt$. Then the divisor associated to $f$ on $\Bbb A^1\times\Bbb A^1=\Bbb A^2$ is the union of the axes, which is a type one divisor plus a type two divisor. There are worse examples, but this is one very accessible one; conversely, you might be able to fix this most of the time. Digging in to the stuff required to make it just a type II divisor is more work than you need to do here because we already know we're surjective on type I divisors, and so we don't have to figure it out to finish the proof.

Part 5 could make an interesting problem to think about further (when can one guarantee the divisor associated to $f$ is just a type II divisor?), but it's not necessary to finish this proof.


问:
Restricted closed subscheme and reduced induced closed subscheme
$(X,\mathcal O_X)$ is a scheme, and $Y \subset X$ is a closed subset. So on $Y$ we can define a canonical structure sheaf, i.e. the restriction structure sheaf as $(Y,\mathcal O_X|_Y)$. I guess it is not hard to see this is a closed subscheme of $(X,\mathcal O_X)$. However, we know that,(Hartshorne Ex 3.2.6) there are many different closed subschemes with underlying space $Y$ but different structure sheaves. In that Ex 3.2.6, it defines reduced induced closed subscheme which is "smaller" (universal) than any other closed subschemes with space $Y$.

So my question is for those two specific constructions of closed subschemes, are they the same or if not, related in some way? For example, if $X=Spec \, A$, then we know that the reduced induced closed subscheme $Y=Spec \, A/I$, where $I$ is the largest ideal for which $V(I)=Y$ (Also in Ex 3.2.6), then what will be the ideal $I$ corresponding to the restricted subscheme $Y, \mathcal O_X|_Y$??

Thank you so much for any help!

答: Here's a recap of the discussion from the comments and chat: the locally-ringed space $(Y,\mathcal{O}_X|_Y)$ is not in general a scheme, because it is not the case that every point has an open neighborhood isomorphic as a locally ringed space to $\operatorname{Spec} A$ for some ring $A$ (this is the defining property for schemes among locally ringed spaces). In particular, taking $X=\operatorname{Spec} k[t]$ for $k$ a field and $Y$ to be the set consisting of the single closed point $(t)\in X$, we have that $(Y,\mathcal{O}_X|_Y)$ is the locally ringed space which is a point with the structure sheaf the constant sheaf with value $k[t]_{(t)}$. If this were to be a scheme, then it must be an affine scheme, and it would need to be isomorphic to the affine scheme $\operatorname{Spec} k[t]_{(t)}$. But this is impossible, because this has two points while $Y$ is a singleton.


问:
Partial derivatives of $v(x,y) = x f(x+y) + y g(x,y) $
For $v(x,y) = x f(x+y) + y g(x+y)$ where $f$ and $g$ are twice differentiable, I need to prove that
$$ v_{xx} - 2v_{xy} + v_{yy} = 0 $$
(it is understood that $v_{xx} = \frac{\partial^2 v}{\partial x^2}$ and $v_{xy} = \frac{\partial^2 v}{\partial x \partial y}$.)

My attempt:
$$v_{xx} = 2f_x(x+y) + xf_{xx}(x+y) + yg_{xx}(x+y),$$
$$v_{yy} =xf_{yy}(x+y) + 2g_y(x+y) + yg_{yy}(x+y),$$
$$-2v_{xy} = -2(f_y(x+y)+ xf_{xy}(x+y)+g_x(x+y)+yg_{xy}(x+y) ).$$

Only the problem is, these terms don't add up to zero! I can't see why. Your help will be appreciated!

答:
$f$ and $g$ are functions of a single variable. There are no partial derivatives w.r.t. $x$ and $y$ for these functions. If you replace $f_x$ by $f'$, $f_{xy}$ by $f''$ etc you will see that the given equation is indeed satisfied.

$$v_{xx} = 2f'(x+y) + xf''(x+y) + yg''(x+y),$$
$$v_{yy} =xf''(x+y) + 2g'(x+y) + yg''(x+y),$$
$$-2v_{xy} = -2(f'(x+y)+ xf''(x+y)+g'(x+y)+yg''(x+y) ).$$


问:
How to project a portion of a sphere to a plane preserving distance?
Stereographic projection preserves angles but not distances. I've been lead to believe that it isn't possible to go from a sphere to a plane preserving distances between points, but is it possible to do so with a segment of a sphere?

For context, I'm planning on using locations in the United States as inputs, and so I only would need to project the area of a sphere corresponding to the continental United States and Alaska (I'd likely just eyeball the numbers for Hawaii given it's distance from the rest of the states). Is there a way to preserve distance in this transformation of a subset of the sphere?

答:
It's not possible to do it even for a segment of a sphere. The reason is essentially the same: a segment of the sphere is intrinsically curved, just as the whole sphere is.  If the segment is small relative to the whole sphere, the distortion will be relatively small; the approximation gets worse as the segment grows.  This is why local maps give approximately correct distances. For a region as large as the continental United States, the distortion will be substantial no matter what you do.

For exact distances, the best you can do is the various equidistant projections.  The azimuthal equidistant projection  preserves the distances along any diameter of the map.  In particular, this means that the distance from the center point $P$ to any other point is correct.

In the equidistant conic projection, distances are correct along two particular parallels.


Title: How to integrate $\int_{0}^{1}(y^2+y)\sqrt{1+(2y+1)^2}dy$

Body: I'm trying to solve this integral and I can't figure out how to do it using the basic methods. Can someone help out?</p>

Body:
Hint:let $f(y)=1+{(2y+1)}^2=4y^2+4y+2$

$f'(y)=8y+4$

Hence we write $$y^2+y=l(4y^2+4y+2)+m(8y+4)+n$$

Now find $l,m,n$ and integrate.....


问:
Compute the scalar line integral $\int_{C}^{}x^2+2y-zds$ where C is the line segment from (0,1,0) to (1,0,1).
I'm having trouble with this question. I tried to make a relation between x, y, and z using the points given, and I came up with $x=z=-y+1$, which led to trying to solve the integral

$$\int_{0}^{1}(y^2+y)\sqrt{1+(2y+1)^2}dy$$

But I don't know if this is the right approach.

I was assisted in evaluating the integral, but I'm just not sure.

How to integrate $\int_{0}^{1}(y^2+y)\sqrt{1+(2y+1)^2}dy$

答:
Your integral is not the correct integral that you came up with. The formula you should be using is:

$$\int_C f(x,y,z) ds = \int_a^{b} f(x(t),y(t),z(t))\sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2}dt$$

With the parametric representation of the line segment. You should get the parameterization as:

$$x = t, y = 1-t, z = t, 0\leq t \leq 1$$

Integral then becomes

$$\int_C (x^2 + 2y - z)ds = \int_0^{1} (t^2+2(1-t)-t)\sqrt{1^2+(-1)^2+1^2}dt$$

$$=\sqrt{3}\int_0^{1} t^2 - 3t +2dt = \frac{5\sqrt{3}}{6}.$$

Let me know if you have any questions. Cheers.


答:
Your path for line integral is the line segment from $(0, 1, 0)$ to $(1, 0, 1)$.

So the points on the line will be given by $\, c(t) = (0, 1, 0) + t(1-0, 0-1, 1-0) = (t, 1 - t, t)$.

$(0 \leq t \leq 1)$

$|c'(t)| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt3$

So your line integral becomes $I = \displaystyle \int_{0}^{1} \Bigg(t^2 + 2 (1-t) - t\Bigg) \, |c'(t)| \, dt$

Can you take it from here?

Title: Compute the scalar line integral $\int_{C}^{}x^2+2y-zds$ where C is the line segment from (0,1,0) to (1,0,1).

Body:
I'm having trouble with this question. I tried to make a relation between x, y, and z using the points given, and I came up with $x=z=-y+1$, which led to trying to solve the integral

$$\int_{0}^{1}(y^2+y)\sqrt{1+(2y+1)^2}dy$$

But I don't know if this is the right approach.

I was assisted in evaluating the integral, but I'm just not sure.

How to integrate $\int_{0}^{1}(y^2+y)\sqrt{1+(2y+1)^2}dy$


问:
Any hints for a proof by mathematical induction?
$$x^{\left(n-1\right)}+x^{\left(n-2\right)}c+x^{\left(n-3\right)}c^{2}+...+c^{\left(n-2\right)}+c^{\left(n-1\right)}=\frac{x^{n}-c^{n}}{x-c}$$ Where n is a positive integer

So step one is to test when $n=1$. I cannot even do this step. The $RHS = 1$ but I am struggling to get the LHS to also be one. I tried factorizing out $2$ after substituting since the end $x$ and $c$ become $1$, but I'm this doesn't really work. Is this a geometric progression?

Also is writing it out in this summation notation correct (I'm not too familiar with this notation, hopefully its right):

$$\sum _{n=n-1}^0\:x^n\:\cdot \:\sum _{n=0}^{n-1}c^n\:$$

答:
The base case $n=1$ is obviously true. Now suppose that it is true for some $r\in\Bbb{Z}^+,r\ge1$. The LHS can be written as
$$\sum_{k=1}^{n}x^{n-k}c^{k-1}$$
We made the assumption that
$$\sum_{k=1}^{r}x^{r-k}c^{k-1}=\frac{x^r-c^r}{x-c}$$
and we have to prove that
$$\sum_{k=1}^{r+1}x^{r-k+1}c^{k-1}=\frac{x^{r+1}-c^{r+1}}{x-c}$$
Now
\begin{align}
\sum_{k=1}^{r+1}x^{r-k+1}c^{k-1}&=c^r+\sum_{k=1}^{r}x^{r-k+1}c^{k-1}\\
&=c^r+x\sum_{k=1}^{r}x^{r-k}c^{k-1}
\end{align}
we had made the assumption that
$$\sum_{k=1}^{r}x^{r-k}c^{k-1}=\frac{x^r-c^r}{x-c}$$
So
\begin{align}
c^r+x\sum_{k=1}^{r}x^{r-k}c^{k-1}&=c^r+x\frac{x^r-c^r}{x-c}\\
&=c^r+\frac{x^{r+1}-xc^r}{x-c}\\
&=\frac{xc^r-c^{r+1}+x^{r+1}-xc^r}{x-c}\\
&=\frac{x^{r+1}-c^{r+1}}{x-c}
\end{align}
And hence the statement is true for $r+1$ if the statement is true for $r$. So by the principle of induction, the statement is true.


问:
The 68-95-99.7% rule gives approximate probabilities of a Normal r.v. being within 1, 2, and 3 standard deviations of its mean ...
The 68-95-99.7% rule gives approximate probabilities of a Normal r.v. being within 1, 2, and 3 standard deviations of its mean. Derive analogous rules for the following distributions.

a) Unif(0, 1).

b) Expo(1).

For part A, I got $σ = \frac{1}{sqrt(12)}$, $μ = \frac{1}{2}$, X ~ Uni(0, 1)

$P(|U - μ|≤ nσ) = P(|U - \frac{1}{2}|≤ \frac{n}{srt(12)}) = P(\frac{1}{2} - \frac{n}{srt(12)}≤U≤ \frac{1}{2} + \frac{n}{srt(12)})$

$P(|U - μ|≤ nσ) = F_U (\frac{1}{2} + \frac{n}{srt(12)}) - F_U (\frac{1}{2} - \frac{n}{srt(12)})$

Then I plugged in, 1, 2, and 3 and got 57.73%, 100%, and 100%

For part B, I got $σ = \frac{1}{ λ}$, $μ = \frac{1}{ λ} = 1$

$P(|E_1 - μ|≤nσ) = P(|E_1 - 1|≤n) = P(1 - n≤E_1≤1 + n)$

$P(|E_1 - μ|≤nσ) = F_{E_{1}}(1 + n) - F_{E_{1}}(1 - n)$

Then I plugged in, 1, 2, and 3 and got 86.47%, 95.02%, and 98.17%

However I feel like my answers are off I'm just not exactly where my work is off. Please any help would be very much appreciated!

答:
Your answers are correct.

The precise values are as follows.  For the uniform distribution, the probability of being within $1$ standard deviation of the mean is $1/\sqrt{3}$.  The probability of being within $2$ standard deviations is $1$, because $2/\sqrt{3} > 1$.  The same is true of $3$ standard deviations.

For the exponential distribution with mean $1$, the standard deviation is also $1$, therefore being within $1$ standard deviation of the mean implies that $0 \le E_1 \le 2$.  This is simply $1 - e^{-2}$.  To be within two standard deviations, we must have $0 \le E_1 \le 3$, or $1 - e^{-3}$.  And to be within three standard deviations, the probability is $1 - e^{-4}$.

Note that in both cases, we encounter situations where one or more of the endpoints of the interval are truncated due to the support of the random variable not being infinite.


问:
The stalk of the image sheaf on a normalization curve
My question is based on an exercise (Ex II.6.9) in  Hartshorne's GTM52.

Suppose that $X$ is a projective curve over $k$ (an algebraically closed field). Let $\tilde{X}$ be its normalization and $\pi: \tilde{X}\rightarrow X$ be the projection map. For each point $P\in X$, let $\mathscr{O}_P$ be its local ring and $\tilde{\mathscr{O}_P}$ be the integral closure of $\mathscr{O}_P$. Then my question is:

Is there a relationship between the stalk of $\pi_*\mathscr{O}_\tilde{X}$ at $P$ and the local ring $\tilde{\mathscr{O}_P}$? Are they equal? If it's true, how to prove it?

My attempt.

In the affine case, suppose $X=\mathrm{Spec}A$ for some domain $A$. I know that $\pi$ is just the ring extension $A\hookrightarrow\tilde{A}$, where $\tilde{A}$ is the integral closure of $A$ in its field of fractions. And $P$ corresponds to a prime ideal $\mathfrak{p}$ of $A$. Then $\tilde{\mathscr{O}_P}$ is the integral closure of $A_\mathfrak{p}$. The stalk of $\pi_*\mathscr{O}_\tilde{X}$ at $P$ can be finally expressed as a direct limit of local rings $\{\tilde{A}_f\}$, where $f$ is taken over all elements in $A\setminus\mathfrak{p}$. So the question is translated into a problem of commutative algebra. That is, is it true for  $\tilde{A_\mathfrak{p}}=\varinjlim\tilde{A}_f$?

Maybe we need something given by the assumption on $X$, i.e. $A$ is noetherian and of dimension $1$. But I still have no strategy to deal with it. Could anyone help me? Thanks in advance!

答: I think your attempt is right. $\tilde{A_\mathfrak{p}}=\varinjlim\tilde{A}_f$ is true because you are working in an integral domain(curve implies integral), so everything is included in the function field, and taking colimit is the same as taking the union.


问:
How can I increase Validation Accuracy when Training Accuracy reached 100%
Initial Situation

I have a Classification Model which I train on a Dataset consisting of 1400 samples where train on a training set (80%) and validate on another validation set (20%). As Model I use a Neural Network. During training I plot the train- and validation-accuracy curves. Both accuracies grow until the training accuracy reaches 100% - Now also the validation accuracy stagnates at 98.7%.

My Assumptions

I think the behavior makes intuitively sense since once the model reaches a training accuracy of 100%, it gets "everything correct" so the failure needed to update the weights is kind of zero and hence the modes "does not know what to further learn". Nonetheless the validation Accuracy has not flattened out and hence there is some potential to further increase the Validation Accuracy.

My Question

What can I possibly do to further increase the validation accuracy? As a side note: I still implement slight Data Augmentation (slight noise, rotation) on the training set (not on the validation set).

Thanks for your suggestions :)

答:
It appears that your network very quickly learns how to classify the data. To check your train/validation errors are not just anomalies, shuffle the data set repeatedly and again split it into train/test sets in the 80/20 ratio as you have done before. If you continue to observe the same behaviour, then it is indeed possible that your model learns very quickly and would continue to improve if only it had more data.

To test that, do a Leave-One-Out-Crossvalidation (LOOC). That is, for each $i=1,\ldots, 1400$ take your test set to be the $i$-th sample, and your training set to be the other $1399$ samples. If the average training accuracy over these $1400$ models is $100$% and the average test accuracy is again very high (and higher than $98.7$%) then we have reason to suspect that even more data would help the model.

Another way to improve the model, related to what you said about your model not knowing "what further to learn" once the training accuracy reaches $100$%, is to add a regularisation term into your error function, so that even when a set of weights gives a training accuracy of $100$%, you can continue to find even simpler weights which also do the same, instead of stagnating.


问:
Problems in solving Number Theory problems (specifically Congruences)
I am very new to Number Theory, I understood the basic concepts and theorems but what I am struggling at is that I am not able to apply them on questions, and basically I am not able to process the problem in 'number theory' theorems rather I am thinking from an absolute manner.

I have solved around 15-20 divisibility problems of Number Theory and I still have no idea how you deal with congruences in a real problem, for example:

Prove that for positive integer $ n$ we have $169| 3^{3n+3}-26n-27$.

My approach to this problem was by simplifying the expression as much as I can but it involved congruences, which didn't even strike me. Even when I am reading congruence questions' answers, I am not able to simply understand

Now I have a number of questions:

first, does it take time to develop that 'intuition' while solving problems?

second, can elementary number theory be mastered through sheer practice?

third, is there a good resource to study and practice congruences?

Body:
Note that $3^{3n+3}=(26+1)^{n+1}$

Can you use binomial expansion from here?

Hint

$3^{3n+3}=(26+1)^{n+1} \\
 \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }={{n+1} \choose {n+1}}(26)^{n+1}+{{n+1} \choose {n}}(26)^{n}+\dots+{{n+1} \choose {2}}(26)^{2}+{{n+1} \choose {1}}(26)+{{n+1} \choose {0}}
\\
 \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }={{n+1} \choose {n+1}}(26)^{n+1}+{{n+1} \choose {n}}(26)^{n}+\dots+{{n+1} \choose {2}}(26)^{2}+(n+1)(26)+1
\\
 \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }=\color{green}{{{n+1} \choose {n+1}}(26)^{n+1}+{{n+1} \choose {n}}(26)^{n}+\dots+{{n+1} \choose {2}}(26)^{2}}+26n+27 $

Can you figure out the expression in $\color{green}{green}$ is divisible by $(13)^2$


问:
Non-isomorphic connected graphs question
How do I find all of the non-isomorphic connected graphs with the degree sequence 233345?</p>

答:
By the Handshaking Lemma, there are $\tfrac12(2+3+3+3+4+5)=10$ edges in these graphs.  We can exhaustively enumerate the $6$-vertex $10$-edge graphs with vertex degrees between $2$ and $6$ using geng which comes with nauty:

```<code>geng 6 10:10 -d2 -D5
</code>```

The 12 graphs generated can be viewed using showg and we can exclude the ones that do not have the desired degree sequence.  The two that remain are:

where the vertices are ascribed their degrees.

Body:
By the Handshaking Lemma, there are $\tfrac12(2+3+3+3+4+5)=10$ edges in these graphs.  We can exhaustively enumerate the $6$-vertex $10$-edge graphs with vertex degrees between $2$ and $6$ using geng which comes with nauty:

geng 6 10:10 -d2 -D5

The 12 graphs generated can be viewed using showg and we can exclude the ones that do not have the desired degree sequence.  The two that remain are:

where the vertices are ascribed their degrees.


问:
Application of closed graph theorem: proving continuity of $F : H \to H$ satisfying $\langle{Fx, y\rangle} = \langle{x, Fy\rangle}$
Let $H$ be a hilbert space and let $F: H \to H$ be any map. Prove by using the Closed Graph Theorem that if $\langle{Fx, y\rangle} = \langle{x, Fy\rangle}$ for all $x, y \in H$, then $F$ is continuous and linear.

答:
First, to show that $F$ is linear,
$$
     \langle F(\alpha x + \beta y),z\rangle=\langle \alpha x+\beta y,Fz\rangle \\
       = \alpha \langle x,Fz\rangle+\beta\langle y,Fz\rangle \\
       = \alpha\langle Fx,z\rangle+\beta\langle Fy,z\rangle \\
       = \langle \alpha Fx + \beta Fy,z\rangle.
$$
Because this holds for all $z$, then $F(\alpha x+\beta y)=\alpha Fx+\beta Fy$. So $F$ is linear.

To prove that $F$ is closed, suppose that $\{x_n\}$ converges to $x$ and $\{Fx_n\}$ converges to $y$. Then we must show that $Fx=y$. To do this, note that the following holds for all $z$:
$$
       \langle Fx,z\rangle = \langle x,Fz\rangle=\lim_n \langle x_n,Fz\rangle=\lim_n\langle Fx_n,z\rangle = \langle y,z\rangle.
$$
Therefore $Fx=y$. So $F$ is closed. It follows that $F$ is continuous by the Closed Graph Theorem.


问:
Solutions of $f(x)=f'(x)$ where $f$ is continuous under certain conditions
This particular question was part of my analysis quiz and I was having a hard time figuring it out.

Let $f:[0,1] \longrightarrow \mathbb R$ be a fixed continuous function such that $f$ is differentiable on $(0,1)$ and $f(0)=f(1)=0$. Then how many solutions of equation $f(x)=f'(x)$ are on interval $(0,1)$?

Rolle's theorem implies that there exists a $c \in (0,1)$ such that $f'(c)=0$. But neither I am able to use it in the question nor think about an alternative approach.

So, Can you please help?

答:
The substitution $g(x) \colon = e^{-x} \cdot f(x)$ is quite all right, since
$$g'(x) = e^{-x}( f'(x) - f(x)$$

and so $f'(x) = f(x)$ if and only if $g'(x) = 0$.

So now $g(0) =g(1)=0$, so in the interval $(0,1)$ there exists at least one solution of $g'(x) = 0$.  That's all one can say, otherwise their number is not determined.


问:
Why is $\sin(4\pi/5)$ equal to $\sin(\pi/5)$
I am trying to understand why $$\sin(4\pi/5) = \sin(\pi/5)$$
Is there any formula that could help me? Thank you

答:
By the identity $\sin(\pi-x) = \sin x$, we have,

$\sin(\frac{\pi}{5}) = \sin(\pi - \frac{\pi}{5})= \sin\frac{4\pi}{5}$


问:
Which is the better approximation to $e$?
Let $a_n = (1+1/n)^n$
and $b_n = (1+1/n)^{n+1}$.
Both $a_n \to e$
and $b_n \to e$,
and
$a_n < e < b_n$.

A better approximation to $e$
is known to be
$c_n = (1+1/n)^{n+1/2}
= \sqrt{a_n b_n}
$,
the geometric mean aof $a_n$ and $b_n$.

My question is:
how about $d_n = (a_n+b_n)/2$,
the arithmetic mean of $a_n$ and $b_n$?

Is $d_n$ a better approximation to $e$
than $c_n$?

More precisely,
let $r_n = \frac{c_n-e}{d_n-e}$.
Does $\lim_{n \to \infty} r_n$
exist? If so, is it $0$,
$\infty$, or a finite value?
If the limit is finite, what is it?

Is there some other mean of $a_n$ and $b_n$
(such as the harmonic mean)
which does better than either?

No, I haven't tried to solve these yet.
I thought they would be interesting questions.

Extra points if the answer
does $not$ use the expansions of
$\ln(1\pm x)$ or $e^x$.

答:
This is not an answer but some numerical insight. From the figure, we have
$$\lim_{n \to \infty} r_n = 0.4$$
The red line is the line at an height of $0.4$. The limit is almost surely $0.4$, since the value of $r_{50}$ is $\approx 0.39999$. Hence, it is pretty clear that the geometric mean does a better job than the arithmetic mean, but only by a factor. Also, as Ittay Weiss has already mentioned in his comment, the convergence of both the series is actually pretty pretty slow; even for $n=50$, both the sequences, $c_n$ and $d_n$, are only accurate to the third digit.

Similarly, if we were to define $h_n = \dfrac2{\dfrac1{a_n} + \dfrac1{b_n}}$ and look at the sequence $$l_n = \dfrac{h_n-e}{a_n-e}$$ from the figure below we see that
$$\lim_{n \to \infty} l_n = -0.2$$ The black line is at a height of $-0.2$. The limit is almost surely $-0.2$, since the value of $l_{50}$ is $\approx -0.19999$.
Also, we have
$$d_n > c_n > e > h_n$$


问:
Isomorphism $f$ preserves the order of an element?
Let's say $f$ is an isomorphism $f:G  \rightarrow G'$, where $G$ and $G'$ are multiplicative groups.

Then for $x\in G$, if $f(x) = x' \in G'$. Do we have always have $\text{ord}(x) = \text{ord}(x')$? Why?

答: The claim holds also for infinite $\langle g\rangle$. In fact, by the injectivity, $\langle g\rangle\cong f(\langle g\rangle)$; moreover, $f(\langle g\rangle)\le\langle f(g)\rangle$ generally holds for any homomorphism. If $\langle g\rangle$ is infinite, then $\langle g\rangle\cong f(\langle g\rangle) \cong \Bbb Z$, and hence $\langle f(g)\rangle\cong \Bbb Z$. Therefore, $|\langle g\rangle|=|\langle f(g)\rangle|$.


问:
solving a system of cubic equations with real roots
Consider the following system of equations:

\begin{align}
\frac{1}{8} (\alpha +2 x)^2 \left((\alpha +2 x)^2-12\right)+\\ +\frac{4}{9} y^2 \left((\alpha
   +2 x)^2+2\right)+\frac{32 y^4}{81}&=0\\
\alpha ^3-6 \alpha +8 x^3-4 \alpha  x^2-2 \left(\alpha ^2+2\right) x&=0
\end{align}

I would like to eliminate the $\alpha$ value in order to have an equation of the form $f(x,y)=0$.

Using Cardano's formula or Mathematica, I end up with imaginary values. However if I solve them  numerically I obtain a nice shape $f(x,y)=0$

Due to the numerical plots I obtain I am thinking that there must be a nice (analytical) solution for these equations. How could I derive such an expression for $f(x,y)=0$?

答:
The second equation write
$$\alpha^3-2x\alpha^2-2(3+2x^2)\alpha+4x(2 x-1)=0$$ The discriminant is
$$\Delta=32 \left(128 x^4+18 x^2+27\right) > 0 \qquad \forall x$$
Using the trigonometric method for three real roots leads to
$$\alpha_k=\frac{2}{3} \left(x+ \sqrt{2(8 x^2+9)} \cos \left(\frac{1}{3} \left(2 \pi  k-\cos
   ^{-1}\left(\frac{\sqrt{2} x \left(27-16 x^2\right)}{\left(8 x^2+9\right)^{3/2}}\right)\right)\right)\right)$$ with $k=0,1,2$.

Replace in the first equation to have $f(x,y)=0$ for each value of $k$. This would lead to
$$c_0+c_1\cos(A)+c_2\cos(2A)+c_3\cos(3A)+c_4\cos(4A)=0$$ with
$$c_0=32 \left(70 x^4+3 x^2 \left(4 y^2+9\right)+y^4\right)+27 \left(8 y^2-9\right)$$
$$c_1=128 \sqrt{2} x \sqrt{8 x^2+9} \left(7 x^2+y^2\right)$$
$$c_2=2 \left(8 x^2+9\right) \left(112 x^2+8 y^2-9\right)$$
$$c_3=16 \sqrt{2} x \left(8 x^2+9\right)^{3/2}$$
$$c_4=\left(8 x^2+9\right)^2$$

$$A=\frac{1}{3} \left(2 \pi  k-\cos
   ^{-1}\left(\frac{\sqrt{2} x \left(27-16 x^2\right)}{\left(8 x^2+9\right)^{3/2}}\right)\right)$$


问:
How do I go about proving a bi-conditional statement?

I'm taking a discrete math class and am kind of confused on how to prove a bi-conditional statement. I understand that your supposed to first prove p implies q and then prove q implies p. However, I don't know how to practically apply that to the problems I've been given, so here I am. In the image link is the problem I'm having trouble with, if anyone could give me an example of working through something similar I would be grateful.

I've tried finding videos on proving bi-conditionals, but they always keep the problems they work rather simple and aren't very helpful.

答: To prove a statement of the form $A \iff B$, you prove both $A \implies B$ and $B \implies A$. So, in your case, you have to prove both the statement $\exists x[P(x) \lor Q(x)] \implies \exists x[P(x)] \lor \exists x[Q(x)]$ and the statement $\exists x[P(x)] \lor \exists x[Q(x)] \implies \exists x[P(x) \lor Q(x)]$.


问:
Determine bases and the dimensions of $Ker(f)$ and $Im(f)$. Complete these bases to bases of $\mathbb {R}^4$ and $\mathbb {R}^5$, respectively.
Let $f:\mathbb {R}^4→\mathbb {R}^5$ be an $\mathbb {R}$-linear map defined by $f(x, y, z, t) = (t, x+ 2y, y+z, z−2x, t+z+y)$.

Determine bases and the dimensions of $Ker(f)$ and $Im(f)$. Complete these bases to bases of $\mathbb {R}^4$ and $\mathbb {R}^5$, respectively.

my take on the exercise:

$(t, x+ 2y, y+z, z−2x, t+z+y) = (0,x,0,-2x,0) + (0,2y,y,0,y) + (0,0,z,z,z) + (t,0,0,0,t) + (0,0,0,0,0) $

$= x(0,1,0,-2,0) + y(0,2,1,0,1) + z(0,0,1,1,1)+ t(1,0,0,0,1)$

$= <(0,1,0,-2,0), (0,2,1,0,1), (0,0,1,1,1), (1,0,0,0,1)>$

so the dim $Im(f)=4$

is it correct so far?

$Ker(f)$ = {$(x,y,z,t)$ ∈ $\mathbb {R}^4$ | $f(x,y,z,t)$ = $(0,0,0,0,0)$}

$Ker(f)$ = {$(x,y,z,t)$ ∈ $\mathbb {R}^4$ | $(t, x+ 2y, y+z, z−2x, t+z+y)$ = $(0,0,0,0,0)$}

$Ker(f)$ = {$(0,0,0,0,0)$ ∈ $\mathbb {R}^5$}

so the dim $Ker(f)=1$

I have trouble with completiting these bases to bases of $\mathbb {R}^4$ and $\mathbb {R}^5$

答:
You can compute $\ker f$ by solving the system$$\left\{\begin{array}{l}t=0\\x+2y=0\\y+z=0\\z-2x=0\\t+z+y=0.\end{array}\right.$$Its only solution is indeed $(0,0,0,0,0)$ and therefore $\dim\ker f=0$. So, a basis of $\ker f$ is $\emptyset$. And, since $\ker f=\{(0,0,0,0,0)\}$, $f$ is injective and therefore $\dim\operatorname{Im}f=4$ and$$\{f(1,0,0,0),f(0,1,0,0),f(0,0,1,0),f(0,0,0,1)\}\tag1$$is a basis of $\operatorname{Im}f$.

Now, in order to get a basis of $\Bbb R^4$ which contains $\emptyset$, you can just take the standard basis. And it turns out that if you add $(1,0,0,0,0)$ to $(1)$, then you get a basis of $\Bbb R^5$.


问:
Let $\{v1, v2, . . . , vn\}$ span $V$, and let $T:V→W$ be a linear transformation. Prove that $\{T(v_1),..., T(v_n)\}$ spans $W$ and $image(T)$.
I got the proof for $\{T(v_1), T(v_2), . . . , T(v_n)\}$ spanning $W$:

Since the $v_i$ spans $V$, given any $v ∈ V$, we can write $v = a_1v_1 + . . . + a_nv_n$ for some scalars $a_i ∈ F$. Applying $T$ to both sides, we have $T(v) = T(a_1v_1 + . . . + a_nv_n) = a_1T(v_1) + . . . + a_nT(v_n)$, by linearity of $T$.

Now $T(v) = w$ for some arbitrary vector $w ∈ W$. Thus we have $w = T(v) = a_1T(v_1) + . . . + a_nT(v_n)$, for any arbitrary vector $w ∈ W$, so $(T(v_1), . . . , T(v_n))$ spans $W$.

But I'm confused about how to do it for $image(T)$. I know that the image of a linear transformation is the span of the vectors of the linear transformation, but I'm not sure how to use that.

答:
Only the second part is correct.  Consider the zero transformation, or one into a space of higher dimension.  Say, for instance, $T:\Bbb R^4\to \Bbb R^5$.  Then $T$ can't be surjective.

To prove the second part is easy by linearity and the fact that $\{v_1,\dots,v_n\}$ spans $V$.  For the image of $T$ is nothing but $T(V)$.


问:
Determine the outer measure induced by a given set function (which counts the number of integers in a subset of $\mathbb{R}$)
On the collection $\mathcal{S}$ of all subsets of $\mathbb{R}$, define the set function $\mu: \mathcal{S} \rightarrow \mathbb{R}$ by setting $\mu(A)$ to be the number of integers in $A$. Determine the outer measure induced by $\mu$ and the $\sigma$-algebra of measurable sets.

I was wondering if the following is true: any $E$ is a subset of $\mathbb{R}$ (i.e., $E \in \mathcal{S}$), which implies that $\mu^*(E)=\mu(E)$. For any set $A$, $\mu^*(A)=\mu^*(A \cap E)+\mu^*(A \cap E^{\complement})$ always holds.

答:
$\mu =\sum_{n \in \mathbb Z} \delta_n$ is a measure  so the  outer measure induced by it $\mu$ itself.

For any sequence of measures $\mu_n$ the sum $\sum_n \mu_n$ is countably additive.


问:
There are 15 balls named from A to O.
Continuing the question: They are thrown in boxes 1, 2, 3 uniformly and at random such that each box gets 5 balls. Calculate probabilitity that A and B are in the same box.

I understand that we may need to do some form of complementary counting where the total number of outcomes are $\binom{15}{5} \times \binom{10}{5} \times \binom{5}{5}$. But I am confused over how to go ahead from this step.

答: Rare probability problem where counting is the wrong approach.  Box with A has 4 empty slots out of 14.  Chance is 4/14.

Title: There are 15 balls named from A to O.


问:
Problem of linear Algebra.
F is a 2 * 2 matrix that defines a transformation that reflects across three different lines in R2 and then rotates by an angle of 90 degree clockwise. Is F invertible?</p>

答: Yes: the inverse is the matrix representing the transformation which rotates $90^\circ$ counterclockwise, then reflects across the third line (which is the inverse of reflecting across said line), then across the second (same as before), then across the first (idem).

Title: Problem of linear Algebra.


问:
Monotone convergence theorem for a decreasing recursive sequence from $n>n_0$
Let a recursive sequence:
$$a(0)=a\in\mathbb{R};\\
a(n+1)=f(a(n));\\$$
if the sequence is decreasing but not for all $n\in\mathbb{N}$, e.g $a(n+1)\leq a(n) \,\, \forall \, n\geq 2$, it holds the monotone convergence theorem? So I can say that exists the limit of the sequence (even if the decrease holds from $n=2$)?

答: It is required that the sequence is $\textit eventually$ monotonic, since it would still mean that $\textit eventually$ all the terms in the sequence are less than $\epsilon$ for every $\epsilon$. However, it is also required by the monotonic convergence theorem that the sequence is bounded. It doesn't look like you have given how $f$ is defined.


问:
"Optical Illusion" in 4D
(Apologies in advance for the wordiness - not very mathematical, I know!)

I open my book of Escher optical illusions and look at the 2D page.  "Aha!" my brain says - "That image doesn't make sense when I try to interpret it as a representation of a 3D scene.  Isn't Escher clever?"

Conceptually, I could throw a Flatlander into the Escher picture and he could explore the 2D page.  The shapes on the page wouldn't make much sense to him but he could move around them (and possibly peek inside them to explore their interiors, by cutting them open) all the same.  Patiently I could explain that there's this third dimension unknown to him and if only he look at the shapes from this third dimension, he might understand the cleverness of Escher and how the perfectly-valid 2D shapes combine to make an optical illusion for 3D beings.  If he's bright enough, he might actually be able to conceptualise the mathematics of viewing things in three dimensions.

So, by analogy, a hypothetical 4D being could open her "book" of optical illusions and look at the 3D page.  Her brain would be suitably fooled and amused as she examined the 3D objects and tried to interpret them as a 4D scene.

If this statement is correct (and perhaps it isn't - perhaps an "optical illusion" in 4D is provably impossible?), then it should be possible for us to construct some 3D solids (possibly with non-trivial interiors) and arrange them in a way which we could convince ourselves would imply an optical illusion if only we were to look at them from a 4D point-of-view.

Would it be possibly to construct and arrange such 3D shapes, do you think?  How we would go about proving that they'd form an "optical illusion" in 4D?

PS I'm not interested using time as a fourth dimension - strictly spatial dimensions, please.

答:
There are different definitions (or rather, categories) of what is an "optical illusion", but the example you are talking about is a special case of ambiguous information. In general, there is an infinite number of 3D interpretations from a 2D image. The basic reason is that you are missing the information on the "depth" direction. Some illusions result in a "local" structure that doesn't make sense when you try to interpret it in a global way, but even with that knowledge it is almost impossible to change the perceived local structure (but that is a limitation of our visual system because it not always look for a globaly consistent solution) .
Our brain, or any 3D recovering algorithm tries to give the 2D image the most likely interpretation, based on past experience or on some a priori probability distribution. So yes, any 3D object has an infinite number of interpretations as a 4D object or scene, and depending on the observer's visual algorithms, even some that are inconsistent at a global level. </p>

Title: "Optical Illusion" in 4D


答:
If you do have Mathematica, use the command

GroebnerBasis[{pol1, pol2}, {x,y},{a}]

where poli are the two expressions that equal $0$. This will eliminate the variable $a$.

On WolframAlpha I got a polynomial in $x$, $y$ with fairly large coefficients.  Check this link here.

Hope there were no mistakes in the input.

$\bf{Added:}$. You will get the resultant of the two expressions considered as polynomials in $a$. The Groebner command  is just more general, can eliminate some variables ( the ones in the second group) from a system of equations.

Title: solving a system of cubic equations with real roots

Body:
Consider the following system of equations:

\begin{align}
\frac{1}{8} (\alpha +2 x)^2 \left((\alpha +2 x)^2-12\right)+\\ +\frac{4}{9} y^2 \left((\alpha
   +2 x)^2+2\right)+\frac{32 y^4}{81}&=0\\
\alpha ^3-6 \alpha +8 x^3-4 \alpha  x^2-2 \left(\alpha ^2+2\right) x&=0
\end{align}

I would like to eliminate the $\alpha$ value in order to have an equation of the form $f(x,y)=0$.

Using Cardano's formula or Mathematica, I end up with imaginary values. However if I solve them  numerically I obtain a nice shape $f(x,y)=0$

Due to the numerical plots I obtain I am thinking that there must be a nice (analytical) solution for these equations. How could I derive such an expression for $f(x,y)=0$?


问:
IMO 2005 problem hint.
Determine all positive integers relatively prime to all the terms of the infinite sequence
$a_n = 2^n + 3^n + 6^n − 1,\ n \ge 1$. ~IMO 2005 P4

I was solving a number theory book in which this was an example question. I know I can find this solution anywhere but I would like a hint as to how to solve this problem coz i don't want to see the solution just yet. I think this may use Fermat's little theorem so I was looking at $a_{p-1}$ . Any hint would be appreciated! Cheers.

答: Modulo a prime $p\ge5$, $a\cdot2^{p-1}+b\cdot3^{p-1}+c\cdot6^{p-1}\equiv a+b+c$ by Fermat's Little Theorem. Choose $a,\,b,\,c$ so you can divide this by $6$ to prove $p$ is a factor of some $a_n$. What about $p\in\{2,\,3\}$?

Title: IMO 2005 problem hint.


问:
$\frac{(x-y)^3}{x+y}\neq g(f(x)-f(y))$?
$$h(x,y)=\frac{(x-y)^3}{x+y}$$

Prove that there does not exist 1D real functions $f,g$ such that $h(x,y)=g(f(x)-f(y))$.

The problem seems really really easy because it is obvious that $x+y\neq f(x)-f(y)$. But I only have a really complicated approach outlined below, which is not necessarily correct. Do you have a simple proof?

My try:

By contradiction. suppose that $h=g(f(x)-f(y))$.

If we further assume that  $h(a,b)\geq h(\alpha,\beta)$ and $h(b,c)\geq h(\beta,\gamma)$,

we must have: $h(a,c)\geq h(\alpha,\gamma)$.

However, since

$$h(x,y)=\frac{(x-y)^3}{x+y}$$

Let $a=1$, $b=2$, $c=3$;

$\alpha=95.95$, $\beta=100$, $\gamma=103.44$.

We have: $h(\alpha,\gamma)=2.10$, which is less than $h(a,c)=4$

$h(a,b)=1/3$, $h(b,c)=1/5$, which are less than $h(\alpha,\beta)=0.339$, $h(\beta,\gamma)=0.2001$

Here is another approach but assuming the differentiability of $f$ and $g$. Assume $h=g(f(x)-f(y))$

The partial derivatives:
$h_x=g'f_x$

$h_y=g'f_y$

$h_x/h_y=f_x/f_y$

Therefore

$$h_x/h_y=\frac{x+2y}{-2x-y}=f_x/f_y$$

答:
I can't follow your attempts.Fn. 1  Here's a correct solution, although I can't exactly call it simple.  This is a surprisingly hard problem!

Generalities

Note that $$f(x)-f(y)=(f(x)-f(0))-(f(y)-f(0))$$  Thus we may take $f(0)=0$ without loss of generality (replace $f$ with $x\mapsto f(x)-f(0)$).

$f$ has constant sign

Unless $x=0$, \begin{equation}
g(f(x))=h(x,0)=x^2\geq0
\end{equation}  Likewise, $$g(-f(y))=h(0,y)=-y^2\leq0$$  So $\mathrm{rng}(f)\cap\mathrm{rng}(-f)\subseteq\{0\}$, i.e. $f$ is either always nonnegative or always nonpositive.

Inverting $g$

Equation (1) also has another, more surprising, property: $g\circ f$, considered as a map $\mathbb{R}^+\to\mathbb{R}^+$, has an inverse.  To see this, note that $g(f(\sqrt{x}))=x$; likewise $\sqrt{g(f(x))}=x$.  Thus $g$ is injective and surjective: that is, invertible.

Likewise $g(-f(x))=-x^2$.  Thus $g$ is invertible on $\mathbb{R}^-$ too; the only place it might not be invertible is $0$.  But that last case is easily taken care of: for any $x\neq0$, we have $g(0)=h(x,x)=0$; whereas $g(f(x))=h(x,0)\neq0$.  Thus $g(0)=0$ is the only root of $g$, and $g$ is invertible on all of $\mathbb{R}=\mathbb{R}^+\sqcup\{0\}\sqcup\mathbb{R}^-$.

A recursion for $f$

Now, $h$ is homogenous of degree $2$: $h(a,1)x^2=h(ax,x)$  Thus $$g(f(a)-f(1))=g(f(ax)-f(x))$$ for any $x\neq0$ and $a\neq1$  Since $g$ can be inverted, $$f(a)-f(1)=f(ax)-f(x)$$ again for any $x\neq0$ and $a\neq1$  In particular, \begin{equation}
f(2)-f(1)=f(2x)-f(x)
\end{equation}  Let $C=f(2)-f(1)$ and fix some $x_0>0$; then $f(2^kx_0)=kC+f(x_0)$ by iterating (2).

A contradiction at last

Now restrict to only positive values of $x$; by our remarks above, this forces $f$ to be all of one sign (either positive or negative).  But iterated (2) conflicts with our conclusion that $f$ is all of one sign.  If $f(x_0)$ is positive, then there is some large $k$ such that $f(2^{-k}x_0)=f(x_0)-kC<0$; if $f(x_0)$ is negative, then $f(2^kx_0)=kC+f(x_0)>0$.  Thus the proposed decomposition is impossible.

Footnote

Fn. 1  Where does your proposed monotonicity formulas for $h$ come from?  What should I conclude from the ratio of partial derivatives?

Title: $\frac{(x-y)^3}{x+y}\neq g(f(x)-f(y))$?