问:
First derivative of $\sqrt[\large 5]{\frac{t^3 + 1}{t + 1}}$
I have yet another derivative I need help with. I have to differentiate :

$$\sqrt[\uproot{3}{\Large 5}]{\frac{t^3 + 1}{t + 1}}$$

with respect to $t$.

I had two thoughts about this, use the chain rule then the quotient rule and multiply out, but then I am left with a mess of:

$$\left(\frac{t^3+1}{t+1}\right)^{-4/5}\cdot\frac{0.6t^3+0.6t^2-0.2t^4+0.2t}{(t+1)^2}$$

This is turning into a real mess and the answer I should get is:

$$\frac{2t-1}{5(t^2-t+1)^{4/5}}$$

Am I going the correct way about this or should I try a different route?

Thanks

答:
The process you describe one way to approach this: you can  use both the chain rule and the quotient rule. But it looks to me as though you misunderstand the quotient rule:

$$h(x) = \frac{f(x)}{g(x)}$$

$$h'(x) = \frac{f'(x)\cdot g(x) - f(x)\cdot g'(x)}{g^2(x)}$$

where the factors in the numerator are multiplied, not added. (It also looks like you multiplied the coefficients by $1/5$?)

In your case, the rational function within the 5th root is $$h(x) = \frac{f(x)}{g(x)} = \frac{t^3 + 1}{t+1}$$

$$f(x) = t^3 + 1, \implies f'(x) = 3t^2;\quad g(x) = t+ 1\implies g'(x) = 1$$

$$h'(x) = \frac{f'(x)\cdot g(x) - f(x)\cdot g'(x)}{g^2(x)} $$
$$h'(x) = \frac{3t^2(t+1) - (t^3 +1)\cdot 1}{(t+1)^2}\;\;=\;\;\frac{2t^3 + 3t^2 - 1}{(t+1)^2}$$

Now, multiply $h'(x)$ with the result you got for the first factor: $$\frac 15\left(\frac{t^3+1}{t+1}\right)^{-4/5}= \left(\frac{t+1}{t^3 + 1}\right)^{4/5}$$

Which gives us:

$$\frac 15\left(\frac{t+1}{t^3 + 1}\right)^{4/5}\cdot \frac{2t^3 + 3t^2 - 1}{(t+1)^2}$$

Now, we can simplify.  (The Chaz's method greatly simplifies the process of taking the derivative of your function. But I think it is a good idea to ensure you understand the quotient rule.)

Chaz's tip will also simplify the left factor above to $\left(\dfrac{1}{t^2 - t + 1}\right)^{4/5}$. After that, there is really no need to simplify. Answers can be correct, yet not match the "textbook's solution" perfectly, simply based on how we choose to simplify.

$$\left(\frac{2t - 1}{5(t^2 + t - 1)^{4/5}}\right)$$


问:
Degree of the minimal sufficient statistic for $\theta$ in $U(\theta-1,\theta+1)$ distribution
Suppose $X_1,X_2,...,X_n$ is a random sample from the Uniform distribution over the interval $(\theta-1,\theta+1)$. By the factorization theorem, it is clear that the order statistics $Y_1=X_\left(1\right)$ and $Y_n=X_\left(n\right)$ are joint sufficient statistics for $\theta$. Now, because $$\theta-1 < Y_1 < Y_n < \theta+1 $$ implies that $Y_n -1 < \theta < Y_1 + 1$, choosing $\hat\theta \in (Y_n-1,Y_1+1)$ will force the likelihood function to achieve its maximum; that is, $$L(\hat\theta)=\left(\frac{1}{2}\right)^n, \forall \hat\theta\in(Y_n-1,Y_1+1)$$

Typically, statisticians takes the value for $\hat\theta$ to be the midrange of $Y_1,Y_n$: $\hat\theta=\frac{Y_1+Y_n}{2}$.

Now, because such a value for $\hat\theta$ maximizes the likelihood function, it follows that $\hat\theta$ is an mle for $\theta$ of in $Unif(\theta-1,\theta+1)$. It is noted in the Hogg, McKean, Craig textbook "Introduction to Mathematical Statistics" ($7^{th}$ edition), however, that even though $\hat\theta$ is an mle of $\theta$ and a function of the joint sufficient statistics $Y_1,Y_n$ for $\theta$, it does not represent a minimal sufficient statistic; this is because $\hat\theta$ is itself not a sufficient statistic for theta.

What I'm wondering here is $why$ $\hat\theta = \frac{Y_1+Y_n}{2}$ is not a sufficient statistic for $\theta$. By the definition of a sufficient statistic and using the factorization theorem, $Y_1,Y_n$ being the joint sufficient statistics for $\theta$ imply that the likelihood function can be written as $$\prod_1^nf(x_i;\theta) = K_1(Y_1,Y_n;\theta)*K_2(x_1,x_2,...,x_n)$$ where $K_1(Y_1,Y_2;\theta)$ depends on $x_1,...x_n$ only through $Y_1,Y_n$ and$K_2(x_1,x_2,...,x_n)$ does not depend upon $\theta$; in our case, $$K_1(Y_1,Y_n;\theta)=\left(\frac1{2}\right)^n \cdot \mathbf 1_{(\theta-1,\theta+1)}(Y_1) \cdot \mathbf 1_{(\theta-1,\theta+1)}(Y_n) $$

Now, although it is clear that we cannot reduce the product of the two indicator functions down any further to any single order statistic $Y_i$ such that an equality remains between $K_1$ and our new function, why is it not the case that $$\mathbf 1_{(\theta-1,\theta+1)}(\frac{Y_1+Y_2}{2}) \neq \mathbf 1_{(\theta-1,\theta+1)}(Y_1) \cdot \mathbf 1_{(\theta-1,\theta+1)}(Y_n)$$ is a valid equality?

In proceeding further in my reasoning, let's reference this question. For an overview, this question revolves around the sufficient statistics for the uniform distribution $Unif(-\theta,\theta)$. In this problem, one can reduce their argument from the joint sufficient statistics $X_\left(1\right),X_\left(n\right)$ down to the single sufficient statistic $max\{X_\left(1\right),X_\left(n\right)\}=Y^*$.

Does it happen to be so that the only thing stopping us from reducing our two-dimensional joint sufficient statistic(s), $\mathbf Y=(Y_1,Y_n)$, for $\theta$ down into the single-dimensional ("minimal") sufficient statistic, $Y=Y^*$, for $\theta$ the fact that our Uniform distribution is not distributed symmetrically about the Y-axis? I could see this argument being the case, but I'd like to be sufficiently (no pun intended) sure of that before proceeding on in my studies. It took me quite a bit of investigating to begin to understand how to work through various operations when working with the indicator function, so I still might a little lost when it comes to the finer intricacies of manipulating this function.

答:
$(X_{(1)},X_{(n)})$ is a minimal sufficient statistic for $\theta$.

Proof:

The joint density of $X=(X_1, \cdots,X_n)$ is $$\mathbb{1}(\theta-\frac{1}{2}\leq X_{i}\leq \theta+\frac{1}{2}),\forall i$$
This is equivalent to say
$$\mathbb{1}(\theta\leq X_{(1)}+\frac{1}{2}\text{ and } \theta\geq X_{(n)}-\frac{1}{2})$$
Thus $(X_{(1)},X_{(n)})$ uniquely determines the area where the p.d.f is one, which means $(X_{(1)},X_{(n)})$ uniquely determines the shape of the log-likelihood function.
We can conclude $(X_{(1)},X_{(n)})$ is a minimal sufficient statistic for $\theta$.

ContentLicense: CC BY-SA 4.0


问:
Name and application of a nonlinear ODE
Is there a name for an ODE taking form:
\begin{equation}
\left(\frac{dy}{dx}\right)^2 + a y = 0,
\end{equation}
and if there is, what is the constant `a' called either generally or in certain applications if there are any?
I've searched online for equations of this form, and it seems a specific form of Chrystal's equation
\begin{equation}
\left(\frac{dy}{dx}\right)^2 + bx \frac{dy}{dx} + ay + cx^2 = 0
\end{equation}
takes this form when $b=c=0$.

I would also like to know if there are any physical applications that use this particular type (or a vectorized version) of this ODE.

答:
Let $a= -1/f$; the equation
$$\left(\dfrac{dy}{dx}\right)^2 + a y; \quad  \left(\dfrac{dy}{dx}\right)^2 - \dfrac{y}{f} = 0 \tag1 $$

is called simple harmonic motion and can be  be parametrized
$$ x= 2 ft, y=f t^2$$
so that $a$ is related to max curvature of a parabola at its vertex.
$$ \dfrac{1}{R} =\dfrac{1}{2f}=\dfrac{-a}{2}\tag2 $$

Title: Name and application of a nonlinear ODE


问:
Is $x=n^x$ solvable where $n>1$?
Since the graphs of $y=x$ and $y=2^x$ (or $y=n^x$ for that matter, where $n>1$) do not intersect, is $x=2^x$ unsolvable? Or is there some kind of a really clever way to find a root?

P.S. I'm not too sure if this is the right tag to use.

答:
$$x=n^x=e^{x\ln n}$$
$$(-x\ln n)e^{-x\ln n}=-\ln n$$
$$-x\ln n=W(-\ln n)$$
$$x=-\frac{W(-\ln n)}{\ln n}$$
Since $W$ is real-valued on at least one branch when its argument is a small negative number (down to $-\frac1e$), $x=n^x$ has real solutions when $1<n\le e^{1/e}$.

Title: Is $x=n^x$ solvable where $n>1$?


问:
Is it possible to separate Pi from Eulers constant with rational constants?

$\pi^2 \cdot e^2 \sim \frac{10000}{137}$, is quite accurate, although I have no idea what this "fine structure constant" $\frac{1}{137}$  from physics really is.

For any prime numbers $p_1<p_2<p_3<p_4$,  $(p_1 p_2 p_3 p_4)^9$  has always exactly $10000$ divisors  and the powers of $2$ do relate to the powers of $10$ in some way dependent of their divisors, which also seems to relate in some form to $\pi$.

So my question is, is it possible to separate $\pi$, from $e$ in the first equation without using recursions or irrational numbers and how does $\pi$ relate to the $2^{10} = 1024, 10^3 = 1000$  coincidence?

答:
This does not answer the question but relates the approximation in the question to another one.

Since $137\pi^2e^2 \approx 9991$, the accuracy of the following approximation is better.

$$\pi^2e^2 \approx \frac{100^2-3^2}{137}$$

Now if we bother to write $$\frac{100^2-3^2}{\pi^2e^2} \approx 136.999899991$$
this is closer to an integer than the nice

$$e^\pi-\pi \approx 19.999099979$$

And the ratio of the distance of both expressions from an integer is close to yet another integer.

$$\frac{e^\pi-\pi-20}{\frac{100^2-3^2}{\pi^2e^2}-137} \approx 9$$

I believe that the following expression is also worth mentioning.
$$\frac{1}{\alpha}\approx137.035999 \approx \frac{34259}{250}=5^3+12+\frac{3^2}{2·5^3}=5^3+\frac{5^2}{2}-\frac{1}{2}+5^{-2}-\frac{5^{-3}}{2}$$

In other words, the polynomial $(2n+1)^3+3n^2$ takes the value $137$ for $n=2$. Equivalently, $(4n+1)^3+3(2n)^2$ gives $137$ for $n=1$. Here is an illustration of $137$ as a three-dimensional shape.

Only the faces seen have additional cubes, shown in red. Those extra cubes could be halved and shared among the seen and unseen faces for another symmetry.

Regarding the $2^{10} \approx 10^3$, since $10=2·5$ we can simplify it down a bit to the following.

$$2^{10} \approx 10^3 = 2^35^3$$
$$2^7 \approx 5^3$$

Update

A simpler fourth power $(9^4)$ yields a better approximation:
$$\pi \approx \frac{27}{e\sqrt{10}} \approx 3.1410097$$

compared to
$$\pi \approx \frac{100}{e\sqrt{137}} \approx 3.143006$$

However, from your approximation the following almost-integer is obtained.

$\frac{10^4}{137\pi^2e^2} \approx 1.0009000 $

This happens to match the gap in $e^\pi-\pi$, so combining both gives

$$e^\pi-\pi+\frac{10^4}{137\pi^2e^2} \approx 21.000000058$$


问:
Use quaternion to represent rotation matrix
I know the unit quaternion can represent the 3D rotation. For example, $Rp=q*p*q^{-1}$ where $R$ is the rotation matrix of body frame with respect to inertial frame, $q$ is the unit quaternion, $*$ is the quaternion multiplication, $p$ is pure quaternion whose first element is $0$ and rest elements are a $3 \times 1$ vector.

Right now I have a formula like $RDR^{T}$ where $R$ is the rotation matrix of body frame with respect to inertial frame, $D$ is a $3 \times 3$ coefficient matrix. Could I use quaternion $q$ to represent this formula in a similar way that I showed in my above example?

答:
Given a pure rotation matrix $\mathbf{R}$, its inverse is $\mathbf{R}^{-1} = \mathbf{R}^T$.  Let
$$\mathbf{R} = \left[ \begin{matrix}
X_x & X_y & X_z \\
Y_x & Y_y & Y_z \\
Z_x & Z_y & Z_z \\
\end{matrix} \right] , \quad \mathbf{D} = \left[ \begin{matrix}
x_x & x_y & x_z \\
y_x & y_y & y_z \\
z_x & z_y & z_z \\
\end{matrix} \right]$$
Then,
$$\mathbf{R} \mathbf{D} \mathbf{R}^{-1} = \left[ \begin{matrix} \chi \\ \gamma \\ \zeta \end{matrix} \right]$$
where
$$\begin{array}{rclclcl}
\chi   & = & x_x X_x X_x & + & x_y Y_x X_y & + & x_z Z_x X_z \\
 ~     & + & y_x X_x Y_x & + & y_y Y_x Y_y & + & y_z Z_x Y_z \\
 ~     & + & z_x X_x Z_x & + & z_y Y_x Z_y & + & z_z Z_x Z_z \\
\gamma & = & x_x X_y X_x & + & x_y Y_y X_y & + & x_z Z_y X_z \\
 ~     & + & y_x X_y Y_x & + & y_y Y_y Y_y & + & y_z Z_y Y_z \\
 ~     & + & z_x X_y Z_x & + & z_y Y_y Z_y & + & z_z Z_y Z_z \\
\zeta  & = & x_x X_z X_x & + & x_y Y_z X_y & + & x_z Z_z X_z \\
 ~     & + & y_x X_z Y_x & + & y_y Y_z Y_y & + & y_z Z_z Y_z \\
 ~     & + & z_x X_z Z_x & + & z_y Y_z Z_y & + & z_z Z_z Z_z \\
\end{array}$$

The $*$ in $q * p * q^{-1}$ refers to Hamilton product, which is very different from matrix product, and for unit quaternion $q = q_r + q_i\mathbf{i} + q_j\mathbf{j} + q_k\mathbf{k}$ (i.e. where $q_r^2 + q_i^2 + q_j^2 + q_k^2 = 1$) and a general quaternion $p = p_r + p_i\mathbf{i} + p_j\mathbf{j} + p_k\mathbf{k}$,
$$\begin{aligned}
q * p * q^{-1} &= p_r ( q_r^2 + q_i^2 + q_j^2 + q_k^2 ) \\
  ~            &~+ \biggr( p_i ( q_r^2 + q_i^2 - q_j^2 - q_k^2 ) - 2 p_j ( q_r q_k - q_i q_j ) +2 p_k ( q_r q_j + q_i q_k ) \biggr) \mathbf{i} \\
               &~+ \biggr( p_j ( q_r^2 - q_i^2 + q_j^2 - q_k^2 ) + 2 p_i ( q_r q_k + q_i q_j ) - 2 p_k ( q_r q_i - q_j q_k ) \biggr) \mathbf{j} \\
               &~+ \biggr( p_k ( q_r^2 - q_i^2 - q_j^2 + q_k^2 ) - 2 p_i ( q_r q_j - q_i q_k ) + 2 p_j ( q_r q_i + q_j q_k ) \biggr) \mathbf{k} \\
\end{aligned}$$
Because $p$ has at most four components, you cannot simply use $q * D * q^{-1}$.

We can, however, examine how we can decompose $D$ into vectors, so we can rotate and sum those.  Since $\mathbf{R}\mathbf{D}\mathbf{R}^{-1} = (\mathbf{R}\mathbf{D})\mathbf{R}^{-1} = \mathbf{R}(\mathbf{D}\mathbf{R}^{-1})$,
$$\mathbf{D} \mathbf{R}^{-1} = \left [ \begin{matrix}
x_x X_x + y_x Y_x + z_x Z_x \\
x_y X_y + y_y Y_y + z_y Z_y \\
x_z X_z + y_z Y_z + z_z Z_z = d\\
\end{matrix} \right ]$$
we have
$$\mathbf{R} \mathbf{D} \mathbf{R}^1 = \mathbf{R} \mathbf{d}$$
and we can express the same vector using $q d q^{-1}$ (with $d$ and the result quaternions with real component zero) with $q$ corresponding to the rotation matrix $\mathbf{R}$.

Unfortunately, an unit quaternion represents a rotation or orientation, via $q = \cos(\varphi/2) + x \sin(\varphi/2)\mathbf{i} + y \sin(\varphi/2)\mathbf{j} + k \sin(\varphi/2) \mathbf{k}$, where $(x, y, z)$ is the rotation axis unit vector ($x^2 + y^2 + z^2 = 1$) and $\varphi$ is the rotation angle, and recovering the basis vector components ($X_x$, $X_y$, $X_z$, and so on) is non-trivial.

So, the answer to OP's question, if $\mathbf{R}\mathbf{D}\mathbf{R}^{-1}$ can be represented via a quaternion $q$ corresponding to the rotation matrix $\mathbf{R}$, the answer is not really, the expression would be very, very long and complicated.

Another approach to examining the question is to realize that Hamilton product between two quaternions $q = q_r + q_i \mathbf{i} + q_j \mathbf{j} + q_k \mathbf{k}$ and $p = p_r + p_i \mathbf{i} + p_j \mathbf{j} + p_k \mathbf{k}$ can be expressed as a matrix product, via
$$q * p = \left [ \begin{matrix} q_r & q_i & q_j & q_k \end{matrix} \right ] \left [ \begin{matrix}
p_r & p_i & p_j & p_k \\
-p_i & p_r & -p_k & p_j \\
-p_j & p_k & p_r & -p_i \\
-p_k & -p_j & p_i & p_r \\
\end{matrix} \right ] = \left[ \begin{matrix}
q_r & q_i & q_j & q_k \\
-q_i & q_r & q_k & -q_j \\
-q_j & -q_k & q_r & q_i \\
-q_k & q_j & -q_i & q_r \\
\end{matrix} \right] \left[ \begin{matrix} p_r \\ p_i \\ p_j \\ p_k \\ \end{matrix} \right ]$$
and for an unit quaternion $q$ and a "vector" p ($p = x \mathbf{i} + y \mathbf{j} + z \mathbf{k}$, a quaternion with real component zero),
$$q p q^{-1} = \left[ \begin{matrix}
q_r & q_i & q_j & q_k \\
-q_i & q_r & q_k & -q_j \\
-q_j & -q_k & q_r & q_i \\
-q_k & q_j & -q_i & q_r \\
\end{matrix} \right] \left[ \begin{matrix} 0 & x & y & z \end{matrix} \right ] \left[ \begin{matrix}
q_r & -q_i & -q_j & -q_k \\
q_i & q_r & -q_k & q_j \\
q_j & q_k & q_r & -q_i \\
q_k & -q_j & q_i & q_r \\
\end{matrix} \right]$$
whereas for a rotation matrix $\mathbf{R}$ (so $\mathbf{R}^{-1} = \mathbf{R}^T$) we have
$$\mathbf{R} \mathbf{D} \mathbf{R}^{-1} = \left[ \begin{matrix}
X_x & X_y & X_z \\
Y_x & Y_y & Y_z \\
Z_x & Z_y & Z_z \\
\end{matrix} \right] \left[ \begin{matrix}
x_x & x_y & x_z \\
y_x & y_y & y_z \\
z_x & z_y & z_z \\
\end{matrix} \right] \left[ \begin{matrix}
X_x & Y_x & Z_x \\
X_y & Y_y & Z_y \\
X_z & Y_z & Z_z \\
\end{matrix} \right]$$
So, the similarity is in the notation, not in the operations done.


问:
If a cubic has a constructable root $a$, then $a$ is either rational or in some number field $Q(\sqrt{r})$ For some $r$ belongs to the rationals.
I know that if a polynomial has a constructible root, then it must have a rational root. Thus I try to prove it in two cases:
If $a$ is the rational root, then we are done. However, if $a$ is not the rational root, how do I show that $a$ must be in $Q(\sqrt{r})$ rather than $Q(\sqrt{r})(\sqrt{s})$ etc.

答:
If a cubic polynomial $x^3+ax^2+bx+c$ has a root it must have a minimal polynomial less then $3$.

The constructible numbers only contain minimal polynomials of degrees which are powers of $2$.

the only two powers of two under $3$ are $1,2$.

Option 1:

if the polynomial can be reduced into 3 linear factors then we are done.

Option 2:

if the polynomnial can be factored into a linear and quadratic factor then the quadratic factor is the root of $x^2+ax+b$, $\frac{-a\pm \sqrt{a^2-4b}}{2}$ which is a part of $\mathbb Q[\sqrt{a^2-4b}]$.


问:
Combinatorial proof of ${2n\choose n}n!\le (2n)^{n}$
Initially I was supposed to prove the proposition $n^n\ge \prod_{r=1}^{r=n}(2r-1)\forall n\in\mathbb{N}$. So I have basically simplified this question down to proving ${2n\choose n}n! \le (2n)^{n}$ by carrying out the following operations.

$$\begin{aligned}n^{n}\prod_{r=1}^{n}2r&\ge \prod_{r=1}^{n}(2r-1)\prod_{r=1}^{n}2r\\ 2^{n}n^nn!&\ge(2n)!\\ (2n)^{n}&\ge {2n\choose n}n!\end{aligned}$$

Any ideas on how to prove this inequality. The right side of the inequality represents the arrangement of $n$ objects chosen out of $2n$ object. And the left side represents the arrangement of $n$ objects into $2n$ vacancies. Any hints are appreciated. Thanks.

答:
Let $X = [n] = \{1,2,\ldots,n\}$, and $Y = [2n] = \{1,2,\ldots ,2n\}$.

Observe that $(2n)^n$ is the number of functions $f:X\to Y$. Indeed, for each $i\in X$, you have $2n$ choices to assign $f(i)$.

Next, note ${2n \choose n}n!$ is the number of injective functions $f:X\to Y$. Indeed, for an injective (one-to-one) function, we first pick an image for $f$, which is a size $n$ subset of $Y$. There are $2n\choose n$ many ways. Now with the image chosen, we have $n!$ ways to assign them to $f(i)$.

Since the set of all injective functions is a subset of all functions $X\to Y$, we have the desired inequality.

Note: In general ${|Y| \choose |X|} |X|!$ is the number of injective functions $X\to Y$ (whenever sensible), while $|Y|^{|X|}$ is the total number of functions $X\to Y$.


问:
Conditional expectation property: independence versus conditional independence properties
My question is about those two properties of conditional expectation:

-If $\mathcal{H}$ is independent of $\sigma(\sigma(X), \mathcal{G}),$ then
$$
\mathrm{E}[X \mid \sigma(\mathcal{G}, \mathcal{H})]=\mathrm{E}(X \mid \mathcal{G}), \quad \text { a.s. }
$$
In particular, if $X$ is independent of $\mathcal{H},$ then $\mathrm{E}(X \mid \mathcal{H})=\mathrm{E}(X), \quad$ a.s.

-If $X, Y$ are conditionally independent given $Z$, then $P(X \in B \mid Y, Z)=P(X \in B \mid Z)$ a.s (equivalently, $\left.E\left(1_{\{X \in B\}} \mid Y, Z\right)=E\left(1_{\{X \in B\}} \mid Z\right) a.s \right)$

Is one of those properties more general than the other ? (in the sense does one imply the other)

答:
The following is true:

If $Y$ is independent of $(X,Z)$, then $X$ and $Y$ are conditionally independent given $Z$, i.e., $$P(X\in A\mid Y,Z)=P(X\in A\mid Z).\tag{*}$$

This is easy to see when $(X,Y,Z)$ have a joint density $f(x,y,z)$, since the hypothesis is saying
$$f(x,y,z)= f(y)f(x,z).$$
This implies that
$$
f(y,z)=\int_x f(x,y,z)dx\stackrel{(1)}=\int_x f(y)f(x,z)dx=f(y)f(z),$$ hence $f(y\mid z)=f(y)$ so that
$f(x,y,z)=f(x,z)f(y\mid z)$ which is equivalent to
$f(x\mid y,z)=f(x\mid z)$, the desired conclusion.

For the general measure-theoretic proof, write $U:=P(X\in A\mid Z)$. By definition of conditional probability, $U$ is $\sigma(Z)$-measurable and
$$ P(X\in A, Z\in C)=\int_{(Z\in C)} UdP\tag1$$ for every $C$.
To prove (*), it is clear that $U$ is $\sigma(Y,Z)$ measurable, so to apply the definition of conditional probability it remains to show
$$
P((X\in A)\cap F)=\int_F UdP
$$ for every set $F\in \sigma(Y,Z)$. It's enough to do so for sets of the form $F:=(Y\in B, Z\in C)$. For such $F$ we have
$$P(X\in A, Y\in B, Z\in C)=P(Y\in B)P(X\in A, Z\in C)\stackrel{(1)}=EI_B(Y)E\left[I_C(Z)U\right]
$$ where the first equality follows from independence. Recall $U$ is $\sigma(Z)$-measurable, so by independence again this last equals
$$E[I_B(Y)I_C(Z)U]=\int_{(Y\in B,Z\in C)}UdP,$$ as was to be shown.


问:
How do you find the independent term in $(x + 1 + x^{-1})^4$?
I know you could expand it all out and take cases but is there a way to group it such that there is a general term? I've been trying to group it in order to establish a general term ie something like $((x + 1) + x^{-1})^4$ to make a binomial. Any hints?

答: An alternative way is to note that $\frac1x+1+x=\frac1x\cdot\frac{1-x^3}{1-x}$ so we are looking for the coefficient of $x^0$ in$$\frac1{x^4}\left(\frac{1-x^3}{1-x}\right)^4$$or the coefficient of $x^4$ in$$\begin{align*}\left(\frac{1-x^3}{1-x}\right)^4&=(1-x^3)^4(1-x)^{-4}\\&=\left[1+\binom41(-x^3)+...\right]\left[1+\binom{-4}1(-x)+...+\binom{-4}4(-x)^4+...\right]\end{align*}$$ which is $\binom{-4}4+\binom41\binom{-4}1=35-16=19$.


问:
Ring of polynomials in $\mathbb F_2$ with roots at a (specific) root of unity
WLOG I'm going to use $\zeta = \sqrt[12]1$ in this example. I'm trying to find some way to work with sums of distinct powers of $\zeta$, specifically so that I can analyze sums that are equal to zero (such as $1 + \zeta + \zeta^2 + \zeta^5 + \zeta^6 + \zeta^8 + \zeta^9$) (EDIT: I want the coefficients in the sum to be either $0$ or $1$, so each distinct power is only included once at most). I initially tried to work in the quotient ring $\mathbb F_2[x] / ( x^{12} + 1)$, but it quickly became clear that while I expected $P=1 + x+ \cdots + x^{11}$ to be equal to $0$ (since $Px = P$), it was not. I'm not sure what the flaw in my logic here was (though I suspect it has something to do with $1$ being a root of $x^{12}+1$), but moving to $\mathbb F_2[x]/P$ obviously fixed that issue.  My problem is that I can't find a sense in which the polynomial (for example) $1 + x +  x^2 + x^5 + x^6 + x^8 + x^9 = 0$, even though that's what I should expect to see if $x = \zeta$. Where am I going wrong?

Body:
$\zeta_{12}\,$ has minimal poly $\, \Phi_{12}(x) = x^4-x^2+1\mid x^6+1\,$ (by $(x^6)^2 = 1,\ x^6\neq 1\Rightarrow x^6= -1)\,$
so by the method of simpler multiples, to reduce $\!\bmod \Phi\,$ we can first reduce $\!\bmod x^6+1,\,$ where we can apply the simple rewrite rules $\,\color{#c00}{x^6 \equiv -1},\,$ so $\,\color{#0a0}{x^8 \equiv -x^2},\,$ and  $\,\color{#90f}{x^9 \equiv -x^3},\,$ e.g. in your case

$$\begin{align}&1+x+x^2+x^5+\color{#c00}{x^6}+\color{#0a0}{x^8}+\color{#90f}{x^9}\\[.1em]
\equiv\ & \color{#c00}{1}+x+\color{#0a0}{x^2}+x^5\color{#c00}{-1}\,\color{#0a0}{-\,x^2}\,\color{#90f}{-x^3}\\[.1em]
\equiv\ &\ \ \ \ \ \ \ x\ \ \ \,+\ \ \ \ \ x^5\ \ \color{#90f}{-\ \ x^3}\\[.1em]
\equiv\ &\ \  x\ (1\ \ \ \ +\,\ \ \ \ x^4\ \ -\ \ x^2)\\[.1em]
\equiv\ &\ \  x\,\Phi_{12}\equiv\, 0
\end{align}\qquad\qquad$$


答: Hint: simplify the radicand to $t^2 -t + 1$ before differentiating.

ContentLicense: CC BY-SA 3.0

Title: First derivative of $\sqrt[\large 5]{\frac{t^3 + 1}{t + 1}}$

Body:
I have yet another derivative I need help with. I have to differentiate :

$$\sqrt[\uproot{3}{\Large 5}]{\frac{t^3 + 1}{t + 1}}$$

with respect to $t$.

I had two thoughts about this, use the chain rule then the quotient rule and multiply out, but then I am left with a mess of:

$$\left(\frac{t^3+1}{t+1}\right)^{-4/5}\cdot\frac{0.6t^3+0.6t^2-0.2t^4+0.2t}{(t+1)^2}$$

This is turning into a real mess and the answer I should get is:

$$\frac{2t-1}{5(t^2-t+1)^{4/5}}$$

Am I going the correct way about this or should I try a different route?

Thanks


问:
If $f_n$ is the $n$th Fibonacci number then $\lim_{n\to \infty} (f_{n-2}+{1-\sqrt 5\over 2}f_{n-1})$?
If $f_n$ is the $n$th Fibonacci number then what is  the value of $$\lim_{n\to \infty}\left(f_{n-2}+{1-\sqrt 5\over 2}f_{n-1}\right)?$$
With direct calculation I saw the first four terms are approximately $(-0.62,0.39,-0.24,0.15)$. It seems that the limit approaches $0$. But how to show it analytically?

Body:
Using Binet's Formula,

for $a+b=1, ab=-1$ with  $a>b$

we need $$\lim_{n\to\infty}\left(\dfrac{a^{n-2}-b^{n-2}}{a-b}+b\cdot\dfrac{a^{n-1}-b^{n-1}}{a-b}\right)$$

$$=\dfrac1{a-b}\lim_{n\to\infty}\left(a^{n-2}-b^{n-2}+b(a^{n-1}-b^{n-1})\right)$$

$$=\dfrac1{a-b}\lim_{n\to\infty} (a^{n-2}(1+ab)+2b^{n-1})$$

Now  $1+ab=0$

and as $|b|<1,\lim_{n\to\infty}b^{n-1}=0$


问:
function and its derivative converges compactly on a open set of $R^2$ implies uniform convergence of derivative

Let $(f_n)$ be a sequence of continuously differentiable functions on an open set $U \subseteq \mathbb{R}^2$. Suppose the three sequences $f_n, \frac{\partial f_n}{\partial x}$ and $\frac{\partial f_n}{\partial y}$ converges compactly (i.e. converges uniformly on every compact subset of domain) to functions $g, h_1$ and $h_2$, respectively. Prove that $\frac{\partial g}{\partial x} = h_1$ and $\frac{\partial g}{\partial y} = h_2$.

Rudin (Theorem 7.17) states that the pointwise convergence of $f_n$ and the uniform convergence of $f_n$'s partial derivatives are enough to conclude this (if I am not reading that wrong), so it is enough to show that $\partial_x f_n \to h_1$ uniformly.

So for every $x \in U$, by openness we can find a open ball $\overline{B(x, R)}$ such that it falls in $U$, so we see that by compact convergence, $\partial_x f_n \to h_1$ uniformly on this compact set.

The problem remains is that, how should I extend it to the whole $U$?

答: To prove that $\frac {\partial g} {\partial x}=h_1$ you take any point in $U$ and work in a neighborhood of that point. So there is no loss of generality in assuming that $U$ is bounded.


问:
How many positive integers $n$ are there such that $(7n + 1)$ is a perfect square and $(3n + 1) < 2008$?

How many positive integers $n$ are there such that $(7n + 1)$ is a perfect square and $(3n + 1) < 2008$ ?

What I Tried: We have $(3n + 1) < 2008$
$\rightarrow n < 669$ .

Now, $(7n + 1) = k^2$ for some $k$ .
$$\rightarrow 7n = (k + 1)(k - 1)$$
Now I have to count all the $n$ so that after multiplied by $7$ , gives a product same as the product of $2$ consecutive numbers with a gap of one. I know that $n = 5,9$ are two solutions, but there are more values so that when multiplied by $7$ gives a valid solution, like $k = 24,32$ .

The problem I am facing is, is there any elegant way here to count all the solutions?

Can anyone help me?

答:
Since $n\le 668$, we know that $7n+1\le 4677$. $\left\lfloor\sqrt{4677}\right\rfloor=68$, so we want to know how many of the integers $1,2,\ldots,68$ have squares of the form $7n+1$. Suppose that $7n+1=k^2$ for some integer $k$; then $k^2\equiv 1\pmod 7$, and it’s easy to check that this is the case precisely when $k\equiv 1\pmod 7$ or $k\equiv -1\pmod 7$, i.e., when $k=7m\pm 1$ for some integer $m$.

There are $9$ positive multiples of $7$ less than $68$, each of which provides two such values of $k$.

E.g., $3\cdot 7=21$ means that $k=20$ and $k=22$ will work: $20^2=400=7\cdot 57+1$, and $22^2=484=7\cdot 69+1$.

Thus, there are altogether $2\cdot9=18$ such integers $n$.


答:
As you mentioned, $7n = (k + 1)(k - 1)$

So $7$ is a factor of either $(k+1)$ or $(k-1)$

which means $k = 7 a -1$ or $7a+1$.

If $k = 7a - 1$, $n = 7a^2 - 2a \lt 669$ as you observed.

Taking $a = 1, 2, ...$ you get the values but if you only have to count the number of solutions, it is easy to see that $a \leq 9$ given the constraint $n \lt 669$.

Similarly take $k = 7a + 1$ and see how many solutions given the constraint of $n \lt 669$.

Title: How many positive integers $n$ are there such that $(7n + 1)$ is a perfect square and $(3n + 1) < 2008$?

Body:
How many positive integers $n$ are there such that $(7n + 1)$ is a perfect square and $(3n + 1) < 2008$ ?

What I Tried: We have $(3n + 1) < 2008$
$\rightarrow n < 669$ .

Now, $(7n + 1) = k^2$ for some $k$ .
$$\rightarrow 7n = (k + 1)(k - 1)$$
Now I have to count all the $n$ so that after multiplied by $7$ , gives a product same as the product of $2$ consecutive numbers with a gap of one. I know that $n = 5,9$ are two solutions, but there are more values so that when multiplied by $7$ gives a valid solution, like $k = 24,32$ .

The problem I am facing is, is there any elegant way here to count all the solutions?

Can anyone help me?


答:
You just have to find the number of solutions.

$k$ such that $(k+1)$ or $(k-1)$ is divisible by $7$ and $k>1$.

Note that $7n+1 < 7\cdot 669+1=4684$

Therefore, $k<69$

Every $k$ such that $(k+1)$ or $(k-1)$ is divisible by $7$ and $69>k>1$ will give you a solution.

Therefore, $k = 6,13,20,...62$ These are $\frac{62-6}{7}+1 = 9$ numbers

or $k = 8,15,22,...,64$ These are $\frac{64-8}{7}+1 = 9$ numbers

$9+9=18$ solutions


问:
Real-life, every day applications of number theory?
Are there any day to day situations in which number theory could be applied to? For example, I was trying to make a mosaic and I realized that diophantine equations could be really useful if I needed to find how many pieces of certain dimensions needed to be used. Are there any other real life situations in which number theory would be useful?</p>

答: The most important application of number theory is that it is the key foundation of cryptography. Our strong encryption algorithms and systems have developed because of the impetus provided by number theory. For example, your data cannot be easily accessed by anyone because of the strong encryption system. Moreover number theory is useful in the study of binary codes and other related concepts. This is the main real life applications of number theory. Other than those, there are myriad of applications which are useful for applied mathematicians and physicists. For example, the q series is extremely useful in the study of strings. Then supersymmetric functions like the mock theta functions and theta functions are used for a large number of advanced purposes. Over-all number theory is extremely useful for applied mathematicians and physicists. But to understand those applications you must have a sound knowledge in both.</p>


问:
Prove equality in triangle inequality for complex numbers
We need to show that
$$ |z_{1}+z_{2}+\cdots+z_{n}|=|z_{1}|+|z_{2}|+\cdots+|z_{n}|$$
if and only if $z_{1},z_{2},\dots,z_{n}$ have the same argument (i.e. $z_{j}=r_{j}e^{i\theta}$ for $j=1,\dots,n$).

This way [$\Longleftarrow$] is easy, but for the other way around, I get stuck proving it for the sum of two complex numbers, even though the induction itself isn't that big a problem.

Does anyone happen to know how to prove this for two complex numbers?

答: Suppose $z_1,z_2\ne0$ (otherwise, it's trivial). Then $$\begin{align}|z_1+z_2|^2 &= (z_1+z_2)\overline{(z_1+z_2)}\\ &= (z_1+z_2)\left(\overline{z_1}+\overline{z_2}\right)\\ &= z_1\overline{z_1}+z_1\overline{z_2}+\overline{z_1}z_2+z_2\overline{z_2}\\ &= |z_1|^2+z_1\overline{z_2}+\overline{z_1\overline{z_2}}+|z_2|^2\\ &=|z_1|^2+2\Re(z_1\overline{z_2})+|z_2|^2\\ &= \bigl(|z_1|+|z_2|\bigr)^2+2\Re(z_1\overline{z_2})-2|z_1||z_2|.\end{align}$$ Thus, you must show that $\Re(z_1\overline{z_2})=|z_1||z_2|$ if and only if $z_1,z_2$ have the same argument.</p>


问:
Fatou's Lemma and Counting Measure
I have a vague problem in a Measure and Integration book here. They ask me to consider $\mu$ the counting measure in $\mathbb{N}$ and interpret Fatou's lemma, monotone and dominated convergence theorems as statements about infinite series. I thought it would be easy but got stuck really quick...

If we consider a sequence of non negative functions $f_n\in L^1$, Fatou's lemma says that $$\int_\mathbb{N}\liminf f_n \ d\mu \leq \liminf\int_\mathbb{N}f_n \ d\mu.$$

The real problem comes when I try to understand what are this integrals. Starting with a simple function $\Phi =\sum_{i=1}^kx_i\mathcal{X}_{E_i}$, we have that $\int_\mathbb{N}\Phi \ d\mu = \sum_{i=1}^kx_i\mu(E_i) = \sum_{i=1}^kx_i|E_i|$, where $|E_i|$ is the number of elements in $E_i$. Each function $f_n$ is the $\sup$ of simple functions, but it's not clear how I should use all this together. Even if I consider a sequence $0\leq \Phi_1\leq\Phi_2\leq\ldots\leq f_n$ converging to $f_n$, they are not partial sums.

To make things worse, there is infinite $f_n$ to consider and the $\liminf$ after that. Any help is welcome to interpret all this.

Thank you.

答:
I would like to provide a clearer proof to the post. As a start, we recall Fatou's lemma, that is, $$\int_{\Omega}\liminf\:h \:d\mu \leq \liminf \int_{\Omega}h\:d\mu $$

We first show the sketch of the proof:

$\textbf{First}$, we show that for any $h$: $(\mathbb{N},P(\mathbb{N})) \rightarrow (\mathbb{R}^{+},B(\mathbb{R}^+))$ (i.e. nonnegative measurable function), we have
\begin{equation}
\sum_{k=1}^{\infty}h(k)=\int_{\mathbb{N}}h(k)\:d\mu_{c}(k), \tag{1}
\end{equation}

where $\mu_{c}$ is a counting measure, $\mu_c$ is defined on $(\mathbb{N},P(\mathbb{N}))$, where $P(\mathbb{N})$ is the powerset of natural numbers.

$\textbf{Second}$, let us define $f(k) \equiv \liminf_{n\rightarrow \infty}\:f_n(k) \;\forall k\in\mathbb{N} $. Since this newly defined $f$ is such that $f$: $\mathbb{N} \rightarrow \mathbb{R}^+$ and is measurable, then from (1), $$\sum_{k=1}^{\infty} \liminf_{n\rightarrow\infty}f_n(k) = \int_{\mathbb{N}}\liminf f_n(k)\;d\mu_c \leq \liminf \int_{\mathbb{N}}f_n \;d\mu_c = \liminf \sum_k f_n (k),$$ where the inequality follows from fatou's lemma above, and the last equality follows from (1) above since $f_n$: $\mathbb{N} \rightarrow \mathbb{R}^+$ and is measurable. $\textbf{Our job is therefore reduced to proving (1)}$

$\textbf{Proof of (1)}$:

For any nonnegative measurable function $h$: $(\mathbb{N},P(\mathbb{N})) \rightarrow (\mathbb{R}^{+},B(\mathbb{R}^+))$, we define $h_n \equiv \chi_{[1,n]} h$. Note first that $h_n \rightarrow h$ as $n\rightarrow \infty$. In addition, $h_n$ forms a sequence of nonnegative increasing function.

\begin{align}
\int_{\mathbb{N}} h_n \; d\mu_c =
 \int_{[1,n]} h\; d\mu_c
&= \int_1 h\; d\mu_c + \int_2 h\; d\mu_c + ... + \int_n h\;d\mu_c \\
&= h(1)+h(2)+...+h(n)=\sum_{k=1}^{n} h(k)\\
\end{align}

Then $\lim_{n\rightarrow\infty} \int_{\mathbb{N}}h_n\;d\mu_c = \lim_{n\rightarrow\infty} \sum_{k=1}^{n} h(k) = \sum_{k=1}^{\infty} h(k) \tag{2}$

However, by $\textbf{monotone convergence theorem}$, we know that $$\lim_{n\rightarrow\infty} \int_{\mathbb{N}}h_n\;d\mu_c = \int_{\mathbb{N}} h\;d\mu_c\tag{3},$$ since $0\leq h_n \uparrow h$. Equating (2) and (3) yields (1)

$ \square$

Title: Fatou's Lemma and Counting Measure


问:
Issue with modulo.
In the proof of the theorem of inverse modulo. I encountered this problem and unable to comprehend it.

```<code>sa + tm ≅ 1 (mod m)
</code>```

Then why it is equal to

```<code>sa ≡ 1 (mod m)
</code>```

答: $\equiv$ is a congruence relation and $m \equiv 0 \pmod m$, so $$sa + tm \equiv sa + t\cdot 0 \equiv sa \pmod m.$$

Title: Issue with modulo.


问:
Extremal points of $L^\infty$ unit ball
Consider a compact set $K\subseteq\mathbb{R}^N$, and $L^p=L^p(K)$. Let $B_p$ the unit ball in $L^p$, denote by $c$ the convex hull and by $e$ the set of extremal points of a set.

Can we write $\overline{c(e(B_\infty))}=B_\infty$?

I think yes: $B_\infty$ is (isometric isomorphic to) the unit ball of the dual of $L^1$, and thanks to the compactness of $K$ this space is separable. Thus we can apply the Krein-Milman theorem to get that $\overline{c(e(B_{{L^1}^*}))}^{w^*}=B_{{L^1}^*}$.

But what about $e(B_{{L^1}^*})$? I would appreciate any hint on how to go on from here.

答:
The extreme points of $B_\infty$ are functions $u\in L^\infty(K)$ such that $|u(x)|=1$ for almost all $x$.

It is straightforward to check that such functions are extreme points, since $u(x)$ is an extreme point of  $[-1,+1]$ for almost all $x$.

Let $u\in B_\infty$. Denote by $\chi_{\{u>0\}}$ the characteristic function of the set $\{x: \ u(x)>0\}$. Then
$$
u = \chi_{\{u>0\}} \left( \frac12(+1) + \frac12(2u-1)\right)
+ \chi_{\{u\le0\}} \left( \frac12(-1) + \frac12(2u+1)\right)\\
= \frac12\left( \chi_{\{u>0\}} -\chi_{\{u\le0\}} \right)
+ \frac12\left( \chi_{\{u>0\}}(2u-1)+\chi_{\{u\le0\}}(2u+1) \right),
$$
which writes $u$ as a convex combination of two elements of $B_\infty$. Then $u$ is an extreme point only if $u=\chi_{\{u>0\}} -\chi_{\{u\le0\}}$, which is equivalent to $|u(x)|=1$ for almost all $x$.

With a similar proof one can prove the claim also for complex-valued functions.

Title: Extremal points of $L^\infty$ unit ball


问:
Solutions of the functional inequality $f(x)\geq f(y)(1+x-y)$
Consider the functional inequalities:

$$
(1)\; f(x)\geq f(y)(1+x-y) \; ; \; x,y\in \mathbb{R},
$$
$$
(2)\; \frac{f(x)}{f(y)}\geq 1+x-y \; ; \; x,y\in \mathbb{R}.
$$

It is obvious that all functions of the form $f(t)=ce^t$ with $c\geq 0$ (resp. $c\neq 0$) satisfy
$(1)$ (resp. $(2)$).

Are there any other solutions for them?

Thnaks in advance

答:
We deal with (2). For any given $t,\delta\in\mathbb R$, we have
$$\frac{f(t+n\delta)}{f(t)}=\prod_{i=0}^{n-1}\frac{f(t+(i+1)\delta)}{f(t+i\delta)}\geq (1+\delta)^n,$$
so
$$\frac{f(x)}{f(y)}\geq \lim_{n\to\infty}\left(1+\frac{x-y}{n}\right)^n=e^{x-y}.$$
Furthermore, as $e^v\geq 1+v$ for all real $v$, this is sufficient for (2). However,
$$\frac{f(y)}{f(x)}\geq e^{y-x};$$
these two inequalities multiply to $1\geq 1$ (and we can multiply them since everything is positive), and so equality must hold; as a result, $f(x)=ce^x$ for some constant $c$.

For (1), the process is very similar, and the answer is only slightly different, in essentially the way that the solutions you found differ (values of $0$ are acceptable; negative values are not).


问:
Eliminate $\theta$ and prove $x^2+y^2=1$
We have:$${ \begin{cases}{2x=y\tan\theta+\sin\theta} \\ {2y=x\cot\theta+\cos\theta}\end{cases} }$$

And want to prove $x^2+y^2=1$

My works:

I multiplied first equation by $\cos\theta$ and second one by $\sin\theta$ and get:

$${ \begin{cases}{2x\cos\theta=y\sin\theta+\sin\theta\cos\theta} \\ {2y\sin\theta=x\cos\theta+\sin\theta\cos\theta}\end{cases} }$$
By extracting $\sin\theta\cos\theta$ we get: $$2x\cos\theta-y\sin\theta=2y\sin\theta-x\cos\theta$$
$$x\cos\theta=y\sin\theta$$
But I don't know whether this helps or not.

答: You got $x=y\tan\theta$. Now substitute back into the original equations to get $y=\cos\theta$ and $x=y\tan\theta=\sin\theta$.


问:
Finding triangle’s area using Cartesian plane
Question 19

Hi,

I am a high school student and I've been trying to solve the problem shown in the image for an hour or so now. The question requires me to basically find the area of triangle using the coordinates of straight lines. If helpful, the unit has been focusing on equations of straight lines. I would really appreciate some guidance on how I might tackle this question.

答:
The slope of the line $L$ thru $A=(-3,12)$ and $B=(6,0)$ is $S=\frac {12-0} {(-3)-6}=-4/3.$ So the slope of the line $M, $ perpendicular to $L$ and passing thru $A,$ has slope $\frac {-1}{S}=3/4.$

$M$ meets the $x$-axis at the un-labelled point that I will call $C.$

As you travel  from $C$ to $A$ along $M,$ your $y$-coordinate's change, which is $12,$ is $3/4$ of your $x$-co-ordinate change because $M$'s slope is $3/4.$ So the $x$-co-ordinate change from $C$ to $A$ is $\frac{12}{3/4}=16.$ So $C=((-3)-16,0)=(-19,0).$

So in $\triangle ABC$ the length of the base $AC$ is $6-(-19)=25.$ And the height of the point $B$ above $AC$  is $12.$ So the area of $ABC$ is $(25)(12)/2=150.$

Now  $\triangle A'B'C',$ with $A'=(0,0), B'=B=(6,0),$ and with $C'$ being the intersection of $AB$ with the $y$-axis, is similar to $\triangle ABC,$ as they have equal sets of interior angles. We have length $A'B'=6.$  We have length $AB=\sqrt {(6-(-3))^2+(0-12)^2}=\sqrt {3^2} \sqrt {3^2+4^2}=15.$

So $A'B'/AB=6/15.$

So the area of $A'B'C'$ is $(6/15)^2$ times the area of $ABC.$ That is, the area of $A'B'C'$ is $(6/15)^2\cdot 150 =(4/25)\cdot 150=24.$

So the shaded area  is $150-24=126.$


答: We have from your last step $3x\cos(\theta)=3y\sin(\theta)$ thus $x=y\tan(\theta)$ putting this in (1) we get $$2y\tan(\theta)=y\tan(\theta)+\sin(\theta) \implies y=\cos(\theta)$$ from this we get $x=\sin(\theta)$ now note the identity that $\sin^2(\theta)+\cos^2(\theta)=1$

Title: Eliminate $\theta$ and prove $x^2+y^2=1$

Body:
We have:$${ \begin{cases}{2x=y\tan\theta+\sin\theta} \\ {2y=x\cot\theta+\cos\theta}\end{cases} }$$

And want to prove $x^2+y^2=1$

My works:

I multiplied first equation by $\cos\theta$ and second one by $\sin\theta$ and get:

$${ \begin{cases}{2x\cos\theta=y\sin\theta+\sin\theta\cos\theta} \\ {2y\sin\theta=x\cos\theta+\sin\theta\cos\theta}\end{cases} }$$
By extracting $\sin\theta\cos\theta$ we get: $$2x\cos\theta-y\sin\theta=2y\sin\theta-x\cos\theta$$
$$x\cos\theta=y\sin\theta$$
But I don't know whether this helps or not.


答:
From your solution,
\begin{gather*}
x\cos \theta =y\sin \theta \\
y=x\cot \theta \\
\end{gather*}
Substitute the corresponding value of y into any of the given two equations, and you should be able to get the values of x any y both.
Let's see what we get by substituting it  into the first equation:

\begin{gather*}
2x=y\tan \theta +\sin \theta \\
2x=x+\sin \theta \\
x=\sin \theta \\
y=x\cot \theta =\cos \theta \\
\\
\end{gather*}
Can you now prove what is required?


问:
Show a certain group is contained in a Sylow p-group.
Statement: Let G be a group and p a prime that divides $|G|$. Prove that if $K\le G$ such that $|K|$ is a power of p, K is contained in at least one Sylow p-group.

I just started studying Sylow p-groups, so although I'm familiar with Sylow theorems and a couple of corollaries, I don't know how to get started with this problem. Any hint is more than welcome.

PS: I looked for something related here at Math.SE but didn't find anything. Sorry if it's a duplicate.

答:
Here's another approach. You want to show for any $p$-subgroup $K$, $K\subseteq aPa^{-1}$ where $P$ is a Sylow $p$-subgroup and $a\in G$ . Let $X=\{aP \mid a\in G\}$ be the set of left cosets of $P$ in $G$. Now let $K$ act on $X$ in the following manner $k \cdot(aP)=(ka)P$ .

Let $|G|=p^n m$ where $p \nmid m$ , we know that $|P|=p^n$ since $P$ is a Sylow $p$-subgroup. Note that $[G:P]=|X|= \displaystyle\frac {|G|}{|P|}=m$, then $p \nmid |X|$. We also know that if a $p$-group acts on a set $X$ then $p\mid |X|-|X_f|$ where $X_f$ is the fixed set under the action (I can supply the proof if needed). But $p \nmid |X|$ then $p \nmid |X_f|$ (otherwise $p \mid |X|-|X_f|+|X_f|=|X|$ which is a contradiction) then $|X_f|\not= 0$ then there is an element $aP\in X$ such that $k(aP)=P \ \forall\ k\in K$. $kaP=aP \implies a^{-1}(ka) \in P \implies k\in aPa^{-1}$ for all $k\in K$ hence $K \subseteq aPa^{-1}$ .

As a side note you can use this as a lemma to prove the second Sylow theorem.


答:
The system
$${ \begin{cases}{2x\cos\theta=y\sin\theta+\sin\theta\cos\theta} \\ {2y\sin\theta=x\cos\theta+\sin\theta\cos\theta}\end{cases} }$$

is a linear system of two (independent) equations in two variables, and it is readly checked that $x=\sin\theta, y=\cos\theta$ is a solution. Therefore it is the unique solution. Now $x^2+y^2=1$ follows.


问:
Why do we take the derivative as zero in this question?
Today, I was solving this KVPY-SA 2017 Question:

In an isosceles trapezium, the length of one of the parallel sides, and the lengths of the non-parallel sides are all equal to $30$. In order to maximize the area of the trapezium, the
smallest angle should be:

A) $π/6$

B) $π/4$

C) $π/3$

D) $π/2$

So, I took the angle as $θ$ and after that I found the Area by using the $sin$ and $cos$ functions like this:

$$A=900\sin\theta(\cos\theta + 1)$$

And after that I found the derivative of area like this:

$$\begin{align*}\frac{dA}{d\theta} &= \frac{d}{d\theta}[900(\sin\theta)(1+\cos\theta)]\\
&= 900\left[(\sin\theta)\frac{d}{d\theta}(1+\cos\theta)+(1+\cos\theta)\frac{d}{d\theta}(\sin\theta)\right]\end{align*}$$

And after many steps I reached here where I put the derivative as equal to $0$:
$$\frac{dA}{d\theta} =900[\sin\theta\cdot(-\sin\theta)+(1+\cos\theta)\cdot(\cos\theta)]=0$$

I don't know why I should put it equal to $0$ but that's how you find the answer. When I searched for answers on google I found this solved paper by Byju's. In their answer, you would see that they've written:

Differentiating both sides with respect to $\theta$,
$$\frac{dA}{d\theta} = 900[{\sin\theta\cdot(–\sin \theta) +(1 + \cos\theta)\cdot\cos\theta }] = 0\text{ (for critical points)}$$

What did they mean by differentiation = $0$ for critical points?

You know, I'm very bad at calculus so please forgive me if this is a dumb question.

Edit:

What I did after putting the derivative=$0$ :-

$$-sin²\theta+cos\theta+cos²\theta=0$$

$$⇒cos²\theta-1+cos\theta+cos ²\theta=0$$

$$⇒2cos²\theta+cos\theta-1=0$$

$$⇒2cos²\theta+2cos\theta-cos\theta-1=0$$

$$⇒2cos\theta(cos\theta+1)-1(cos\theta+1)=0$$

$$⇒(2cos\theta-1)(cos\theta+1)=0$$

答:
If a function is continuous on a closed interval a maximum either is interior (0 first derivative, negative second derivative) or must lie on the boundary of the domain.

We need to compare the value of the local maximum and the values on the boundaries of the domain.

A local maximum has 0 first derivative because intuitively the next point in infinitessimal difference will not be any different.

A local maximum has negative second derivative because intuitively the next point will have a negative first derivative (the second next point will attain even less)


问:
Non-affine transform for visual detail
I'm designing a transformation that I think needs to be non-affine by definition. Technically it's a chain of two transformations:

The first transforms from the frequency space to a logarithmic chromatic music note space, in units of cents,  via

$$ x = 100 \cdot 12 \log_2 \frac f {55} - t $$

where $t$ is some known, desired tuning note.

The second is an antisymmetric non-affine transformation $y(x)$ for the purposes of graphing with the following requirements:

$$y(0) = 0$$
$$y(600) = 600$$
$$y(x) = -y(-x)$$
$$ \text{For all } 0 \le x \le 600, \frac {\partial y}{\partial x} > 0$$
$$ \text{For all } 0 < x < 600, y(x) < x $$

The purpose is to zoom into $x=0$ while still showing $x = -600, x=600$. The function must be invertible and the inverse will be used about as frequently as the function itself, due to the way that matplotlib is written.

The two candidates I thought of are a cubic and a tangent. The cubic implementation is

$$ y = a x^3 + b x , b > 0$$
$$ a = \frac {s - 100} {s(600^2 - s^2)} $$
$$ b = 1 - 600^2 a $$

where $s$ is a scaling parameter, interpreted as the value of $x$ where $y = 100$.

This works well enough; here is a $1 - \left| \frac x {600} \right|$ triangle shown with this horizontal zoom for $s = 320$:

This formulation has two big disadvantages:

Values of $s > 330$ are not possible without making $b < 0$, in turn giving the curve undesired inflection points

The inverse $x(y)$ is truly monstrous and I won't include it here unless requested

Is there a different function that will meet these requirements but is more easily invertible?

答:
An easy way to get what you're looking for is using the equation for the motion of a particle in a velocity field, $v$:

$\dot x(t) = v(x(t))$

In your case, $x(t)$ is $y$, $x(0)$ is $x$, and $t$ is $s$. Rewriting with these variable names,

$\dot y(s) = v(y(s))$

$y(0) = x$

This will produce smooth uniform zooming, moving the view of the original function at the same rate, no matter what the value of $s$ is. It is a first order ODE, so $y(x)$ is guaranteed to be invertible. Where $v=0$, the velocity is zero and the action is stationary. Where $\frac{dv}{dx} > 0$, the effect will be a zoom inward, and vice versa with $\frac{dv}{dx} < 0$. So, it is very easy to control the zoom using $v$. This also is a very easy ODE to solve, with a judicious choice of $v$.

First, let's just scale the problem from $\pm600$ to $\pm1$. It is trivial to scale the problem back again afterward. If you choose $v(x) = \frac 1 2 x(1 - x)(1 + x)$, you get the desired zooming in at $x=0$, while keeping $x=0$, $x=1$, and $x=-1$ stationary, as required. (The factor of 1/2 is there to simplify the math.) Math, math, math, and

$$y(x, s) = \frac x {\sqrt{e^{-s} + (1 - e^{-s}) x^2}}$$

Well, that's a pretty function! In $x$, it's just a straight line divided by a hyperbola!

There are other choices of $v$ that give similar results. $v(x) = \sin(\frac \tau 2 x)$ should work, though the integral may be messier. $v(x) = x(1 - |x|)$ will give even simpler results than the cubic I just did—a tanh function will result from it—but the zooming may be more unnatural because of the singularity in $v$ at $x=0$.

Title: Non-affine transform for visual detail


问:
Find integer solutions of $x^2 -px +q=0$, where $p$ and $q$ are prime
Quick number theory question that I have just come across, was wondering if anyone could shed some light on it.

So $p$ and $q$ are given to be prime numbers, and we are told that the equation $x^2 -px +q=0$ has two integer solutions.  How can might one find them?

Thanks for your help.

答: Suppose the equation $x^2 - px + q$ has two integer solutions $a$ and $b$. Then $a + b = p$ and $ab = q$. Since $q$ is prime, we can assume $a = 1$, $b = q$. Therefore $q + 1 = p$ is also prime. The only possibility is $p = 3, q = 2$ and the solutions are 1 and 2. Similarly is $a = -1$, $b=-q$ and $p = 2, q = 3$.</p>


问:
DU decomposition
My professor said that we can't express a square matrix as a product of upper triangular matrix and lower triangular although it can be expressed as a product of lower triangle matrix and an upper triangular matrix. I'm not convinced with her answer. Is she right or not?</p>

答:
It is in fact always possible. This can be seen by understanding that one can obtain a lower triangular matrix from an upper triangular one, and vice versa, by simply reversing the order of indexing.

Suppose $A$ is the matrix in question, and let $\tilde{A}$ be the same matrix but with indices reordered in reverse, in particular $\tilde{A} = PAP$ where $P$ is the permutation matrix that reverses the order of the indices; $P$ is a square matrix consisting of all zeroes except for $1$s on the anti-diagonal (the diagonal of the matrix that goes from bottom-left to upper-right). Then $\tilde{A}$ has an $LU$ decomposition. But now simply reverse the indices again, and this gives a decomposition of $A$ into $\tilde{U} \tilde{L}$, where $\tilde{U} = PUP$ and $\tilde{L} = PLP$, with $L,U$ being the $LU$ factors of $\tilde{A}$.

Title: DU decomposition


问:
Basic topology definition: open sets as arbitrary unions but only finite intersections?
Why is it that in topology one can obtain an open set by taking the union of arbitrarily many open sets, but if one wishes to obtain an open set from intersections, one is restricted to taking only finitely many intersections?

I saw the answers to this question but I'm still confused:
On definition of a topology

Thank you very much.

答:
Perhaps an example will help. Consider the standard $\mathbb{R}$. You may be aware that intervals of the form $(a,b)$ are open. But intervals of the form $[a,b]$ are not, these are closed. In particular singletons $\{a\}$ are not open.

So consider now open intervals $U_n=(-1/n, 1/n)$ for any $n\in\mathbb{N}$. You can then check that $\bigcap_{n=1}^\infty U_n=\{0\}$ which is not open.

Now of course you can invent a new definition such that an arbitrary intersection of open subsets is again open. And in fact we do consider such situations. This is also known as the Alexandrov topology. So $\mathbb{R}$ is not Alexandrov, but for example every topological space that is finite as a set is Alexandrov. But this situation is rather rare in practice and less useful then the classical definition of topology. For example $T_1$ Alexandrov space is automatically discrete. And 99% of spaces we encounter "in the wild" are non-discrete $T_1$.

We can invent definitions however we want. But the rule of thumb is: a definition is as it is, because it is useful.


问:
Let $a_n=|nx-1|$, $n\in\mathbb{Z}^+$ and $x\in\mathbb{R}$. Find the least possible value of $\sum_{n=1}^{696}a_n$
Let $a_n=|nx-1|$, $n\in\mathbb{Z}^+$ and $x\in\mathbb{R}$. Find the least possible value of $\displaystyle\sum_{n=1}^{696}a_n$

My attempt:
\begin{align*}
S_{696}&=\sum_{n=1}^{696}|nx-1|\\
&=|x|\sum_{n=1}^{696}|n-\frac{1}{x}|\\
\end{align*}

答:
Denote $a_n(x) = \vert nx -1 \vert$ and $$f_m(x)= \sum_{n=1}^m a_n(x)$$

You're looking for the minimum of $f_{696}$.

The minimum of $f_m$ is attained for one of the values:

$$1, 1/2, 1/3, \dots, 1/m.$$

And for $1 \le k \le m$ you have

$$f_m\left(\frac{1}{k}\right) = \frac{2k^2 - 2(m+1)k + m(m+1)}{2k}.$$

Looking at the minimum of the map

$$g : x \mapsto \frac{2x^2 - 2(m+1)x + m(m+1)}{2x},$$ you find (by differentiation) that the minimum is achieved at

$$x_m = \sqrt{\frac{m(m+1)}{2}}.$$

Therefore the minimum of $f_m$ is attained at $$\lfloor 1/x_m \rfloor \text{ or } \lceil 1/x_m \rceil.$$ You just have to compute $f_m$ at those two values to get the minimum. For example the minimum of $f_{696}$ is attained at $1/492$ and its value is 288.

A python program to illustrate the question.

```<code>from math import sqrt

def abs_nx(n, x):
    return abs(n * x - 1.0)

def f(m, x):
    y = 0.0
    for n in range(1, m + 1):
        y += abs_nx(n, x)
    return y

def g(m, k):
    return (k * (k - 1) + (m - k) * (m - k + 1)) / (2.0 * k)

def min_f(m):
    xm = sqrt(m * (m + 1) / 2.0)
    k, kp1 = round(xm), round(xm) + 1
    fm_k, fm_kp1 = f(m, 1.0 / k), f(m, 1.0 / kp1)
    km = k if fm_k < fm_kp1 else kp1
    return km

def main():
    m = 10
    for k in range(1, m + 1):
        print(f"f(10, {1.0 / k: .2f}): {f(m, 1.0 / k): .2f}, g(10, {k: 2d}): {g(m, k): .2f}")
    km = min_f(m)
    print(f"minimum: {f(m, 1.0 / km): .2f} attained at: 1/{km: 2d}")

    k696 = min_f(696)
    print(f"minimum 696: {f(696, 1.0 / k696): .2f} attained at: 1/{k696: 2d}")

main()
</code>```

Body:
Denote $a_n(x) = \vert nx -1 \vert$ and $$f_m(x)= \sum_{n=1}^m a_n(x)$$

You're looking for the minimum of $f_{696}$.

The minimum of $f_m$ is attained for one of the values:

$$1, 1/2, 1/3, \dots, 1/m.$$

And for $1 \le k \le m$ you have

$$f_m\left(\frac{1}{k}\right) = \frac{2k^2 - 2(m+1)k + m(m+1)}{2k}.$$

Looking at the minimum of the map

$$g : x \mapsto \frac{2x^2 - 2(m+1)x + m(m+1)}{2x},$$ you find (by differentiation) that the minimum is achieved at

$$x_m = \sqrt{\frac{m(m+1)}{2}}.$$

Therefore the minimum of $f_m$ is attained at $$\lfloor 1/x_m \rfloor \text{ or } \lceil 1/x_m \rceil.$$ You just have to compute $f_m$ at those two values to get the minimum. For example the minimum of $f_{696}$ is attained at $1/492$ and its value is 288.

A python program to illustrate the question.

from math import sqrt

def abs_nx(n, x):
    return abs(n * x - 1.0)

def f(m, x):
    y = 0.0
    for n in range(1, m + 1):
        y += abs_nx(n, x)
    return y

def g(m, k):
    return (k * (k - 1) + (m - k) * (m - k + 1)) / (2.0 * k)

def min_f(m):
    xm = sqrt(m * (m + 1) / 2.0)
    k, kp1 = round(xm), round(xm) + 1
    fm_k, fm_kp1 = f(m, 1.0 / k), f(m, 1.0 / kp1)
    km = k if fm_k < fm_kp1 else kp1
    return km

def main():
    m = 10
    for k in range(1, m + 1):
        print(f"f(10, {1.0 / k: .2f}): {f(m, 1.0 / k): .2f}, g(10, {k: 2d}): {g(m, k): .2f}")
    km = min_f(m)
    print(f"minimum: {f(m, 1.0 / km): .2f} attained at: 1/{km: 2d}")

    k696 = min_f(696)
    print(f"minimum 696: {f(696, 1.0 / k696): .2f} attained at: 1/{k696: 2d}")

main()


答: If the roots are two integers $a$ and $b$, then $p=a+b$ while $q=ab$. Since $q$ is prime, without loss of generality we may assume $a=\pm 1$. So suppose $a=1$, then $b=q$ and $p=q+1$ which forces $q=2,p=3$ (since both $p$ and $q$ are prime and differ by one). <s> Similarly, if $a=-1$, we must have $p=2,q=3$</s>. No! When $a=-1,~b=-q$ and $p=-q-1<0$ cannot be prime, so there is no solution.</p>

Title: Find integer solutions of $x^2 -px +q=0$, where $p$ and $q$ are prime

Body:
Quick number theory question that I have just come across, was wondering if anyone could shed some light on it.

So $p$ and $q$ are given to be prime numbers, and we are told that the equation $x^2 -px +q=0$ has two integer solutions.  How can might one find them?

Thanks for your help.


问:
One definite integral problem

Let $f$ be a continuous function, and $\int_0^1 f(1-x^2) dx=\sqrt{3}$. Find $\int_0^2xf(2x-x^2)dx$.

I tried this:
Let $x=t+1$, then
$$\int_0^2xf(2x-x^2)dx = \int_{-1}^1 (t+1)f(1-t^2) dt$$.
Also, $\int_{-1}^1f(1-x^2) dx = 2\int_{0}^1f(1-x^2) dx $.

But I don't know what to do next. Both substituting and integrating by parts does not seem to work...

答: $\int_{-1}^1 (t+1)f(1-t^2) dt=\underbrace{\require{cancel}\cancelto{0}{\int_{-1}^1 tf(1-t^2) dt}}_\text{integrand is an odd function}+\int_{-1}^1 f(1-t^2) dt=2\sqrt3$

Title: One definite integral problem


问:
Can I find inverse normal in Casio calculator fx- 991ms?
The mean and standard deviation of a normal distribution is given(mean=18 and standard deviation=2.5).
We have to find the value of k for which P(X<k)=0.2578.
It's exhausting to look for the value of z for which the area is 0.2578 from the normal integral table. So how can I find inverse normal in calculator (fx-991ms)?</p>

答:
It's exhausting to look for the value of z for which the area is 0.2578 from the normal integral table

really? it is enough to find the quantile of $1-0.2578=0.7422$ and take the opposit value:

Result:-0.65

(quicker than pressing a complicated key's combination on a calculator...and works also in a battery low situation)


问:
How few $(42^\circ,60^\circ,78^\circ)$ triangles can an equilateral triangle be divided into?
This is the parallel question to this other post with many answers already, in the sense that the $(42^\circ,60^\circ,78^\circ)$-similar triangles form the only non-trivial rational-angle tiling of the equilateral triangle (and the regular hexagon), modulo a real conjugation of the coordinate field (a subfield of $\mathbf{Q}(\zeta_{60})$) which transforms between $(42^\circ,60^\circ,78^\circ)$-similar triangles and $(6^\circ,60^\circ,114^\circ)$-similar triangles. (Reference: M. Laczkovich's Tilings of triangles.)

My attempt has been the following:

Since $\sin(42^\circ)$ and $\sin(78^\circ)$ have nested radicals, I tried to get rid of them by restricting my basic tiling units to only the $60^\circ$-angled isosceles trapezoids and parallelograms that are a single row of the triangular tiles. They have shorter-base-to-leg ratios of the form
$$m\cdot\frac{9-3\sqrt{5}}{2}+n\cdot\frac{11-3\sqrt{5}}{2}\quad\left(m,n\ge 0\right)$$
which are automatically algebraic integers. Any potential tiling of the equilateral triangle from these quadrilateral units corresponds to some integer polynomial relation of the above algebraics, whose polynomial degree correlates with the number of quadrilateral pieces in the tiling.

Unfortunately all the above algebraics have large norms, so a blind search for the desired polynomial is out of the question, and I had to reduce the pieces' proportions again to the rationals. I was able to find a $60^\circ$-angled isosceles trapezoid with shorter-base-to-leg ratio of $10$ using $79$ tiles, and a $60^\circ$-angled parallelogram with neighboring sides' ratio of $11$ using $80$ tiles. Thus a few more tiles produce a $60^\circ$-angled rhombus, and another few more tiles produce a $60^\circ$-angled isosceles trapezoid with shorter-base-to-leg ratio $1$, three of which tile an equilateral triangle, using a total of $121\,170$ triangular tiles. While I was at it, I found this less related post that might reduce my number of tiles to a bit below a hundred thousand.

Meanwhile, I also did a quick computer search through some conceptually simple configurations that attempt to tile the equilateral triangle using less than about $50$ tiles, and I found nothing at all.

I get this feeling that about a hundred thousand tiles is not the optimal amount for such a tiling, so I'm asking to see if people have better ideas. I'm unable to provide cash incentive as the parallel post did, but anyone who tries this puzzle will sure have my gratitude.

Edit suggested by RavenclawPrefect:

To get to the quadrilateral tiling units that I used, the first thing is to de-nest the radicals as I mentioned above. As $\mathbf{Q}(\zeta_{60})$ is Galois over $\mathbf{Q}(\sqrt{3})$ (the base field here should not be $\mathbf{Q}$ but instead the coordinate field of the equilateral triangle), if we can geometrically construct any length $\ell$ (or techinically, ratio $\ell$), such that when we perform the same geometric construction but with all the $42^\circ$ angles and $78^\circ$ angles swapped with each other, we still provably construct the same $\ell$, then it must hold that $\ell\in\mathbf{Q}(\sqrt{5})$, so that $\ell$ doesn't contain any nested radicals.

There were a couple of ideas on what $\ell$ should specifically be, most of them parallel ideas that can all be found in the parallel question for the square. I settled on the above $\mathbf{Q}(\sqrt{5})$-quadrilaterals (the ones that are a single row of triangular tiles) because they had the smallest numerator norms among others. As a non-example, there was a double-decker idea using $9$ tiles that resulted in a trapezoid with ratio a rational multiple of $889-321\sqrt{5}$, yuck. There were also some non-triviality in which way the triangles should be oriented when being put into a single row, but some more calculation showed that the above $(m,n)$ form are all we really get. More precisely, a trapezoid also can't have $m=0$, and a parallelogram also can't have $n=0$.

After all that work, the rest has really been a matter of trial-and-error. Among all the $(m,n)$ form, I picked a parallelogram with the smallest norm, which is an $(m,n)=(0,1)$ parallelogram with $4$ tiles, and rotated it so that it becomes a $\frac{11+3\sqrt{5}}{38}$-parallelogram. Then $19$ of those make a $\frac{11+3\sqrt{5}}{2}$-parallelogram with $76$ tiles, and obviously I combined it to a $(1,0)$-trapezoid and a $(0,1)$-parallelogram to get to the rational quadrilaterals.

So the process was more like "I frankly don't know what else to do" rather than "I see potential simplifications but I don't know the optimum". It's also why I'm seeking for completely new ideas (see above) that aren't found in the parallel question about the square.

RavenclawPrefect also asked a well-motivated question for if the same tiling could be performed but with congruent tiles. M. Laczkovich proved this is impossible in a subsequent paper Tilings of Convex Polygons with Congruent Triangles.

答:
From the OP, I'm using the fact that we can use $79$ triangles to tile a trapezoid with side lengths $11,1,10,1$ and angles of $60$ and $120$ degrees, as well as the parallelogram with side lengths $1$ and $11$ with $80$ triangles. This means that we can tile a "diamond" (the union of two edge-connected equilateral triangles) using $11\cdot80=880$ triangles.

We can then fit all these pieces onto a triangular grid: the trapezoid takes up $21$ triangles, the skinny parallelogram $22$, and the diamond-shaped region just $2$ (but at a great cost). Of course, any of them can be scaled up by some integer factor and still lie on the grid.

Using some code I wrote to solve tiling problems plus some manual modifications, I have found the following packing of an isosceles trapezoid with base-to-leg ratio $1$ (in this case, scaled up on the triangular grid by a factor of $12$ in each dimension):

It uses $12$ trapezoids and $19$ diamonds (the latter of varying sizes). Thus, tiling an equilateral triangle with three copies of this shape will use $3\cdot(12\cdot79+19\cdot880)=\textbf{53004}$ tiles.

Edit by nickgard:

A smaller tiling of the same trapezoid using $10$ long trapezoids and $12$ diamonds.

$3\cdot(10\cdot79+12\cdot880)=\textbf{34050}$ tiles.

(End of edit)

EDIT (RavenclawPrefect): I've found some improved ways to tile parallelograms, which can be used along with nickgard's solution to reduce the number further.

Here is a tiling of a $1\times 2$ parallelogram with seven $1\times 11$ parallelograms (contrast with the $22$ it would take by joining two rhombi together):

In general, one can tile a $1\times n$ parallelogram for $n=1,\ldots,9$ with $11,7,6,6,6,6,6,6,7$ skinny parallelograms; these values arise from taking a tiling of an $11\times n$ rectangle by squares (see A219158 on OEIS) and applying an appropriate affine transformation.

For the $1\times 7$, using $6$ skinny parallelograms gives us $6\cdot 80$, but we can also use $6$ trapezoids as described in Edward H's comment on this answer for $6\cdot 79$ tiles, which offers a slight improvement.

Using these more efficient packings, I can fill in the "staircase" shape in nickgard's answer as follows:

This uses a total of $4874$ tiles in the staircase, $4874+10\cdot79 = 5664$ in the trapezoid, and $\textbf{16992}$ in the triangle.

Edit 2 (RavenclawPrefect): After lots of fiddling around with decomposing the "staircase" shape into nice axis-aligned parallelograms, I realized that I could just apply an affine transformation, turning the whole staircase into a very tall polyomino of size ${10\choose 2}\cdot 11=495$ with "steps" of height $11$, and try to tile the resulting thing with squares directly.

This resulted in a substantial improvement, giving a tiling with $46$ squares (hence, $1\times 11$ parallelograms once transformed back); the resulting image would not embed well due to its height, but I have uploaded it to imgur here. Update: I have slightly improved this tiling to a $45$-square solution, seen here.

This results in $3\cdot(45\cdot80+10\cdot79)=\textbf{13170}$ tiles.

Ways this might be improved:

Trying for a better packing of this $495$-omino by squares - my search was not exhaustive, and I think there's at least a $30\%$ chance it can be tiled more efficiently.

Finding a better packing of some trapezoid or equilateral triangle with these same methods - I certainly haven't optimized things as much as I could.

Finding a more efficient "base" packing of either of the seed shapes used in this tiling, or generating new relatively simple polyiamonds that can be efficiently tiled with $42-60-78$ triangles.


问:
About trivial restrictions of a vector bundle
Let $M$ be a smooth manifold of dimension $n$. Let $U \subset M$ be an open subset such that the restriction $TM|_U$ is trivial. Let $N$ be a closed submanifold of $M$ contained inside $U$. Does it follow then that $TM|_N$ is trivial as well?

I thought this was a trivial result (simply take $n$ linearly independent sections over $U$ and restrict to $N$), until I played around with Stiefel-Whitney classes and got a contradictory answer.

In particular take $N$ to have codimension $1$, and $N$ and $M$ to be orientable. Let $N$ be not parallelizable. Then $TM|_N \cong TN \oplus \nu$ where $\nu$ is the normal bundle of $N$ inside $M$, and $\oplus$ denotes the Whitney sum. It can be shown that since $\nu$ is a line bundle and $N$ and $M$ are orientable, $\nu$ is trivial. Hence $w_i (TM|_N) = w_i (TN \oplus \nu) = w_i(TN)$.

But $TN$ is not trivial, so not all $w_i$ are zero for $i>0$ so $TM|_N$ cannot be trivial, a contradiction.

Clearly I made a mistake somewhere, either at the beginning with the $n$ sections or when working with the Stiefel-Whitney classes, but I can't spot it. Any help in pointing out my mistake would be appreciated.

EDIT: Correction from the discussion in the comments, in the second part, $N$ should have some nonzero Stiefel-Whitney class instead of being just not parallelizable. In this case, there cannot be such an open set $U$ containing it.

答:
The first argument is correct. In addition to the discussion in the comments, a nice example to think about explicitly here is the usual codimension-$1$ embedding of $N = S^2$ into $M = \mathbb{R}^3$. They're both orientable, $N$ is not parallelizable, the normal bundle is trivial, $TM$ is trivial (so we can take $U$ to be, say, a sufficiently large open ball containing $N$, or just all of $M$), and so the pullback of $TM$ to $N$ is also trivial.

This embedding shows that the tangent bundle $TS^2$ is stably trivial (it is trivializable after adding a trivial line; the same is true for every sphere, using the embedding $S^n \hookrightarrow \mathbb{R}^{n+1}$), and in particular all of its Stiefel-Whitney classes vanish. But $TS^2$ is not trivial (by the hairy ball theorem; in terms of characteristic classes, its Euler class does not vanish, and the same is true for every even-dimensional sphere).


Title: One definite integral problem

Body:
Let $f$ be a continuous function, and $\int_0^1 f(1-x^2) dx=\sqrt{3}$. Find $\int_0^2xf(2x-x^2)dx$.

I tried this:
Let $x=t+1$, then
$$\int_0^2xf(2x-x^2)dx = \int_{-1}^1 (t+1)f(1-t^2) dt$$.
Also, $\int_{-1}^1f(1-x^2) dx = 2\int_{0}^1f(1-x^2) dx $.

But I don't know what to do next. Both substituting and integrating by parts does not seem to work...

Body: $\int_{-1}^1 tf(1-t^2)dt+2\int_0^1f(1-t^2)dt=0+2\sqrt 3$ . [Let $g(t)=tf(1-t^2)$. Then $g(-t)=-g(t)$].


答: Hint: by the Rational Root Test (or Factor Theorem) the roots must be factors of the prime $q$.</p>


问:
Why is this diagonal matrix not possible over the reals
Let $\alpha = \begin{pmatrix}
7 &3  &-4 \\
 -2&-1  &2 \\
 6&2  &-3
\end{pmatrix}$ over the reals.

Show that there does not exist a invertible real matrix $\beta$, so that $\delta = \beta ^{-1} \alpha \beta$ is a diagonal matrix

My "proof"

the characterisitc polynomial is $-\lambda ^3 + 3\lambda ^2 -\lambda + 3=-(\lambda - 3)(\lambda ^2 +1)$. The solutions would then be $3,i,-i$. Now, we have that $\delta = \beta ^{-1} \alpha \beta$ where the diagonal values of $\delta$ would then be eigenvalues of $\alpha$. if both $\alpha , \beta$ are real $3 \times 3$-matrices then $\delta$ must also be a real $3 \times 3$-matrix. But that is not possible since the eigenvalues of $\alpha$ are  $3,i,-i$ and thus $\beta$ must be complex.....

Now I am stuck since I would think I would have to show that the diagonal of $\delta$ must be the eigenvalues of $\alpha$. Is this a generel fact or should this be proven? If so - how?

答: If two matrices $A$ and $B$ are similar, $A-\lambda\operatorname{Id}$ and  $B-\lambda\operatorname{Id}$ are similar too, and therefore they have the same determinants. In other words, $A$ and $B$ have the same characteristic polynomials. But, if $B$ is a diagonal matrix and the entries of the main diagonal are $b_1,b_2,\ldots,b_n$, then the characteristic polynomial of $B$ is $(b_1-\lambda)(b_2-\lambda)\cdots(b_n-\lambda)$, whose roots are $b_1,b_2,\ldots,b_n$. But $b_1,b_2,\ldots,b_n$ are the eigenvalues of $B$ and therefore of $A$.


问:
$G_1$ of a nodal projective curve.
Let $C$ be the projective nodal curve over the field $k$ i.e. the projective line with two points identified. I'm having trouble calculating the $G$-theory of $C$, especially $G_1$. Writing the localization sequence wrt closed subscheme the singular point we get: $$G_1(k)\rightarrow G_1(C) \rightarrow G_1(k[t,t^{-1}])\rightarrow G_0(k)\rightarrow G_0(
C)\rightarrow G_0(k[t,t^{-1}])\rightarrow 0$$
This reduces to: $$k^{\times}\rightarrow G_1(C) \rightarrow \mathbb{Z}\oplus k^{\times}\rightarrow \mathbb{Z}\rightarrow G_0(C)\rightarrow \mathbb{Z}\rightarrow 0$$
The question is whether $G_0(k)\rightarrow G_0(C)$ is injective or not?

答: The morphism $G_0(C) \to G_0(k)$ induced by the pushforward via the structure morphism $C \to \mathrm{Spec}(k)$ splits the morphism $G_0(k) \to G_0(C)$, hence the latter is injective.

Title: $G_1$ of a nodal projective curve.


问:
Some horrid integrals I am confused with
$$(1) ~~~~ \int_{1/e}^{\infty} \sqrt{\frac{\ln{x} + 1}{x^3 + 1}}~~dx$$

Speaking as an example on (1).

The only thing I could do from here is only to do a u-sub:

$$ u = \ln(x) + 1 \\ x = e^{u-1} \\ du = \frac{1}{x} dx \\ dx = x (du) \Rightarrow dx = e^{u-1} (du) $$

And so this becomes:

$$ \int \sqrt{ \frac{u}{e^{3u-3} + 1} } ~~ du \\ $$

And from here it's basically a deadend.. I would appreciate if you could give some insights..

Interestingly, these similar integrals have interesting solutions:

$$(2) ~~~~ \int_{1/e}^{\infty} \sqrt{\frac{\ln{x} +1}{x^3}}~~dx =  \sqrt{2 e \pi}$$

$$(3) ~~~~ \int_{1/e}^{\infty} \sqrt{\frac{\ln{x}}{x^3 + 1}}~~dx = \text{a complex number ?!}$$

(2) - where and why does the pi comes into play? this looks like an interesting solution..

(3) - why is the solution a complex number if the area is right infront my eyes, and it is real just like every other areas? What is happening here?

Mystery solved about the complex number - the lower bound should be $1$ and not $1/e$ my bad! But this is not the main mystery ;)

Thank you :-)

答:
Concerning $(3)$ $$\int_{\frac 1e}^{\infty} \sqrt{\frac{\log{(x)}}{x^3 + 1}}\,dx $$ it is normal to get a complex result.

Around $x=\frac 1e$ we have
$$\sqrt{\frac{\log{(x)}}{x^3 + 1}}=i\, \sum_{n=0}^\infty a_n \left(x-\frac{1}{e}\right)^n$$ and the first coefficients are
$$a_0=\frac{ e^{3/2}}{\left(1+e^3\right)^{1/2}}\qquad   a_1 =-\frac{e^{5/2} \left(4+e^3\right)}{2 \left(1+e^3\right)^{3/2}}\qquad   a_2 =\frac{e^{7/2} \left(22-4 e^3+e^6\right)}{8 \left(1+e^3\right)^{5/2}}$$ So
$$\int_{\frac 1e}^{1} \sqrt{\frac{\log{(x)}}{x^3 + 1}}\,dx = i\, \sum_{n=0} ^\infty  \frac {(e - 1)^{n + 1} } {(n+1) e^{n+1} }a_n$$ using the given terms, this leads to
$$\int_{\frac 1e}^{1} \sqrt{\frac{\log{x}}{x^3 + 1}}\,dx \sim 0.373129 \,i$$

In French, we have an expression which says "This, Sir, is the cause of your daughter's being dumb". In French, this simply means "This explains that !"

Edit

Consider
$$\sqrt{\frac{\log{(x)}}{x^3 + a}}=\sum_{n=0}^\infty \binom{-\frac{1}{2}}{n} x^{-3 n-\frac{3}{2}} \sqrt{\log (x)} \,a^n$$  This gives
$$\int\sqrt{\frac{\log{(x)}}{x^3 + a}}\,dx=$$
$$\sum_{n=0}^\infty \binom{-\frac{1}{2}}{n} \left(\frac{\sqrt{2 \pi } \text{erf}\left(\sqrt{3
   n+\frac{1}{2}} \sqrt{\log (x)}\right)}{(6 n+1)^{3/2}}-\frac{2 x^{-3
   n-\frac{1}{2}} \sqrt{\log (x)}}{6 n+1}\right)\,a^n$$

$$\int_1^\infty\sqrt{\frac{\log{(x)}}{x^3 + a}}\,dx=\sqrt{2\pi}\sum_{n=0}^\infty \frac{\binom{-\frac{1}{2}}{n}}{(6 n+1)^{3/2}}\,a^n$$

For $a=1$, computing the partial sums
$$S_p=\sqrt{2\pi}\sum_{n=0}^p\frac{\binom{-\frac{1}{2}}{n}}{(6 n+1)^{3/2}}$$
$$\left(
\begin{array}{cc}
p & S_p \\
10 & 2.45333 \\
 20 & 2.45303 \\
 30 & 2.45297 \\
 40 & 2.45294 \\
 50 & 2.45293 \\
 60 & 2.45293 \\
 70 & 2.45292
\end{array}
\right)$$
which is the result given by numerical integration (this number is not recognized by inverse symbolic calculators).

This sum is alternating and we have
$$\frac{S_{p+2}}{S_p}=1-\frac{4}{p}+\frac{51}{4 p^2}+O\left(\frac{1}{p^3}\right)$$ which explains the rather slow convergence.


问:
How to prove the following integral inequality with exponential function
It is known that $\alpha \geqslant1$,prove
$$\frac{1}{\alpha+\frac{1}{2}}<\int_{0}^{1}e^t(1-t)^{\alpha}dt<\frac{1}{\alpha}$$
I try to use $e^x>1+x$,but I can only prove it $\geqslant\frac{1}{\alpha+1}$.Then I thought about variable substitution, but I still couldn't get the right result,it is
$$\int_{0}^{1}e^t(1-t)^{\alpha}dt=\int_{0}^{1}e^{1-t}t^{\alpha}dt$$I want to know how to start to prove.Thanks!

答:
This is a partial answer.

$$\int_0^1 e^t(1-t)^\alpha> \int^1_0 (1+t)(1-t)^\alpha = \frac{1}{\alpha+1} +\frac{1}{\alpha^2+3\alpha+2}>\frac{1}{\alpha+\frac{1}{2}}$$
(the latter can be proved by simple algebriac manipulation)


问:
Determining whether the set of eigenvectors form an orthonormal basis
I have a question concerning this question.

In that case, can we say that the orthonormalized eigenvectors of the normal, compact operator T form an orthonormal basis for $H$? If not, under what conditions can we usually guarantee that the eigenvectors of an operator acting on an infinite-dimensional Hilbert space are to form an orthonormal basis? Thanks.

答:
The orthonormalized eigenvectors of a normal compact operator $T$ form an orthonormal basis for $H$:

Since $T$ is normal, eigenvectors of distinct eigenvalues are orthogonal. For the same eigenvalue, they can be orthonormalized.

Since $T$ is compact, there is a countable set of eigenvectors with non-zero eigenvalues. Each eigenvalue has a finite number of independent eigenvectors, hence the non-zero eigenvalues give a countable set of orthonormal eigenvectors.

The answer to the referenced question shows that the perpendicular space is then the eigenspace of the zero eigenvalue (unless it is trivial). This space need not be finite dimensional or even have a countable basis. However any Hilbert space, here $\ker T$, has an orthonormal basis of some cardinality.


问:
Property of modulus of a number raised to the power of a complex number
I had been trying to find the relationship between $|x^s|$ and $x^p$, where $s$ is a complex number expressed as $s=p+iq$ and $p>0$. We know that in general, $|x^s|\leq x^{|s|}$. Can I say that since $p\leq |s|$, so $x^p \leq x^{|s|}$? Please tell me if I am wrong.

答: If $x >0, s=p+iq$ with $p,q$ real then $|x^{s}|=|e^{s \ln x}|=e^{p \ln x}=x^{p}$. From this you get $|x^{s}| \leq x^{|s|}$ provided $x\geq 1$. For $0<x<1$ your inequality is false. I will let you find  counter-example with $x=\frac1 2$ and $q=0$.


问:
Why demonstrations are important in mathematics?
Good evening, I'm studying math and would like to know how important are mathematical proofs in the world and particularly in a school of mathematics

Thanks for your help

答: Most of the deep proven results in abstract mathematics were proven with mathematical proofs that could not be derived by intuition alone.        </p>


答:
What you said is a general fact and your proof is correct.

If any matix $\alpha$ is similar to a diagonal matrix $\delta=\text{diag}[\delta_{11},\delta_{22},\ldots,\delta_{nn}]$ i.e. $\exists$ an invertible matrix $B$ such that $\delta=B^{-1}\alpha B$, the diagonal entries of $\delta$ must be the eigenvalues of $\alpha$. You can prove this fact easily.

We have $\alpha B=B\delta$. Let the columns of $B$ be $B_i$. Then$$\alpha B=\alpha[B_1~B_2~\ldots B_n]=[\alpha B_1~\alpha B_2\ldots\alpha B_n]$$and$$B\delta=[\delta_{11}B_1~\delta_{22}B_2\ldots\delta_{nn}B_n]$$Equating the columns we get $\alpha B_i=\delta_{ii}B_i$, i.e. $B_i$ are eigenvectors of $\alpha$ and $\delta_{ii}$ their corresponding eigenvalues.

Title: Why is this diagonal matrix not possible over the reals

Body:
Let $\alpha = \begin{pmatrix}
7 &3  &-4 \\
 -2&-1  &2 \\
 6&2  &-3
\end{pmatrix}$ over the reals.

Show that there does not exist a invertible real matrix $\beta$, so that $\delta = \beta ^{-1} \alpha \beta$ is a diagonal matrix

My "proof"

the characterisitc polynomial is $-\lambda ^3 + 3\lambda ^2 -\lambda + 3=-(\lambda - 3)(\lambda ^2 +1)$. The solutions would then be $3,i,-i$. Now, we have that $\delta = \beta ^{-1} \alpha \beta$ where the diagonal values of $\delta$ would then be eigenvalues of $\alpha$. if both $\alpha , \beta$ are real $3 \times 3$-matrices then $\delta$ must also be a real $3 \times 3$-matrix. But that is not possible since the eigenvalues of $\alpha$ are  $3,i,-i$ and thus $\beta$ must be complex.....

Now I am stuck since I would think I would have to show that the diagonal of $\delta$ must be the eigenvalues of $\alpha$. Is this a generel fact or should this be proven? If so - how?


答: The characteristic polynomial is satisfied by $\alpha$. $$(\alpha-3I)(\alpha^2+I)=0$$ This is still true if $\alpha$ is replaced by the diagonal matrix $\delta$ $$(\delta-3I)(\delta^2+I)=\beta^{-1}(\alpha-3I)(\alpha^2+I)\beta=0$$ If $\delta$ consists of the diagonal terms $x$, then $$(x-3)(x^2+1)=0$$ Over the reals, this only gives $x=3$ as a solution, so $\delta=3I$ and $\alpha=\beta(3I)\beta^{-1}=3I$, which is not the case.


问:
How to get quartic polynomial function given graph, 4 points, and 1 root
I'm doing a math assignment right now, which is due in more or less four and a half hours. What I need to find is a polynomial function given this graph this graph and the points on it. Here are the points:

0,15

7,-1

22,-7

26,0

28,14

The other two points marked on the graph were just marked for another question; I'm not exactly sure if they are x intercepts because I can see that they are a few pixels above or below. Thanks in advance :)

Edit: So far, using the point 7,-1 and the x- and y-intercepts, I tried substituting and transposing the terms in the polynomial, so from y=ax^4+bx^3+cx^2+dx+e, I got 2401a + 343b + 49c + 7d + 16 = 0, and I'm stumped from there. I've also tried doing it in factored form, so y = a(x-26)(x-b)(x-c)(x-d), and I don't really know what to do next.

答:
Given a set $S=\{x_1,...,x_{n+1} \}$ of $n+1$ distinct values:

For $1\le j\le n+1$ let $S_j=S\setminus \{x_j\}.$

The polynomial $$P_j(x)=\frac {\prod_{s\in S_j}(x-s)}{\prod_{s\in S_j}(x_j-s)}$$ is $1$ when $x=x_j$ and is $0$ when $x_j\ne x\in S.$

Given a sequence $(y_1,...,y_{n+1}),$ the polynomial $$Q(x)=\sum_{j=1}^{n+1}y_jP_j(x)$$ satisfies $Q(x_j)=y_j$ for all $j\in \{1,...,n+1\}.$


问:
Are 2 rare events that happen simultaneously likely dependent?
Assuming there are two extremely rare events in a given time period A and B, and given both events have happened in the same time period, can this be an indication that these two events are likely not independent?

Intuitively, if these are the only two events, then certainly, P(A)P(B) < P(A)P(B|A). However, if there are more events, C,D,E etc, then it seems pretty likely that at least two of them will happen simultaneously.

I am being curious due to a real life situation I encountered. After running a nightly test suite on a software I am working on, two rare problems, never seen before, appeared. So we had a discussion with one of my colleagues, on whether it is wise to assume the two are related, or it can be coincidence.

So can someone please clarify this, also is it applicable to real life situations like the one above?

答:
Suppose that $A$ and $B$ are disjoint events, each with positive probability. Can they be independent? No. This follows since $P(A)P(B) \gt0$ yet $P(AB)=P(\emptyset)=0$.

So, if $P(AB)=0$, then $P(A|B) = P(B|A) = 0$. Thus, you might ask yourself if once you've lived a rare event $A$ or $B$, then, after a period of time, rare events $B$ or $A$ can occur, respectively. This is, ¿Are $P(A|B) = P(B|A) = 0$? If the anser is yes, then them both are dependent.

However, it can't be assumed they are independent otherwise.


问:
Are there any practical implications of Infinite Monkey Theorem?
I've recently come across the Infinite Monkey Theorem which loosely states that if you gave a monkey a typewriter and an infinite amount of time, it would almost surely type out any given finite text (say, a Shakesperian sonnet).

I understand also that the "monkey" is a placeholder for some notion of an abstract machine that continually produces random strings of symbols (say, letters and numbers).

What are the implications of this theorem? I find the theorem fascinating, but am really not sure how (if at all) it has contributed to other branches of mathematics, real world applications, or academia in general. Any insight would be appreciated!

答:
The "Infinite Monkey Theorem" is a thought experiment to illustrate the Borel 0-1 law/Borel-Cantelli Lemma. This Lemma says that for a sequence of events $(A_n)_n$ living in a single probability space, we have

i) If $\sum_{n} \mathbb{P}(A_n)< \infty$, then $\mathbb{P}(\limsup_n A_n)=0$

ii) If $\sum_{n} \mathbb{P}(A_n)= \infty$, and the events $A_n$ are independent, then $\mathbb{P}(\limsup_n A_n)=1$.

The $\limsup_n A_n$ is just the event in which a infinite number of $A_n$'s happening. Notice that the "infinite monkey" refers to events $\mathbb{P}(A_n)=p>0$ and we assume independence as well, so it is a rather simple case of the duality presented above.  I don't think it is fair to ask if such example itself is useful.

However, the Borel 0-1 Law itself is rather important result in Probability Theory. A professor of mine told me "this is one of the only 2 or 3 things you need to understand probability". In fact, there are many applications where we want to show that a property holds almost surely by translating it into a sequence of simpler events. We then prove examine $\sum_{n} \mathbb{P}(A_n)$ to conclude
whatever you need. True, it is more common to use the statement i) instead of ii) (as often such simpler events are not necessarily independent), but I think it still falls under the same big picture.

As for an example, I suggest the proof of the Kolmorogov Continuity Theorem, which is required to prove that the Brownian Motion is a continuous function almost surely. The Brownian Motion is one of the most important objects in probability theory and has a wide range of applications, from financial maths to physics.


问:
The parametric equation of the shortest curve along a sphere between any 2 points on sphere
So, I am trying to find the parametric equation of the shortest curve on a sphere between two arbitrary given points on a sphere.

Starting from cartesian coordinate system,

My plan so far is,

find the normal vector from two vectors $v_1, v_2$ made by the given two points on a sphere and the origin of the sphere.

find angles between the normal and $z$-axis by using inner product with two vectors which made by making the normal's $x,y$ components $0$.

find a 3d rotation matrix $A$ with respect to the angles found.

find inverse of A, and rotate $v_1, v_2$ to $xy$-plane.

use parametric equation of a circle $\vec r(t) = \langle r\cos(t), r\sin(t)\rangle$, and find corresponding range of $t$, $(a\leq t\leq b)$ for the angle of rotated $v_1, v_2$.

finally, the parametric equation of the shortest curve is $A\vec r(t)$, $(a\leq t\leq b)$

So my question is :

Is this right method?

Is there any simpler method?

what would be the general function of it to generate points of the curve as a code, python or any pesudo..

答:
You start off okay.
Define two vectors from the center of the sphere to the the identified points (lets call them $ u, v$).
You would use the cross product to find the normal vector (not the inner product).
But, what you really want are two orthogonal vectors in this plane, rather than the normal to the plane.  You can choose either $ u$ or $ v$ to be one of those vectors.

Lets start by normalizing these vectors.

$\hat u = \frac { u}{\| u\|}\\
\hat v = \frac {v}{\|v\|}$

The subtract the projection to find a vector in the plane orthogonal to $mathbf u.$

$w = \hat v - (\hat u\cdot \hat v)\hat u$

And normalize $w$

$\hat w = \frac {w}{\|w\|}$

Parametric equations....

$(x,y,z) = (r\cos \theta) \hat u + (r\sin\theta) \hat w$


问:
First derivative of $\sqrt[\large 5]{\frac{t^3 + 1}{t + 1}}$
I have yet another derivative I need help with. I have to differentiate :

$$\sqrt[\uproot{3}{\Large 5}]{\frac{t^3 + 1}{t + 1}}$$

with respect to $t$.

I had two thoughts about this, use the chain rule then the quotient rule and multiply out, but then I am left with a mess of:

$$\left(\frac{t^3+1}{t+1}\right)^{-4/5}\cdot\frac{0.6t^3+0.6t^2-0.2t^4+0.2t}{(t+1)^2}$$

This is turning into a real mess and the answer I should get is:

$$\frac{2t-1}{5(t^2-t+1)^{4/5}}$$

Am I going the correct way about this or should I try a different route?

Thanks

答:
I would prefer to take the fifth power of both sides and get
$$y^5(t+1)=t^3+1.$$
Differentiate both sides, using implicit differentiation on the left.
We get
$$y^5+5(t+1)y^4 y'=3t^2.$$
Now solve for $y'$.

For more complicated examples of a similar kind, I would suggest logarithmic differentiation. Take the (natural) logarithm of both sides. We get
$$\log y=\frac{1}{5}\log(1+t^3)-\frac{1}{5}\log(1+t).$$
Differentiate both sides with respect to $t$. On the left we get $\frac{y'}{y}$. Differentiation on the right is straightforward.


问:
Difference between countable unions and union of any two arbitrary sets in the context of sigma fields.
One of the criteria for a collection of events to be a sigma field is that it should be closed under all "countable unions" as the textbook says. However almost all the places i referred have countable unions defined on a sequence of sets that belong to the collection. So what exactly is the criteria saying ? That if any two sets belong to the collection their unions must also belong, or that the unions of all possible sequences that can be formed from the collection must also be in the collection ?</p>

答: That if $\{S_i\}_{i\in I}$ is a collection of sets from $\mathcal A$ (the to-be sigma algebra) and $I$ is a countable set (i.e., is either finite or infinite of cardinality $\aleph_0$), then $\bigcup_{i\in I}S_i\in \mathcal A$. This is equivalent to saying that the union of any sequence of sets from $\mathcal A$ is in $\mathcal A$.


问:
Number of ways to construct a 4 digit pin code, under some conditions
Given the digits $0,1,...,9$, the numer of possibilities to construct a PIN code is $10^4$ (zero can be at the beginning since it's a code and not a number).

I need to find the number of PIN codes that have at least two of the same digits.

I tried using two ways to calculate, but got different results and I could really use some help figuring out why it happend.

First solution:

The number of PIN codes with all distinct digits is $10\times9\times8\times7$. So the number of PIN codes with at least two of the same digits should be $10^4 - 10\times9\times8\times7 = 4960$.

Second solution:

If a PIN code has at least two of the same digits, it has exactly two, three or four of the same digits.

For having exactly two of the same digits, I have $10$ choices for what digits should be the same. I then choose their position in the code, ${4 \choose 2}$ choices for that , and then there are $9\times8$ choices for the other two digits. All in all, there are $10\times{4 \choose 2}\times8\times9=4320$ possibilities for that.

For having exactly three of the same digits, again there are $10$ choices for what digits should be the same. For choosing their position in the code there are ${4 \choose 3}$ ways, and $9$ ways to pick the fourth digit. So there are $10\times{4 \choose 3}\times9 = 360$ possibilities for that.

Lastly, the number of possible ways to construct a PIN code with exactly 4 of the same digits is $10$, which is trivial $(0000,1111,...,9999)$.

Summing up all of the above, we get that the number of possibilities is $4320+360+10 = 4690$.

I could really use some help figuring out why one (or both) of my methods is wrong.

Thanks!

答:
Ok, so thanks for the help guys, it turns out I wasn't counting pairs of the same digit.
The way I calculated it:

We have $10\times9$ choices for the two pairs of identical digits. Next, we need to find out how many possibilities are there to divide them to pairs. This is much like dividng $2n$ people to pairs, when $n=2$.

All in all we get: $10\times9\times\frac{4!}{2^{2}\times2!} = 270$.

Summing this with the above we get $270+4690=4960$, which is the correct answer.


问:
Sigmoid Curve function calculation
So I am learning about Sigmoid Curves and in this article it says the formula is:
$$\tag{1}
    \frac{1}{1+e^{-\beta_0-\beta_1x}}
$$
Then it gives an example problem: Let’s say you take $ \beta_0 = -15$ and $\beta_1 = 0.065$. Now, what will be the probability of diabetes for a patient with sugar level $220$?

And the answer is given as follows:

The probability of diabetes for a person with sugar level $x$ is given by $\mathrm{P}(\mathrm{Diabetes})=\frac{1}{1+e^{-\beta_0-\beta_1x}}$. Now, taking $\beta_0=−15$ and $\beta_1=0.065$, the probability of diabetes for a person with sugar level $220$ will be given by

$$\mathrm{P(Diabetes)}= \frac{1}{1+e^{15-0.065\cdot 220}}\approx 0.33$$

So the answer is $0.33$. But I am not sure how they reached this result. I know that $-15+ 0.065\cdot 220 = -0.7$, but I dont know how the whole equation computes to $0.33$

Can anyone help in understanding? My math knowledge is very poor so I appreciate any help!

答:
I'll just go ahead and post this as an answer. In Excel, you can write:

```<code>=1/(1+EXP(-(-15+0.065*220)))
</code>```

which returns $\approx 0.3318$. Another way to evaluate the expression, in Wolfram Alpha, is here: https://www.wolframalpha.com/input/?i=1%2F%281%2Be%5E%28-%28-15%2B0.065*220%29%29%29

So the lesson to take home from this, is that in Excel you just need to use the EXP function to calculate the exponentiation. And remember to use proper parentheses.

Title: Sigmoid Curve function calculation

Body:
I'll just go ahead and post this as an answer. In Excel, you can write:

=1/(1+EXP(-(-15+0.065*220)))

which returns $\approx 0.3318$. Another way to evaluate the expression, in Wolfram Alpha, is here: https://www.wolframalpha.com/input/?i=1%2F%281%2Be%5E%28-%28-15%2B0.065*220%29%29%29

So the lesson to take home from this, is that in Excel you just need to use the EXP function to calculate the exponentiation. And remember to use proper parentheses.


问:
Gaussian integral with complex linear component
I want to prove this gaussian integral is equivalent to left side using cauchy's theorem:
$$I=\int^\infty_{-\infty} dx \exp(-ax^2+bx)=\sqrt{\frac{\pi}{a}}e^{b^2/4a} $$
with $a \in\mathbb R$, $a>0$ and $b=\beta + i \nu$.

答:
$$\int^\infty_{-\infty}e^{-ax^2+bx} =\int^\infty_{-\infty}e^{-(\sqrt ax-\frac{b}{2\sqrt a})^2-\frac{b^2}{4a}}$$
Using the substitution $u=\sqrt ax-\frac{b}{2\sqrt a}$,$du=adx$
$$\frac{e^{-\frac{b^2}{2a}}}{\sqrt a}\int^\infty_{-\infty}e^{-u^2}du=\sqrt{\frac{\pi}{a}}e^{-\frac{b^2}{2a}}$$