问:
Linear Dependence Preserved after Transformation.
Let $V$ and $W$ be vector spaces and let $T: V\to W$ be a linear transformation. Let $\{v_1,v_2,\ldots,v_p\}$ be a linearly dependent set of vectors in $V$. Show that $\{Tv_1,Tv_2,\ldots,Tv_p\}$ is also linearly dependent.

Here's what I have, I need someone to tell me if they think this works.

Since $\{v_1,\ldots,v_p\}$ is linearly dependent, we know that there are scalars that are not all zero (lets this scalar be an) that make $a_1v_1+\ldots+a_nv_n = 0$

Once we apply the transformation we get $T(a_1v_1+\ldots+a_nv_n)=T(0)$

Because are told that this thing is a linear transformation, we know its closed under addition and scalar multiplication. We can use this to change the set to.

$T(a_1v_1)+\ldots+T(a_nv_n) = T(0)$

and we can also pull those scalars out since its dependent.

$a_1T(v_1)+...+a_nT(v_n) = T(0)$

the scalars are still not all zero since we just factored them out of the transformation. Therefore, we have a set here that has scalars, not all zero that gives us $T(0)$ which is $0$. This set with the transformation applied is linearly dependent which is what we needed to show.

The only question I have here is about the end where I said that $T(0) = 0$. Can I make that assumption or am I missing something?

答:
By linearity of $T$ you have

$T(\alpha u+\beta v)=\alpha T(u)+\beta T(v)$ for scalars $\alpha,\beta$ belonging to underlying field and $u,v\in V$.

Put $\alpha=\beta=0$ to see $T(0)=0$.


问:
PQ subtends an angle of $45^{\circ}$ to the origin. Find the locus of the mid point of a chord PQ to circle $x^2+y^2-2ax-2by=0$
A chord PQ of the circle $x^2+y^2-2ax-2by=0$ subtends an angle of $45^{\circ}$ to the origin.

Given that M is the mid point of PQ. Find the equation of the locus of M.

MY ATTEMPT
Centre of the given circle $=
\frac{-1}{2}[(-2a),(-2b)]=C(a,b)$

Let, $M(x,y)$

Slope of $CM$=$(\frac{b-y}{a-x})$

$CM\perp PQ$

$\therefore$ Slope of $PQ$=-$(\frac{a-x}{b-y})$

Equation of $PQ$ gives locus of $M(x,y)$

$\implies$$y_{PQ}=m_{PQ}x+c$

$\implies$$y=-(\frac{a-x}{b-y})+c$

But my approach above is wrong since the textbook answer is as below:

$x^2+y^2-2ax-2by+\frac{1}{2}(a^2+b^2)=0$

YOUR HELP WILL BE HIGHLY APPRECIATED

答:
Hint: $O$ the origin is on the circle, $\angle POQ = 45^\circ$ means $\angle PCQ = 90^\circ$.

$M$ is on a circular arc centred at $C$ with shorter radius.

ContentLicense: CC BY-SA 4.0


问:
Function of the height of a burning sphere-shaped candle
While looking for examples non-linear functions, I thought: If a "normal", cylinder shaped candle burns down at a constant pace (which means its height can be modelled by a linear function), then what function describes the height of a burning sphere-shaped candle with radius r that burns down completely in t minutes? Since its narrower at the top and bottom, the candle burns down faster in the beginning and in the end and slower in the middle. A graph of the function can be seen in this pdf:

http://groolfs.de/klasse9pdf/FunktionenFortsetzung.pdf

Any advice would be appreciated! And in the same vein, what is the function if the candle is cone-shaped with radius r and height h and burns down completely in t minutes?

答:
The best model is to assume that equal volumes of wax melt in equal times. Let the rate at which the wax melts be $v$. Consider a candle of height $H$ with a cross sectional area of $A(h)$ for $h\in[0,H]$. The volume of wax $\mathrm{d}V$ in a strip $[h,h+\mathrm{d}h]$ is then simply
$$\mathrm{d}V=A(h)\mathrm{d}h$$
Since we stated before that the rate at which wax melts is $v$, this means that
$$-v=\frac{\mathrm{d}V}{\mathrm{d}t}$$
Hence,
$$\frac{\mathrm{d}V}{\mathrm{d}t}=-v=A(h)\frac{\mathrm{d}h}{\mathrm{d}t}$$
So what we have is a first order autonomous ODE (IVP):
$$\frac{\mathrm{d}h}{\mathrm{d}t}=\frac{-v}{A(h)}~~|~~h(0)=H$$
So we can express the solution implicitly as
$$-vt+c=\int A(h)\mathrm{d}h$$
In general you'll need to invert the expression obtained to get a solution for $h(t)$ - typically this will be quite hard, even in simple cases like a spherical or conical candle.

EDIT: An illustration of the spherical candle.

Imagine we have a spherical candle of radius $1$. Let's say for arguments sake $v=1$. Let $h(t)$ be the position of the flame, with $0$ at the ground and $2$ at the top of the candle. The area function is
$$A(h)=\pi \cdot r(h)^2$$
And of course geometry tells us that
$$r(h)=\sqrt{1-(h-1)^2}=\sqrt{2h-h^2}$$
Verify here. So our area function is
$$A(h)=\pi h\left(2-h\right)$$
Now our IVP is
$$\dot{h}=\frac{-1}{\pi h\left(2-h\right)}~~|~~h(0)=1.999$$
Our initial condition is theoretically $h(0)=2$, but because of the singularity in the RHS at $h=2$, we have to pick some value just under $2$. Not bothered to work out any integrals, here is a plot of the numerical solution:

Numerical techniques show that $h(4.189)\approx 0$.

Here is also the conical case. Let the cone have height $1$ and radius $1$ and again let $v=1$. In this example $r(h)=1-h$ So
$$A(h)=\pi (1-h)^2$$
So our IVP is
$$\dot{h}=\frac{-1}{\pi (1-h)^2}~~|~~h(0)=0.999$$

Once again numerical techniques show that $h(1.0472)\approx 0$.


答:
The given circle of center $O(a,b)$ and radius $r=\sqrt{a^2+b^2}$ contains the origin. Then, the chord $PQ$ subtends a right-angle to $O$ and the distance from $O$ to $PQ$ is $\frac1{\sqrt2}r$.

Thus, the locus is on a circle of center $(a,b)$ and radius $\frac1{\sqrt2}r$, which determines its equation as
$$(x-a)^2+(y-b)^2=\frac{1}{2}(a^2+b^2)$$

Title: PQ subtends an angle of $45^{\circ}$ to the origin. Find the locus of the mid point of a chord PQ to circle $x^2+y^2-2ax-2by=0$

Body:
A chord PQ of the circle $x^2+y^2-2ax-2by=0$ subtends an angle of $45^{\circ}$ to the origin.

Given that M is the mid point of PQ. Find the equation of the locus of M.

MY ATTEMPT
Centre of the given circle $=
\frac{-1}{2}[(-2a),(-2b)]=C(a,b)$

Let, $M(x,y)$

Slope of $CM$=$(\frac{b-y}{a-x})$

$CM\perp PQ$

$\therefore$ Slope of $PQ$=-$(\frac{a-x}{b-y})$

Equation of $PQ$ gives locus of $M(x,y)$

$\implies$$y_{PQ}=m_{PQ}x+c$

$\implies$$y=-(\frac{a-x}{b-y})+c$

But my approach above is wrong since the textbook answer is as below:

$x^2+y^2-2ax-2by+\frac{1}{2}(a^2+b^2)=0$

YOUR HELP WILL BE HIGHLY APPRECIATED


问:
Help solving a PDE similar to the heat equation
In my course on PDEs, I was given this PDE similar to the heat equation, which we solve for $t \geq 0$ and $0 \leq x \leq 1$

$u_t=\alpha u_{xx} + u$ with $\alpha$ a positive constant. The conditions provided are $u_x(0,t)=u_x(1,t)=0$ and $u(x,0)=\sin^2(\pi x)$.

I was not told how to solve this equation. This looks similar to the heat equation which I know can be solved using separation of variables. I am not very familiar with separation of variables or Fourier series, so I was wondering, can this be solved with separation of variables? I found myself unable to solve this equation using separation of variables, probably due to lack of familiarity. Can anyone please show me how to solve this? I thank all helpers.

答:
Separation of variables definitely works.
$$f(x)g'(t)=\alpha f''(x)g(t)+f(x)g(t)$$
Divide through by $f(x)g(t)$:
$$\frac{g'(t)}{g(t)}=\alpha \frac{f''(x)}{f(x)}+1=\text{constant}=-K$$
So you have some differential equations
$$g'(t)=-K~g(t)$$
$$ f''(x)=-\frac{K+1}{\alpha} f(x)$$
More work...
$$g(t)=b e^{-K t}$$
$$f(x)=a_1 \sin\left(\sqrt{\frac{K+1}{\alpha}}x\right)+a_2 \cos\left(\sqrt{\frac{K+1}{\alpha}}x\right)$$
Meaning our $x$ partial derivative is
$$\partial_x u(x,t)=be^{-Kt}\sqrt{\frac{K+1}{\alpha}}\left(a_1 \cos\left(\sqrt{\frac{K+1}{\alpha}}x\right)-a_2 \sin\left(\sqrt{\frac{K+1}{\alpha}}x\right)\right)$$
So our boundary conditions are
$$\partial_x u(0,t)=0\implies a_1 \cos\left(\sqrt{\frac{K+1}{\alpha}}\cdot 0\right)-a_2 \sin\left(\sqrt{\frac{K+1}{\alpha}}\cdot 0\right)=0$$
$$\implies a_1=0$$
Let's rename $b\cdot a_2 \to C$
$$u(x,t)=C\cdot e^{-Kt} \cos\left(\sqrt{\frac{K+1}{\alpha}}x\right)$$
Can you perhaps apply the second boundary condition now?
EDIT:
The second boundary condition yields
$$\sin\left(\sqrt{\frac{K+1}{\alpha}}\right)=0\implies \sqrt{\frac{K+1}{\alpha}}=n\pi$$
So we have separation constants
$$K_n = \alpha n^2\pi^2 -1$$
Meaning our full solution is an arbitrary linear combination (for choices of $\{C_n\}$ that converge)
$$u(x,t)=\sum_{n=0}^\infty C_n e^{-(n^2\pi^2-1)t}\cos\left(n\pi x\right)$$
Meaning
$$u(x,0)=\sum_{n=0}^\infty C_n \cos\left(n\pi x\right)$$
But also
$$u(x,0)=\sin^2(\pi x)=\frac{1}{2}-\frac{1}{2}\cos(2\pi x)$$
So $C_0=1/2$, $C_2=-1/2$, and all others are $0$.


问:
$V=W_1\oplus\cdots\oplus W_k$ if and only if $\dim(V)=\sum{\dim(W_i)}$
If $W_1,\dots, W_k$ are subspaces of a finite dimensional vector space $V$ such that $W_1+\cdots+W_k=V$, and I want to show that $V=W_1\oplus\cdots\oplus W_k$ if and only if $\dim(V)=\sum{W_i}$, then will what's displayed below suffice?

$$V=W_1\oplus\cdots\oplus W_k$$
$$\iff$$
$$V=W_1+\cdots+W_k~\text{and}~W_i \cap (W_1 + \ldots + W_{i-1} + W_{i+1} + \ldots + W_k) = \{0\}$$
$$\iff$$
$$\text{The subspaces $W_i$ are independent; that is, no sum $w_1+\cdots+w_k$ with $w_i$ in $W_i$ is zero except the trivial sum.}$$
$$\iff$$
$${\scr{B}}=\{\beta_1,\dots,\beta_k\}~\text{is a basis for $V$, where $\beta_i$ is a basis for $W_i$}$$
$$\iff$$
$$\dim{V}=\dim{(W_1+\cdots+W_k)}=\dim{W_1}+\cdots+\dim{W_k}=~\mid\beta_1\mid+\cdots+\mid\beta_k\mid=k$$
$$\iff$$
$$\overset{\text{Does this belong here?}}{\dim{\scr{B}}=k}$$

答:
Your first equivalence is false: when $k \geq 3$, $\bigcap_{i=1}^k W_k = \{0\}$ is too weak to imply that the spaces $W_1,\ldots,w_k$ are independent, i.e., that $W_1 + \ldots + W_k = W_1 \oplus \ldots \oplus W_k$.  For example, take $V = \mathbb{R}^2$, $W_1 = \langle (1,0) \rangle$, $W_2 = \langle (0,1) \rangle$, $W_3 = \langle (1,1) \rangle$.  This is one of two or three subtle traps in linear algebra that even professional mathematicians can fall into if they're not careful.

Any of the following is an acceptable definition of independent subspaces (and they are equivalent):

(i) For all $1 \leq i \leq k$, $W_i \cap (W_1 + \ldots + W_{i-1} + W_{i+1} + \ldots + W_k) = \{0\}$.

(ii) If for all $i$ we choose a nonzero $v_i \in W_i$, then $\{v_1,\ldots,v_k\}$ is a linearly independent set.

(iii) If for all $i$ we choose a linearly independent set $S_i \subset W_i$, then
$S = \bigcup_{i=1}^k S_i$ is a linearly independent set.

Note that if you believe that (iii) is equivalent to the sum being a direct sum then you've got the equivalence you're asking about, so I suggest you concentrate instead on showing these conditions are equivalent to your given definition of internal direct sums.

ContentLicense: CC BY-SA 3.0


问:
Probability of getting two heads in k tosses
You toss a fair coin until you get a total of two heads. What is the probability that you made
k tosses?

My answer was $P(2h, K) = Pr_{2}^K/Vr_{2}^K$

$P(2h, K) = \frac{P_{K}}{(K-2)!2!}/2^k$

$P(2h, K) = \frac{K!}{(K-2)!2!}/2^k$

But the teacher told me it was wrong. What am I not seeing?

答:
Your answer isn't quite right, as the teacher wants the probability that you got the second head on toss $k$. You worked out the probability that among the first $k$ tosses, exactly two are heads.

To solve the question, just note that

The $k$-th toss must be heads, which has probability $1/2$.

Exactly one of the first $k-1$ throws is heads.

The latter event has probability $(k-1)\cdot(1/2)^{k-1}$, so the final answer is $\frac{k-1}{2^k}$.


问:
Is the variance of sample median lower than the variance of any individual observation?
Let $X_1,...,X_n$ be random variables with common variance $\sigma^2$. It is easy to see that irrespective of our knowledge about the distribution or dependence structure of $X_i$'s, the variance of any weighted sample average, $\sum_{i=1}^n w_iX_i$ where $w_i\in[0,1],\sum_{i=1}^n w_i=1$, is at most $\sigma^2$, and hence any weighted average is better (in variance sense) than any individual $X_i$.

But what can we say about the variance of another commonly used measure of central tendency, the sample median? Remember that I am not assuming any dependence structure among $X_i$'s. Can we say that the variance of the sample median is at most $\sigma^2$? Assume for simplicity that each $X_i$ is symmetric if you want.

答:
Comment continued with simulation: Assume you have $n = 100$ observations from $\mathsf{Norm}(\mu=100,\sigma=15),$ so that $Var(X_i) = 15^2 = 225$ and population median is $\eta = 100.$

```<code>set.seed(122)
h = replicate(10^5, median(rnorm(100, 100, 15)))
mean(h);  var(h);  sd(h)
[1] 100.003
[1] 3.490295
[1] 1.868233
</code>```

So the variance $3.49$ of the median of 100 observations
is smaller that the variance $225$ of an individual observation.

Notes: (1) However $n = 100$ is not large enough
to achieve the asymptotic variance quoted in my comment.

(2) The variance of the sample mean $\bar X_{100} = \sigma^2/n = 225/100 = 2.25 < 3.49, $ the variance of the sample median of 100 observations.

Body:
Comment continued with simulation: Assume you have $n = 100$ observations from $\mathsf{Norm}(\mu=100,\sigma=15),$ so that $Var(X_i) = 15^2 = 225$ and population median is $\eta = 100.$

set.seed(122)
h = replicate(10^5, median(rnorm(100, 100, 15)))
mean(h);  var(h);  sd(h)
[1] 100.003
[1] 3.490295
[1] 1.868233

So the variance $3.49$ of the median of 100 observations
is smaller that the variance $225$ of an individual observation.

Notes: (1) However $n = 100$ is not large enough
to achieve the asymptotic variance quoted in my comment.

(2) The variance of the sample mean $\bar X_{100} = \sigma^2/n = 225/100 = 2.25 < 3.49, $ the variance of the sample median of 100 observations.


问:
Find all $f(x)$ so that $[y(x)]^2\cos x + y(x)f(x)y'(x)=0$ is exact
Finde all functions $f(x)$, so that the ODE is exact...
$$[y(x)]^2\cos x + y(x)f(x)y'(x)=0$$
First I rewrote the ODE slightly...
\begin{align*}
    &[y(x)]^2\cos x + y(x)f(x)y'(x)=0 \\
    \Longrightarrow \quad & [y(x)]^2\cos x + y(x)f(x)\frac{dy}{dx}=0 \\
    \Longrightarrow \quad & ([y(x)]^2\cos x)dx + y(x)f(x)dy=0 \\
\end{align*}
Then I identified $P$ and $Q$...
\begin{align*}
    &P_x(x,y) = [y(x)]^2\cos x\\
    &Q_y(x,y) = y(x)f(x) \\
    \\
    \Longrightarrow \quad &P_{x,y}(x,y) = 2y(x)\cos x \\
    \Longrightarrow \quad &Q_{x,y}(x,y) = y'(x)f(x) + y(x)f'(x) \\
\end{align*}
Since $P_{x,y}$ must be equal to $Q_{x,y}$ for the ODE to be exact, I put them equal to find out $f(x)$ but following equations doesn't look solveable...
\begin{align*}
    y'(x)f(x) + y(x)f'(x) \overset{!}{=}2\cos(x) y(x)
\end{align*}
Any suggestions where I went wrong?

答:
At your fourth step :

$[y^2\cos x] dx+[y f(x)]dy=0$

This implies $M(x,y)=y^2\cos x$ and $N(x,y)=yf(x)$ both are functions of $x$ and $y$. Intuitively,  $N$ is the direct output for inputs $x$ and $y$.

If you still consider $y$ as an output from $x$ then it results in function composition $N\circ y(x)$ which suggests you to compute the ordinary derivative $\left (N\circ y(x)\right )'=\frac{dN}{dy}\times\frac{dy}{dx}$ using chain rule rather than the actually required partial derivative $\frac{\partial M}{\partial x}=yf'(x)$.


问:
A functional equation while attempting to integrate a function differently: $ f(x)-f(y) = z \bigl( f(xz) - f(yz) \bigr) $
Here is a functional equation :

For $f : \mathbb (0,\infty) \to \mathbb R$ continuous non-constant, $f$ has the property that for every $b,x,y>0$, $f(1) = -1$  and
$$
f(x) - f(y) = b \bigl( f(bx) - f(by) \bigr) \text .
$$

I want to show that the only possible solutions are $f(x) = a-\frac {a+1}x$ for some constant $a$, without using any form of calculus(with the exception of limits, if required).

For an approach, there are too many free variables, so I tried to set $b$ to a certain value, say $b=2$. Then I get:
$$
f(x)-f(y) =  2 \bigl( f(2x) - f(2y) \bigr) \text .
$$

but I don't know how to proceed : I can't use this to find a particular value of any single $x$, because the value of $x$ depends on the value of $2x$, and so on, so there's no way of proceeding I can think of. This also makes me feel that there is missing information, but that doesn't seem likely because in the presence of three constraining variables and domain, some kind of uniqueness must hold, right?

Setting $g(x) = xf(x)$ and trying to prove that it is constant has also proved futile, for the same reason as before.

Context : This post contains the question I am trying to answer.

Let $f : (0,\infty) \to \mathbb R$ be a function satisfying $f(x)-f(y) = \int_y^x \frac{\mathrm dt}{t^2}$ for all $y,x>0$. Under the fix $f(1) = -1$, we know that there is a unique function $f$ satisfying these conditions, given by $f(x) = -1 + \int_1^x \frac{1}{t^2}\,\mathrm dt = -\frac 1x$. I attempted to locate such an $f$ without using calculus, but rather using functional equations.

The logic was as follows : let $b>0$ be arbitrary, and in the expression $\int_x^y \frac{\mathrm dt}{t^2}$, make the change of variables $bt =u$, so that $b\,\mathrm dt = \mathrm du$. Accordingly, we get:
$$
f(x)-f(y) = \int_x^y \frac{\mathrm dt}{t^2} = \int_{bx}^{by} \frac{b^2 \,\mathrm du}{u^2} = b \int_{bx}^{by} \frac{\mathrm du}{u^2} = b \bigl( f(bx)-f(by) \bigr) \text .
$$

and thus the functional equation comes out. Now if I can use non-calculus techniques to prove that this functional equation is satisfied only by $-\frac 1x$ then I will have obtained the antiderivative of $\frac 1{x^2}$.

答:
Suppose we have a functional equation
$$ x(f(x)-f(x\,y)) = z(f(z)-f(z\,y)) \tag{1}$$
which holds for all $\,x,y,z>0.\,$
We want solutions for $\,f(x)>0.$ Since this
equation holds for all $\,x,z>0\,$ both sides depend only on $\,y.\,$ call it $\,g(y).\,$
Thus if $\,z=1\,$ we define
$$ g(x) := f(1)-f(x). \tag{2} $$
Equations $(1)$ and $(2)$ imply that
$$ f(x) - f(x\,y) = g(y)/x. \tag{3} $$
Add equations $(2)$ and $(3)$ to get
$$ f(1) - f(x\,y) = g(x) + g(y)/x. \tag{4} $$
Rewrote the left side of equation $(4)$ using equation $(2)$ to get
$$ g(x\,y) = g(x) + g(y)/x. \tag{5} $$
Eliminate the denominator in equation $(5)$ by defining
$$ h(x) := x\,g(x). \tag{6} $$
Rewrite equation $(5)$ using equation $(6)$ to get
$$ h(x y) = h(x)y + h(y). \tag{7} $$
Define the limit
$$ c := \lim_{x\to 0} h(x). \tag{8} $$
The limit as $\,x\to 0\,$ of equation $(7)$ is
$$ c = c\,y + h(y). \tag{8} $$
Solve equation $(8)$ for $\,h\,$ to get
$$ h(y) = c - c\,y. \tag{9} $$
Using equations $(9)$, $(6)$, and $(2)$ to get
$$ f(x) = f(1) + c - c/x. \tag{10} $$
The original equation in the question was
$$ f(x)-f(y)=b(f(b\,x)-f(b\,y)). \tag{11} $$
Replacing $\,y\,$ by $\,x\,y\,$ and
$\,b\,$ with $\,z/x\,$ in this equation
is equivalent to our equation $(1)$.

EDIT: This version is simpler than my previous one
from Jan 22, 2021.


问:
How to solve for $x$ in the equation, $4\sqrt{x-3} - \sqrt{6x-17} = 3$, with two square root terms?
So guys, my girlfriend is taking a college algebra class this summer and I figured I would help her study for her upcoming final because I am an engineering major and this kind of math would be easy for me. As we were doing problems, we came about one that I have no idea how to solve. It seems to be a straight forward "here's an equation, solve for $x$." Just one problem, I ended up not being able to solve for $x$, making me feel embarrassed since I was trying to help her and I am supposed to be the one who is "good at math." Anyway, it would be a grreat help if anyone can help me out on this. The equation is....

$$4\sqrt{x-3} - \sqrt{6x-17} = 3$$

The answer comes out to be $x = 7$ because I did it on my calculator. I first tried to just square each side to get rid of the square roots. But once I "FOILed" the left side, there were still square roots and things didn't look good for me. By the way, I am sorry I don't know how to format this equation correctly.

答:
There are more creative subsitutions. For the square roots to be defined, $x ≥ 3$ and $x ≥ \frac{17}{6}$, so the substitution $x = 3 \sec^2 u$ is valid as its range is $[3, \infty)$, so for every value of $x$, there is a corresponding value of $u$. Doing this gives:

$$4 \sqrt{3(\sec^2 u - 1)} - \sqrt{18 \sec^2 u - 17}  = 3$$
$$\Rightarrow - \sqrt{18 \sec^2 u - 17} = 3- 4 \sqrt{3} \tan u$$
$$\Rightarrow 18 \sec^2 u - 17 = 9- 24 \sqrt{3} \tan u+48\tan^2 u$$

$$\Rightarrow 18(\tan^2 u+1) - 17 = 9 - 24 \sqrt{3} \tan u+ 48 \tan^2 u$$
$$\Rightarrow 30 \tan^2 u- 24 \sqrt{3} \tan u+8=0$$
$$\Rightarrow 15 \tan^2 u - 12 \sqrt{3} \tan u + 4 = 0$$

and now using the quadratic formula, $\tan u = \frac{2}{5 \sqrt3}, \frac{2}{\sqrt3}$, so $x = 3 \sec^2 u = 3(\tan^2u+1) = \frac{79}{25}, 7$. However, substituting back into the original equation, $x = \frac{79}{25}$ does not work but $x = 7$ does. Hence $x = 7$ only.


问:
Does two minimum of a continous function imply at least one maximum?
Can we prove that for a continous function (not necessarily smooth), if it has two local minimum in the confined domain, then it must has at least one local maximum.

This seems to be obvious for one dimensional function, but not for multivariable functions.

答:
Consider function $f(x,y) = sin(x) + y^{2}$. It has infinitely many local minimums and no local maximum. Intuitively (that is geometrically) it is easily to see why - at any point where $sin(x) = 1$ and $y = 0$ we have a saddle i.e. along x-axis the graph looks like we have a local maximum at the point and, at the same time along y-axis it looks like local minimum.
At the same time each point where $sin(x) = -1$ and $y=0$ we have local minimum.
One can confirm this by computing derivatives of course.


问:
Can a quadratic equation not equal to zero
Well I wanted to know whether or not $y = x^2 + x + 7$ is a quadratic equation since the general form is $ax^2 + bx + c = 0$ here the equation
$y=x^2+x+7$. Isn't equal to zero so I'm a bit confused

答:
The general form is not $ax^2+bx+c=0$, but $y = ax^2+bx+c$. To have a function, you must be able to put one number in ($x$ in this case), and output one number out ($y$).

$ax^2+bx+c =0$ when $y = 0$, so when you write this, you are trying to find which $x$ values the function outputs $0$ (or the roots). This doesn't help you figure out if $y=x^2+x+7$ is a quadratic equation.


问:
$e^{-1/f(x)}$ convexity
Given:

$$g(x)=e^{-1/f(x)}, \textbf{dom} g=\{x|f(x)<0\}$$

where $f$ is convex.

How can I prove that $g(x)$ is convex, concave or neither?

I have tried to find a counter example and failed. So based on those examples I've tried, I think  $g$ is convex but not sure how to prove it.

Tried using the definition of convexity where $ 0 \le \theta \le 1$ and I know that:
$$f(x_0) < 0, f(x_1) < 0$$
$$f(\theta x_0 + (1-\theta) x_1) \le \theta f(x_0) + (1-\theta) f(x_1)$$

and need to show that:
$$g(\theta x_0 + (1-\theta) x_1) \le \theta g(x_0) + (1-\theta) g(x_1)$$
$$e^{-1/f(\theta x_0 + (1-\theta) x_1)} \le \theta e^{-1/f(x_0)} + (1-\theta) e^{-1/f(x_1)}$$

But I am stuck here. Taking the log of both sides didn't get me very far.

答:
Let $\phi(t) = e^{-{1 \over t}}$ for $x <0$ and $+\infty$ otherwise, it is straightforward to check that $\phi$ is convex and non decreasing.

The composition of a non decreasing convex function with a convex function is convex.

Here $g= \phi \circ f$.

Title: $e^{-1/f(x)}$ convexity


问:
Does every set in ZFC have a value when interpreted as a misère game?
A misère game is one where a player with no moves wins. The object of the game is to be presented with a board position in which you cannot do anything.

If we imagine a two-player game where the players take alternate turns, then we can think of a game as being isomorphic to a set of games that we can choose to present to our opponent. This means we can represent any game fitting our criteria as a set.

Suppose the value to the player is $+1$ if they win and $-1$ if the opponent wins.

The game $\{\}$ has value $+1$ because the player cannot make any moves and therefore wins.

The game $\{\{\}\}$ has value $-1$ because the player must pick the game $\{\}$ and give it to the opponent, who then wins.

The game $\{\{\}, \{\{\}\}\}$ has value $+1$ because the player picks the game $\{\{\}\}$ (which has value -1), to give to the opponent.

Does every set in ZFC have a value as a misère game? I think the axiom of regularity, guarantees that there must be some kind of "trace" through any set which has a value in some sense, but I don't know whether it's possible to extend this to a value for the entire set.

答:
This is mainly about set theory; the misère game business is a red herring. Note that the outcome of a misère game can be calculated recursively: It's $\mathcal N$, which you called "value $+1$", if there are no moves or there's a move to a $\mathcal P$-position. It's $\mathcal P$, which you called "value $-1$", otherwise. Therefore, all that matters is whether play in the set ends in finite time, so that the induction can get to the base case.

As you suspected, the axiom of regularity/foundation proves no infinite descending sequence of sets exists.

(As an aside, instead of relying on regularity/foundation, the graduate textbook "Combinatorial Game Theory" by Aaron N. Siegel just defines the class of long games inductively using ordinals, so that it doesn't matter if regularity/foundation holds.)


问:
Which polygon tile grids allow convex polygons to be formed from multiple tiles?
If I have a grid made of equilateral triangles, I can easily form larger convex polygons as a set of tiles in that grid. I believe this holds for some (but not all) tilings of non-equilateral triangles.

The same for quadrilaterals - it's obvious for squares, rectangles and parallelograms, but I believe it also holds for some other (but not all) tilings of quadrilaterals.

I'm pretty sure tiling a flat area purely with convex pentagons is impossible, so skip that case.

With hexagons, a tiled grid is possible, but the only way to make a convex polygon from those convex hexagonal tiles seems to be to use only one tile - no multi-tile convex polygons seem possible.

My speculation is that it's only possible to form a multi-tile convex polygon from convex polygon tiles if all those tiles have interior angles at all vertices of 90 degrees or less, at least for those vertices that are at a vertex or edge of that larger convex polygon.

Is that speculation correct? Is there a proof or disproof?

答: Are you looking for items like the Hirschhorn Medallion?


问:
Number of subsets from a set of size $2n$ with no two elements differing by $n-1$
Let $f_r(n)$ = #$\{S \subset [2n]$ , s.t. there are no $x, y \in S$ for which $x − y = r\}$.
$[2n]$ denotes the set of size $2n$

Is there a formula for $f_{n-1}(n)$?

For example,

$f_1(2) = 8$

since the possible subsets from the set of size $4$ $\{1,2,3,4\}$ are $\{\phi,1,2,3,4,[1,3],[1,4],[2,4]\}$

If I'm not wrong, $f_2(3)=25$

I've already figured $f_n(n)=3^n$ using bijection, but I don't know how to solve for $f_{n-1}(n)$. Any idea would be appreciated.

答:
Assume $n\ge3$. Partition $[2n]$ into two groups of three – $1,n,2n-1$ and $2,n+1,2n$ – and $n-3$ groups of two. Each pair where $x-y=n-1$ fits into exactly one such group.

There are $5$ subsets of each group of three such that there is no pair with $x-y=n-1$, and $3$ such subsets for each group of two. Hence for $n\ge3$, $f_{n-1}(n)=25\cdot3^{n-3}$.


问:
Prove $(x-1)(y-1)>(e-1)^2$ where $x^y=y^x$, $y>x>0$.

Let $x,y$ be different positive real numbers satisfying $x^y=y^x$.
Prove $(x-1)(y-1)>(e-1)^2$.

We may suppose $a=\frac{y}{x}>1$. Then we obtain $x=a^{\frac{1}{a-1}},y=a^{\frac{a}{a-1}}.$ But how to go on?

答:
Sketch of a proof:

Fact 1: Let $f(u) = \frac{\ln u}{u}$. Then, $f(u)$ is strictly increasing on $(0, \mathrm{e})$,
and strictly decreasing on $(\mathrm{e}, \infty)$.

WLOG, assume $y < x$. From $x^y = y^x$, we have $\frac{\ln x}{x} = \frac{\ln y}{y}$.
By Fact 1, it is easy to prove that $1 < y < \mathrm{e} < x$.

We need to prove that $y > 1 + \frac{(\mathrm{e} - 1)^2}{x-1} \triangleq y_0$. Clearly, $y_0\in (1, \mathrm{e})$.
By Fact 1, it suffices to prove that $f(y) > f(y_0)$
or $\frac{\ln x}{x} > f(y_0)$. It suffices to prove that
$$\frac{\ln x}{x} > \frac{\ln\left( 1 + \frac{(\mathrm{e} - 1)^2}{x-1}\right)}{1 + \frac{(\mathrm{e} - 1)^2}{x-1}}, \ \forall x > \mathrm{e}.$$
It is true. But my proof is not nice. (Hint: Use Fact 2, then use derivative.)

Fact 2: $\frac{\ln v}{v} \ge \frac{(2\mathrm{e} + 10)v - 16\mathrm{e} + 4\mathrm{e}^2}
{(3\mathrm{e} - 3)v^2 + (16\mathrm{e} - 4\mathrm{e}^2)v - 19\mathrm{e}^2 + 7\mathrm{e}^3}$ for all $v \ge \mathrm{e}$.
(Hint: Take derivative.)

Remark 1: I propose the following problem:

Let $0 < y < x$ such that $\frac{\ln x}{x} = \frac{\ln y}{y} = a$ ($a>0$). Prove that
$$(x-1)(y-1) > \left(\mathrm{e} - 1 + \frac{5}{8}\left(\frac{1}{\mathrm{e}} - a\right)\right)^2.$$

By the way, there are many similar problems in MSE or AoPS. These problems has the following description:
Let $f(z)$ be a unimodal function. Let $f(x) = f(y) = a$. Then $g(x, y) \ge h(a)$.

See my closed post:
Inequalities involving roots of some functions (e.g., $\frac{\ln x}{x}$, $x\ln x$)

For example,
Estimate the bound of the sum of the roots of $1/x+\ln x=a$ where $a>1$,

let $f(x) = (x-1)\ln x$, and given $0 < a < b$. If $f(a) = f(b)$, prove that $\frac{1}{\ln a}+\frac{1}{\ln b} < \frac{1}{2}$


问:
marginal distribution of two random variables
I have a question about joint distributions and i'm really struggling to find a solution:

$X$ and $Y$ are two independent random variables with $X \sim Exp(\lambda)$ and $Y \sim Exp(\mu)$. Suppose that we don't know the value of $X$ and $Y$, instead we observe the random variables $Z$ and $W$, where
$Z = \min(X, Y)$ and  W=\begin{cases}
1,  & \text{if  $Z=X$} \\
0, & \text{if $Z=Y$}
\end{cases}

I find the joint distribution for $\mathbb P$($Y>X>z$) = ${\frac{\lambda}{\lambda+\mu}}\cdot{\mathrm e^{-(\lambda+\mu)z}}$ and for $\mathbb P$$(X\gt Y\gt z)={\frac{\mu}{\lambda+\mu}}\cdot{\mathrm e^{-(\lambda+\mu)z}}.$
Now i have to find the marginal distribution of $Z$ and $W$ but i don't know how to build the integral for it, especially for $Z$.
I appreciate any help. Thank you in advance

答: On many occasions, it is quite helpful to find equivalent events. For instance, $W=1$ simply means that $X<Y$, so $\Pr\{W=1\}=\Pr\{X<Y\}$, which you already have evaluated. Or if you want to evaluate the joint distribution of $W$ and $Z$, if $W=1$ (with probability that you already know), then the distribution of $Z$ is the distribution of $X$, which you already know. Sometimes the key is not to get caught up in the calculus and look for equivalent events, hopefully those that are known, or simple to demonstrate.


问:
What is the probability that at least $N$ players out of $M$ flip heads $R$ times in a row?
Let's say $M$ people gather to play heads or tails. They each flip their coin $R$ times and count the heads. What is the probability that at least $N$ of them (of course $N < M$) hit heads $R$ times (perfect score) ?

I can do the combinatorics by brute force for small values of $M$ and $R$, and $N = 1$, but have trouble generalizing it to a formula tractable even for large numbers and any value of $N$. Any hint is appreciated.

Background: I realize this looks like straight out from a textbook assignment (^_^;)
The motivation is I had a discussion with people who take the existence of US bellwether counties and the fact that 18 out of 19 selected the loosing candidate in the recent presidential election as a proof of massive fraud, on grounds of the improbability they could be right for 40 years and then wrong. Instinctively, I don't find it improbable that 19 counties out of 3243 have a "winning streak" of 10 "successes", which would mean there is nothing special about those counties and their existence can be pure dumb luck, but can't make formally sure of it, which has bugged me for a few days now.

答:
$\displaystyle \sum_{i=N}^M {M \choose i} (.5^T)^i (1-.5^T)^{M-i}$

Explanation: i people hit heads T times in a row. The probability of hitting heads T times in a row is .5^T, and this happens i times. The rest of them don't hit heads T times in a row, so it's 1 minus that probability. This happens for the remaining M-i people. And you need to choose which of the M people hit heads T times in a row. Sum this for each i from N to M.


问:
How to find the equation of diameter of a circle that passes through the origin?
So this was a question that I was solving that got me stuck. Its as follows:
Q. Find equation of diameter of the circle $x^2 + y^2 - 6x + 2y = 0$ which passes through the origin.

Now I have tried the following,

From the equation I found,
$g=(-3)$ and $f=1$ and subsequently found the radius as $r=\sqrt{10}$ and hence diameter as $d=2\sqrt{10}$.

But I cant seem to understand how to progress, help would be greatly appreciated.

答:
Let centre of the given circle be $C(x_{1},y_{1})$

Hint Diameter passes through the centre of the circle as well as the origin

$C(x_{1},y_{1})$= $\frac{-1}{2}$$(-6,2)$

$\implies$=$(3,-1)$

Origin, $O(x_{0},y_{0})$=$(0,0)$

$M_{OC}$=slope of diameter=$\frac{(0- -1)}{(0-3)}$

$\implies$$M_{OC}$=$\frac{-1}{3}$

Equation of diameter

$Y=M_{OC}(X-X_{0})+Y_{0}$

$y=\frac{-1}{3}(x-0)+0=\frac{-1}{3}x$


问:
dividing integrals to derive the equations
I'm trying to derive an equation for which I have a relationship between a constant and its integral -- namely, $$W = 2 \int_0^\infty f(L)\,dL$$ -- but also have another definition for $W$ -- $$W = -L\ln(1-P)$$  $f(L)$ itself is $f(L) = P_d$, a value related to $P$ but not the same ($P$ is the value of $P_d$ integrated from $0$ to $L$ somehow, though my sources don't actually give the $P_d$ equation).  So what I need to do is solve $$2\int_0^\infty f(L)\,dL = -L\ln(1-P)$$ so that $P_d$ is the $y$ value and $L$ is the $x$ value, but I'm flubbing this one as I keep coming up with $P_d = \frac{W}{2L}$, which is clearly wrong (partly because it should go to $P_d = 1$ at $L = 0$ rather than infinity).  I'm an experimentalist and out of my depth so long after taking calculus!

答:
The first thing you should note is that $\int_0^\infty f(L)\,dL$ is a constant. It is just a number. The only thing that it depends upon is the function $f$. If you change what function you are integrating, you will get a different value. But it does not depend on a value of the variable $L$. The $L$ in $\int_0^\infty f(L)\,dL$ is a dummy variable - it only exists to support the integral notation, and it does not exist outside that integration. You could equally well use any other variable:

$$\int_0^\infty f(L)\,dL = \int_0^\infty f(q)\,dq = \int_0^\infty f({\frak L})\,d{\frak L}$$
And since $L$ is also a variable in use outside this expression, you definitely should use a different dummy variable. Then you don't get confused thinking that the variable $L$ inside the notation means the same thing as the variable $L$ outside the notation.

But since $\int_0^\infty f(q)\,dq$ is just a constant, that means $W$ is a constant. It doesn't change with $L$. So the equation you need to solve is
$$W = -L\ln(1 - P)\\-\dfrac WL = \ln(1 - P)\\e^{-W/L} = e^{\ln(1-P)}\\e^{-W/L} = 1 - P\\P = 1 - e^{-W/L}$$
Now you say that $P = \int_0^L P_d(q)\,dq$. By the fundamental theorem of calculus, that would mean $P_d$ is the derivative of $P$:
$$P_d = \frac{dP}{dL} = -\frac{W}{L^2}e^{-W/L}$$

But alas, this cannot be true. If $P = \int_0^L P_d(q)\,dq$, then $P = 0$ when $L = 0$, but instead $P=1 - e^{-W/L}$ approaches $1$ as $L$ approaches $0$ from above.

So unfortunately something you've said here has been garbled too much to be able to figure out the answer.


问:
Prove that $A^T \cdot A$ equal $I$ for an orthonormal matrix $A$ directly using matrix multiplication
Apparently it is always true that $A \cdot A^T=A^T \cdot A=I$ for an orthonormal matrix $A$. I know this can be proven using theorems regarding the rank and invertibility of matrices, but I would like to show it directly with matrix multiplication.

Suppose that
$A =
\begin{bmatrix}
u_1 &v _1 \\
u_2 &v_2 \\
\end{bmatrix}
$ is an orthonormal matrix.
Then we know that $\begin{bmatrix} u_1\\u_2 \end{bmatrix}^T \begin{bmatrix} u_1\\u_2 \end{bmatrix}=1$
and $\begin{bmatrix} v_1\\v_2 \end{bmatrix}^T \begin{bmatrix} v_1\\v_2 \end{bmatrix}=1$, and also that $\begin{bmatrix} u_1\\u_2 \end{bmatrix}^T \begin{bmatrix} v_1\\v_2 \end{bmatrix}=0$

However, we must have that $A \cdot A^T = \begin{bmatrix}
u_1^2+v_1^2 & u_1u_2+v_1v_2 \\
u_1u_2+v_1v_2  & u_2^2+v_2^2 \\
\end{bmatrix}
$, and I do not see why $u_1^2+v_1^2 =1$ or why $u_1u_2+v_1v_2=0.$ It seems like $A \cdot A^T = I$ is not necessarily true.

Where is my mistake?

答:
Per VHarisop's comment, orthonormal columns imply orthonormal rows, implying the extra relations
$$\begin{bmatrix}u_1\\v_1\end{bmatrix}^T\begin{bmatrix}u_1\\v_1\end{bmatrix}=1$$
and so on. This should allow you to get $I$.


问:
Show that $\langle X\rangle$ is a subgroup of $G$
Let $G$ a group and $X\subseteq G$ show that $\langle X\rangle=\left\{x_{1}^{r_{1}} x_{2}^{r_{2}} \ldots x_{n}^{r_{n}} \mid n \geq 1, x_{i} \in X, r_{i} \in \mathbb{Z}\right\}$ is a subgroup of G

How $0 \in \mathbb{Z}$ then exist $x\in \langle X\rangle$ such that $x=x_{i}^0=e$ then $e\in\langle X\rangle$

Let $x=x_{1}^{r_1}...x_{n}^{r_n}$ and  $y=y_{1}^{s_1}...x_{k}^{s_k}$ elements in $x\in \langle X\rangle$

then $xy=x_{1}^{r_1}...x_{n}^{r_n}y_{1}^{s_1}...x_{k}^{s_k}$ but i dont know  how to justify xy is in $\langle X\rangle$

答:
The subgroup criterion is that $S\ne\emptyset$ and $x,y\in S\implies xy^{-1}\in S$.

We need $X\ne\emptyset$.

For the first part, let $x\in X$. Then $e=x^0\in \langle X\rangle$.

Secondly, given $x,y\in\langle X\rangle$, we have $x=x_1^{r_1}\dots x_n^{r_n},y=y_1^{s_1}\dots y_m^{s_m}$. Then $xy^{-1}=x_1^{r_1}\dots x_n^{r_n}y_m^{-s_m}\dots y_1^{-s_1}\in\langle X\rangle$.

$\langle X\rangle$ is the smallest subgroup containing $X$.


问:
How to calculate $\lim_{n \to \infty}\frac{r^n}{n}$
I'm a bit rusty on calculus and I'm not able to solve this rather simple limit:
$$\lim_{n \to \infty}\frac{r^n}{n}$$
In my case $r = -1$, and "just by looking at it" I'd guess that for $\left|r\right| = 1, n \to \infty, \frac{\pm1^n}{n} \to 0$. But I was wondering if there's a general rule, since given values like $0.1, e, -2$ for $r$ the limit does not seem so obvious for me.</p>

答:
Hint: For $|r|\le 1$ we have

$$-\frac{1}{n}\le \frac{r^n}{n}\le \frac{1}{n}$$

and for $|r|>1$ the values $\frac{r^n}{n}$ are unbounded.


问:
Exercise of Stein manifold and analytic subset
Let  $\Omega$ be a Stein manifold and $V\subset \Omega$ an analytic set. Show that para every $z\in \Omega\setminus V$ there is a holomorphic function $f$ on $\Omega$ vanishing on $V$ and $f(z)\neq0$.

If $\Omega$ is Stein, it is holomorphically convex, I think that by taking some point out of a compact set I can get a function.

Could you give me any hints?

Thank you!

答:
Since $\Omega$ is Stein, by Cartan's Theorem A, ideal sheaf $I_V$ is globally generated, namely there is a surjection

$$\mathcal{O}_{\Omega}^{I}\twoheadrightarrow I_{\Omega}.$$

So $\Omega$ is zero loci of a family of holomorphic functions $\{f_i\}_{i\in I}$ defined on $\Omega$.

In particular, for any $x\in \Omega\setminus V$, there is an $i\in I$ such that $f_i(x)\neq 0$.


问:
Computing Bayesian Nash Equilibrium
I recently made a model to solve a conflict problem in sports.

The model is based on Bayesian Games: a complete definition can be found here.

The thing is that I am now looking to compute the equilibria of the system, but I don't know any algorithm to do so.

Do you know of any?

Which language do you think I should use? I was a priori going to use R.

Does anybody have any experience with this kind of problem?

答:
For finite Bayes games, i.e., both the action and the type space are finite, you can transform them into either the agent form (see Theorem 9.51 of Game Theory) or the induced normal form (see section 6.3.3 of Multiagent Systems). After you do the transformation, they becomes game in normal forms and you can compute the corresponding Nash Equilibrium (NE). Then recover BNE from NE.

If you further have only two players, then you can form a math program of a bi-linear objective function and linear constrains to solve it (see Theorem 1 of this paper). You can use any nonlinear solver such as Baron to solve it. I would recommend use matlab with the toolbox Yalmip. Note that no nonlinear solver can guarantee global optimal, but you can judge whether BNE is achieved by checking whether the objective function equals 0 (see Theorem 1 of this paper for details. Basically, the objective function of the bilinear program should achieve 0 at BNE. The value closer to 0 represents a smaller error, which can serve as a metric of the solution quality. If the value is eps not equals 0, then you obtain an eps-BNE). We can directly extend the above math programming method to N-player games.

Title: Computing Bayesian Nash Equilibrium


问:
Minimum number of numbers needed to uniquely define a plane
As title says, what would be the minimum number of real numbers needed to uniquely define any plane in $\mathbb{R}^3$?

What I mean by that is, if you define a plane with 3 non-colinear points, then that would count as 9 real numbers (3 real numbers per point).

The minimum (I think) I have found so far is 5 real numbers (the angle along x-axis, the angle along y-axis and a point in the plane), but I don't know what math I could use to prove that, I do not know what to search for, what terms to use (there must be some area in mathematics about that stuff, right?) and cannot find a related question to this in here.

What I would like is pointers to resources, names of related area in mathematics, etc.

答:
The minimum number is actually three: $ax+by+cz=d$
defines a plane with unit normal $(a,b,c)$ (which can be parametrised by two angles), whereupon $d$ defines how far the origin is from the plane.


问:
Extracting distinct sequence
Let $(x_n)$ be a non-constant sequence in $\mathbb{R}$ and $x_n\rightarrow p$ for some $p\in \mathbb{R}$. Can I always extract a subsequence $(x_{n_k})$ whose all elements are distinct? (of course $x_{n_k}\rightarrow p$). Thanks in advance.

By non-constant I mean $\forall a\in\mathbb{R}, x_k=a$ for finitely many $k\in \mathbb{N}$

答: Rigorously, you can define your sequence as follows.  Let $n_1 = 1$, and then for all $k \ge 2$, let $n_k$ be a number such that $x_{n_k} \ne x_{n_1}, \cdots ,x_{n_{k-1}}$.  Such a number always exists, because there are only finitely many elements of the sequence equal to $x_{n_1}$, and only finitely many equal to $x_{n_2}$, etc.</p>

Title: Extracting distinct sequence


问:
Closed form of $\frac{e^{-\frac{\pi}{5}}}{1+\frac{e^{-\pi}}{1+\frac{e^{-2\pi}}{1+\frac{e^{-3\pi}}{1+\ddots}}}}$
It is well known that
$$\operatorname{R}(-e^{-\pi})=-\cfrac{e^{-\frac{\pi}{5}}}{1-\cfrac{e^{-\pi}}{1+\cfrac{e^{-2\pi}}{1-\cfrac{e^{-3\pi}}{1+\ddots}}}}=\frac{\sqrt{5}-1}{2}-\sqrt{\frac{5-\sqrt{5}}{2}}$$
where $\operatorname{R}$ is the Rogers-Ramanujan continued fraction:
$$\operatorname{R}(q)=\cfrac{q^{\frac{1}{5}}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\ddots}}}},\, q=e^{\pi i\tau}.$$

But I'm interested in $\operatorname{R}(e^{-\pi})$. Numerically, I checked that it agrees with the root of the following octic equation near $x=\frac{1}{2}$ to at least $16$ decimal places:
$$x^8+14x^7+22x^6+22x^5+30x^4-22 x^3+22 x^2-14x+1=0;$$
the root turns out to be equal to
$$\frac{\sqrt{5}-1}{2}\frac{\sqrt[4]{5}+\sqrt{2+\sqrt{5}}}{\sqrt{5}+\sqrt{2+\sqrt{5}}}.$$

So is it true that
$$\frac{e^{-\frac{\pi}{5}}}{1+\cfrac{e^{-\pi}}{1+\cfrac{e^{-2\pi}}{1+\cfrac{e^{-3\pi}}{1+\ddots}}}}=\frac{\sqrt{5}-1}{2}\frac{\sqrt[4]{5}+\sqrt{2+\sqrt{5}}}{\sqrt{5}+\sqrt{2+\sqrt{5}}}?$$
The $2$'s and $5$'s under the square roots seem very suggestive of the nature of $\operatorname{R}$.

Also, how could it be proved in that case?

Using
$$\frac{1}{\operatorname{R}(q)}-\operatorname{R}(q)=\frac{\left(q^{\frac{1}{5}};q^{\frac{1}{5}}\right)_{\infty}}{q^{\frac{1}{5}}(q^5;q^5)_{\infty}}+1$$
and
$$\frac{\eta (e^{-\pi\sqrt{n}})}{\eta \left(e^{-\frac{\pi}{\sqrt{n}}}\right)}=n^{-\frac{1}{4}},\, n\gt 0$$
(where $\eta (q)=q^{\frac{1}{12}}\prod_{n\ge 1}(1-q^{2n})$ is the Dedekind eta function),
I've been able to evaluate $\operatorname{R}(-e^{-\pi})$ and $\operatorname{R}(e^{-2\pi})$, but I don't know how could it be used to evaluate $\operatorname{R}(e^{-\pi})$. Perhaps something else is necessary.

I was inspired by Ramanujan's first letter to Hardy, where Ramanujan's $7\text{th}$ theorem states that
$$\cfrac{1}{1+\cfrac{e^{-\pi\sqrt{n}}}{1+\cfrac{e^{-2\pi\sqrt{n}}}{1+\cfrac{e^{-3\pi\sqrt{n}}}{1+\ddots}}}}$$
can be exactly found for any $n\in\mathbb{Q}^{+}$.

答:
Another approach is to use the identity $$\frac{1}{R(q)}-1-R(q)=\frac{f(-q^{1/5})}{q^{1/5}f(-q^5)}\tag {1}$$ where $$f(-q) =\prod_{n=1}^{\infty} (1-q^n)\tag{2}$$ Also note that the above is essentially Dedekind eta function $$\eta(q) =q^{1/24}\prod_{n=1}^{\infty}(1-q^n)\tag{3}$$ Thus in order to evaluate $R(q) $ for $q=e^{-\pi} $ we need the values of $\eta(e^{-\pi/5})$ and $\eta(e^{-5\pi})$. In fact what is needed is the ratio of their values and not the individual values.

We can note that if $k$ is the elliptic modulus corresponding to $q$ and $K$ is the corresponding complete elliptic integral of first kind then $$\eta(q) =2^{-1/6}\sqrt {\frac{2K}{\pi}}k^{1/12}k'^{1/3}\tag{4}$$ Also it should be noted that if $k$ is replaced by $k'$ then $K$ is replaced by $K'$ and $q$ is replaced by $q'$ where $(\log q) (\log q') =\pi^2$. Clearly the values $q=e^{-\pi/5},q'=e^{-5\pi}$ satisfy this equation and hence using a similar equation for $\eta(q') $ we get $$\frac{\eta(q)} {\eta(q')} =\sqrt{\frac{K} {K'}} (k'/k) ^{1/4}$$ and $\pi K'/K=-\log q=\pi/5$. Thus we get $$\frac{\eta(q)} {\eta(q')} =\sqrt{5}\cdot(k'/k)^{1/4}$$ where $k$ is the modulus corresponding to $q=e^{-\pi/5}$. It is well known from the value of Ramanujan class invariant $G_{25}=\phi$ (golden ratio) that $$2k'^2=1-\sqrt{1-\phi^{-24}}=1-12\sqrt{161\sqrt{5}-360}$$ and $$2k^2=1+\sqrt {1-\phi^{-24}}=1+12\sqrt{161\sqrt{5}-360}$$ and hence $$\frac{\eta(e^{-\pi/5})} {\eta(e^{-5\pi})} =\sqrt{5}\left(\frac{1-12\sqrt{161\sqrt{5}-360}}{1+12\sqrt{161\sqrt{5}-360}}\right)^{1/8}=A\text{(say)}\tag{5}$$ It should now be observed that if $q=e^{-\pi}$ then the value of right hand side of equation $(1)$ is $A$. Thus $R(e^{-\pi}) $ is the root of the equation $$\frac{1}{x}-1-x=A$$ or $$x^2+(A+1)x-1=0$$ so that $$R(e^{-\pi}) =x=\frac{-(A+1)+\sqrt{(A+1)^2+4}}{2}$$ where $A$ is given by equation $(5)$.

It requires some patience to get a simple expression for the above root though. Fortunately Ramanujan provided a denesting formula which helps us here. I reproduce it from one of my previous questions here: $$ \sqrt[8]{1\pm\sqrt{1-\phi^{-24}}} = \frac{-1+\sqrt{5}}{2}\cdot \frac{\sqrt[4]{5}\pm 1}{\sqrt{2}}\tag{6}$$ and thus $$A=\sqrt{5}\cdot\frac{\sqrt[4]{5}-1}{\sqrt[4]{5}+1}$$ It is now easy to verify the desired form for $x$ using a little symbolic algebra with the symbol $a$ given $a>0,a^4=5$.

Another curious fact is that if we change the value of $A$ to $A'=5/A$ then the corresponding $x$ gives the value of $R(e^{-4\pi})$. This is generalized by Ramanujan as a relation between the values of $R(e^{-2\alpha}) $ and $R(e^{-2\beta})$ where $\alpha, \beta$ are positive with $\alpha\beta=\pi^2$: $$\left(\frac{\sqrt {5}+1}{2}+R(e^{-2\alpha}) \right) \left(\frac{\sqrt{5}+1}{2}+R(e^{-2\beta})\right)=\frac{5+\sqrt{5}}{2}\tag{7}$$


问:
Positive integers $(x,y)$ satisfying $x^2-y^2= 12345678$?
I saw this question in one of the previous year Olympiad questions.

Let $S$ be the set of all ordered pairs of positive integers $(x,y)$ satisfying the condition : $x^2-y^2= 12345678$

Then what can be said about the number of elements in set $S$ ?

The solution of this question doesn't seem to be correct. Look at it :

Can someone provide me a reasonable approach ?

答:
$x^2 - y^2 = 12345678$ which is even.

That means that $x^2, y^2$ must both be even or both be odd.

If they are both even then $x,y$ are both even $x^2 -y^2$ would be divisible by $4$ which $12345678$ is not.

If they are both odd then we have $x^2 - y^2 = (x-y)(x+y)$. And both $x-y$ and $x+y$ are both even, so again $x^2-y^2$ is divisible by $4$ (actually it is divisible by $8$).

.....

I'm not sure why the answer claimed that if $x^2 - y^2$ is even that $x,y$ should both be odd (was it assuming $x,y$ were relatively prime??? I don't know.)

But other than that the explanation is perfectly correct.


问:
Is it possible to draw all regular polygons on a double grid formed by a rational and an irrational grid, superimposed?
Is it possible to draw any regular polygon in such a way that its vertices be on the double grid, made of a rational and an irrational grid, superimposed, like in the image below?

If yes, what would be the step of the irrational grid?

(It is, evidently, possible to find such a double grid for drawing squares and equilateral triangles, as can be seen in the image below.)

Example of a double grid, rational and irrational (the step of the blue grid was taken $\sqrt 3$).

答:
What I actually ended up doing below was merely to cover the case where certain pairs of vertices had to be at grid points of one of the grids.
Allowing one vertex of the pair to be on one grid and one on the other opens up some possibilities; allowing the use of intersections of lines from different grids, along with some arbitrary offset of the grids' origins, opens up others. There might still be a way to make an argument like the following stick, perhaps using a polygon of more than ten sides.
So I'll leave this here, but it is unfinished and possibly unproductive.

I interpret your question to be whether it is true for every $n \geq 3$ that there exists a regular polygon of $n$ sides that has all vertices on grid points.

Note that if $P,$ $Q,$ $R,$ and $S$ are grid points of a regular rectangular grid
and the lines $PQ$ and $RS$ are parallel,
then the ratio of distances $PQ:RS$ is rational.
(Note that this result is true for any orientation of the grid that allows each line to intersect multiple grid points, not just the case where the lines are parallel to a line of the grid.)

Given a regular decagon $ABCDEFGHIJ$,
consider the lines $AF,$ $BE,$ and $CD.$
These lines are all parallel.
But the distances $BE$ and $CD$ are in the ratio
$$\frac{BE}{CD}
 = \frac{\csc\left(3\pi/10\right)}{\csc\left(\pi/10\right)}
        = \frac12(3 - \sqrt5),$$
which is irrational, so the points $B$ and $E$ cannot be on the same grid
as the points $C$ and $D.$

The distances $AF$ and $BE$ are in the ratio
$$\frac{AF}{BE}
 = \frac{\csc\left(5\pi/10\right)}{\csc\left(3\pi/10\right)}
        = \frac14(1 + \sqrt5),$$
so $A$ and $F$ cannot be on the same grid as $B$ and $E.$

Finally, the distances $AF$ and $CD$ are in the ratio
$$\frac{AF}{BE}
 = \frac{\csc\left(5\pi/10\right)}{\csc\left(\pi/10\right)}
        = \frac14(\sqrt5 - 1),$$
so $A$ and $F$ cannot be on the same grid as $C$ and $D.$

In short none of the three pairs of points can be grid points of the same grid as either of the other two pairs. You need at least three grids in order to be able to place all the points of a regular decagon on regular rectangular grids.

So if the pairs of vertices must fall on a regular grid point of one grid or the other, the answer is no, there is no mere pair of two grids that can provide vertices for regular polygons of all numbers of sides.


答:
You can apply the following theorem. Let $a_n$ be a positive sequence and let$$\ell = \lim _{n\rightarrow +\infty} \dfrac{a_{n+1}}{a_n}.$$
Then: $$\ell \in [0,1) \implies a_n\rightarrow0, $$

and $$\ell \in (1,+\infty) \implies a_n \rightarrow +\infty.$$
Note that the case $\ell=1$ doesn't tell you nothing. Examples: $$a_n=1,\qquad a_n=n.$$

Title: How to calculate $\lim_{n \to \infty}\frac{r^n}{n}$

Body:
I'm a bit rusty on calculus and I'm not able to solve this rather simple limit:
$$\lim_{n \to \infty}\frac{r^n}{n}$$
In my case $r = -1$, and "just by looking at it" I'd guess that for $\left|r\right| = 1, n \to \infty, \frac{\pm1^n}{n} \to 0$. But I was wondering if there's a general rule, since given values like $0.1, e, -2$ for $r$ the limit does not seem so obvious for me.</p>


问:
Can I say only when $g(t)=0$ the above condition holds? Why?
$$\int_0^\theta g(t) n t^{n-1} \theta^{-n} dt =0, 0<t<\theta$$

Can I say only when $g(t)=0$ the above condition holds for all $\theta$? Why?

I can see $n$ and $\theta^{-n}$ above are constants (n is finite and fixed, but positive integer). So I can throw away them. Then the integral becomes
$$\int_0^\theta g(t)  t^{n-1}  dt =0$$

答:
Apply the fundamental theorem of calculus (take the derivative of both sides with respect to $\theta$).

$g(\theta)\theta^{n-1}=0$


答: Actually three numbers suffice, since you can always multiply by a constant.  You can make the length of the normal $=1$, so you only need two other numbers.  Alternatively, as long as one of the four numbers is not zero, you can divide it out.

Title: Minimum number of numbers needed to uniquely define a plane

Body:
As title says, what would be the minimum number of real numbers needed to uniquely define any plane in $\mathbb{R}^3$?

What I mean by that is, if you define a plane with 3 non-colinear points, then that would count as 9 real numbers (3 real numbers per point).

The minimum (I think) I have found so far is 5 real numbers (the angle along x-axis, the angle along y-axis and a point in the plane), but I don't know what math I could use to prove that, I do not know what to search for, what terms to use (there must be some area in mathematics about that stuff, right?) and cannot find a related question to this in here.

What I would like is pointers to resources, names of related area in mathematics, etc.


问:
What does this equation equivalent to?
I have encountered the following *expression and while it looks very familiar, I cannot figure out what it simplifies to or what it represents. My initial guess is that it will be equal to the expected value of something. I'd appreciate any help. Thanks!

$$
\binom{n}{l}p^{n-l}(1-p)^{l}
$$

答:
It is just an expression, not an equation: an equation says that two quantities or expressions are equal. It is the general term of the summation that appears in an application of the binomial theorem in probability:

$$1=1^n=\big((1-p)+p\big)^n=\sum_{\ell=0}^n\binom{n}\ell(1-p)^\ell p^{n-\ell}\,.$$

Specifically, if you have a coin that comes up heads with probability $p$, $\binom{n}\ell(1-p)^\ell p^{n-\ell}$ is the probability that you will get $\ell$ tails if you toss the coin $n$ times.

Title: What does this equation equivalent to?


问:
Let $I$ be a nilpotent left ideal of $R$. Prove that for each left $R$-module $M$, $IM\ll M$
Let $I$ be a nilpotent left ideal of $R$. Prove that for each left $R$-module $M$, $IM\ll M$.  ($\ll$ means superfluous)

If $IM + N = M$, then $I^2M + IN + N = M$

I know we will use this information in somewhere but I dont know how to prove.

答:
Don’t quit now!

From where you started notice  $M=I^2M + IN + N = I^2M+N$.

Repeating that again, $M=I^3M+N$.

You see where this is headed?


问:
How can I solve this kind of systems: $\left\lbrace \begin{array}{rcl} 3x-y'-2y&=& 8t\\ x'-2x+y &=&16e^{-t} \end{array}\right.$
How can I solve this kind of systems: $$\left\lbrace \begin{array}{rcl} 3x-y'-2y&=& 8t\\ x'-2x+y &=&16e^{-t} \end{array}\right.$$

I don't know how to approach it. I tried:  $$\left\lbrace \begin{array}{rcl} y'&=& -8t+3x-2y\\ x' &=&16e^{-t}+2x-y \end{array}\right.$$ But since $y,y' x$ are mixed seems difficult, to continue.

答:
This is a Linear Nonhomogeneous Systems of Differential Equations with Constant Coefficients.

You have

$$X'(t) = \begin{pmatrix}x'\\y'\end{pmatrix}= AX+f(t) = \begin{pmatrix}2&-1\\3&-2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}+\begin{pmatrix}16e^{-t}\\-8t\end{pmatrix}$$

Next, follow Variation of Paramters or many different methods to solve such a system.

Can you proceed?

You should get

$$x(t)= \dfrac{1}{2} e^{-t} \left(-16 \left(e^t+1\right) t+c_1 \left(3 e^{2 t}-1\right)-c_2 e^{2 t}-24+c_2\right)\\y(t)=\frac{1}{2} \left(-32 t-3 e^{-t} (16 t+8+c_1-c_2)+(3 c_1-c_2) e^t+16\right)$$


问:
Smooth Quartics in $\Bbb{P}^3$
Algebraic category. Ground field $\Bbb{C}$.

This is a naive question: are all smooth quartic surfaces in $\Bbb{P}^3$ isomorphic ?

The answer is NO if and only if there is a smooth quartic in  $\Bbb{P}^3$ containing some (-1)-curve.

答:
A smooth quartic in $\mathbb P^3$ is a K3 surface, and they are typically non-isomorphic.  If you quotient out by the obvious projective symmeteries,
you get a 19-dimensional family, and the K3's honestly depend on these 19 moduli.
(There is a Torelli theorem to this effect, I think.)

[But, by adjunction, there are no -1 curves on a K3.  I don't really understand
your (-1)-curve remark.  Details: Adjunction says that $2g - 2  = C \cdot C - C\cdot K,$ but on a K3 we have $K = 0$, so $2g-2 = C \cdot C.$  The left hands side
is even, and so a curve on K3 cannot have self-intersection $-1$.]

Title: Smooth Quartics in $\Bbb{P}^3$


问:
Calculating number of free variables in rotation matrices
In Modern Robotics, there is a line that says, because we have 3 unitary constraints, and 3 orthogonality constraints for rotation matrices, summarized by the equation $R^T R = I$, and $R$ is 9-dimensional, that means $R$ has 3 free variables.

My understanding of free variables comes from linear algebra, where the number of "free variables" is essentially the dimensionality of the null space.

In order to argue that the previous setup has a 3-dimensional nullspace, the 3 unitary and 3 orthogonality constraints must be linearly independent constraints.

However, looking at these constraints:

These constraints are quadratic in terms of the rotation matrix variables. Therefore, linear independence does not make sense as a concept here.

How do I rigorously argue that $R$ has 3 "free" variables in this setting? How do you define "free" variables in a general way with non-linear constraint equations, instead of using the nullspace, which is a concept for linear constraint equations?

答:
The appropriate "rigorous" notion to apply here is that $O(3)$, the set of rotations in three dimensions, is a three-dimensional Lie Group, which in particular means that it is a three-dimensional Smooth Manifold. This means essentially that $O(3)$ is "locally" described by three smooth coordinate functions, but this takes a bit of work to make precise.

The standard way to show that $O(3)$ is indeed a manifold is through the preimage theorem, which states that, given a smooth map $f:M\to N$ between manifolds, and $x\in M$ is a regular value, then $f^{-1}(\{x\})$ is a smooth manifold of dimension $\dim f^{-1}(\{x\})=\dim M-\dim N$ (and in fact a smooth submanifold of $M$).

Here, we can identify the space of $3\times 3$ matrices with $\mathbb{R}^9$ and the set of symmetric $3\times 3$ matrices with $\mathbb{R}^6$, and consider the map $f:\mathbb{R}^9\to\mathbb{R}^6$ defined by $f(A)=A^TA$ under this identification, so that $O(3)$ is then the level set $f^{-1}(I_3)$. The level set theorem then states that $O(3)$ is a smooth manifold if $I_3$ is a regular value of $f$, i.e. if the Jacobian of $f$ is surjective on all of $O(3)$.


问:
Proving that the complementary of a surface is connected
I must prove that, being $S$ a differentiable surface of dimension $d$, which is a closed subset of $\mathbb{R}^k$, where $k\geq d+2$, one have $\mathbb{R}^k-S$ connected.

This is trivial for $d=1$, because so $S$ is (a subset of) one line in a space at least $\mathbb{R}^3$, so one can deviate of this line and obsviouly the complementary is connected.

However, I have doubts to formalize this argument in such a way as to make it extendable.

I'd appreciate some ideas. However, for ethical reasons, I ask you to be kind enough not to post detailed answers.

答:
For any compact, locally contractible, nonempty, proper subspace of $K$ of $n$ dimensional sphere $\mathbb{S}^n$, Alexander Duality says: $\tilde{H}_i(\mathbb{S}^n-K;\mathbb{Z})\simeq\tilde{H}^{n-i-1}(K;\mathbb{Z})$ for all $i$.

Thinking $\mathbb{S}^n$ as one point compactification of $\mathbb{R}^n$ and applying Alexander Duality to the cases when $K$ is compact $\mathbb{R}^n$ and non-compact in $\mathbb{R}^n$ gives the result since $\tilde{H}_0(\mathbb{S}^n-\overline{K};\mathbb{Z})\simeq 0$ where $\overline{K}$ is the closure of $K$ in $\mathbb{S}^n$. This means we have path-connectedness for $\mathbb{S}^n-\overline{K}$.  Hence connectedness for $\mathbb{S}^n-\overline{K}$. Now you need to argue removing one point back does not change the connectedness.


问:
If two multiplicative functions are close, are their Dirichlet inverses necessarily close?
Let $f$ be a multiplicative arithmetic function such that $\sum_{n = 1}^{\infty} |f(n)|$ converges.

Given some $\epsilon > 0$, can we necessarily choose some $\delta$ such that for all arithmetic functions $g$ that obey the same above criteria as $f$, then:

$$\bar{\rho}(f, g) < \delta \ \Rightarrow \ \bar{\rho}(f^{-1}, g^{-1}) < \epsilon$$

where $h^{-1}$ is the Dirichlet inverse of $h$, and $\rho(f, g) = \sup \{ \min(|f(n) - g(n), 1) \ | \ n \in \mathbb{N}\}$ (it is the uniform metric).

答:
Try with $$\sum_{n\ge 1} f(n)n^{-s}=\sum_{n\ge 1} f^{-1}(n)n^{-s}=1,\qquad \sum_{n\ge 1} g_r(n)n^{-s}=1-\delta \sum_{n\ge 1} r^n n^{-s}$$
with $\delta>0$ and $1-r>0$ small.

$$\frac1{G_r(s)}=1+\sum_{k\ge 1}\delta^k (\sum_{n\ge 1} r^n n^{-s})^k$$
Since everything is non-negative we get that $$g_r^{-1}(n)\ge \delta^2\sum_{d|n} r^d r^{n/d} \qquad \implies \sup_{r<1}\sup_n |f^{-1}(n)-g_r^{-1}(n)|=\infty$$

If you assume additionally that $\sum_n |f^{-1}(n)|=C$ converges and $\sum_n |g(n)-f(n)|<\delta<1/C$ then $\sup_{\Re(s)\ge 0} |F(s)|\ge 1/C$ so that $$\sum_{n\ge 1} (g^{-1}(n)-f^{-1}(n))n^{-s}=\frac1{F(s)-(F(s)-G(s))}-\frac1{F(s)}=\sum_{k\ge 1} \frac{(F(s)-G(s))^k}{F(s)^{k+1}}$$
This implies that
$$\sum_{n\ge 1} |g^{-1}(n)-f^{-1}(n)| \le \sum_{k\ge 1} (\sum_n |f(n)-g(n)|)^k (\sum_n |f^{-1}(n)|)^{k+1}\le \sum_{k\ge 1} \delta^k C^{k+1}$$


问:
Why is it okay to swap the integral with the average here?
In these notes, Pg. 23, they show: $$\text{avg}_U \left(\int_{S^{n-1}} \frac{1}{\|U\theta\|^n_K\|\theta\|^n_K} d\sigma(\theta) \right) = R^{2n}$$

Here's the complete argument, and the line marked $(\star\star\star)$ is the one that I'm talking about:

The group of orthogonal transformations carries an invariant probability measure. This means that we can average a function over the group in a natural way. In particular, if $f$ is a function on the sphere and $θ$ is some point on the sphere, the average over orthogonal $U$ of the value $f(Uθ)$ is just the average of $f$ on the sphere: averaging over $U$ mimics averaging over the sphere:
$$\text{avg}_U f(Uθ) = \int_{S^{n-1}}f (\phi) dσ(\phi)$$
Hence,
\begin{align*}
    \text{avg}_U \left(\int_{S^{n-1}} \frac{1}{\|{U\theta}\|^n_K\|{\theta\|}^n_K} d\sigma(\theta) \right) &= \int_{S^{n-1}} \left( \text{avg}_U\frac{1}{\|{U\theta}\|^n_K} \right)\frac{1}{\|{\theta}\|^n_K}d\sigma(\theta)\ (\star\star\star)\\
    &= \int_{S^{n-1}} \left( \int_{S^{n-1}} \frac{1}{\|{\phi}\|^n_K} d\sigma(\phi) \right)\frac{1}{\|{\theta}\|^n_K}d\sigma(\theta) \\
    &= \left(\int_{S^{n-1}}\frac{1}{\|{\theta}\|^n_K}d\sigma(\theta)\right)^2 = R^{2n}
\end{align*}

How did they take the average inside the integral? Is that a mistake? If not, why? Can we always swap averages and integrals like this?

I'd appreciate any help with understanding the above work!

答: An average is an integral, so Fubini-Tonelli would be the most applicable theorem. In this case, the integrand is non-negative, so we can interchange the order of integration.</p>


问:
How can i show the relationship between a sequence in a space in norm E with a genuinely increasing sequence of natural numbers?
Supposing i have a sequence $(a_n)_{n\epsilon\mathbb{N}}$ in a space with norm E .

How can i show that a sequence $(x_n)_{n\epsilon\mathbb{N}}$ exists in E , so that $ \lim_{n\to\infty} x_n = 0 $ and a genuinely increasing sequence $(k_n)$ of natural numbers so that $a_n = x_0 + x_1 +x_2 +x_3+...+x_{k_n} \forall n\epsilon\mathbb{N}$

Body:
Hint

You need
$$
x_{k_{n}+1}+...+ x_{k_{n+1}}=a_{n+1}-a_n
$$
Try to pick inductively $k_{n+1}$ such that
$$
\|\frac{a_{n+1}-a_n}{k_{n+1}-k_n}  \| < \frac{1}{n}
$$
and set
$$
x_{k_{n}+1}=...= x_{k_{n+1}}=\frac{a_{n+1}-a_n}{k_{n+1}-k_n}
$$


问:
Can a vector space with infinite basis be represented as a countable union of proper subspaces?
I am familiar with the case where the basis is not infinite (it becomes the case that it CANNOT be represented as such). However, many sources suggest that this is not the case for an infinite basis. How could one go about proving this?</p>

答: Suppose that $V$ is infinite dimensional and fix a basis $B$. Then, by assumption, we have $|B| \geq \aleph_0$, and thus there exists a countable subset $S \subset B$. For each $s \in S$, consider $V_s = \langle B \setminus \{s\} \rangle$. Then $V = \bigcup_{s \in S}V_s$.


答:
Hint

From the first equation, you have
$$x=\frac 13(y'+2y+8t)\tag 1$$ Replace in the second to have
$$y''-y=16 t+48 e^{-t}-8\tag 2$$ which not difficult.

When solved for $y$, go back to $(1)$ to obtain $x$.

Title: How can I solve this kind of systems: $\left\lbrace \begin{array}{rcl} 3x-y'-2y&=& 8t\\ x'-2x+y &=&16e^{-t} \end{array}\right.$

Body:
How can I solve this kind of systems: $$\left\lbrace \begin{array}{rcl} 3x-y'-2y&=& 8t\\ x'-2x+y &=&16e^{-t} \end{array}\right.$$

I don't know how to approach it. I tried:  $$\left\lbrace \begin{array}{rcl} y'&=& -8t+3x-2y\\ x' &=&16e^{-t}+2x-y \end{array}\right.$$ But since $y,y' x$ are mixed seems difficult, to continue.


问:
Probability that at least 2 people chose the same number
Each person in a group of n people is requested to select a whole number between 1 to k. Describe the probability that at least 2 people chose the same number.

My answer was

$P(x)=favorable\ cases / posible\ cases$

$P(x) = C_2^nC_{n-2}^K\ /\ Vr_{K}^n$

But the teacher told me it was wrong. What am I not seeing?

答:
For $k\ge n$, it's just 1- the probability that everyone chooses different numbers.

$1-\displaystyle \frac{k*(k-1)*...*(k-n+1)}{k^n}$


问:
If G is a nonabelian group with order of $p^3$ which p is a prime number then prove $G’=Z(G)$
I have an exam in two hours and this is one of the exercises in my textbook that I have problem with, I tried solving it using the main theorems but i couldn’t figure it out.
The question goes:

If G is a nonabelian group with order of $p^3$ which p is a prime number then prove $G’=Z(G)$

A hint or an answer would be appreciated .

答:
Since it is a $p$-group, the center is nontrivial. The center cannot have prime index, since $G/Z(G)$ cannot be cyclic nontrivial; and the center cannot be of order $p^3$, because then $G$ would be abelian. Thus, $|Z(G)|=p$.

Now, $G/Z(G)$ is of order $p^2$ and hence abelian. Thus, $[G,G]\leq Z(G)$. But $Z(G)$ is of prime order, so either $[G,G]$ is trivial or $[G,G]=Z(G)$. But $G$ is nonabelian, so $[G,G]\neq\{e\}$. Thus, $[G,G]=Z(G)$.


答:
Below are some general remarks on inductive proofs that are too long for comments. The primary point that I'd like to emphasize is that when you are first learning how to present rigorous proofs by mathematical induction it is important to be careful to avoid composing circular proofs. Let's consider a specific example. In another answer here a proof of the desired result, that $\rm\ 10^n\equiv (-1)^n\ (mod\ 11)\:,\:$ was proposed based upon the following identity, which is true for all integers $\rm\:a,b\:$ and all naturals $\rm\:n\:$

$$\rm a^{n+1}-b^{n+1}\ =\ (a-b)\ (a^n + a^{n-1}b + a^{n-2}b^2 + \cdots + ab^{n-1}+b^n) $$

The sought result follows immediately upon specializing $\rm\: a = 10,\: b = -1\:$ in the above identity.

Does the above constitute a valid proof by induction of the sought result?  Probably your instructor would accept it as a valid proof if you first proved by induction said identity and then derived the sought result as a corollary, by specializing the identity as indicated above. However, if you don't already have available a rigorous proof that the above identity holds true for all $\rm\:n\:$ then it would not be correct to invoke the identity in your proof since, in effect, you would be assuming true a more general result, that itself requires a rigorous proof by induction. Even though you may have learned this identity in your prior studies, before you learned how to give rigorous inductive proofs, probably at that time you did not have available the tools to give such a rigorous proof. Perhaps you may even have learned more general results, such as the Factor Theorem $\rm\ x-y\ |\ f(x)-f(y)\:$ in $\rm\:\mathbb Z[x,y]\:$ for all polynomials $\rm\:f(x)\in\mathbb Z[x]\:.\:$ But this too requires a rigorous inductive proof, e.g. by induction on $\rm\: deg\ f\:.\:$

Thus if you propose to tackle an exercise on induction by "passing the inductive buck" to a more general result, be sure that you have already given a rigorous inductive proof of that result. Since the point of such exercises is to provide experience to help strengthen your inductive intuition, it is essential to experience the induction - whether in a specific form or a more general form.

Once you have gained some experience with inductive proofs it is important to introspect a bit at a meta-level - to abstract out various common patterns of induction. For example, many of my posts here emphasize the fact that many common inductive proofs are simply special cases of telescopy. By noticing this telescopic structure, one can often reduce the induction to a very trivial normalized form, e.g. to the trivial induction that $\rm\: 1^n = 1\:.\:$ Such reduction of inductive complexity frees up our mental faculties so that we can concentrate on more essential aspects of the problem.

Title: Complete induction of $10^n \equiv (-1)^n \pmod{11}$

Body:
To prove $10^n \equiv (-1)^n\pmod{11}$, $n\geq 0$, I started an induction.

It's $$11|((-1)^n - 10^n) \Longrightarrow (-1)^n -10^n = k*11,\quad k \in \mathbb{Z}. $$

For $n = 0$:
$$ (-1)^0 - (10)^0 = 0*11 $$

$n\Rightarrow n+1$
$$\begin{align*}
 (-1) ^{n+1} - (10) ^{n+1} &= k*11\\
(-1)*(-1)^n - 10*(10)^n &= k*11
\end{align*}$$
But I don't get the next step.


问:
Prove that in every tree, any two paths with maximum length have a node in common.
Prove that in every tree, any two paths with maximum length have a node in common. This is not true if we consider two maximal (i.e. non-extendable) paths.

What does this even mean?

答:
If we have a path we can measure its length (how many vertices, or edges, it goes through).

The length is a natural number, the question asks to show that given two paths which have the longest possible length (i.e. the length is maximum possible) then they have a common node; however if we only take maximal paths (which are paths which we just cannot extend any further) then it might not be true if the length of these paths is not the maximum from the first part.


问:
the (seeming) loop caused by defining Natural numbers through inductive sets.
I'm a first year student in mathematics, and we recently defined Natural numbers as following:

"For all numbers, a number is an element of the set Natural numbers, if and only if it is an element of every inductive set".
An inductive set is defined as "A subset of the Real numbers that contains 1 and for each element, that element + 1 is also an element of the set'.

One's natural instinct is to ask why not define $\mathbb{N}$ = {1, 2, 3 ...} which is easy to dismiss because we cannot define an infinite set in Roster notation when there is no set builder notation. We have not yet defined what natural numbers are and therefore these numbers are not related by any means (The only property that they share is being Natural which yet has no definition in our stack. even if one wishes to define this set inductively, through something call a successor function if I'm not mistaken, the domain of the said function must be the natural numbers which...defeats the purpose? Someone might argue for a definition through sequences, which...can one have sequences before defining Natural numbers?).

I suppose the reason why we are being so rigorous in defining natural numbers in such way is that we wish to prove things by induction. When we prove through induction we basically define a set like M to be the subset of $\mathbb{N}$ of all numbers who satisfy P(n) which P(n) is usually our the algebraic conclusion in our theorem. We prove 1 $\in$ M and $\forall n (n+1 \in M)$. From this we conclude M = $\mathbb{N}$.

This triviality is in need of proving, so we naturally (no pun intended) resort to making a more rigorous definition of $\mathbb{N}$ in order to be able to use induction.
We define inductive sets, then define a Natural number as said above, see that according to definition $\forall \hspace{1mm} inductive A, \hspace{1mm}(\mathbb{N} \subseteq A)$.

So it is sufficient to know a subset of Natural numbers is inductive to prove they are equal (which we apply in M for the above). It is easy now to prove that 1, 2, 3 and so on are natural numbers individually, and I don't know if I'm being nitpicky about this, but I have the following concern: when we prove things through induction we essentially have in mind that we want to prove that a theorem holds for {1, 2, ...} and that happens to be called $\mathbb{N}$. We can't have {1, 2, 3, ...} because of the reasons above so we take $\mathbb{N}$ as nothing but a name and define it as we did above. we can prove that 1 $\in \mathbb{N}$ and so on, but that is a stretch of it being equal to the infinite set of {1, 2, 3, ...}. Instantly, it seems ridiculous to call it a stretch for it not being equal to something that we cannot define or is defined through something we earlier defined. does it suffice to say {n| n $\in \mathbb{N}$} = {1, 2, 3, ...}, or we should find a separate way to define the algebraic form of Natural numbers and then prove it to be equal to the $\mathbb{N}$ we defined above through inductive sets.
It also seems that we can prove 1 $\in \mathbb{N}$, 2 $\in \mathbb{N}$ and ... . But it seems that proving {1, 2, ...} to be equal to $\mathbb{N}$ is (in addition to needing to define {1, 2, ..} in a way it makes sense) in itself in need of induction for proving.

I would be most thankful if someone would help clarify my perspective.

答:
There is a serious problem in the definition you have been given! Basically, it is circular!! One cannot construct $ℝ$ unless one already has a set $ℕ$ with constants $0,1∈ℕ$ and binary operations $+,·$ on $ℕ$ such that $(ℕ,0,1,+,·)$ satisfies PA, and PA includes induction!  For an actually proper mathematical construction of $(ℝ,0,1,+,·)$ from $(ℕ,0,1,+,·)$, you can consult Spivak's "Calculus". So whoever gave you this circular 'definition' of natural numbers clearly does not know that it actually does nothing from a proper foundational perspective. And if foundations are not the point, then there is absolutely no reason to talk about inductive sets!

Patrick has hinted at, but not really explained, the true meaning of the idea of inductive sets. This is a matter of set theory. Historically, some people did not like having to axiomatize $ℕ$ via all the axioms of PA, and sought to reduce everything to set theory. They did so via introducing the notion of an inductive set. Formally, a set $S$ is inductive iff $∅∈S ∧ ∀x{∈}S\ ( \ x∪\{x\}∈S \ )$. The problem is, there is no way to prove the existence of an inductive set without using some assumption of the same kind! So they decided to simply add an axiom (to the set theory they were building) that just baldly asserts the existence of an inductive set. Though such an assumption cannot be justified non-circularly, it does get us somewhere, because from that and other set-theoretic axioms we can prove that there is an inductive set that satisfies induction!
$
\def\ind{\text{Ind}}
$

Namely, define $\ind(S)$ to denote that $S$ is inductive. Let $I$ be such that $\ind(I)$ (by the axiom that an inductive set exists). Let $N = \{ x : x∈I ∧ ∀S\ ( \ \ind(S) ⇒ x∈S ) \ \}$. Then we can prove that $∀S\ ( \ \ind(S) ⇒ N⊆S \ )$. From this we get $∀S{⊆}N\ ( \ \ind(S) ⇒ S=N \ )$, which is what we call induction in set theory. Note that the specific initial element and successor operation we use in the definition of an inductive set is not essential, except that it must be definable. In the above definition we used initial element $∅$ and successor operation $( \ x ↦ x∪\{x\} \ )$. However, if you study set theory later you will see that it is most convenient to use this particular choices, and one reason is that you can define for every $k,m∈N$ that $k<m$ iff $k∈m$.


问:
Compute $\det A$ for the following elementary matrices
$ A = P_2^4 =
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 1 & 0 \\
0 & 1 & 0 &  0
\end{bmatrix} $

$ A = Q_2^4(-4) =
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & -4\\
0 & 0 & 1 & 0 \\
0 & 0 & 0 &  1
\end{bmatrix} $

$ A = S_2(-2) =
\begin{bmatrix}
1 & 0 & 0 \\
0 & -2 & 0 \\
0 & 0 & 1
\end{bmatrix} $

I know it's a stupid question but I started learning matrices and I can't find $ S , Q $ and $P$ in theory can you tell me what that means?

答:
Hint:

The identity matrix has determinant $1$.

Exchanging two rows of a matrix multiplies the determinant by $-1$.

Adding a multiple of one row to another does not change the determinant.

Multiplying a row by a non-zero number multiplies the determinant by that number.


问:
Summing a (partial) harmonic series
Given that a is a real number greater than 1, what is the value of

$$S = \lim\limits_{n\rightarrow\infty} S_n$$ where $S_n = \displaystyle \sum\limits_{i=1}^n \dfrac{1}{an-i}$, i.e.,
$$S_n = \frac{1}{an - 1} + \frac{1}{an - 2} + \ldots + \frac{1}{an - n}$$

I have chosen that if $a > 1$, S converges to a finite number, but I can't seem to arrive at an exact answer (which is what I need).

Thanks for any insights you can give.

Ron

答:
We have
$$S_n(a) = \sum_{i=1}^n \dfrac1{an-i} = \dfrac1n \sum_{i=1}^n \dfrac1{a-\dfrac{i}n} $$
Note that the above is nothing but a Riemann sum. Hence,
$$S(a) = \lim_{n \to \infty} S_n(a) = \lim_{n \to \infty} \dfrac1n \sum_{i=1}^n \dfrac1{a-\dfrac{i}n} = \int_{0}^{1} \dfrac{dx}{a-x} = -\left. \log(\vert a-x \vert) \right \vert_{0}^1= \log(\vert a \vert) - \log(\vert a-1 \vert)$$
Hence, we have
$$S(a) = \log\left(\left \vert \dfrac{a}{a-1}\right \vert\right)$$</p>

Title: Summing a (partial) harmonic series


问:
Can a quadratic equation not equal to zero
Well I wanted to know whether or not $y = x^2 + x + 7$ is a quadratic equation since the general form is $ax^2 + bx + c = 0$ here the equation
$y=x^2+x+7$. Isn't equal to zero so I'm a bit confused

答:
When you write $y=x^2 + x+7$, that is not generally considered a "quadratic equation" in the commonly used sense. Most of the time, that is taken to mean a functional relationship between two variables, namely $y$ and $x$. Because the right hand side takes the form of a quadratic polynomial, you are justified in calling it a "quadratic function" of $x$. When writing the relationship between two variables in this form, you're looking to answer questions like: what is the value of $y$ for a given value of $x$? What does a plot of $y$ against $x$ look like? And so forth.

A quadratic equation is (already, or can easily be rearranged into) something of the canonical form $ax^2 + bx +c =0$. The last term on the left hand side and is a constant term while the right hand side is zero.

So these are quadratic equations:

$x^2 + x +7 =0$ (already in the canonical form)

$x^2 + x + 7 = 2$ (can be immediately rearranged into the canonical form)

$x^2 + x + 7 = k$ (where $k$ is specified as a constant, even if it's not a known constant, allowing rearrangement into the proper form)

Note that the quadratic functional relationship $y = x^2 + x +7$ can be made into a quadratic equation if we ask and try to answer questions like:

What value(s) of $x$ makes $y = 10$? In this case $x^2 + x +7 =10$, which is a quadratic equation with two real roots, so you have your two possible $x$ values.

Does the curve $y = x^2 + x +7$ intersect the $x$ axis? The answer is 'no' because the quadratic equation $x^2 + x + 7 =0$ has no real roots, only complex ones.

I hope I've understood the essence of your question and answered clearly enough.


答:
Let's to suppose that $y'=y'(t)$ and $x'=x'(t)$ with $t\in I$ and $I$ an open interval, then
note that $$\begin{cases} 3x-y'-2y&=& 8t\\ x'-2x+y &=&16e^{-t} \end{cases} \iff \begin{cases}-y'&=&-3x+2y+8t\\ x'&=&2x-y+16e^{-t} \end{cases} \iff \begin{cases}y'&=&3x-2y-8t\\ x'&=& 2x-y+16e^{-t} \end{cases}$$

Finally, note that the system of ordinary differential equations can be written as

$$\underbrace{\begin{pmatrix} x'(t) \\ y'(t) \end{pmatrix}}_{X'(t)}=\underbrace{\begin{pmatrix}2 & -1 \\ 3 &-2 \end{pmatrix} }_{A(t)}\underbrace{\begin{pmatrix} x(t) \\ y(t) \end{pmatrix}}_{X(t)}+\underbrace{\begin{pmatrix} 16e^{-t} \\ -8t \end{pmatrix}}_{F(t)} $$

Since you have a linear nonhomogeneous systems of differential equations with constant coefficients, so you can first to find the solution for $F(t)\equiv 0$ and then you can find the solution with $F(t)=\begin{pmatrix} 16e^{-t} \\ -8t \end{pmatrix}$.

First Part: Let's to find the complementary solution, since that $$A(t)=\begin{pmatrix}2 & -1 \\ 3 &-2 \end{pmatrix}$$
It's easy to see that the eigenvalues are given by $$\boxed{\lambda_{1}=-1} \quad \text{and} \quad \boxed{\lambda_{2}=1}$$
and yours corresponding eigenvectors are given by $$\boxed{v_{1}=\begin{pmatrix} 1 \\ 3 \end{pmatrix}} \quad \text{and} \quad \boxed{v_{2}=\begin{pmatrix} 1 \\ 1 \end{pmatrix}} $$
so, since that $\lambda_{1}\not=\lambda_{2}$ and $\lambda_{1},\lambda_{2}\in \mathbb{R}$ then we can conclude that $$\color{blue}{\boxed{X_{c}(t)=c_{1}\begin{pmatrix} 1 \\ 3 \end{pmatrix} e^{-t}+c_{2}\begin{pmatrix}1 \\ 1 \end{pmatrix}e^{t}}}$$

Remember that if $\lambda_{1},\lambda_{2},\ldots,\lambda_{n}$ be reals eigenvalues and differents of the coefficients matrix $A(t)$ of system $X'(t)=A(t)X(t)$ and if $v_{1},v_{2},\ldots,v_{n}$ be correspondent eigenvectors. Therefore, the general solution of $X'(t)=A(t)X(t)$ on the interval $(-\infty,\infty)$ is given by $$\color{blue}{\boxed{X(t)=c_{1}v_{1}e^{\lambda_{1}t}+c_{2}v_{2}e^{\lambda_{2}t}+\cdots+c_{n}v_{n}e^{\lambda_{n}t}}}$$ with $c_{1},c_{2},\ldots,c_{n}$ constants.

Second Part: In general this is the tricky part, you mainly have two methods here: the indeterminate coefficients method and the parameter variation method. Of the methods I mentioned, the parameter variation method is the strongest; although not necessarily the simplest. I will use parameter variation, to solve this problem, to illustrate the method.

To use the parameter variation method the idea is to construct a matrix that I will call a fundamental matrix $\Phi(t)$  it is defined in the following way $$\color{blue}{\boxed{\Phi(t):=\begin{pmatrix} \uparrow & \uparrow \\ v_{1} & v_{2} \\  \downarrow & \downarrow \end{pmatrix}}}$$
where $v_{1}$ and $v_{2}$ are the eigenvectors of matrix $A(t)$, then we find the inverse of the fundamental matrix that I will call $\color{blue}{\boxed{\Phi^{-1}(t)}}$ and as a final step, we use the well-known formula from variation parameter method to find the particular solution given by $$\color{blue}{\boxed{X_{p}(t)=\Phi(t)\int \Phi^{-1}(t)F(t)\operatorname{dt}}}$$

Note that it is not necessary to add a constant of integration to the above identity for the same reason that you do not add the constant of integration in the parameter variation method for higher order ODE.

Returning to our problem, we can see that $$\Phi(t)=\begin{pmatrix} 1 & 1 \\ 3 & 1 \end{pmatrix} $$
and $$\Phi^{-1}(t)=\begin{pmatrix}-\frac{1}{2} & \frac{1}{2} \\ \frac{3}{2} &  -\frac{1}{2}\end{pmatrix}$$

Finally, you can find $$X_{p}(t)=\begin{pmatrix} 1 & 1 \\ 3 & 1 \end{pmatrix} \cdot \int \begin{pmatrix}-\frac{1}{2} & \frac{1}{2} \\ \frac{3}{2} &  -\frac{1}{2}\end{pmatrix}\cdot \begin{pmatrix} 16e^{-t} \\-8t \end{pmatrix}\operatorname{dt}$$

I'll let you figure that part out.

Finally when you have solved that part you found $X_{p}(t)$ and so the general solution for your problem is given by $$\color{blue}{\boxed{X(t)=X_{c}(t)+X_{p}(t)}}$$

I hope this detailed explanation will help you in your studies and future readers of MathSE.


问:
Help me factorise
Please help me factorize $9x^2-4a^2+4ay +y^2$. I took out "-" as a common factor from $(- 4a^2+4ay +y^2)$ but I cannot factorize it further using difference of 2 squares.

Title: Help me factorise

Body:
I believe the exact question is :

$$9x^2-4a^2+4ay-y^2$$ which is

$$=9x^2-(4a^2-4ay+y^2)=(3x)^2-(2a-y)^2=?$$


答:
We proceed by using the hints given by @twosigma.

$\det{P_2^4} = -1$ because $P_2^4$ is obtained by swapping the second and forth rows of the identity matrix.

$\det{Q_2^4} = 1$ because $Q_2^4$ is obtained by from the identity by adding a multiple $-4$ of the forth row to the second row.

$\det{S_2} = -2$ because $S_2$ is obtained by multiplying $-2$ to one row of the identity.

Title: Compute $\det A$ for the following elementary matrices

Body:
$ A = P_2^4 =
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 1 & 0 \\
0 & 1 & 0 &  0
\end{bmatrix} $

$ A = Q_2^4(-4) =
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & -4\\
0 & 0 & 1 & 0 \\
0 & 0 & 0 &  1
\end{bmatrix} $

$ A = S_2(-2) =
\begin{bmatrix}
1 & 0 & 0 \\
0 & -2 & 0 \\
0 & 0 & 1
\end{bmatrix} $

I know it's a stupid question but I started learning matrices and I can't find $ S , Q $ and $P$ in theory can you tell me what that means?


问:
Show that there exists a partition and sample points such that the Riemann sum equals any intermediate value
Let $f:[0, 1]\rightarrow \mathbb{R}$ be a bounded function. Show that for any c such that $f(a) < c < f(b)$ where $a,b \in [0, 1]$ there exists a partition $P$ and sample points such that the Riemann sum is equal to c.

I tried proving this by making 2 or 3 subintervals containing $a$, $b$, and an endpoint such that the Riemann sum will be c but I cannot seem to figure it out.

答:
$c$  may be written as $c=tf(a)+(1-t)f(b)$ with $0<t<1$. This is a Riemann sum with partition $P=\{0 ,t, 1\}$  and sample points $a,b$ if $a\leq t\leq b$.
(This is only a begining.)


问:
How to get the integral of a function with respect to another variable
I want to make a graph of the integral of a function (on the y-axis) where its integration is with respect to one of its variables, with respect to another one of its variables (on the x-axis). Please pardon me, the previous sentence is rather confusing because I am pretty confused myself; this is my first time dealing with more than two variables.

Here are the specifics:
I want to get the area under the curve of the Maxwell-Boltzmann distribution for chemical reactions and see how that area changes as absolute temperature, another one of its variables changes. The equation is given below:

$f_E(E) = \left(\frac{1}{\pi k_BT}\right)^{3/2} 2\pi \cdot E^{1/2} e^{-\frac{E}{k_B T}}$

I want to see how the integration of the above equation changes with temperature.
how this

$f_E(T) =\int_{E_a}^{\infty
} \left(\frac{1}{\pi k_BT}\right)^{3/2} 2\pi \cdot E^{1/2} e^{-\frac{E}{k_B T}}dE$

changes with T, temperature.

With my limited exposure and knowledge, I've tried a number of 3D graphing software including grapher and geogebra but I've only gotten as far as making a 3D graph of the normal distribution curve versus temperature, not its integral. For context, my current lesson in high school are derivatives of trigonometric functions.

Any help would be appreciated.

Edit: (for graphing the integral vs. T on Desmos)
I graphed the following, where T = x.

$ f(x)=\frac{m}{kx}\left(1-erf \left(\frac{E_a \sqrt{m}}{\sqrt{2kx}}\right) \right) $

Where k is the Boltzmann constant, $1.38064852 × 10^{-23}$

$m = 4.022 \times 10^{-25}$

$E_a = 101kJmol^{-1}$

I got the equation from here
https://physics.stackexchange.com/questions/366804/kinetic-energy-in-maxwell-boltzmann-distribution
and https://en.wikipedia.org/wiki/Maxwell%E2%80%93Boltzmann_distribution

答:
This coming from statistical thermodynamics, it should be $n(T)$ (as already said in comments).

$$n(T)=4\pi\left(\frac{m}{2\pi kT}\right)^{\frac32}\int_{E_a}^{\infty
} e^{-\frac{mx^2}{2kT}}dx$$ and we face a gaussian integral. So,
$$\int_{E_a}^{\infty} e^{-\frac{mx^2}{2kT}}dx= \sqrt{\frac{\pi k T}{2m}}\,\text{erfc}\left(\frac{E_a \sqrt{m}}{ \sqrt{2k T}}\right)$$
$$n(T)=\frac{m}{k T}\,\text{erfc}\left(\frac{E_a \sqrt{m}}{ \sqrt{2k T}}\right)$$
Compute the deivative $n'(T)$


问:
How can I solve this kind of systems: $\left\lbrace \begin{array}{rcl} 3x-y'-2y&=& 8t\\ x'-2x+y &=&16e^{-t} \end{array}\right.$
How can I solve this kind of systems: $$\left\lbrace \begin{array}{rcl} 3x-y'-2y&=& 8t\\ x'-2x+y &=&16e^{-t} \end{array}\right.$$

I don't know how to approach it. I tried:  $$\left\lbrace \begin{array}{rcl} y'&=& -8t+3x-2y\\ x' &=&16e^{-t}+2x-y \end{array}\right.$$ But since $y,y' x$ are mixed seems difficult, to continue.

Body:
$$\left\lbrace \begin{array}{rcl} 3x-y'-2y&=& 8t\\ x'-2x+y &=&16e^{-t} \end{array}\right.$$
Sum both DE:
$$x-y'+x'-y= 8t+16e^{-t}$$
$$(x-y)+x'-y'= 8t+16e^{-t}$$
$$u+u'= 8(t+2e^{-t})$$
Where $u=x-y$.


问:
What is the convex hull of $\text{conv}(u_1,u_2,\cdots,u_p)+\text{conv}(v_1,v_2,\cdots,v_s)$?
Let $u_i, i= 1,\cdots,p$ and $v_j, j= 1,\cdots,s$ be finitely many vectors in $\mathbb{R}^n$. Show that

$$
\text{conv}(u_1,u_2,\cdots,u_p)+\text{conv}(v_1,v_2,\cdots,v_s)=\text{conv}\{u_i+v_j \mid i= 1,\cdots,p, \,\, j= 1,\cdots,s\}
$$

We need to show

$$
x+y \in \text{conv}\{u_i+v_j \mid i= 1,\cdots,p, \,\, j= 1,\cdots,s\}
$$

where $x \in \text{conv}(u_1,u_2,\cdots,u_p)$ and $y \in \text{conv}(v_1,v_2,\cdots,v_s)$. Also, we need to show

$$
z \in \text{conv}(u_1,u_2,\cdots,u_p)+\text{conv}(v_1,v_2,\cdots,v_s)
$$

where $z \in \text{conv}\{u_i+v_j \mid i= 1,\cdots,p, \,\, j= 1,\cdots,s\}$.

I have tried the following for the first one:

Let $x \in \text{conv}(u_1,u_2,\cdots,u_p)$ so $x=\sum_{i=1}^p\lambda_iu_i$ where $\sum_{i=1}^p\lambda_i=1$. Also, Let $y \in \text{conv}(v_1,v_2,\cdots,v_s)$ so $x=\sum_{j=1}^s\mu_jv_j$ where $\sum_{j=1}^s\mu_j=1$.

Summing them

$$x+y=\lambda_1u_1+\lambda_2u_2+\cdots+\lambda_pu_p+\mu_1v_1+\mu_2v_2+\cdots+\mu_sv_s.$$

Now the question is how we can get something in the form of $\text{conv}\{u_i+v_j \mid i= 1,\cdots,p, \,\, j= 1,\cdots,s\}$?

答:
For two sets
$$\operatorname{co}(S_1 \times S_2) = \operatorname{co}(S_1) \times \operatorname{co}(S_2)$$

the inclusion from left to right is clear, for the other way notice
the equality
$$(\sum \lambda_i u_i, \sum \mu_j v_j) = \sum \lambda_i \mu_j (u_i , v_j)$$
if $\lambda_i$, $\mu_j$ positive with sum $1$. Now apply the affine map $+\colon V\times V \to V$.


问:
$e^{-1/f(x)}$ convexity
Given:

$$g(x)=e^{-1/f(x)}, \textbf{dom} g=\{x|f(x)<0\}$$

where $f$ is convex.

How can I prove that $g(x)$ is convex, concave or neither?

I have tried to find a counter example and failed. So based on those examples I've tried, I think  $g$ is convex but not sure how to prove it.

Tried using the definition of convexity where $ 0 \le \theta \le 1$ and I know that:
$$f(x_0) < 0, f(x_1) < 0$$
$$f(\theta x_0 + (1-\theta) x_1) \le \theta f(x_0) + (1-\theta) f(x_1)$$

and need to show that:
$$g(\theta x_0 + (1-\theta) x_1) \le \theta g(x_0) + (1-\theta) g(x_1)$$
$$e^{-1/f(\theta x_0 + (1-\theta) x_1)} \le \theta e^{-1/f(x_0)} + (1-\theta) e^{-1/f(x_1)}$$

But I am stuck here. Taking the log of both sides didn't get me very far.

Title: $e^{-1/f(x)}$ convexity

Body: $$g(x)=e^{-\frac{1}{f(x)}}\implies g''(x)=\frac {e^{-\frac{1}{f(x)}} } {[f(x)]^4}\Big[f(x)^2 f''(x)+(1-2 f(x)) f'(x)^2\Big]$$


问:
Integration of elementary function $\int \frac {\log x}{(1+x)^3}\,{\rm d}x $

The question is to find the integral
  $$\int \frac {\log x}{(1+x)^3}\,{\rm d}x $$

It can be easily be solved by integration by parts, but I want to solve it without using integration by parts.

答:
Integrate by parts

\begin{align}
\int \frac {\log x}{(1+x)^3}\,dx
& = \int \frac {\log x}2d\left(1-\frac{1}{(1+x)^2}\right)\\
&=\frac12{\log x}\left(1-\frac{1}{(1+x)^2}\right)-\frac12\int\left(\frac1{1+x}+\frac1{(1+x)^2}\right)dx\\
&= \frac12\left( \frac{x(x+2)\log x}{(1+x)^2}-\log(1+x)+\frac1{1+x}\right)+C
\end{align}


问:
Higher Order Euler Differential Equations
What is the pattern for writing the indicial equation for higher order Euler Ordinary Differential Equations?

For first order Euler ODEs, the indicial equation coefficients mirror the ODE coefficients exactly, i.e., $xy' + 5y = 0$ yields $s + 5 = 0$.  Define this pattern as $[0, 0]$, as nothing is added to either term.

For second order Euler ODEs, $1$ must be subtracted from the $xy'$ coefficient, i.e., $x^2 y'' + 9xy' - y = 0$ yields $s^2 + 8s - 1 = 0$.  This is the pattern $[0, -1, 0]$, as $-1$ is added to the middle term.

These patterns can be computed by converting a first order constant coefficient linear ODE to the associated Euler ODE via the substitution $x = e^t$.  However, checking the patterns by doing this conversion for higher order ODEs would require applying a mess of third/fourth/fifth derivative chain rules and product rules.

What is the pattern for higher order Euler ODEs?  My guess is that something related to binomial coefficients ($[0, -1, -1, 0]\ [0, -1, -2, -1, 0]$, perhaps up to some sign oscillation) or another collection of numbers famous in Combinatorics appears.

答:
For the ODE
\begin{align}
\sum_{n=0}^m\alpha_nx^n\frac{\mathrm d^ny}{\mathrm dx^n}=0,
\end{align}
assume the solution $y=x^r$, so
\begin{align}
x^r\sum_{n=0}^m\alpha_n(r)_n=0,
\end{align}
where $(r)_n$ is the falling factorial. Equating the coefficients to $0$ gives
\begin{align}
\sum_{n=0}^m\alpha_n(r)_n=0,
\end{align}
writing out the falling factorial in terms of stirling numbers gives
\begin{align}
\sum_{n=0}^m\alpha_n \sum_{k=0}^n s(n,k)r^k=0.
\end{align}
Not sure how to concisely do this, but I manually wrote out and added like terms of $r^k$ to get that the $k^{\text{th}}$ coefficient is
\begin{align}
c_k=\sum_{n=0}^m\alpha_n s(n,k).
\end{align}

To actually answer your question, to find the change from the original $k^{\text{th}}$ constant you could simply compute
\begin{align}
c_k-\alpha_k=\sum_{n=0}^m\alpha_n s(n,k)-\alpha_k,
\end{align}
which is indeed shorter than computing a $7^{\text{th}}$ order chain rule. . . .


问:
Next number in the sequence $7,31,211,2311,...$
I am trying hard to figure out next number in $7,31,211,2311,...$

I tried as $31=7*4+3$

$211=31*6+25$

$2311=211*11+10$

I also tried as, sum of the digits of each number is $7,4,4,7$ respectively. So the next number has the sum of the digits $4$.

Body:
$$2*3 +1=7$$
$$2*3*5 +1=31$$
$$2*3*5*7+1=211$$
$$2*3*5*7*11+1=2311$$
$$2*3*5*7*11*13+1=30031$$
Can you identify the pattern?


问:
Homework - Getting Started - Optimization
I am starting up my Masters after being out of school for awhile and this is my first week back. I am recalling how to complete a problem and I am wondering if anyone can help me get started. I do not need a final answer, just steps needed to complete.

Problem: Solve the following optimization problem using Lagrange multiplier. Express your answer in terms of $A, b,$ and $y$.

minimize $x$, $||x-y||^2$
subject to $Ax=b$

Attempt

$$L(x,\lambda) = ||x-y||^2 + \lambda^T(Ax -b)$$
$$L(x,\lambda) = ||x-y||^2 + \lambda^TAx -\lambda^Tb$$

Now, do I take a partial derivative with respect to $x$? For clarity, below is an image of the problem.

答:
I'm not 100% sure, but the derivative of the langrangian with respect to x is $2(x-y)+A^\intercal \lambda=0$ and the derivative with respect to $\lambda$ is $Ax-b=0$.

This solution assumes that y is given and at some point that A is a square matrix, but it should work even if A is not square if the last few steps are omitted.

From the first equation, $x=\frac 1 2 (-A^\intercal\lambda+y)$. Putting this in the second equation,$\frac 1 2 A(-A^\intercal\lambda+y)-b=0$. So $-\frac 1 2 AA^\intercal\lambda+\frac 1 2Ay=b$. So $\lambda=-2(AA^\intercal)^{-1}(b-\frac 1 2 Ay)$. Putting this back in what we got from the first equation, we eventually get $x=\frac 1 2(-A^\intercal(-2)(AA^\intercal)^{-1}(b-\frac 1 2 Ay)+y)=\frac 1 2(2A^{-1}b-y+y)=A^{-1}b$


问:
(Possibly Lebesgue) integral over an indicator function
I have trouble understanding the following integral:

\begin{equation*}
\int_{0}^{1} \textbf{1}_{[x\geq b]}db
\end{equation*}

I don't know any way to integrate it "normally" like doing a Riemann integral. But I don't know how to analyze it as a Lebesgue integral, either. For example, $b$ is supposed to be a measure on a $\sigma$-algebra (presumably Borel sets) over $[0,1]$, yet the function $\textbf{1}_{[x\geq b]}$ implies that $b$ is a variable.

I appreciate any insights explaining the integral and solving my confusion.

答: Here $b$ is the variable of integration and $x$ is fixed. Let $f(b)=1$ if $b \leq x$ and $0$ otherwise. Then the given intergal is $\int_0^{1}f(b)db$ which is equal to the measure of $\{b\in [0,1]: b \leq x\}$. So the value is  $x$ if $0 \leq x \leq 1$, $1$ if $x >1$ and $0$ if $x<0$.


问:
Power series expansion of function involving logarithm and power function
It is well-known that the power series
$$\log\left(\frac{2}{1+\sqrt{1-x}}\right)=\sum_{n=1}^\infty \frac{\binom{2n}{n}}{n4^n}x^n\quad (x\in[-1,1]).$$ However, for the following function
$$\log\left(1+(1-x)^{1/p}\right)\quad (p>2)$$
Does it have a simple power series expansion? More general,
$$\log\left(1+c_1(a_1-x)^{1/{p_1}}+c_2(a_2-x)^{1/{p_2}}+\cdots+c_r(a_r-x)^{1/{p_r}}\right)$$
Does it have a simple power series expansion? Here $p_1,\ldots,p_r\ge 2$ are positive integers, $a_1,\ldots,a_r\in [-1,1]$ and $c_1,\ldots,c_r$ are real numbers.

Body:
I think that, because of the required composition of Taylor series, it could be difficult to find the general term
$$(1-x)^{\frac{1}{p}}=1-\frac{1}{p}x+\frac{(1-p)}{2 p^2} x^2+\frac{ (1 - p) ( 2 p-1)}{6
   p^3}x^3+$$
$$\frac {(1-p) (2 p-1) (3 p-1) }{24p^4}x^4+\frac {(1-p) (2 p-1) (3 p-1)(4p-1) }{120p^5}x^5+O\left(x^6\right)$$
$$\log \left(1+(1-x)^{\frac{1}{p}}\right)=\log (2)-\frac{1}{2 p}x+\frac{(1-2 p) }{8 p^2}x^2+\frac{(3-4 p) }{24
   p^2}x^3+\frac{(1-4 p) \left(6 p^2-4 p-1\right)}{192 p^4} x^4-\frac{\left(48 p^3-50 p^2+5\right) }{480 p^4}x^5+O\left(x^6\right)$$

Edit

Let $k=\frac 1p$ and we have
$$\log \left(1+(1-x)^k\right)=\log (2)+\sum_{n=1}^\infty \frac {a_n}{b_n}x^n$$ where the $b_n$'s make the sequence
$$\{2,8,24,192,480,5760,13440\}$$ and the $a_n$'s are polynomials of degree $n$ in $k$; the first are given below
$$\left(
\begin{array}{cc}
 n & a_n \\
 1 & -k \\
 2 & (k-2) k \\
 3 & k (3 k-4) \\
 4 & -(k-4) k \left(k^2+4 k-6\right) \\
 5 & -k \left(5 k^3-50 k+48\right) \\
 6 & (k-6) (k-2) k \left(2 k^3+16 k^2+19 k-40\right) \\
 7 & k \left(14 k^5-245 k^3+1176 k-960\right)
\end{array}
\right)$$