问:
Confused about a proof involving concave functions
This proof is from a book about linear programming; it is about interior point methods, but I think that might not be important for my question.

We have:

$$f_{\mu}(\mathbf{x}) = \mathbf{c}^T\mathbf{x} + \mu\sum_{i=1}^{m}\log(b_{i} - \mathbf{a}_{i}\cdot\mathbf{x}) $$

For vectors $\mathbf{c},\mathbf{x},\mathbf{a}_i$ for $i = 1,2,3,...,m$ and $\mu > 0$. The book claims that $f_{\mu}$ has a unique maximum inside the polytope defined by $A\mathbf{x} < \mathbf{b}$ where the $\mathbf{a_i}$'s are the rows (not the columns) of $A$; we also assume that this polytope is bounded. $A$ is an $m \times n$ real matrix of rank $m$; so the polytope is in $\mathbb{R}^n$.

I don't understand a line in the proof of the uniqueness claim. The books states that if there were two different maxima $\mathbf{x}$ and $\mathbf{y}$ then $f_{\mu}$ must be constant along the line segment $\mathbf{xy}$ because the polytope is convex so all of the line segment is inside the polytope and $f_{\mu}$ is concave.

The book then states that "since the logarithm is strictly concave, this can happen only if $A\mathbf{x} = A\mathbf{y}.$" This is the part I don't understand; why does the logarithm being concave imply that $A\mathbf{x} = A\mathbf{y}$? I don't have much experience with convexity.

答:
$A:= \begin{bmatrix}
\mathbf a_1^T\\
\vdots \\
\mathbf a_m^T\\
\end{bmatrix}$

Assuming a maximum exists, I'll give the uniqueness proof.  But it is prudent to exploit linearity and consider a simpler function, with a change of variables

in essence define $g = f -\mathbf c^T\mathbf x$  and assume $\mu: = 1$ for now, then simplify notation with

$\mathbf z:=A\mathbf x $  and

$ \mathbf z':=A\mathbf x'$

$g\big(\mathbf z\big) := \sum_{i=1}^m \log\big(b_i - z_i\big)$

$g$ is a strictly negative convex function (you can show this e.g. by differentiating twice and seeing the Hessian is a diagonal matrix with negative entries on the diagonal).  By the definition of strict negative convexity (aka strict concavity)

for any $p\in (0,1)$

$p\cdot g\big(\mathbf z\big)+(1-p) \cdot g\big(\mathbf z'\big)\leq g\big(p\mathbf z+(1-p)\mathbf z'\big)$

with equality iff $\mathbf z = \mathbf z'$

in other words, if two distinct points $\mathbf z$ and $\mathbf z'$ attain a maximum under $g$, then there is an even bigger value in their convex hull-- a contradiction.  (It is useful to recall that a convex hull of finitely many points is a polytope, so considering convex hulls is quite natural.)  This is essentially the proof.  Everything that follows is just book-keeping to show that the above implies the result for the original problem statement.

This is still true for any $\mu \gt 0$, i.e. rescale each side by $\mu$

$\mu\cdot p\cdot g\big(\mathbf z\big)+\mu\cdot (1-p)\cdot g\big(\mathbf z'\big)\leq \mu\cdot g\big(p\mathbf z+(1-p)\mathbf z'\big)$

this implies the function given by

$h_\mu\big(\mathbf x\big) := f_\mu\big(\mathbf x\big) - \mathbf c^T\mathbf x$

cannot have two distinct maximizing x's, $\mathbf x$ and $\mathbf x'$ unless $\mathbf z=\mathbf z'$ i.e. unless $\mathbf x$ and $\mathbf x'$ are equal under the image of $A$, so putting it all together

$p\cdot f_\mu\big(\mathbf x\big) +(1-p) \cdot f_\mu\big(\mathbf x'\big) -\mathbf c^T \big(p\mathbf x+(1-p)\mathbf x'\big) $

$=p\cdot \big\{f_\mu\big(\mathbf x\big) - \mathbf c^T (\mathbf x)\big\}+(1-p)\big\{f_\mu\big(\mathbf x'\big) - \mathbf c^T (\mathbf x')\big\}$

$=p\cdot h_\mu\big(\mathbf x\big)+(1-p)\cdot h_\mu\big(\mathbf x'\big)$

$\leq h_\mu\big(p\mathbf x+(1-p)\mathbf x'\big)$

$= f_\mu\big(p\mathbf x+(1-p)\mathbf x'\big) - \mathbf c^T \big(p\mathbf x+(1-p)\mathbf x'\big) $

with equality iff $A\mathbf x = A\mathbf x'$

this of course simplifies to

$p\cdot f_\mu\big(\mathbf x\big) +(1-p) \cdot f_\mu\big(\mathbf x\big) \leq f_\mu\big(p\mathbf x+(1-p)\mathbf x'\big) $

one awkward point:

$A \in \mathbb R^{m\times n}$ with rank $m$.  If $n\neq m$ it is short and fat and hence surjective, but also has linearly dependent columns.  Thus the optimizing $\mathbf x$ shouldn't be unique since multiple other choices exist for $\mathbf x'$ which take the same value under the image of $A$.


问:
Maximize $f(x,y) = x+y$ subject to $x^2+xy+y^2+y=1$
I'm stuck trying to find the maximum value and at which point it occurs of the function
$$f(x,y) = x + y,$$
subject to the constraint $x^2 +xy +y^2 + y = 1$.

my working so far is

$f_x = \lambda g_x \rightarrow 1 = \lambda \cdot (2x + y)$

$f_y = \lambda g_y \rightarrow 1 = \lambda \cdot (x + 2y + 1) $

so im left with the system.

$\lambda \cdot (2x + y) - 1 = 0$

$\lambda \cdot (x + 2y + 1) - 1 = 0 $

$x^2 + xy + y^2 + y -1 = 0$

I dont know what  I have missed but the maximum value is $1$ at $(1, 0)$, I'm not sure how I get to that.

答:
Your start looks fine except some $1$'s need to be multiplied by $\lambda$.   For the second equation I get $\lambda \cdot (x+2y + 1) - 1 = 0$

Now you have to solve the system of equations. Solve one equation for one variable and substitute.  For example, $\lambda = \frac{1}{2x+y}$ from the first equation.  Plug that into the second.  We get
$$ \frac{1}{2x+y} \cdot(x + 2y + 1) = 1$$
which can be solved for either x or y by multiplying and rearranging to get $y = x-1$.  Plug that into the last equation and solve for the possible values of $x$.

ContentLicense: CC BY-SA 4.0


问:
Given this limit involving T can I imply that T is injective?
Recently, trying to solve a problem, I got stuck in the following fact: Given a linear transformation $T:\mathbb{R}^m \to \mathbb{R}^n$ such that $$\lim\limits_{v \to 0} T \cdot \frac{v}{|v|}=0.$$ Is it true that $T \equiv 0$? I think I was able to prove it by supposing there is some $w \in \mathbb{R}^m \setminus \{0\}$ such that $T \cdot w \neq 0$. Then the sequence $(v_k)_{k \geq 1} \subset \mathbb{R}^m \setminus \{0\}$ given by $v_k=\frac{1}{k} \cdot w$ is such that $\lim\limits_{k \to \infty} v_k=0$ and$$0=\lim\limits_{v \to 0} T \cdot \frac{v}{|v|}=\lim\limits_{k \to \infty}T \cdot \frac{v_k}{|v_k|}=\lim\limits_{k \to \infty} T \cdot \frac{w}{|w|}=T \cdot \frac{w}{|w|}\neq 0.$$ Hence $T \cdot w =0 $ for all $w \in \mathbb{R}^m$ and $T$ must be the zero transformation.

In weakening the hypothesis, I tried to find an answer to the following problem: Given a linear transformation $T:\mathbb{R}^m \to \mathbb{R}^n$, a sequence $(v_k)_{k \geq 1} \subset \mathbb{R}^m \setminus \{0\}$ such that $\lim\limits_{k \to \infty}v_k=0$ and $$\lim\limits_{k \to \infty} T \cdot \frac{v_k}{|v_k|}=0.$$ Is it true that $Ker(T) \neq 0$?
First I tried to find a counterexample, but I couldn't. And since It is easy to prove the converse (i.e. if $Ker(T) \neq 0$ then there is sequence $(v_k)_{k \geq 1} \subset \mathbb{R}^m \setminus \{0\}$ satisfying the above conditions) I thought this might be true but have no idea how to prove it. Any help with a proof or counterexample is appreciated.

答: Let $\ S\ $ be the set of cluster points of $\ \big\{\frac{v_k}{|v_k|}\big\}_{k=1}^\infty\ $.  The kernel of $\ T\ $ must contain  $\ S\ $, and $\ S\ $ must contain a non-zero element.  Therefore $\ Ker(T)\ne\{0\}\ $.


答:
Great question! I think it is true. A somewhat analytical way to show the contrapositive would roughly be "$|T(v)|$ must attain a minimum on the unit sphere in $\mathbb R^m$ because $T$ is continuous and the unit sphere is compact. This minimum can't be $0$ by injectivity, so if $v_k$ is a nonzero sequence then $T(v_k / |v_k|) \not \to 0$, as this second sequence is bounded below in magnitude.

A somewhat different (but secretly the same) angle: Effectively what you're asking is "if $T(u_i) \to 0$ for some sequence of unit vectors, can $T$ be injective?" Again by compactness $u_i$ has a convergent subsequence, and by continuity its limit is in the kernel of $T$ (and of course the limit can't be $0$).

Title: Given this limit involving T can I imply that T is injective?

Body:
Recently, trying to solve a problem, I got stuck in the following fact: Given a linear transformation $T:\mathbb{R}^m \to \mathbb{R}^n$ such that $$\lim\limits_{v \to 0} T \cdot \frac{v}{|v|}=0.$$ Is it true that $T \equiv 0$? I think I was able to prove it by supposing there is some $w \in \mathbb{R}^m \setminus \{0\}$ such that $T \cdot w \neq 0$. Then the sequence $(v_k)_{k \geq 1} \subset \mathbb{R}^m \setminus \{0\}$ given by $v_k=\frac{1}{k} \cdot w$ is such that $\lim\limits_{k \to \infty} v_k=0$ and$$0=\lim\limits_{v \to 0} T \cdot \frac{v}{|v|}=\lim\limits_{k \to \infty}T \cdot \frac{v_k}{|v_k|}=\lim\limits_{k \to \infty} T \cdot \frac{w}{|w|}=T \cdot \frac{w}{|w|}\neq 0.$$ Hence $T \cdot w =0 $ for all $w \in \mathbb{R}^m$ and $T$ must be the zero transformation.

In weakening the hypothesis, I tried to find an answer to the following problem: Given a linear transformation $T:\mathbb{R}^m \to \mathbb{R}^n$, a sequence $(v_k)_{k \geq 1} \subset \mathbb{R}^m \setminus \{0\}$ such that $\lim\limits_{k \to \infty}v_k=0$ and $$\lim\limits_{k \to \infty} T \cdot \frac{v_k}{|v_k|}=0.$$ Is it true that $Ker(T) \neq 0$?
First I tried to find a counterexample, but I couldn't. And since It is easy to prove the converse (i.e. if $Ker(T) \neq 0$ then there is sequence $(v_k)_{k \geq 1} \subset \mathbb{R}^m \setminus \{0\}$ satisfying the above conditions) I thought this might be true but have no idea how to prove it. Any help with a proof or counterexample is appreciated.


问:
How to show this function defined using a homotopy is continuous?
Suppose there exists a homotopy $H$ between continuous functions $f,g:X\longrightarrow Y$, where  $X,Y$ are non-empty topological spaces. Consider the function $h:X\times [-1,1]\times [0,1]\longrightarrow Y\times[-1,1]$ defined by $h(x,s,t)=(H(x,t),s)$. I want to show that this function is continuous, but I am not sure how to.

I tried to show that such a map is "induced" (and is therefore automatically continuous) by the continuous map $f':X\times [-1,1]\longrightarrow Y\times [-1,1]$ given by $f'(x,t)=(f(x),t)$, under some identification map from $X\times [-1,1]$ to $X\times [-1,1]\times [0,1]$, but I am not sure if such an identification map exists.

Or is there a simpler argument along the lines of "$h$ is continuous because it is defined using $H$, which is continuous because its a homotopy"? I feel this is not sufficient mainly due to the fact that we 'swap' the order of arguments $s$ and $t$ in some way.

Can anyone provide some clarity on this problem? Thanks

答:
You only need to know the following fact about the product topology:

A map $\phi : A \to P  = \prod_{j \in J} X_j$ is continuous if and only if all component functions $p_j \circ \phi : A \to X_j$ are continuous. Here, $p_j : P \to X_j$ denotes projection onto the $j$-th component.

Note that this statement includes the continiuity of the $p_j$. Simply take $\phi =id_P$ which is clearly continuous. In fact, the product topology is introduced as the coarsest topology on $P$ making all $p_j$ continuous.

Thus your function $h$ is continuous iff $h_1(x,s,t) = H(x,t)$ and $h_2(x,s,t) = s$ are continuous. But $h_2$ is the projection onto the second coordinate, hence continuous. Moreover, $h_1 = H \circ p$, where $p : X\times [-1,1]\times [0,1] \to X\times [0,1], p(x,s,t) = (x,t)$. But $p$ is continuous by the above statement. Thus $h_1$ is continuous.


问:
Evaluating Large Expression Concerning the Roots of a Cubic
A few months ago I was doing a math competition and I came up with a very roundabout way of finding the answer. This ended up in me having to evaluate
$$r=\frac{(1-c_i+c_i^2)^3}{(1-6c_i+3c_i^2+c_i^3)^2}$$
where $c_i$ is any root of $z^3-3z^2+1$. Through approximating a root around 2 to 5 digits using a calculator, the answer came to be 1/4.

Is there any way to even begin evaluating this by hand?

Some information: If $z_i$ is a root of $z^3-6z^2+3z+1$, then $\frac{-z_i+2}{1-2z_i}=c_i$, so reversing the coefficients to reciprocate the roots could help. If $P(x)=z^3+1$, then $z^3-3z^2+1=P(x)-P'(x)$, which is probably not helpful.

I suppose that you could assume that the value of $r$ could be found by assuming that it is independent on which root of $P$ you pick, and then add/multiply the equal values of $r$, and then expand and use Vieta's, but that sounds unlikely. I'm betting that there's probably no easy way, but maybe there is.

答: Hint: Start by replacing $c_i^3=3c_i^2-1$ in the denominator.


问:
3 different ways to express $S(x)=(\sum_{n=0}^{\infty}x^n)^2$
Let the function $S(x)$ be a sum, such that $S(x)=(\sum_{n=0}^{\infty}x^n)^2$ and $|x|<1$. Which of the following is true: (can be multiple options)

$S(x)$ can be expressed as $\sum_{n=0}^{\infty}x^{2n}$

$S(x)$ can be expressed as $\sum_{n=0}^{\infty}(n+1)x^n$

$S(x)$ can be expressed as $\frac{1}{(1-x)^2}$

(1)This one seems to be applying the square to the sum. I think this is one of the properties of these series, so I think it is true.

(2)This one can be rewritten as $\sum_{n=0}^{\infty}nx^n+x^n$, and the square is gone too, after some testing, I declare this one to be true with values $S(0.5)$

(3)This one seems sus, I have no idea how this one is derived, and intuitively it seems false. It's not even a sum and expressed as a fraction as well, which is vastly different from the rest

答:
Hint 1: $(a+b)^2=a^2+2ab+b^2\neq a^2+b^2.$

Hint 2: See hint 3. Then perform Taylor series expansion.

Hint 3: Geometric series!!!


问:
Change in the price of an item based on a group's need for the item.
Okay so this is likely to be confusing to read, please bear with me.

The base price of an item is 1440. There are only 6 of these items and there is a population of 20.  10 of the population want or need the item.
Is it possible to up the price by a percentage based on a the need of the population?

For instance if the 10 who don't want the item wouldn't pay over the base cost, but the 10 who wanted it would be willing to pay up to 5% more. (or some other number, I'm just throwing out ideas now.)

I hope I worded this in a way that can be understood. What I'm trying to do is set up a makeshift economy for a game I'm making, I'd like the prices to fluctuate based on the needs of the population at any given time, I'm just at a loss for where to begin.

Any help is greatly appreciated.

答: Since this is for a game, simply set up a demand curve that is downward sloping. Depending on the game specifications, you can set up the demand curve to be linear or nonlinear as along as the slope is negative.</p>

ContentLicense: CC BY-SA 3.0


问:
Can we always replace $\underset{\delta\rightarrow 0}{\lim}$ with $\underset{k\delta\rightarrow 0}{\lim}$ for $k$ constant?
Is it in general acceptable to  replace $\underset{\delta\rightarrow 0}{\lim}$ with $\underset{k\delta\rightarrow 0}{\lim}$ for constant $k$?

What I mean by this is, if we have a real-valued function $f$  and we know that, say $\underset{\delta \rightarrow 0}{\lim} f(\delta)=L$ for real constant $k$, is it then valid to say that  $$\underset{\delta \rightarrow 0}{\lim} f(k \delta)=\underset{k\delta \rightarrow 0}{\lim} f(k \delta)=\underset{\delta \rightarrow 0}{\lim} f(\delta)=L\:?$$

(I feel like this should always be the case, because for any constant $k$, $(\delta\rightarrow 0) \Longleftrightarrow (k\delta \rightarrow 0)$ but maybe I'm overlooking something)

答:
Yes you can!

$\forall k \in \Bbb{R}\backslash \{0\}$,

$$\underset{\delta \rightarrow 0}{\lim} f(\delta)=\underset{k\delta \rightarrow 0}{\lim} f(k \delta)=\underset{t \rightarrow 0}{\lim} f(t)=L$$

where $t=k\delta$

Sometimes be careful with the sign of $k$. It may change a RHL to LHL or vice versa and this creates a problem when the function is not continous


问:
Induced map of kernels is an isomorphism for a pullback diagram in an Abelian category.
In this stacks project proof, we have the following diagram (with $w$ pullback):

$$\begin{array}{ccccccccc}  \text{ker}(f) & \xrightarrow{i} & w & \xrightarrow{f} & y\\
\downarrow \uparrow & \nearrow & \downarrow && \downarrow \\
 \text{ker}(k) & \xrightarrow{j} & x & \xrightarrow{k} & z \end{array}$$
The vertical arrows are $e:\text{ker}(f) \to \text{ker}(k)$, $u:\text{ker}(k) \to \text{ker}(f)$, $g:w\to x$ and $h:y \to z$.

The diagonal arrow $t:\text{ker}(k) \to w$ is the unique arrow such that $ft = 0$ and $gt=j$ (obtained from the universal property of pullback $w$). Also $iu = t$ and $je=gi$.

It is said that

It follows $giue=gte=je=gi$ and $fiue=0=fi$, hence $iue=i$.

Q: How did the author conclude $iue=i$?

答: This is from the universal property of the pullback.  Letting $a=gi$, we have $ka=kgi=hfi=0$, so there is a unique arrow $b:\ker(f)\to w$ such that $gb=a$ and $fb=0$.  One such arrow is $b=i$, and another is $b=iue$, so they are equal.


问:
Instrument to make sphere in space
I checked one of my old math books (for about $5$ years ago) and one of the problems is as follow:

$a)$ Why compass draw a circle?

$b)$ Can you give an example of an instrument to make sphere in space ?

Part $(a)$ is easy to answer. but I'm eager about knowing the answer of second part. I searched on internet but couldn't find such instrument.

答:
Rotating a circle around its diameter produces a sphere.

If they want a physical device, a gyroscope can be pointed in any direction and spins.  An ice cream scoop carves out parts of a sphere.

Not sure what else they might be asking for.  It's not as if we have a 3d piece of paper to draw on.

Title: Instrument to make sphere in space


问:
Compute $\int_0^1\frac{x}{\sqrt{x(1-x)}}dx$
I haven't integrated in a while... so I'm a bit rusty. I have to integrate $\int_0^1\frac{x}{\sqrt{x(1-x)}}dx$. Should I apply some change of variables? If I take $u=x-1/2$, I get something of the form $\int\frac{1}{\sqrt{u^2+1/4}}du$ (along with another integral, which I can integrate), but I don't know how to deal with this one.

答:
Apply another change of variables to get

$$\int\frac1{\sqrt{v^2+1}}dv,$$

and from there, try $v = \sinh x$. Use the following identities:

$$\frac{d}{dx}\sinh x = \cosh x\\
\cosh^2x - \sinh x^2 = 1$$

Everything should cancel out nicely; you'll end up with just $\int dx$, which is to say $\sinh^{-1}v$.


问:
How to show that $\mathbb{E}[X\mathbb{1}_{\lbrace X\le a \rbrace}]=\mathbb{P}(X\le a)$?
Let $X$ be a positive countinuous random variable. How do I show that $\mathbb{E}[X\mathbb{1}_{\lbrace X\le a \rbrace}]=\mathbb{P}(X\le a)$?
It seems that I have missed something important.

$\mathbb{E}[X\mathbb{1}_{\lbrace X\le a \rbrace}]=\int\limits_{-\infty}^{\infty}x\cdot f_{X\mathbb{1}_{\lbrace X\le a \rbrace}}(x)dx$ and i want to justify it equals to $\int\limits_{-\infty}^{a}x\cdot f_X(x)dx$.

答:
That is not quite correct.  It is that : $\mathsf E(\mathbf 1_{X\leqslant a})=\mathsf P(X\leqslant a)$

What $\mathsf E(X\,\mathbf 1_{X\leqslant a})$ equals is $\int\limits_0^a\mathsf P(y\lt X\leqslant a)\,\mathrm d y$

The key is that $X$ is a positive continuous random variable; meaning that what ever the support for $f$ may be, it does not intercept the negative reals.  So for any positive $a$...

$$\begin{align}\mathsf E(X\,\mathbf 1_{X\leqslant a})&=\int_0^a x~f(x)\,\mathrm d x\\&=\int_0^a \left(\int_0^x \mathrm d y\right)~f(x)~\mathrm d x\\&=\int_0^a\int_y^a f(x)\,\mathrm d x\,\mathrm d y\\&=\int_0^a\mathsf P(y\lt X\leqslant a)~\mathrm d y\end{align}$$


答:
Using $x=\sin^2 \theta$, then
\begin{align}
\int^1_0 \sqrt{\frac{x}{1-x}}\ dx =&\ \int^\frac{\pi}{2}_0\sqrt{\frac{\sin^2\theta}{1-\sin^2\theta}} d(\sin^2\theta)\\
=&\ \int^{\pi/2}_0 \tan\theta\ d(\sin^2\theta)\\
=&\ 2\int^{\pi/2}_0 \sin^2\theta\ d\theta = \frac{\pi}{2}
\end{align}

Title: Compute $\int_0^1\frac{x}{\sqrt{x(1-x)}}dx$

Body: I haven't integrated in a while... so I'm a bit rusty. I have to integrate $\int_0^1\frac{x}{\sqrt{x(1-x)}}dx$. Should I apply some change of variables? If I take $u=x-1/2$, I get something of the form $\int\frac{1}{\sqrt{u^2+1/4}}du$ (along with another integral, which I can integrate), but I don't know how to deal with this one.


问:
Finding angle in a triangle with one median and an isosceles triangle in it

In the following figure $AF=BD=DC$ and $AE=EF$. Find the angle $\alpha$.

I tried so many different things (constructing triangles, drawing parallel lines, using Ceva theorem in triangles $ABD$ and $CBE$, ...). Any help for solving this will be much appreciated.

A geomtrical solution (without trigonometry) would be very nice.

答:
Here is a purely geometric proof without any trigonometry.

Draw the three lines $a, \, b, \, c$ such that:
$$A \, \in \, a \,\,\text{ and } \,\, a \, || \, BC$$
$$B \, \in \, b \,\,\text{ and } \,\, b \, || \, CA$$
$$C \, \in \, c \,\,\text{ and } \,\, c \, || \, AB$$
Furthermore, let
$$B^* \, = \, a \, \cap \, c \,\,\, \text{ and } \,\,\, A^* = b \, \cap \, c$$
Then, $ABA^*C$ and $ABCB^*$  are parallelograms and furthermore
$$AB = CA^* = CB^*$$ as well as
$$AB^* = BC$$
Apply Menelaus' theorem to the triangle $BCE$ intersected by the line that passes through the collinear points $A, \, F,\, D$:
$$\frac{AB}{AE} \cdot \frac{EF}{CF} \cdot \frac{DC}{BD} = 1$$
Since by assumption $AE = EF$ and $BD = CD$, one concludes that
$$AB = CF$$
Combining this latter fact with the earlier conclusions, one observes that
$$CF = AB = CA^* = CB^*$$ which is possible if and only if triangle $\Delta \, A^*B^*F$ has $90^{\circ}$ angle at vertex $F$, i.e. $\angle \, A^*FB^* = 90^{\circ}$. But that means that:
$$\angle \, AFB^* = 180^{\circ} - \angle \, A^*FB^* = 180^{\circ} - 90^{\circ} = 90^{\circ}$$
Recall that by assumption $AF = BD = DC = \frac{1}{2}\,BC$. But by construction, $ABCB^*$ is a parallelogram and $AB^* = BC$ so
$$\Delta \, AFB^*  \, \text{ is a triangle with the properties that } \, \angle \, AFB^* = 90^{\circ} \,\, \text{ and } \,\, AF = \frac{1}{2}\, AB^* $$

which is possible if and only if $$\angle \, FAB^* = 60^{\circ}$$
However, $AB^*$ is parallel to $BC$ and consequently, by the properties of a pair of parallel lines intersected by a third line,
$$\angle \, ADB = \angle \, DAB^* = \angle \, FAB^* = 60^{\circ}$$


问:
Looking for reference on triangulating convex (or regular) $n$-gons
Looking for reference on triangulating convex (or regular) $n$-gons.

There are some basic interesting facts on them, like the number of triangulations of an $n$-gon is the $n-2$ Catalan number. I'm looking for a more in depth exposition, and google is failing me.

Thanks!

P.S. I have access to Springer's electronic database of books through my university

答: See Chapter 1 of the wonderful book Discrete and Computational Geometry by Devadoss and O'Rourke. That chapter is freely available.</p>


答:
Here is how I would attack the problem.

After you get the fraction, expand the numerator to get

$(1-c_i+c_i^2)^3=1-3c_i+6c_i^2-7c_i^3+6c_i^4-3c_i^5+c_i^6$

Now since $c_i^3=3c_i^2-1$, we may eliminate the sixth power using $c_i^6=3c_i^5-c_i^3$.  This reduces the numerator to $1-3c_i+6c_i^2-8c_i^2+6c_i^4$.  We similarly put in $c_i^4=3c_i^3-c_i$ to eliminate the fourth power, and after that $c_i^3=3c_i^2-1$ to remove the cubic term.  We finally end with the quadratic expression $-9-9c_i+36c_i^2$.

For the denominator we may put in $c_i^3=3c_i^2-1$ at the beginning to get $(-6c_i+3c_i^2)^2=36c_i^2-36c_i^3+9c_i^4$.  Eliminating the higher powers using a procedure similar to that above leads to $-36-36c_i+144c_i^2$.

So our fraction is $(-9-9c_i+36c_i^2)/(-36-36c_i+144c_i^2)$ where the denominator is exactly four times the numerator.

Title: Evaluating Large Expression Concerning the Roots of a Cubic

Body:
A few months ago I was doing a math competition and I came up with a very roundabout way of finding the answer. This ended up in me having to evaluate
$$r=\frac{(1-c_i+c_i^2)^3}{(1-6c_i+3c_i^2+c_i^3)^2}$$
where $c_i$ is any root of $z^3-3z^2+1$. Through approximating a root around 2 to 5 digits using a calculator, the answer came to be 1/4.

Is there any way to even begin evaluating this by hand?

Some information: If $z_i$ is a root of $z^3-6z^2+3z+1$, then $\frac{-z_i+2}{1-2z_i}=c_i$, so reversing the coefficients to reciprocate the roots could help. If $P(x)=z^3+1$, then $z^3-3z^2+1=P(x)-P'(x)$, which is probably not helpful.

I suppose that you could assume that the value of $r$ could be found by assuming that it is independent on which root of $P$ you pick, and then add/multiply the equal values of $r$, and then expand and use Vieta's, but that sounds unlikely. I'm betting that there's probably no easy way, but maybe there is.


问:
How to prove Big O, Omega and Theta asymptotic notations?
I know the definitions of this notations

Big $\mathcal{O} \; : \enspace T(n) \in \mathcal{O}(f(n))$ if and only if $∃ \, c, n_0$, such that $T(n) \leq c \cdot f(n) \enspace \forall \,n \geq n_0$.

Big $\Omega \; : \enspace T(n) \in \Omega(f(n))$ if and only if $∃ \, c, n_0$, such that $T(n) \geq c \cdot f(n) \enspace \forall \,n \geq n_0$.

Big $\Theta \; : \enspace T(n) \in \Theta(f(n))$ if and only if $∃ \, c_1, c_2, n_0$ such that $c_1 f(n) \leq T(n) \leq c_2 f(n) \enspace$ $\forall \,n \geq n_0$.

I'm having trouble manipulating this definitions to prove some notation for example:

Suppose that $f = \Theta(g)$ and $g = \Theta(h)$.
Prove that $f = \Theta(h)$

Suppose that $f$ and $g$ are two non-negative functions such that  $g = \mathcal{O}(f)$.
Prove that $f + g = \Theta(f)$.

Is there anything that could help me? I've read, watch videos but nothing helps me clarifying to prove these notations. Anything helps thanks. Also if you're a tutor and there's a way to set up a meeting it would help.

答:
From scratch then:

Consider some arbitrary function $f$ and the "Big O" of $f$, i.e. $\mathcal{O}(f)$. This $\mathcal{O}(f)$ is a set, consisting of all functions that satisfy a certain condition. This condition somehow includes $f$. You have already given the definition of $\mathcal{O}(f)$, but I will rewrite it now:

$$\mathcal{O}(f) \enspace = \enspace \Big\{ \; g : \mathbb{N} \longrightarrow \mathbb{R} \; \Big| \; \exists \, c > 0 \; \exists \, n_0 \in \mathbb{N} \; \forall \, n \geq n_0 \; : \; g(n) \leq c \cdot f(n) \; \Big\}$$

In other words, the set $\mathcal{O}(f)$ is a set of functions who map from the natural numbers $\mathbb{N}$ to the real numbers $\mathbb{R}$. These functions have to satisfy a certain condition, namely that there is some constant $c$, which is greater than zero, such that $g(n) \leq c \cdot f(n)$. This inequality does not have to hold for every function value of $g(n)$ and $f(n)$. It only has to hold for all function values after a certain "point" $n_0$. Let us look at an example:

Example:
Consider the functions

$$g(n) = n \qquad \text{and} \qquad f(n) = n^2 - n$$

We want to show that $g \in \mathcal{O}(f)$ holds. If we insert $n=1$, we find

$$g(1) = 1 \qquad \text{and} \qquad f(1) = 0$$

But there exists no constant $c > 0$, such that $1 \leq c \cdot 0$. Does this mean, that $g \notin \mathcal{O}(f)$? The answer is NO. Why? Because if we insert $n=2$ or $n=3$ or $n = 4,5,6, \ldots$ we find that for these values the inequality

$$ g(n) \leq c \cdot f(n) $$

is fulfilled when e.g. picking $c = 1$. This means, that when $c = 1$ then for all $n$ that are greater than $n_0 = 1$ we find $g(n) \leq c \cdot f(n)$. However, this is exactly the definition of $\mathcal{O}(f)$. Therefore,

$$g(n) \in \enspace \mathcal{O}(n^2 - n)$$

${}$

Problem 2

Consider now the second problem you have stated. We want to prove that if $g \in \mathcal{O}(f)$ then $f + g \in \Theta(f)$.

How do we do this? First of all, we gather the information we have. In this case, we know that $g \in \mathcal{O}(f)$. By definition, we know that there exists a $c>0$ and a $n_0$ such that for all $n \geq n_0$ the inequality $g(n) \leq c \cdot f(n)$ holds.

Now we simply add $f(n)$ on both sides of the inequality (which we know by assumption to be true). This gives us

$$g(n) + f(n) \leq (c+1) \cdot f(n)$$

This was the first part. Now the second part: $g(n)$ is nonnegative, so we know that $g(n) \geq 0$ and therefore

$$f(n) \leq f(n) + g(n)$$

(because adding something positive to a number always makes the number greater).

We combine these two results and have

$$ f(n) \; \leq \; f(n) + g(n) \; \leq \; (c+1) f(n)$$

This completes the proof that $(f + g) \in \Theta(f)$.


问:
Determine the following subsets of R3 are subspaces of R3. If so, give a proof, and if not, explain why.
U1={(x, y, 1)|x, y are real numbers}

U2={(a, a - b, b)|a, b are real numbers}

U3={(r, s, t)|r, s, t are non-negative numbers}

For U1:

According to the Zero Vector: $\lt x, y, 1\gt \neq \lt 0, 0, 0\gt$

According to Under Addition: $\lt 0, 0, 1\gt + \lt 0, 0, 1\gt = \lt 0, 0, 2\gt$ Not a part of the set

This is not a subspace of $\mathbb{R}^3$

For U2:

According to the Zero Vector: $\lt a, a-b, b\gt = \lt 0, 0, 0\gt$

According to Under Addition $\lt 0, 0, 0\gt + \lt 0, 0, 0\gt = \lt 0, 0, 0\gt$

This is a subspace of $\mathbb{R}^3$

With U3 however, (0,0,0) can work, but in order for the statement to be correct r, s, & t would all have to be greater than or equal to 0. I can use the obvious Zero Vector method, but what about Under Addition or Under Multiplication?

答:
Linear algebra can be a bit tricky when it first comes to learning subsets, but if you understand the axiomatic approach, then it becomes much clearer.

As stated in the comments, a subset $U⊆R^3$ will be a subspace if $a) 0∈U$, and b) whenever $x∈U$ and $y∈U$, we have $x+y∈U$ and c) whenever $λ∈R$ and $x∈U$, we have $λx∈U$.

In other words, $U$ must be equal to trivial solution of $(0,0,0)$ since its in $R^3$. And there must be closure under multiplication and addition by the other two axioms.

All three axioms must be true for $U$ to be a subspace. If you notice that the trivial solution $(0,0,0)$ is not equal to $U$, since $U$ = {$(x,y,1)$ for some $x,y$ real numbers}. Meaning, no $x$ and $y$ can make $U$ = {$(0,0,0)$} And so, $U$ is not a subspace of $R^3$.

After checking the first axiom, check for closure.

Closure under addition states that $m,n∈W$, then $m+n∈W$.

Closure under multiplication states that $m,n∈W$, then $mn∈W$.

By closure under multiplication, you want to see if scalar multiples of $m,n$ will be elements of $W$. So, if any scalar $c$ is multiplied by $m,n$, then you need to check if $cm$ or $cn$ will satisfy the conditions of the set $W$. So, if $W$ has the condition that elements must have non-negative numbers, then it won't be closed under multiplication since $c$ can be any scalar, so it could be negative. And so, multiplication of elements of $W$ with a negative $c$ value will give you a negative number, not satisfying conditions of $W$.

To see if $m+n∈W$ and $mn∈W$ hold, you must check to see the conditions of the set $W$. So, if $W$ has the condition that $m,n∈W$, such that $m$ and $n$ are real numbers, then you need to check if $m+n∈W$ and $mn∈W$ will give real numbers too and thus, be elements of $W$.

Hope this helps you get started. (I haven't done linear algebra in a while so if any one notices any error on my part, then feel free too comment or imporve the post)


问:
Integral inequality, $f$ continuous, increasing function
Let $f$ be a continuous, increasing  function on $[a,b]$, $c$ is the middle of $[a,b]$.

Prove that $\frac{f(a)+f(c)}{2} \le \frac{1}{b-a} \cdot \int _a ^b f(x)dx \le \frac{f(b)+f(c)}{2} $ .

Could you help me with that?

I thought I may use intermediate value theorem, but I didn't come up with anything.

答:
We have
$$\int_a^b f(x)dx=\int_a^c f(x)dx+\int_c^b f(x)dx$$
and since $f$ is increasing function then
$$\int_a^cf(a)dx=\frac{f(a)(b-a)}{2}\leq\int_a^cf(x)dx\leq\int_a^cf(c)dx=\frac{f(c)(b-a)}{2}$$
and
$$\int_c^bf(c)dx=\frac{f(c)(b-a)}{2}\leq\int_c^bf(x)dx\leq\int_c^bf(b)dx=\frac{f(b)(b-a)}{2}$$
Now add the two inequalities.</p>


答: You can do this from scratch: Change $\delta$ to $x$ for clarity. Fix $k\neq 0.$ Suppose $\underset{x \rightarrow 0}{\lim} f(x)=L$. Then, let $\epsilon>0.$ There is a  $\delta>0$ so that if $|x|<\delta$ then $|f(x)-L|<\epsilon.$ Suppose such a $\delta$ has been produced and that  $|kx|<|k|\delta.$  Set $\delta'=|k|\delta$. Then, if $|kx|<\delta'$ we have  $|kx|<|k|\delta\Rightarrow |x|<\delta\Rightarrow |f(x)-L|<\epsilon$. So, with $\delta'$ answering the $\epsilon$-challenge, we have by definition, $\underset{kx \rightarrow 0}{\lim} f(x)=L$.

Title: Can we always replace $\underset{\delta\rightarrow 0}{\lim}$ with $\underset{k\delta\rightarrow 0}{\lim}$ for $k$ constant?

Body:
Is it in general acceptable to  replace $\underset{\delta\rightarrow 0}{\lim}$ with $\underset{k\delta\rightarrow 0}{\lim}$ for constant $k$?

What I mean by this is, if we have a real-valued function $f$  and we know that, say $\underset{\delta \rightarrow 0}{\lim} f(\delta)=L$ for real constant $k$, is it then valid to say that  $$\underset{\delta \rightarrow 0}{\lim} f(k \delta)=\underset{k\delta \rightarrow 0}{\lim} f(k \delta)=\underset{\delta \rightarrow 0}{\lim} f(\delta)=L\:?$$

(I feel like this should always be the case, because for any constant $k$, $(\delta\rightarrow 0) \Longleftrightarrow (k\delta \rightarrow 0)$ but maybe I'm overlooking something)


问:
a real convergent sequence has a unique limit point
How to show that a real convergent sequence has a unique limit point viz. the limit of the sequene?

I've used the result several times but I don't know how to prove it!

Please help me!

答: Hint: Suppose that a sequence $\{x_n\}$ has two different limits $l_1$ and $l_2$. There exists a number $\varepsilon>0$ such that $2\varepsilon<|l_1-l_2|$. Then $\varepsilon$-neighborhoods of the points $l_1$ and $l_2$ are disjoint. Therefore, for sufficiently large $n$ we have that ...</p>


问:
Why is $\cos\theta$ used in three-dimensional geometry in place of $\tan\theta$ or $\sin\theta$?
While going through the introduction of 3D geometry, I'm finding the discussion starts with $\cos\theta$ called "direction cosines".

My query is, in 2D geometry, we find simpler and clear to understand equations of a line as $y= mx+ c$, so

Why all of a sudden we introduce $\cos\theta$ and also take help of a parallel vector which is not a part of the line?

Why do we depend on a vector introduction? Can not the equation of a line be derived using its own points?

Why can not it be used with $\tan\theta$ or $\sin\theta$ notations?

Why is it that the equation $ax+by+cz+ d=0$, which looks like the equation of a straight line, is the equation of a plane?

Due to all this 3D happens to be inaccessible to understand in the introduction itself. Why it becomes so complex?

Can 2D geometry be done using cosine or sine functions? or is $\tan\theta$ fixed for it, and $\cos\theta$ is fixed for 3D geometry?

Thanks.

答:
I count six questions there.  Taking #4, the reason $ax + by + cz = d$ is a plane is because we are in three dimensions. Each linear equation is a constraint, and reduces the dimension of the space by one.  So when you are in two dimensions, one equation reduces it to a line.  When you are in three dimensions, one equation reduces it to a plane.  In higher dimensions one linear equation restricts to an $n-1$ dimensional subspace, and is called a "hyperplane".

If you start with the equation of a line in two dimensions, say $y = 2x + 3$, if you move to three dimensions and write the same equation, that equation has no restriction on z.  So if you take the line in the x-y plane in 3D and move it directly up and down to different values of z, you will get a plane parallel to the z axis.  Tilted planes require the more general $ax + by + cz = d$.


问:
Is there an method similar to FFT for Fractionally Iterated Fourier Transforms?
FFT is one of the 20th Century's greatest inventions, running as $O(n \log(n))$ rather than as $O(n^2)$ as a simple implementation of a discrete Fourier transform would.  But what about half-order Fourier transforms, or arbitrary real order iterated Fourier transforms?

Literature on this topic is quite sparse.

答:
After a little Google searching. I found a webpage
The Fractional Fourier Transform
about a book which seems to be very comprehensive and useful. The website states:

Little need be said of the importance and ubiquity of the ordinary Fourier transform in many areas of science and engineering. As a generalization of the ordinary Fourier transform, the fractional Fourier transform is only richer in theory and more flexible in applications--but not more costly in implementation. This book provides a comprehensive and widely accessible account of the transform covering both theory and applications.

Also on the webpage is a link to
Matlab code for fast computation of the fractional Fourier transform
which may be what you are looking for.


问:
solving "Cascaded measurements are single measurements"

This is an exercise from p. 86 Quantum Computation and Quantum Information by Michael A. Nielsen and Isaac L. Chuang.

Here I think I have to show that $p\left(l\right)p\left(m\right)=p\left(lm\right)$, meaning the probability of getting $l$ and then getting m is equal to the probability of getting $lm$. But I can't prove it... The definition of $p\left(m\right)$ (same for $p\left(l\right)$ of course) is like below.

Could anyone help me how to prove $p\left(l\right)p\left(m\right)=p\left(lm\right)$?

答:
Using ${L_l}$ to measure $|\varphi\rangle$, the probability of resulting $l$ is $\langle{\varphi} |L_l^{\dagger}L_l|{\varphi}\rangle$, and this will make the state collapse to $\frac {L_l|\varphi\rangle}{\sqrt {\langle{\varphi} |{L_l}^{\dagger} L_l|{\varphi}\rangle}}$.

Using ${M_m}$ to measure $\frac {L_l|\varphi\rangle}{\sqrt {\langle{\varphi} |L_l^{\dagger}L_l|{\varphi}\rangle}}$, the probability of resulting $m$ is $\frac {\langle{\varphi} |M_m^{\dagger} L_l^{\dagger}L_l M_m|{\varphi}\rangle}{\langle{\varphi} |L_l^{\dagger}L_l|{\varphi}\rangle}$, and this will make the state collapse to $\frac {L_l M_m|{\varphi}\rangle}{\langle{\varphi} |M_m^{\dagger} L_l^{\dagger}L_lM_m|{\varphi}\rangle}$.

Easy to verify this consecutive process is equivalent to $M_mL_l$


问:
Two circles inscribed in a semicircle
Two circles are inscribed in a semi circle. Given the areas of the shaded triangles, what's the radius of the semicircle?
(Note there's a similar but different question here)

答:
Let $EM=a$ be tangential to both circles at the point $E$ and the angle $\theta$. Then, the area $[ABE] = 33= a^2\sin\theta$ and

\begin{align}
[ABCD]&=153 =[AOD]+[BOC]=[COD] \\
& = \frac12Rr_1\cos\alpha + \frac12Rr_2\cos\beta + \frac12R^2\sin(\alpha+\beta)
\end{align}
Substitute $\sin\alpha = \frac{r_1}{R-r_1}$, $\sin\beta= \frac{r_2}{R-r_2}$ into above equation and, after simplifing
\begin{align}
153=aR\left( \frac1{(1-\frac{r_1}{R})(1-\frac{r_2}{R})}-1\right)\tag1
\end{align}

Note that $r_1 = a \tan\frac{\theta}2$, $r_2 = a \cot\frac{\theta}2$ and $AB=AO +OB$, or
\begin{align}
2a & = \sqrt{(R-r_1)^2-r_1^2} + \sqrt{(R-r_2)^2-r_2^2}
= \sqrt{R^2-2aR\tan\frac{\theta}2}+ \sqrt{R^2-2aR\cot\frac{\theta}2}
\end{align}
which reduces to
$$R = \frac a{\sqrt2-\csc\theta}$$

Then, along with $a^2=33\csc\theta$, substitute $r_1$, $r_2$ and $R$ derived above into (1) to obtain the equation for $\theta$
$$\frac{153}{33} = \frac{\csc\theta}{\sqrt2-\csc\theta}\left(\frac1{3-4\sqrt2\csc\theta+3\csc^2\theta} -1\right)
$$
Solve to get the valid solution $\csc\theta = \frac{17}{12\sqrt2}$ and, in turn, the radius

$$R = \frac{\sqrt{33\csc\theta}}{\sqrt2-\csc\theta} = \frac67\sqrt{187\sqrt2}$$

Title: Two circles inscribed in a semicircle


问:
$\ln k$ where $k>2$
Let $k>2$, is a natural number, and also it's true $\ln k=\int_{1}^{k}\frac{dt}{t}=\int_{1}^{2}\frac{dt}{t}+\int_{2}^{3}\frac{dt}{t}...\int_{k-1}^{k}\frac{dt}{t}$. Then, is $\ln k-\sum_{n=2}^{k}\frac{1}{n}<\sum_{n=2}^{k}\frac{1}{n^2-n}$?For $lnk$ we can use the formula, from before.  Then for the right side, we can use partial fractions? and seperate $n^2,n$. So it's true.

Next questions, is $\lim_{k \to \infty}[\ln k-\sum_{n=2}^{k}\frac{1}{n}]$ exists? So this is calculating for a sequence and $lnk$. But $\frac{1}{n}$ is always diverge, when $k=\infty$, so I think no, it's not exists.

答:
You have

$\begin{array}\\
\ln k-\sum_{n=2}^{k}\dfrac{1}{n}
&\lt \sum_{n=2}^{k}\dfrac{1}{n^2-n}\\
&= \sum_{n=2}^{k}\left(\dfrac{1}{n-1}-\dfrac{1}{n}\right)\\
&=1-\dfrac1{k}\\
&\lt 1\\
\end{array}
$

so the difference is bounded.

By being a little more careful,
you can show that
the limit of the difference exists.
This is called the
Euler-Mascheroni constant,
denoted by 1-$\gamma$,
with a value of about
1-0.57721.

See https://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant
for more information.

Title: $\ln k$ where $k>2$


问:
Evaluate $\lim_{x \to 0}S(x)$
Let the function $S(x)$ be a sum, such that $S(x)=(\sum_{n=0}^{\infty}x^n)^2$ and $|x|<1$. Question is, does $\lim_{x \to 0}S(x)=0$? This is the equivalent of taking the limit of the infinite sum. Since we know that $|x|<1$, the infinite sum can only be decreasing. As $n$ increases. However, this does not mean that the limit is 0. In fact, it should be a positive number. So I think the limit equation is false.

答:
First find the infinite sum. $1+x+x^2 ...$ is a GP with common ratio $x$... So infinite sum is $\dfrac{1}{1-x}$ and lim $x$ goes to zero $\dfrac{1}{(1-x)^2}$ is one.

1.

Title: Evaluate $\lim_{x \to 0}S(x)$


问:
Is it meaningful to define the Dirac delta function as infinity at zero?
I am in discussion with someone online on the subject of the Dirac delta function. This other person wants to say:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \infty & : x = 0 \end{cases}$$

and wants to justify it by saying:

We have that:

$$\delta (x) = \lim_{\epsilon \mathop \to 0} F_\epsilon (x)$$

where:

$$F_\epsilon (x) = \begin {cases} 0 & : x < -\epsilon \\ \dfrac 1 {2 \epsilon} & : -\epsilon \le x \le \epsilon \\ 0 & : x > \epsilon \end {cases}$$

Therefore:

$$\delta (0) = \lim_{\epsilon \mathop \to 0} F_\epsilon (0) = \dfrac 1 {2 \times 0} = \infty$$

and:

$$ \delta {x \ne 0} = \lim_{\epsilon \mathop \to 0} F_\epsilon (x \ne 0) = 0$$

Therefore:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \infty & : x = 0 \end{cases}$$

This comes across as iffy to me. I don't trust $\delta (0) = \infty$ as it is not well-defined exactly what $\infty$ actually means in this context, whereas in fact $\delta$ is precisely defined by means of the limit definition as given above.

I am also seriously unsure about that $\dfrac 1 {2 \times 0}$ in the middle of the exposition, which is at best meaningless and at worst a blasphemous lie.

However, when I consult a number of mathematical works on my shelves and online, there are many of them which give that above definition quite casually, along the lines "As an obvious consequence of the definition: $\delta (0) = \infty$" or some such. Other works heavily italicise the warning that $\delta$ is not a function, and $\delta (0)$ is undefined.

What is the current mode of thought here? I am trying to craft a set of webpages which are mathematically rigorous, but coworkers on the same site are of the mind "It doesn't really matter, we all sort of know what we mean, and hey, $\delta (0) = \infty$ looks really cool!"

When questioning the matter, he is prepared to compromise and say: $\delta (0) \to \infty$ as $x \to 0$, or even:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \to \infty & : x = 0 \end{cases}$$

I am insufficiently mathematically sophisticated as to be able to explain why I hate this approach, but I seriously dislike bandying the $\infty$ sign around, when (once we get past the obvious intuitive meaning we learn in elementary school) we really don't understand what it means.

EDIT:

As requested, I have actually dug up one of my texts which defines the delta function as the limit of the rectangle function as its width goes to zero, as follows. It comes from I.N. Sneddon's "Special Functions of Mathematical Physics and Chemistry", appendix.

If we consider the function
$$\delta_a (x) = \begin {cases} 0 & : |x| > a \\ \dfrac 1 {2 a} & : |x| < a \end {cases}$$
then it is readily shown that
$$\int_{-\infty}^\infty \delta_a(x) d x = 1.$$

...

We now define
$$\delta(x) = \lim_{a \to 0} \delta_a (x).$$
Letting $a$ tend to zero in equations [above] we see that $\delta(x)$ satisfies the relations
$\delta(x) = 0$, if $x = 0$
$$\int_{-\infty}^\infty \delta (x) d x = 1.$$

答:
I think the confusion here may be due to the difference between definition and description.

A good definition of the Dirac delta function is "the function/distribution $\delta(x)$ that satisfies $\delta(x)=0$ for $x\neq0$ and $\int_a^b \delta(x)\ dx = 1$ if $a<0<b$." - one may also define it as a limit of functions.

When describing this function, it is reasonable to state that the value at $x=0$ can be thought of as infinite. And thus, as a description, it is reasonable to say
$$
\delta(x)=\begin{cases} \infty & :x=0\\0 & :x\neq 0 \end{cases}
$$

This allows one to understand that the value is zero away from $x=0$, but at $x=0$, in order for the function to have its other properties, the value at $x=0$ must be infinite. However, this is not sufficient to operate as a definition, as it does not establish the properties that make it the delta function. Indeed, as "infinity" is not, in itself, a well-defined value, the description isn't a definition because it isn't well-defined.

It is conceivable to turn the description into a proper definition, of sorts, if one can somehow introduce the concept of the "special infinity" that has the necessary properties to make the description work as a definition. For example, one might plausibly define $\infty_{dx}$ as an infinite number that can be thought of as the inverse of the width $dx$ that appears in the integration symbol. It would require more careful definition, but with the right structure of the definition (and likely, using some nonstandard analysis), it would become a reasonable concept. And then the concept of $2\infty_{dx}$ is perfectly sane as the value of $2\delta(0)$

But that would be like defining $a\times b=e^{\ln(a)+\ln(b)}$ as the base definition of multiplication (rather than merely an example of a property that can be found). Sure, it's workable, but it's also an overcomplicated solution to a simple problem.


问:
System of ODE's - Positivity on each coordinate
I have a system of ODE's

$$x'(t)=f(x,y,t)\\y'(t)=g(x,y,t)$$

s.t. $f(0,y,t)\equiv 0$ and for an initial condition $x(0)=a>0$, $y(0)=b$ I have solution $x_1(t),y_1(t)$.

Could $x_1(t)$ be negative on some $t^*$?

I think no, because taking $x\equiv 0$ the system becomes

$$y'(t)=g(0,y,t)$$

Let $y_2(t)$ be the solution of this least equation s.t. $y_2(t^{**})=y_1(t^*)$ for some $t^{**}$. So, $x_2(t)\equiv 0, y_2(t)$ is a solution to the system and contains point $(0,y_1(t^*))=(x_1(t^*),y_1(t^*))$, contradiction.

Is it correct?

Thank  you so much.

答: Take $f(x,y,t)=-x^{2/3}t$ and $g(x,y,t)=0$. A solution to your system is $$x(t)=\Bigg(-\frac{t^2}{6}+a^{1/3}\Bigg)^3$$ $$y(t)=b$$ As you can see $x(t)$ will eventually assume negative values.


问:
Probabilities in a game
John plays a game with a die. The game is as follows: He rolls a fair six-sided die. If the roll is a $1$ or $2$, he gets $0$ dollars. If he rolls a $3, 4$ or $5$, he gets $5$ dollars. If he rolls a $6$, he wins $X$ dollars where $X$ is a continuous random variable that is uniform on the interval $(10, 30)$. Let $Y$ be the amount of money John wins by playing the game.

(i) Compute the cumulative distribution function (cdf) of $Y$.

(ii) What is the probability that John rolled a $6$ given that he won less than $15$ dollars?

(iii) Compute $E(Y)$.

My attempt:

(i) If $Y > 5$ then $P(Y \leq y) = P(X \leq y)$. Then $P(Y \leq 5) = P(Y = 0) + P(Y = 5) = P(roll = 1 or 2) + P(roll = 3, 4 or 5) = \frac{1}{3} + \frac{1}{2} = \frac{5}{6}$.

Not sure if this is even correct.

(ii) $P(roll = 6 | Y < 15) = \frac{P(roll = 6 \land Y < 15)}{P(Y<15)}$

But by Bayes theorem $P(roll = 6 | Y < 15) = \frac{P(Y < 15 | roll = 6) \cdot P(roll = 6)}{P(Y<15)}$

We know $P(roll = 6) = \frac{1}{6}$.

Given that he rolled a $6$, the probability that he won less than $15$ dollars is equal to $P(X < 15) = \frac{15 - 10}{30 - 10} = \frac{1}{4}$

Lastly, $P(Y < 15) = 1 - P(Y \geq 15)$. However, since the only situation in which John wins $15$ dollars or more is if he rolls a $6$. We may infer that $P(Y \geq 15) = P(X \geq 15) = 1 - P(X < 15) = \frac{3}{4}$. So $P(Y < 15) = 1 - \frac{3}{4} = \frac{1}{4}$.

So $P(roll = 6 | Y < 15) = \frac{\frac{1}{4} \cdot \frac{1}{6}}{\frac{1}{4}} = \frac{1}{6}$

Not sure how to determine the other probabilities.

(iii) We can check $Y$ conditioned on $X = x$: $E(Y|X=x) = (0 \cdot \frac{1}{3}) + (5 \cdot \frac{1}{2}) + (x \cdot \frac{1}{6}) = \frac{5}{2} + \frac{x}{6} = \frac{x + 15}{6}$. But $E(E(Y|X)) = E(Y)$ by the law of iterated expectation.

Applying it we have $E(Y) = E(E(Y|X)) = E(\frac{X + 15}{6}) = \frac{1}{6} E(X) + \frac{15}{6}$

Since $X$ is uniform on $(10, 30)$ we know $E(X) = \frac{1}{2} (10 + 30) = 20$

Therefore $E(Y) = \frac{35}{6}$.

Is this correct? I have a feeling that my attempt for (i) is incorrect, and I am unsure about (ii) and (iii). Any assistance is much appreciated.

答:
Here is a simulation of a million iterations of your
process. Possibly except for conditional probabilities, it should
be accurate to a couple of decimal places. Also the ECDF
plot of from the simulation is very nearly the CDF of $Y,$
which you have not drawn.

```<code>set.seed(2021)
die =sample(1:6, 10^5, rep=T)
y = runif(10^5, 10, 30)
y[die<=2.1] = 0
y[die>=2.9 & die <=5.1] = 5
[1] 5.838549              # aprx E(Y)
35/6
[1] 5.833333              # exact
mean(die[y<15]==6)
[1] 0.04736204            # aprx P(Die = 6 | Y < 15)

hdr = "ECDF: Mixture of Discrete and Continuous"
plot(ecdf(y), ylab="ECDF", xlab="y", lwd=2, main=hdr)
</code>```

At the 'jumps', the CDF takes the upper value.

Title: Probabilities in a game

Body:
Here is a simulation of a million iterations of your
process. Possibly except for conditional probabilities, it should
be accurate to a couple of decimal places. Also the ECDF
plot of from the simulation is very nearly the CDF of $Y,$
which you have not drawn.

set.seed(2021)
die =sample(1:6, 10^5, rep=T)
y = runif(10^5, 10, 30)
y[die<=2.1] = 0
y[die>=2.9 & die <=5.1] = 5
[1] 5.838549              # aprx E(Y)
35/6
[1] 5.833333              # exact
mean(die[y<15]==6)
[1] 0.04736204            # aprx P(Die = 6 | Y < 15)

hdr = "ECDF: Mixture of Discrete and Continuous"
plot(ecdf(y), ylab="ECDF", xlab="y", lwd=2, main=hdr)

At the 'jumps', the CDF takes the upper value.


问:
Bending invariants in higher dimensions
The determinant of the Gauss map of an embedded surface in 3 space is a bending invariant of the surface.

What about the determinant of the Gauss map for hypersurfaces of n space?

How does one obtain all of the bending invariants of the hypersurface from the Gauss map?

答: I assume what you are interested in is the product of the principal curvatures for hypersurfaces in Euclidean $n$-space for $n\geq 4$.  This quantity is not a Riemannian invariant, i.e. it is not preserved by isometries.  When you say that you want to obtain all the bending invariants from the Gauss map, do you mean sectional curvature?  If so, in each two-dimensional direction determined by a pair of principal directions, the sectional curvature will be the product of the two principal curvatures, just as for Gaussian curvature.</p>

Title: Bending invariants in higher dimensions


问:
How to solve this ODE $(x^3+y^3)\,dx=3x^2y\,dy$
Is there a good way to solve this ODE $$(x^3+y^3)\,dx=3x^2y\,dy$$

I work it until $$\frac{dy}{dx}=\frac13\left(\frac xy+ \left(\frac yx\right)^2\right)$$

and let $u=\frac{y}{x}$

I cannot go on.... It seems difficult to integrate.

Thanks ahead.

答:
$u =\dfrac{y}{x} \implies y = ux \implies y' = u + xu' \implies 3u + 3xu' = \dfrac{1}{u} + u^2\implies 3x\dfrac{du}{dx} = u^2+\dfrac{1}{u}-3u= \dfrac{u^3-3u^2+1}{u}\implies \displaystyle \int \dfrac{udu}{u^3-3u^2+1}= \displaystyle \int \dfrac{dx}{3x} \implies \displaystyle \sum_{\{c: c^3-3c^2+1 = 0\}} \dfrac{\log (u-c)}{c-2}= \ln |x|+K, K:$ constant according to Wolfram Alpha.

Note: If this is a math model to solve an applications, the best method would be the Euler's method.


问:
Inviting 5 friends out of 11 to dinner, with restriction
I'm revising for finals and I have come across this following question:

a) A woman has 11 close friends. Find the number of ways she can invite 5 of them to dinner.

b) Repeat a) but 2 of the friends are married and will not attend separately.

For a) I got ${11 \choose 5} = \frac{11!}{(11-5)! \cdot 5!} = 462$.

I'm completely lost on b). I tried $n-2$ (minus two friends that are married) and then:
${9 \choose 5} = \frac{9!}{(9-5)! \cdot 5!} = 126$, but I'm pretty sure this is wrong,

Could someone explain b) to me?

Thanks in advance!

答:
Originally, the possible invitees consisted of 11 units (each is one friend)
Now,since two of theme are married and will not be attending separately,
the possible invitees consist of 10 units (9 corresponding to one friend, and the last unit is the couple.)
Hence the no, of ways we can select 5 from 10 units is:
10C5, i.e., 210.</p>


问:
Suppose $f:X\to Y$ is continuous at $a$. Can I use $a$ in my $\epsilon-\delta$ proof?
For example, is the following proof valid?

Assume $f:\mathbb{R}\to\mathbb{R}$ is continuous everywhere. Let $a\in\mathbb{R}$ be a designated point of continuity. Let $\epsilon>0$ and find $\delta>0$ such that $|x-a|<\delta$ implies $|f(x)-f(a)|<\min(\epsilon,\frac{\epsilon}{\epsilon+2|f(a)|})$, then
\begin{align*}
|f(x)-f(a)|&<\frac{\epsilon}{|f(x)|+|f(a)|}\hspace{15 mm} b/c\hspace{5 mm} |f(x)|<\epsilon+|f(a)|\\
&\leq\frac{\epsilon}{|f(x)+f(a)|}
\end{align*}
Therefore, $|f^2(x)-f^2(a)|<\epsilon$ and so $f^2$ is continuous on $\mathbb{R}$.

I believe this is good because $a$ is chosen arbitrarily from the outset. I just don't like it. If it is indeed valid, maybe you can help convince me it's not so bad?

答:
Yes, your proof is fine, although your approach seems a little bit convoluted. What I mean by that is that one would start estimating the quantity $\vert f^2(x) - f^2(a) \vert$ and only then see what constants you need to get the correct $\delta-\varepsilon$ estimate. For example, I would approach this problem in the following way. We have that $\vert f^2(x) - f^2(a) \vert = \vert f(x) - f(a) \vert \vert f(x) + f(a) \vert \leq \vert f(x) - f(a) \vert (\vert f(x) \vert + \vert f(a) \vert).$ Let $0 < \varepsilon < 1.$ Since $f$ is continuous at $a$, let $\delta_\varepsilon > 0$ such that $\vert f(x) - f(a) \vert \leq \min\{\frac{\varepsilon}{2 \vert f(a) \vert + 1}, 1\}$ whenever $\vert x - a \vert < \delta_\varepsilon$. Then from the above estimate we infer that $\vert f^2(x) - f^2(a) \vert \leq \frac{\varepsilon}{2 \vert f(a) \vert + 1} (2 \vert f(a) \vert + 1) = \varepsilon$ whenever $\vert x - a \vert < \delta_\varepsilon.$ Therefore, $f^2(\cdot)$ is continuous.

Note that as I said in the beginning of the answer, your proof is also correct, so I am not debating it, rather just sharing what is in my opinion a more natural way to get your message across. After all, it's just a matter of taste. Finally, in regards to your original question, when you are interested in proving the continuity of a function $f$ at a certain point $a,$ you are allowed to use the point $a$ when choosing the $\delta_\varepsilon.$ Similarly, you may use it to assert continuity by replacing $\varepsilon$ with a function involving $\varepsilon$ and $a$ as long as you make sure that you can make the desired bound go to $0$ as this is your ultimate goal. That is because $a$ and $\varepsilon$ are both fixed, so you can play around with them to make your proof look neat. However, if you are trying to prove uniform continuity of a function for example, you are not allowed to use any points in your argument since you are interested in a uniform distance between any two points that allow you to make the distance between their images arbitrarily small. I hope this helps. :)


问:
How to count partitioned subsets of a binomial coefficient
I have been reading Dr. Carl Wagner's book Basic Combinatorics, and I cannot wrap my head around a particular theorem. Someone else has asked a very similar question and accepted an answer, but I'm afraid I still don't understand.

As stated, the theorem runs thus:

For all $ n, \; k \in \mathbb{N} $
$ \displaystyle \sum_{j = 0}^{n} \binom{j}{k} = \displaystyle \sum_{j = k}^{n} \binom{j}{k} = \binom{n + 1}{k + 1} $

There follows a proof of the $\binom{n + 1}{k + 1} $ bit using some handy substitutions. What confuses me is the next part:

#2. Clearly $ \binom{n+1}{k+1} $ counts the class of all (k + 1)-element subsets of [n + 1]. But this class of k+1 subsets may be partitioned into subclasses corresponding to j = k, k + 1, . . . , n as follows.

The subclass of subsets with largest element equal to k + 1 is counted by $ \binom{k}{k} $ (why?), the subclass of subsets with largest element equal to k + 2 is counted by $ \binom{k+1}{k} $ (why?), . . ., and the subclass of subsets with largest element equal to n + 1 is counted by $ \binom{n}{k} $ (why?).

I don't understand why at all! Indeed, it makes no sense to me.

It seems to me that the number of subsets of [n+1] with largest element k+1 would be $\binom{k+1}{k+1}$. Sure, that's the same as $ \binom{k}{k} $  since they both equal one, but that just seems to be a coincidence. We're choosing k+1 numbers from a set that runs from 1 to k+1. There can, naturally, be only one set that fits that bill. Yet to use $ \binom{k}{k} $ to represent that one set seems confusing to me, and every example that follows makes less sense.

What am I missing?

答: The largest element of a set has to be in the set.  If you specify that $j+1$ is the largest element of a $(k+1)$-subset, it means that you have only $j$ choices $\{1,\dots,j\}$ for the remaining $k$ elements.  The three examples given correspond to $j=k$, $j=k+1$, and $j=n$.


问:
How to establish inequality $U_{n+1} \leq U_{n}+\frac{1}{2^{^n}}$
$\begin{cases}
U_{1} = 1
      \\
U_{n+1} = \sqrt{U_{n}²+\frac{1}{2^{^n}}}
   \end{cases}$

How do I establish the following inequality ?

For $n \geq 1$

$U_{n+1} \leq U_{n}+\frac{1}{2^{^n}}$

I thought about proving that the sequence is increasing so that

$U_{n+1} \geq U_{n}$

But it's pretty useless here, I don't know how to advance any further.

Body:
$\begin{cases}
u_{1} = 1
      \\
u_{n+1} = \sqrt{u_{n}^2+\frac{1}{2^{^n}}}
\end{cases}
$

Therefore
$u_{n+1}^2
=u_n^2+\dfrac1{2^n}
$
so
$u_{n+1}^2-u_n^2
=\dfrac1{2^n}
$
so
$u_n > 1$
for $n > 1$.

Therefore
$u_{n+1}-u_n
=\dfrac1{2^n(u_{n+1}+u_n)}
\lt\dfrac1{2^n2}
=\dfrac1{2^{n+1}}
$.


问:
Show that $S = \{(a,a) | a \in A\}$ is an equivalence relation on A. How may equivalence classes does S define in A?
For note $|A| =  n\ge 1$

$aSa$ so S is reflexive.

I'm unsure how you prove symmetry and transitivity when there's only one element. For symmetry, would you say that $aSa$ has the reverse $aSa$? But transitivity still wouldn't work.

There are $n $ equivalence class, $[a] = {a}$, as there is only one element per $a $ for the whole set A.

答: It doesn't matter whether there is only one element or not.  Your assumption only refers to one element at a time.  Go back to the definition.  Symmetry means that $(xSy) \implies (ySx)$  If $x \neq y$ then both clauses are false and the implication is true.  If $x=y$ both clauses are true and the implication is true.  Thus the relation is symmetric.  The proof of transitivity is similar.


问:
Why not to extend the set of natural numbers to make it closed under division by zero?
We add negative numbers and zero to natural sequence to make it closed under subtraction, the same thing happens with division (rational numbers) and root of -1 (complex numbers).

Why this trick isn't performed with division by zero?

答:
Define $0k=1$ for some new element $k$. Certainly $k$ is not a real number, so let us extend $\mathbb{R}$ to a new set $\mathbb{E}$ which includes every $a+bk$ for all $a,b \in \mathbb{R}$. Note that, because of k's property, an existing real number like 60 would not be represented as $60+0k$, but rather as $59+0k$.

We now define addition and multiplication of these numbers as follows:
$$(a_{1}+b_{1}k)+(a_{2}+b_{2}k) = (1+a_{1}+a_{2})+(b_{1}+b_{2})k\\
  (a_{1}+b_{1}k)\times{(a_{2}+b_{2}k)} = \begin{cases} b_{2}\;\;\; & a_{1} = -1, b_1=0\\ b_{1} &a_{2}=-1,b_{2}=0 \\ (a_{1}+a_{2}+a_{1}a_{2})+(b_1+a_2b_1+b_2+a_1b_2+b_1b_2)k& \text{otherwise} \\ \end{cases}$$

An example of addition:
$$2k+(1+k)\\
  =(0+2k)+(1+k)\\
  =(1+0+1)+(2+1)k\\
  =2+3k$$
An example of multiplication:
$$2k\times{5}\\
  =(0+2k)\times{(4+0k)}\\
  =(0+4+(0\times{4}))+(2+(4\times{2})+0+(0\times0)+(2\times0))k\\
  = 4+10k $$
We also define subtraction and division:
$$
x-y=x+(y\times{-1})\\
\text{rec}(a+bk)=\begin{cases} k\;\;\; & a = -1, b=0\\ \frac{1}{a+1} &a\neq{-1},b=0 \\ 0& \text{otherwise} \\ \end{cases}\\
x/y=x\times{\text{rec}(y)}
$$
Properties $(\mathbb{E},+,-,\times)$ satisfies:

$(x+y)+z = x+(y+z)$

$x+y=y+x$

$x+0=x$

$x-x=0$

$x\times{1}=x$

$x\times{(y+z)} = x\times{y}+x\times{z}$

But this extension does not satisfy some other expected properties like $x/x=1$, $k\times{x} = kx$ for $x \in \mathbb{R}$, and $x\times{(y\times{z})} = (x\times{y})\times{z}$.


问:
Showing that $a^{12}\equiv 1\pmod{210}$ if $a$ is coprime to $210$
I ran into this next question:

Show that if $a$ is coprime to $210$:

$a^{12}\equiv1 \pmod{210}$.

This is not a homework question.

Thank you very much in advance,

Yaron

答:
HINT:

Using Carmichael Function,

$$\lambda(210)=\text{lcm}\{\lambda(2),\lambda(3),\lambda(5),\lambda(7)\}=\text{lcm}(1,2,4,6)=12$$

In fact,we can find a larger $n$ such that $a^{12}\equiv\pmod n$ for all $a$ co-prime to $n$

We need $\lambda(n)$ divides $12$

Case $1: $ If $p=2,$ we know, $\lambda(2^r)=\begin{cases} 2^{r-1} &\mbox{if } 1\le r \le 2 \\
2^{r-2} &  \text{ otherwise }\end{cases}$

$2^{r-2}$ needs to divide $12,r-2\le 2\implies r \le 4$

Case $2: $ If $p$ is odd prime, $\lambda(p^r)=\phi(p^r)=p^{r-1}(p-1),$

If $r>1,$ as the only odd divisor$(>1)$ of $12$ is $3^1\implies p=3,r=2$

Else $r=1\implies p-1$ must divide $12\implies p$ can be $2,3,5,7,13$

$2,3$ have already been encountered

So, the highest value of $n$ will be $13^1\cdot7^1\cdot5^1\cdot3^2\cdot2^4$

As $a^{12}\equiv1\pmod {13\cdot7\cdot5\cdot3^2\cdot2^4}$ if $(a,13\cdot7\cdot5\cdot3^2\cdot2^4)=1,$

$a^{12}\equiv1\pmod d$ where $d$ is any divisor of $13\cdot7\cdot5\cdot3^2\cdot2^4$ and $(a,d)=1$ as $(a,13\cdot7\cdot5\cdot3^2\cdot2^4)=1$


问:
Show that $m^{*}([a, b]\backslash G)=b − a- m^{*}{(G)}$
In the proof of Theorem 1.23 in Real Analysis by Bruckner it is used the following without a proof

Let $G$ be an open subset of an interval $[a, b]$ and write $K =[a, b]\backslash G$. Then $m^{*}{(K)}=b − a- m^{*}{(G)}$.

If it were measure, yes $G$ and $K$ are disjoint thus $m(K)=b − a- m(G)$ holds but how to prove the equality for outer measure that is defined by $m^{*}{(E)}=\inf{\{ \sum_{i=1}^{\infty} |I_i| \ :\ E\subset\cup_{i=1}^{\infty}I_i }\}$. I even know $G$ can be written as a countable union of open intervals and $K$ is a compact set but those don't help too. A detailed explanation would be much appreciated.

Added: Actually $m^{*}([a, b]\backslash G)=b − a- m_{*}{(G)}$ holds for every set $G$ and the claim to be proved is $m^{*}{(G)}=m_{*}{(G)}$. The book hasn't introduced much things yet!

答:
It suffices to prove that, if $G$ is an open subset of $[a, b]$, then we have $m_*(G) = m^*(G)$, where $m_*(G)$ is defined as $\sup\{m^*(K): K \subseteq G \text{ compact}\}$.

We write $G$ as a countable disjoint union of open intervals $G = \bigcup_{i = 1}^\infty I_i$. I assume that you know that $m^*(G) = \sum_{i = 1}^\infty |I_i|$.

Since $m_*(G) \leq m^*(G)$ always holds, it suffices to show the other direction.

In the following, we assume that $m^*(G) < \infty$. The case $m^*(G) = \infty$ can be treated similarly.

For a given $\epsilon > 0$, we choose a finite subset $S \subseteq \{1, 2, \dots\}$ such that $m^*(G) - \sum_{i \in S} |I_i| < \frac \epsilon 2$.

Without loss of generality, we may assume that $S$ is the set $\{1, \dots, n\}$. We write $I_i = (a_i, b_i)$.

We then construct a compact set $K = \bigcup_{i = 1}^n [a_i + \frac \epsilon{4n}, b_i - \frac \epsilon{4n}]$. Since $K$ contains the disjoint union $\bigcup_{i = 1}^n (a_i + \frac \epsilon{4n}, b_i - \frac \epsilon{4n})$, we have $$m^*(K) \geq m^*(\bigcup_{i = 1}^n (a_i + \frac \epsilon{4n}, b_i - \frac \epsilon{4n})) = \sum_{i = 1}^n (|I_i| - \frac \epsilon {2n}) = (\sum_{i = 1}^n |I_i|) - \frac \epsilon 2 > m^*(G) - \epsilon. $$

Therefore we have shown that $m_*(G) > m^*(G) - \epsilon$ for any $\epsilon > 0$.


答:
A relation $S$ is reflexive on $A$ exactly when $\forall x\in A~.xSx$.   Which is obviously the case when $S:=\{\langle a,a\rangle: a\in A\}$ .

A relation on $S$ is symmetric in $A$ exactly when $\forall x\in A~\forall y\in A~.(xSy\to yS x)$.   Well when $S=\{\langle a,a\rangle: a\in A\}$ we have that for any $x$ and $y$ in $A$: when $x=y$ then $xSy$ and $ySx$, else when $x\neq y$ then $\lnot(xSy)$ and $\lnot(ySx)$.   Therefore $xSy\to ySx$ is always valued as true, and so we have symmetry.

Likewise for transitivity.  Show that there are no elements $x,y,z$ in $A$ which value $(xSy\land ySz\to xSz)$ as false.

Title: Show that $S = \{(a,a) | a \in A\}$ is an equivalence relation on A. How may equivalence classes does S define in A?

Body:
For note $|A| =  n\ge 1$

$aSa$ so S is reflexive.

I'm unsure how you prove symmetry and transitivity when there's only one element. For symmetry, would you say that $aSa$ has the reverse $aSa$? But transitivity still wouldn't work.

There are $n $ equivalence class, $[a] = {a}$, as there is only one element per $a $ for the whole set A.


问:
Determine convergence of $\sum_{n=1}^{\infty}a_n$
Let $a_n=1-2^{\frac{-1}{n}}$, $b_n=\frac{1}{n^p}$.

Does $\sum_{n=1}^{\infty}a_n$ converge or diverge when compared to $\sum_{n=1}^{\infty}b_n$ if $p < 1$?

My answer to this is that it diverges, since $b_n$ diverges if $p<1$, and by a comparison test $a_n$ does as well.

Is this correct? Thank you.

答:
If $p = 1$, then $\sum_{n=1}^{\infty}b_n$ diverges. Then use the limit comparison test. The limit of the ratios of the terms is $$L = \lim_{n \to \infty } \frac{a_n}{b_n}=\lim_{x \to 0}\frac{1-2^{-x}}{x}$$

Can you finish from here?

Hint for evaluating that limit:

 Use L'hopital's rule (or the Maclaurin series expansion).


问:
Maclaurin expansion of $f(x)=e^{ax^{1/2}}$ with undefined derivative at $0$
I am asked to find the Maclaurin expansion of the following function:

\begin{equation}
 f(x)=e^{ax^{1/2}}
\end{equation}

I have trouble understanding how to do that given that the first derivative of the function (for example) is undefined at $0$.
Any help would be greatly appreciated.

Thank you in advance!

答:
Since
$e^x
=\cosh(x)+\sinh(x)
$,

$e^{ax^{1/2}}
=\cosh(a\sqrt{x})+\sinh(a\sqrt{x})
$.

As $x \to 0$,
$\cosh(a\sqrt{x})$
behaves nicely
but
$\sinh(a\sqrt{x})
\approx a\sqrt{x}
$
which does not have a derivative
at $x = 0$.


问:
biconditional matrices
So my professor gave me this question:

Assuming that $A\in M_{m\times n}$ and $B\in M_{m\times n}$. We will say that $A,B$ are biconditional matrices if there exist invertible matrices $P\in M_{m\times m}$ and $Q\in M_{n\times n}$ such that $PAQ=B$.

First, he asked us to prove that being biconditional defines an equivalence relation. That was very easy.

After that he asked us to find how many equivalence classes in $M_{m\times n}(\mathbb{F})$ are there.

Or basically how many matrices there is in $M_{m\times n}(\mathbb{F})$ so they are not  biconditional matrices.

So how do I do it.

I have no idea. I believe that you could always find such matrices except from the zero matrix which for that it will be biconditional matrices only to itself.

答:
Hints. The following is the key observation (and you should fill in the details).

We can always use elementary row operations and elementary row operations to transform $A$ to a matrix that is simultaneously in reduced row echelon form and reduced column echelon form, i.e. there exist invertible matrices $P$ and $Q$ such that $PAQ$ is a matrix of the following form:
  $$
\left[\begin{array}{cc}I_k&0_{k\times(n-k)}\\ 0_{(m-k)\times k}&0_{(m-k)\times(n-k)}\end{array}\right].\tag{1}
$$
  (When $k=0$, the RHS above is the zero matrix and $I_k$ is void.)

In other words, every equivalence class in $M_{m\times n}(\mathbb{F})$ contains a representative member of the above form. Now the question is: how many different (i.e. inequivalent) representatives of the form $(1)$ are there?

Title: biconditional matrices


问:
Is there a unifying notion of "structure" that can be referenced when saying that a morphism is structure preserving?
I have read the following:

Why is there apparently no general notion of structure-homomorphism?

What does Structure-Preserving mean?

Can we deduce that morphisms in categories of structures should be “structure preserving”

Morphisms has to be structure preserving?

Is every abstract category a concrete category of structures?

They all seem to be attempting to scratch the same itch, and I'm unsatisfied with the answers. It seems to me that in order for someone to answer questions like "do morphisms have to be structure preserving?" we would first need to agree on what structure is.

Question

Is structure in the sense that we might say "FOO is a structure preserving morphism in the category of BAR" formalized? For example, is there consensus on a definition that generalizes the structure preserved by continuous maps in the Top category, the structure preserved by group homomorphisms in the Grp category, and the structure preserved by smooth maps in the Man category?

The wiki article for mathematical structure seems a bit vague.

答:
An "algebraic structure" can be defined as a set together with a collection of formal logical sentences satisfied by the set. These sentences are usually called the axioms of the structure, and structure-preserving maps may be defined as those such that the sentences remain true when applied to the image set of the morphism.

I'm sure you've read about examples, such as the thin category's generalization of the poset, wherein morphisms are not structure-preserving. Such things' ubiquity is why the idea that morphisms are "structure-preserving" is little more than a pedagogical aid; the question of how morphisms act on structure is perhaps the definition of standard abstract algebra.

Working directly with such a definition of "algebraic structure" according to such a definition has a storied history; see for example universal algebra. The basic idea of category theory is to understand such algebraic structure without looking at structures themselves, only the maps between classes of structures. In my opinion, category theory is by far the best way to do such meta-analysis. It actually has predictive power, allowing one to relate wildly distant algebraic structures to one another (an example I'm familiar with is categorical quantum mechanics connecting Hilb to linear logics), whereas other approaches feel more ad-hoc. Furthermore, it is truly universal: one of the first categories one reads about is Cat, which describes the structure of category theory in terms of itself. Furthermore, some current researchers believe categories may end up being used to describe all of mathematics, even formal logic.

TL;DR: while one can can rigorously define structure, the (informal) idea behind category theory is that you don't need to.


答:
If you think it diverges by comparison with $\sum_{n=1}^{\infty}b_n$ which diverges, use the limit comparison test to be sure.

The limit comparison test states

$L = \lim_{n \to \infty } \frac{a_n}{b_n}$

$=\lim_{n \to 0 }\frac{1-2^{-n}}{n}$

$=\frac{\lim _{n\to 0\:\:}\left(1-2^{-n}\right)}{\lim _{n\to 0\:\:}\left(n\right)}$

$=$undefined

And thus, by limit comparison test, both $a_n$ and $b_n$ diverge. Note that if you evaluate the limit for $n \to \infty$ instead of $n \to 0$ (common error), the limit will be $0$. And again, since $L=0$, diverges. For it to converge, $L$ must be greater than $0$ and less than $\infty$.

Hope this helps you out!

Title: Determine convergence of $\sum_{n=1}^{\infty}a_n$

Body:
Let $a_n=1-2^{\frac{-1}{n}}$, $b_n=\frac{1}{n^p}$.

Does $\sum_{n=1}^{\infty}a_n$ converge or diverge when compared to $\sum_{n=1}^{\infty}b_n$ if $p < 1$?

My answer to this is that it diverges, since $b_n$ diverges if $p<1$, and by a comparison test $a_n$ does as well.

Is this correct? Thank you.


问:
If $ab > 1$, can $\frac{3^a - 2^a}{2^{a+b} - 3^a}$ be an integer?
I've been thinking about this positive fraction where $a,b$ are positive integers and whether it can be an integer for $ab > 1$:

$$\frac{3^a - 2^a}{2^{a+b} - 3^a}$$

Clearly, if $a=1, b=1$, then:

$$\frac{3^1-2^1}{2^2 - 3^1} = 1$$

Is there any other value of $a,b$ that results in a positive integer?

I am interested in either finding a solution where $ab > 1$ and it is a positive integer or an approach to show that no such integer exists.

答:
You've reminded me of my least favorite fact about the fundamental theorem of arithmetic, which is that it ruins music.

We need the denominator to be (relatively) small if it's going to be a divisor of anything. But small values in the denominator are going to be hard, if not impossible, to find, as $a$ increases. Obviously we cannot have $2^x = 3^y$ unless $x = y$. We have $3^2 - 2^3 = 1$ from Catalan's Conjecture, but that's in the wrong order.

But then it gets worse. The generalized/extended Catalan Conjecture suggests that prime powers have a tendency to repel one another as they get larger. For instance, there are exactly five instances below $2^64$ where a pair of prime powers fall between one prime and the next. The pairs of powers are $(8,9)$ as above, $(25,27),(121,125),(2187,2197)$, and $(32761,32768)$.

The smallest positive value of $2^{a+b} - 3^a$ for a given $a$ (as found by varying $b$) isn't technically an increasing function, but the decreases are few and far between. And decreasing from $24062143$ to $5077565$ (for $a = 16,17$) isn't what we're looking for. The only "small" values other than your $a = b = 1$ are $5,7,13,47$ for $a = 3,2,5,4$. None of those give an integer quotient (they are $\frac{19}{5}, \frac{65}{7}, \frac{211}{13}, \frac{665}{47}$), and after that most of the values just get exponentially larger.

Essentially, barring an exception to the extended Catalan Conjecture, which I think would be hugely shocking, you can't get an integer out of this. But, technically, it's an open question.


问:
Eigenspace problem
I am trying to solve the following problem:

Let $(\lambda,v)$ be an eigenpar of $A \in \mathbb{R}^{n \times n}$ such that the algebraic multiplicity of $\lambda$ is $2$, but with geometrical multiplicity equal to $1$. Show that $v \in$ Im$(A-\lambda I) $, where Im$(A-\lambda I)$ denotes the image space of the matrix $A-\lambda I$.

My strategy was to suposse that $v \notin$ Im$(A - \lambda I)$, this way i conclude using the Kernel Image Theorem that $\mathbb{R}^{n}$ can be made as the direct sum of Kernel$(A-\lambda I)$ and Im$(A-\lambda I)$. I thought this way i could get some contradiction, but if it is the case i do not noticed how.

答:
Suppose $v\notin \text{Im}(A-\lambda I)$. As you noticed, we have $\mathbb{R}^n=\ker (A-\lambda I)\oplus\text{Im}(A-\lambda I)=K\oplus J.$ Now, observe that $A$ is invariant under $K$ and under $J$. So the restrictions $A|_K:K\rightarrow K$, $A|_J:J\rightarrow J$ are well defined linear maps.

Now choose a basis $\{v_1,\dots,v_{n-1}\}$ of $J$ and let $[A]_B$  be the matrix of $A$ with respect to the basis $B=\{v,v_1,\dots,v_{n-1}\}$. Then $[A]_B$ has the form

$$ [A]_B = \left( \begin{matrix} \lambda & * & \dots & *\\ 0 & a_{11} & \dots & a_{1(n-1)}\\\vdots & \vdots & \ddots & \vdots\\ 0 & a_{(n-1)1} & \dots & a_{(n-1)(n-1)}\end{matrix} \right). $$

Let $p$ be the characteristic polynomial of $A$. We can see by cofactor expansion of the determinant that
$$
p(t)=det(A-t I)=det([A-t I]_B)=det(A|_K-tI|_K)det(A|_J-tI|_J)=p_K(t)p_J(t),
$$
where $p_L$ is the characteristic polynomial of the restriction $A|_L$, for $L=K,J$.

Note that $p_J(\lambda)\neq 0$. If not, then there exists $0\neq x\in \text{Im}(A-\lambda I)$ such that $Ax=\lambda x$, which contradicts the fact that $\ker(A-\lambda I)=\langle v\rangle$. Moreover, as $K$ is one-dimensional vector space, we have $A|_K=(\lambda)$, so $p_K(t)=\lambda-t$, and so $$p(t)=(\lambda-t)p_J(t),\quad p_J(\lambda)\neq 0$$
and this contradicts the fact that $\lambda$ has multiplicity 2 as a root of $p$.

Title: Eigenspace problem


问:
Application of Hahn–Banach theorem
I'd like to prove this statement:

Let $X$ a normed space. If $X$ has a linearly independent subset of cardinal $n$, then so does $X'$.

I know that I have to use the Hahn-Banach theorem to prove the existence of functionals in $X'$ that hold the above mentioned property but I'm not sure how.

Any suggestions are welcome. Thanks in advance.

答: Let $E=\{x_1, x_2, \cdots, x_n\}$ be the linearly independent set. Let $Y:=\textit{span}\{x_1, x_2, \cdots, x_n\}$. Then $E$ is a basis of $Y$. Let $p_i: Y \rightarrow \mathbb{K}$     ($\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$) be the co-ordinate functional, i.e., $p_i(\sum_{i=1}^nc_ix_i)=c_i$. Since $Y$ is finite dimensional each $p_i$ is continuous  and $p_i(x_j)=1$ if $i=j$ and $p_i(x_j)=0$ if $i \neq j$. Now by the Hahn-Banach theorem, $\exists~~~f_1, f_2, \cdots, f_n \in X'$ such that $f_i|_Y=p_i~~~\forall~i$. $\therefore~~~f_i(x_j)=p_i(x_j)~~~\forall~i,j$. It can be easily shown that $f_i$'s are linearly independent.

Title: Application of Hahn–Banach theorem


问:
continuous function on a set of lines through a point
This came up in class once:

Suppose we have a set $R \subset \mathbb{R}^2$ and $x \in
 \mathbb{R}^2$. If we define a function $A$ on all the
  lines $l$ going through $x$ such that $A(l)$ gives how much of the area
  of $R$ is "above" (or to the <s>right</s> left of) the line $l$, must
  $A$ be continuous?

I do think it is continuous. If we take a horizontal line through $x$ and rotate it by $\theta$, we have a continuous function giving the area of the part of $R$ whose pre-image (under this rotation) is the part of $R$ initially under the horizontal line.

However, I'm at loss how to rigorously show continuity using pre-image of open sets is an open set. In particular, how should an open set be defined on the set of lines through $x$, and thus allowing me to rework a proof.

EDIT: What we mean by the ``above'': For a non-vertical line $l$ passing through $x$, we rotate $S$ clockwise until $l$ becomes horizontal. Then $A(l)$ gives the area of $R$ which is above this horizontal line (the image of $l$ after the rotation). If $l$ were vertical, we take the area to the left of $l$,

答: If the line is vertical, do you take the area in the plane to the right?  Apparently for all other lines you take the area "above".  Then the region whose area you are calculating does not depend continuously on the parameter.  Therefore there is no reason to think that this will define a continuous function.</p>


问:
Splitting a circle into 3 equal parts using 2 lines
The picture probably explains my question best.

I need to find a way to divide a circle into 3 parts of equal area with only 2 lines that intersect each other on the outline of the circle.

Also I need to check, if whatever diameter is between those lines, also splits circles with a different diameter into equal parts.

And lastly, and probably the most difficult question: How do I have to calculate the angle between x lines that all intersect in one point, so that the circle is split into x+1 parts with area = 1/(x+1) of the circle?

I tried my best, but couldn't even find a single answer or the right strategy to tackle the question...

1

答:
Given the angle $\theta$, split by the diameter containing $B$, consider the following diagram:

$\overline{BO}$ is the line through the center and $\overline{BA}$ is the chord cutting off the lune whose area we wish to compute.

The area of the circular wedge subtended by $\angle BOA$ is
$$
\frac{\pi-\theta}2r^2\tag1
$$
The area of $\triangle BOA$ is
$$
\frac12\cdot\overbrace{r\sin\left(\frac\theta2\right)}^\text{altitude}\cdot\overbrace{2r\cos\left(\frac\theta2\right)}^\text{base}=\frac{\sin(\theta)}2r^2\tag2
$$
Therefore, the area of the lune is $(1)$ minus $(2)$:
$$
\frac{\pi-\theta-\sin(\theta)}2r^2\tag3
$$
To get the area divided into thirds, we want
$$
\frac{\pi-\theta-\sin(\theta)}2r^2=\frac\pi3r^2\tag4
$$
which means we want to solve
$$
\theta+\sin(\theta)=\frac\pi3\tag5
$$
whose solution can be achieved numerically (e.g. use $M=\frac\pi3$ and $\varepsilon=-1$ in this answer)
$$
\theta=0.5362669789888906\tag6
$$
Giving us

Numerical Details

The iteration from this answer, applied to $\theta+\sin(\theta)=\frac\pi3$, is
$$
\theta_{n+1}=\frac{\pi/3-\sin(\theta_n)+\theta_n\cos(\theta_n)}{1+\cos(\theta_n)}\tag7
$$
Here is the result of this iteration starting at $0$;
$$
\begin{array}{l|l}
n&\theta_n\\\hline
0&0\\
1&0.5\color{#AAA}{23598775598298873077107230547}\\
2&0.5362\color{#AAA}{45321326153808318904236597}\\
3&0.5362669789\color{#AAA}{24456230942633093381}\\
4&0.53626697898889055276\color{#AAA}{1878717471}\\
5&0.53626697898889055276244906787\\
6&0.53626697898889055276244906787
\end{array}
$$


问:
Let $a,b$ be positive integers. Prove that if for all integers $s,t$ we have $sa+tb \neq \gcd(a,b)$ or $s+t=1$ then we have $a=b$
Question: Let $a,b$ be positive integers. Prove that if for all integers $s,t$ we have $sa+tb \neq \gcd(a,b)$ or $s+t=1$ then we have $a=b$.

Attempt:
Well I think for any integers $a,b$ by Bezout's theorem there must exist integers $s,t$ with $sa+tb=\gcd(a,b)$. So then $s+t=1$.

Then $s = 1-t$ and so

$(1-t)a + tb = \gcd(a,b) \implies a - t(a-b) = \gcd(a,b) $

then $t(a-b) <a$ since the gcd is positive.

From here I'm not really sure how to proceed.

Edit: Now that I actually understand this problem, the problem is saying is all of the Bezout coefficients $s,t$ have difference $1$ then $a=b$. To show this you can use the general form of the Bezout coefficients to choose two different representations and reason about $a-b$.

答: Note that if $t a + s b = (a, b),$ then $t_1 = t-b$ and $s_1=s+a$ also work. But $1 =  t_1 + s_1 = t + s + (a-b) = 1 + (a-b),$ so $a-b=0.$


问:
Calculate the limit $\lim_{(x,y)\to (0,0)} \frac{x-\cos(x^2+y^2)-\arctan(x)+1}{x^2+y^2}$
Find the limit, $$\lim_{(x,y)\to (0,0)} \frac{x-\cos(x^2+y^2)-\arctan(x)+1}{x^2+y^2}.$$

My attempt:

I have tried several paths for evaluating the limit. For instance, $y=0$, $x=0$, $y=x$, $x=\tan y$, etc. I always obtain $0$. But I haven't been able to prove that the limit is equal to $0$.

答:
The magic words are: polar coordinate. The limit equals:

$$\lim_{r\to 0} \frac{r \cos \theta - \cos{r^2}  - \arctan{ r \cos \theta} +1}{r^2}.$$

Now $1-\cos r^2 = O(r^4),$ so that part goes to zero, and $\arctan r \cos \theta - r \cos \theta = O(r^3),$ so that goes to zero also.


问:
A question on altering a definition of matrix norm
In the definition of the matrix norm, we have
$$\|A\|=\!\!\sup_{\substack{x\in \mathbb{R}^n\\ ||x||=1}} \!\! \|Ax\|$$
I was wondering if we can replace the condition $\|x\|=1$ with $\|x\|\leq 1$. Will the two conditions give the same definition? If so, then why are we not considering the latter one as it will also give us a convex set along with being compact. Since I am new to norms, I would request you to please suggest to me some good books that can make my concepts clear.
Thank you.

答: Yes, by homogeneity of the norm, you get the same definition (homogeneity: $\| t x\| = t\|x\|.$)


问:
question on T period bargaining(alternating offers).
the following question is a kind of Rubinstein bargaining model:

2 players, A and B, have 100dollars to divide between them.
They agree to spend T days negotiating over this division.

The first day, A will make an offer, B either accepts or comes back with a counteroffer the next day, and on T day, B gets to make one final offer. If they cannot reach an agreement in T days, both players get 0 dollar.

Assuming that both A and B are having the same degree of impatience:
A and B discount payoff in the future at a rate of r per day.

Finally, we assume that if a player is indifferent between two offers, he wil accept the one that is most preferred by his opponent.

This idea is that the opponent could offer some arbitrary small amount that would make the player strictly prefer one choice and that this assumption allows us to approximate such an"arbitrarily small amount" by zero. It turns out that there is a unique subgame perfect nash equilibrium of this bargaining game.

So the question is that, what is the SPNE in this alternating-offer bargaining game when T is even?

Am i making it much clearer?

Thanks for all the advice that has given on modifying this question, thanks for being patient for my stupid question.

答:
I am not sure what you mean by "...the opponent could offer some arbitrary small amount." The first mover has advantage in a Rubinstein-type bargaining. So when it's A's turn to offer, the next-period subgame where B becomes first mover gives B a reservation utility that A must meet.

Replace "100 dollars" by division of a pie of size $1$. An offer is of the form $(x,y)$ where $x + y = 1$. $x$ is A's piece of the pie and $y$ for B.

SPNE by backward induction:

In the subgame starting at the last period $T$, B offers the division $(0, 1)$ and A accepts. (This being the last period, A has reservation utility $0$.)

So in period $T-1$, B has reservation utility $r$. A offers $(1-r, r)$.

In period $T-2$, A has reservation utility $r(1-r)$. B offers $(r(1-r), 1 - r(1-r))$.

In period $T-1$, B has reservation utility $r(1 - r(1-r))$. A offers $( 1-r(1-r(1-r)), r(1-r(1-r) )$.

Continue this way by induction gives the SPNE outcome.


问:
Counting the number of partitions such that every part is divisible by $k$.
I would like to do this using generating functions. I'm comfortable with the generating function for the number of partitions of $n$: $$(1+x+x^2+\dots)(1+x^2+x^4+\dots)(1+x^3+x^6+\dots)\cdot\dots,$$ although I'm not entirely sure how to adapt it to deal with certain restrictions, such as asking that every part is divisible by $k$, or divisible by some pair of numbers, say $k$ and $j$.

I was thinking something like $$(1+x^k+x^{2k}+\dots)(1+x^{2k}+x^{4k}+\dots)(1+x^{3k}+x^{6k}+\dots)\cdot\dots.$$

Any advice or hints to get be going would be greatly appreciated!

答: A partition of $n$ with every part divisible by $k$ correspond bijectively to partitions of $n/k.$


问:
How to count partitioned subsets of a binomial coefficient
I have been reading Dr. Carl Wagner's book Basic Combinatorics, and I cannot wrap my head around a particular theorem. Someone else has asked a very similar question and accepted an answer, but I'm afraid I still don't understand.

As stated, the theorem runs thus:

For all $ n, \; k \in \mathbb{N} $
$ \displaystyle \sum_{j = 0}^{n} \binom{j}{k} = \displaystyle \sum_{j = k}^{n} \binom{j}{k} = \binom{n + 1}{k + 1} $

There follows a proof of the $\binom{n + 1}{k + 1} $ bit using some handy substitutions. What confuses me is the next part:

#2. Clearly $ \binom{n+1}{k+1} $ counts the class of all (k + 1)-element subsets of [n + 1]. But this class of k+1 subsets may be partitioned into subclasses corresponding to j = k, k + 1, . . . , n as follows.

The subclass of subsets with largest element equal to k + 1 is counted by $ \binom{k}{k} $ (why?), the subclass of subsets with largest element equal to k + 2 is counted by $ \binom{k+1}{k} $ (why?), . . ., and the subclass of subsets with largest element equal to n + 1 is counted by $ \binom{n}{k} $ (why?).

I don't understand why at all! Indeed, it makes no sense to me.

It seems to me that the number of subsets of [n+1] with largest element k+1 would be $\binom{k+1}{k+1}$. Sure, that's the same as $ \binom{k}{k} $  since they both equal one, but that just seems to be a coincidence. We're choosing k+1 numbers from a set that runs from 1 to k+1. There can, naturally, be only one set that fits that bill. Yet to use $ \binom{k}{k} $ to represent that one set seems confusing to me, and every example that follows makes less sense.

What am I missing?

答:
Let's say $n = 10$ and you want a subset of $4$ elements, the largest of which you want to be $6$.

Now that is not the same thing as choosing a subset of $\{1,2,3,4,5,6\}$ with $4$ elements because that would not guarantee that the largest element is $6$, only that the largest element is $6$ or less.

Here are all the $4$ element subsets whose largest element is $6$:

$$1236, 1246, 1256, 1346, 1356, 1456, 2346, 2356, 2456, 3456$$

Additionally, here are the $4$ element subsets of $\{1,2,3,4,5,6\}$ which were not listed already:

$$1234, 1235, 1245, 1345, 2345.$$

The important observation is that if you want $6$ to be in your set and you have no choice about the matter, then you are not choosing $6$. You choose $3$ elements of $\{1,2,3,4,5\}$ and then you get an extra $6$ whether you like it or not. And you will agree that we can pair everything in that first list with:

$$123, 124, 125, 134, 135, 145, 234, 235, 245, 345$$

simply by adding or removing the $6$ that is required to be there (not chosen).


问:
Proof linked to linear algebraic equations
Let a, b, c, e, f and g be real numbers such that:

e/a + b/f = 1

f/b + c/g = 1

Prove that abc + efg = 0

I have tried all possible manipulations at my end but am unable to prove the identity. Help or hint appreciated.

答:
if $e=0$ then $b=f$ and $c=0$

otherwise let
$$  \lambda = \frac{b}{f} \; , \; \;  \sigma = \frac{a}{e} \; , \; \; \tau = \frac{c}{g} \; . $$

Then
$$   \sigma = \frac{1}{1-\lambda}  = \frac{-1}{\lambda-1} $$
$$   \tau = \frac{\lambda-1}{\lambda} $$

And
$$  \sigma \tau = \frac{-1}{\lambda} $$
so that
$$ \color{red}{ \sigma \tau \lambda = -1,}$$
which is the desired result


问:
How many verticies, edges and faces (cells) does an nd hypercube have?
So I guess a 1d hypercube is a line segment.  It has 2 verticies and 1 edge.  Not sure how many faces it has?

A 2d hypercube is a square.  It has 4 verticies and 4 edges.  Again not sure how many faces it has?

A 3d hypercube is a cube.  It has 8 verticies and 12 edges and 6 faces.

A 4d hypercube has 16 verticies and not sure how many edges or 3d faces.

We can define the faces of an nd hypercube as being (n-1)d hypercubes.  So the faces of a cube are squares.

An nd hypercube has $2^n$ verticies.  (If we imagine a vertex coordinate $v = (v_1, v_2, ..., v_n)$ then choose some subset and set them to $1$, set the rest to $-1$.  This describes one vertex.  There are $2^n$ subsets of $n$ items.)

How many edges does it have as a function of n?  How do you derive that?

How many $(n-1)$d faces does it have?  How do you derive that?

Update

For the faces I think we can say: pick a dimension $i$, partition the verticies into two sets of $2^{n-1}$ verticies.  One with $v_i = -1$ and the other with $v_i = 1$.  Each of these two sets describes a face.  There are $n$ ways to pick $i$.  So an nd hypercube has $2n$ faces.  So a line segment has 2 faces (which are points).  A square has 4 faces (which are lines).  A cube has 6 faces (which are squares).  A 4d hypercube has 8 faces (which are cubes). and so on.

So just edges remains.

答:
Here are a couple hints. I will consider the vertices to be length-$n$ bitstrings.

An edge connects two vertices if the vertices differ by a single bit flip. How many edges are incident to a given vertex? If you multiply this by the number of vertices, what would you do next to end up with the total number of edges?

A facet (i.e. $(n-1)$-dimensional cell) may be chosen by first picking a coordinate, fixing its bit, and then letting all the other bits be arbitrary. How many ways are there to do this?


问:
Prove that none of the isomorphisms above can be extended to an isomorphism of the ring $\mathbb{Q}$ to itself.
Let $G=\mathbb{Q}^\times$ be the multiplicative group of nonzero rational numbers. If $\alpha=p/q\in G$, where $p$ and $q$ are relatively prime integers, let $\phi: G \to G$ be the map which integer changes the primes 2 and 3 in the prime power factorizations of $p$ and $q$ (so, for example, $\phi(2^4 3^{11} 5^1 13^2) =3^4 2^{11} 5^1 13^2$, $\phi(3/16)=\phi(3/2^4)=2/3^4=2/81$, and $\phi$ is the identity on all rational numbers with numerators and denominators relatively prime to 2 and to 3).

a) Prove that $\phi$ is a group isomorphism.

b) Prove that there are infinitely many isomorphisms of the group $G$ to itself.

c) Prove that none of the isomorphisms above can be extended to an isomorphism of the ring $\mathbb{Q}$ to itself.

I have proven part (a) and (b). And I found that we can generalize $\phi$ by $\phi(p^m q^n ab)=p^n q ^m ab$, where $p,q$ are any primes.

How do I go about proving (c)?

答: Note that under a ring automorphism $\varphi,$  $\varphi(1) = 1.$ So, $\varphi(1+1) = \varphi(1) + \varphi(1) = 2 \neq 3.$


问:
What is the partial of the Ridge Regression Cost Function?
What is the partial of the Ridge Regression Function? Does it provide insight to a gradient descent with ridge Regression? I am using Hands-on Machine Learning with Scikit Learn, Keras, and TensorFlow Geron.

$$\mathbf{Ridge \ Regression\ Equation} => J(\hat\theta) = MSE(\hat\theta) + \alpha\sum_{i=1}^n\theta_i^2 $$
$$\mathbf{ Gradient\ Descent\ Equation} => \frac{\delta}{\delta\theta_j}MSE=\frac{2}m\sum_{i=1}^m(\hat\theta^Tx^{(i)}-y^{(i)})x_j^{(i)}$$

答:
Your ridge term is: $R = \alpha\sum_{i=1}^n\theta_i^2$

Its partial derivative can be computed using the power rule and the linearity of differentiation:

$$
\frac{\delta}{\delta \theta_j}R = 2\alpha\theta_j
$$

You also asked for some insight, so here it is: In the context of gradient descent, this means that there's a force pushing each weight $\theta_j$ to get smaller. If a given weight doesn't have much effect on the loss function, then it will tend to get pushed towards a very small value. This can help to prevent overfitting, where weights with weak effects on the loss are pushed to unrealistically large values, just because it decreases the loss a little bit.


问:
Prove by induction that for all $n \geq 3$: $n^{n+1} > (n+1)^n$
I am currently helping a friend of mine with his preperations for his next exam. A big topic of the exam will be induction, thus I told him he should practice this a lot. As at the beginning he had no idea how induction worked, I showed him some typical examples.

Now he showed me an exercise he was having trouble with, which states that one should prove
$$n^{n+1} > (n+1)^n$$
for all $n \geq 3$. I have to admit that I too have trouble showing this inequality, as in all of my attempts, my lower bound is too low. Also, I have not yet figured out, how one gets the $(n+2)^{n+1}$, primarely the number $2$ is a problem. I think this might be solved using the binomial theorem, however, I don't think they have already seen the binomial theorem in school.

Is there an easy method to show this inequality by induction not using the binomial theorem? If no: How can one show it using the binomial theorem?

Thanks for answers in advance.

答:
In this answer, it is shown by Bernoulli's Inequality, which is in turn shown by induction at the end of that answer, that $\left(1+\frac1n\right)^n$ is an increasing sequence which is term by term less than $\left(1+\frac1n\right)^{n+1}$, which is a decreasing sequence. This means that for all $n\gt0$,
$$
\left(1+\frac1n\right)^n\le\left(1+\frac15\right)^6\lt3
$$
Therefore, for $n\ge3$, we have
$$
n\ge3\gt\left(1+\frac1n\right)^n
$$
which, upon multiplication by $n^n$, gives the requested inequality for $n\ge3$,
$$
n^{n+1}\gt(n+1)^n
$$


问:
Modelling with Higher-order DEs
I would like to know if there are any models that make use of higher-order (5th, 6th and so on...) linear differential equations. The highest that I know is the Euler-Bernoulli beam equation. Do you know any? Or how to model with higher-order ones?

Thanks

答:
I wasn't able to find any high-order linear differential equations but was able to find some examples of nonlinear DEs. A lot of the times when these are used, they are used when approximations of other functions are used, such as in this paper, which looks at the Fermi-Pasta-Ulam model. Equation $(6)$ on page $4$ is: $$\begin{align}\frac{1}{4}\frac{d^2y}{dT^2}&=y_{\xi\xi}+y_\xi y_{\xi\xi}-\mu y^2_{\xi}y_{\xi\xi}+\delta y_{\xi\xi\xi\xi}+2\delta y_{\xi\xi}y_{\xi\xi\xi}+\delta y_{\xi}y_{\xi\xi\xi\xi}-\\ &-4\mu \delta y_{\xi}y_{\xi\xi}y_{\xi\xi\xi}-\mu \delta y_{\xi\xi}^3- \mu \delta y_{\xi}^2y_{\xi\xi\xi\xi}+\frac{2}{5}\delta^2y_{\xi\xi\xi\xi\xi\xi}\end{align}$$

There's also this paper, which models the diffusion of a fluid using the equation $$\frac{\partial h}{ \partial t} = \frac{\partial}{\partial x}\left( h^n \frac{\partial^5h}{\partial x^5} \right)$$

The Korteweg-De Vries equation is a model of "waves on shallow water surfaces", and there's higher order generalizations of it, such as equation $(2)$ on page $3$ of this paper: $$u_t+uu_x+u_{xxxxx}=0$$

That paper also references the Kawahara equation (equation $(4)$). I don't have access to the original paper, but in the last linked one, it says it is $$u_t + uu_x + \alpha u_{xxx}+u_{xxxxx}=0$$ which is also used for modeling shallow water waves.

These last two aren't models per se, but more so examples of when higher-order DEs pop up in physics.

I don't have access to this paper either, but the abstract says that a fifth order equation must be satisfied in order for the equation modeling the fluid to admit a "Lie point symmetry" (I don't understand what that means though).

The last one I'll put here came up in the study of black holes. I understood virtually nothing in this paper, but the equation $H(x)$ , which is the "conformal factor of the two dimensional space" (which has something to do with black holes, I'm guessing), must satisfy a sixth order DE in order to "generate a five dimensional solution". This differential equation (equation $6.19$) is $$\left(H^2H''''\right)'' = 0$$

although it trivially reduces to a $4$th order DE.

Title: Modelling with Higher-order DEs