问:
Can a formula that has free variables be logically equivalent with a sentence?
Let $\phi$ be a formula in F.O.L. that has free variables.

I wondered at first hand:

Can we always find a structure $\mathfrak C$ together with assignments $\alpha,\beta$ such that: $\mathfrak C\vDash\phi[\alpha]$ and $\mathfrak C\nvDash\phi[\beta]$?

I found counterexamples for that in formulas like $x=x$ and $\neg x=x$.

But the fact that in those cases we are dealing with a tautology and a contradiction respectively made me wonder: what if counterexamples of that sort are not allowed?

So I pose the question again, but under the extra condition that structures $\mathfrak A, \mathfrak B$ exists together with assignments $\alpha$ and $\beta$ satisfying $\mathfrak A\vDash\phi[\alpha]$ and $\mathfrak B\nvDash\phi[\beta]$.

If the answer is in that case a "no" then I can draw the conclusion that formulas that are not sentences but are logically equivalent with sentences must be tautologies or contradictions.

I do not know whether that is true (actually I hope so) or not (I do have my doubts), and am quite curious.

Thank you in advance.

Edit:

Let me for compleness make clear what definition  I practicize for logical equivalence of formulas.

Two formulas $\phi$ and $\psi$ are logically equivalent if for every structure $\mathfrak A$ and every $\mathfrak A$-assignment $\alpha$:
$$\mathfrak A\vDash\phi\leftrightarrow\psi[\alpha]$$
or equivalently:$$\mathfrak A\vDash\phi[\alpha]\iff\mathfrak A\vDash\psi[\alpha]$$

答:
Let's fix some theory $T$ in some language $\mathcal{L}$. If you wish to just consider structures you can take $T$ to be the empty theory (i.e. no axioms). We define two formulas $\phi(x)$ and $\psi(x)$ to be logically equivalent (modulo $T$) if
$$
T \models \forall x(\phi(x) \leftrightarrow \psi(x)).
$$
This is indeed equivalent to the definition you mention for the empty theory, and for non-empty theories it just restricts us to the models of $T$. Note that in the above definition $\phi$ and $\psi$ do not need to actually mention the variables $x$.

Your main question then was the following:

Can a formula that has free variables be logically equivalent with a sentence that is not a tautology or a contradiction?

The answer is yes. Take any sentence $\psi$ that is not a tautology or contradiction. Then $\psi \wedge x = x$ is logically equivalent and has free variable $x$.

One example would be to take $\psi$ to be the sentence "there are at least 10 elements" and $T$ to be the empty theory. Then there will be structures (models of $T$) where $\psi$ is true and there will also be structures where $\psi$ is false, so $\psi$ is neither a tautology nor contradiction.

If every sentence is a tautology or contradiction (modulo $T$), so if $T$ is complete, then this can obviously not happen. Simply because there are no sentences that are not a tautology or contradiction.

Your original motivation was the following question:

Given $\phi(x)$, can we find a structure $C$ and $a,b \in C$ such that $C \models \phi(a)$ and $C \not \models \phi(b)$?

You mentioned that if $\phi(x)$ is a tautology or contradiction then this cannot happen. Something more general is actually true: if $\phi(x)$ is logically equivalent to a sentence then this cannot happen. Because in such a case we have $C \models \phi(a)$ if and only if $C \models \psi$ if and only if $C \models \phi(b)$, making use of the fact that $C \models \psi$ does not depend on $a$ or $b$. This does not depend on the fact whether or not $\psi$ is a tautology or contradiction.

Finding structures $A$ and $B$ with elements $a \in A$ and $b \in B$ such that $A \models \phi(a)$ and $B \not \models \phi(b)$ heavily depends on the theory $T$ and $\phi(x)$ (also when $A = B$).


问:
Polynomial inequality
I found the following problem on a website and would be curious to find a solution.

Let $a_1\ge a_2\ge\cdots\ge a_n$ be real numbers such that for all integer $k>0$:
$$a_1^k+a_2^k+\cdots+a_n^k\ge 0$$
Let $p=\max\{|a_1|,\ldots,|a_n|\}$. Show that $p=a_1$, and that
$$(x-a_1)(x-a_2)\cdots(x-a_n)\le x^n-a_1^n$$
for all $x>a_1$.

The first part is rather simple since we can normalize by $p$ and consider the limit $k\to\infty$. However, I can't seem to get the second part right...

POSSIBLE SOLUTION FOR $n\ge 6$:
Without loss of generalization, assume that $a_1=1$. Since $x>|a_i|$ for all $i$, the following expansions converge:
$$\sum_i\ln\left(1-\frac{a_i}{x}\right)=-\sum_{i}\sum_{k=1}^\infty\frac{a_i^k}{kx^k}.$$
Since the sum over $i$ for a given $k$ is non-negative, and  we have
$$\sum_i\ln\left(1-\frac{a_i}{x}\right)\le-\sum_{k=1}^\infty\sum_i\left(a_i^{2k}+a_i^{2k+1}\right)\frac{1}{(2k+1)x^{2k+1}}.$$
Notice that $a_i^{2k}+a_i^{2k+1}\ge0$ since $|a_i|\le1$. Then
$$\sum_i\left(a_i^{2k}+a_i^{2k+1}\right)\ge2.$$
We have
$$\sum_i\ln\left(1-\frac{a_i}{x}\right)\le-\sum_{k=1}^\infty\frac 2{(2k+1)x^{2k+1}}\le-\sum_{k=1}^\infty\frac{2}{(2k+2)x^{2k+2}}.$$
The last inequality then reduces to
$$\sum_i\ln\left(1-\frac{a_i}{x}\right)\le-\sum_{k=2}^\infty\frac{1}{kx^{2k}}=-\sum_{k=1}^\infty\frac{1}{kx^{2k}}+\frac{1}{x^2}.$$
Therefore,
$$\prod\left(x-a_i\right)\le x^n\left(1-\frac{1}{x^2}\right)\exp(1/x^2).$$
We can actually show that this is a sharper bound that the one required in the question, for $n\ge 6$. Consider the functions $f(y)=(1-y)\exp(y)$ and $g(y)=(1-y^3)$ on the interval $[0,1]$. We have $f(0)-g(0)=f(1)-g(1)=0$. Furthermore, $f'(y)-g'(y)=y(-\exp(y)+3y)$, $f'(0)-g'(0)<0$, and let $h(y)=-\exp(y)+3y$. Then, $h(0)<0$, $h(1)>0$, and $h'(y)=-\exp(y)+3>0$. All this implies that $f'(y)-g'(y)$ has a single root on the interval $(0,1)$, which in turn means that $f(y)-g(y)<0$ on the same interval. Therefore,
$$\prod\left(x-a_i\right)\le x^n\left(1-\frac{1}{x^6}\right)\le x^n-x^{n-6}\le x^n-1$$
for $n\ge 6$. As @bgins said earlier, it is easy to show that the above holds for $n=2$ and $3$. However, it seems very tedious to show this for $n=4$ and $5$.

POSSIBLE SOLUTION FOR $n=4$: Assume again $a_1=1$, and let $a_2=a$,$a_3=b$,$a_4=c$. The case where all of these are positive is easy to do by induction, so assume at least $c$ is negative. Then, $$(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ac)x-abc=\\x^3+\frac{a+b+c}{2}(a+b+c-2x)x-\frac{a^2+b^2+c^2}{2}x-abc.$$
Under the assumption that $c<0$, we have $-1\le a+b+c\le 2<2x$, and it follows that
$$\frac{a+b+c}{2}(a+b+c-2x)x\le \frac{-1}{2}(-1-2x)x=x^2+x/2.$$
Therefore,
$$(x-a)(x-b)(x-c)\le x^3+x^2+x+1$$
and this proves it.

答:
Part a) is indeed simple. Let me give a proof for part b). For $k=1$ our condition gives $a_1\ge -\sum_{k=2}^{n}a_k=S.$ Now we apply AM-GM to estimate $$\prod_{k=2}^{n}(x-a_k)\le \left(\frac{(n-1)x-S}{n-1}\right)^{n-1}\le \left(x+\frac{a_1}{n-1}\right)^{n-1}.$$
So we are left to show that $$(x-a_1)\left(x+\frac{a_1}{n-1}\right)^{n-1}\le x^n-a_1^n.$$
Binomial theorem now implies the result, since $$\left(x+\frac{a_1}{n-1}\right)^{n-1}=x^{n-1}+\sum_{k = 2}^{n-2}{\binom{n-1}{k}x^{n-1-k}\left(\frac{a_1}{n-1}\right)^k}+a_1^{n-1}\le \sum_{k=0}^{n-1}x^{n-1-k}a_1^k$$ and $$\binom{n-1}{k}\cdot\frac{1}{(n-1)^k}< 1.$$</p>

ContentLicense: CC BY-SA 3.0

Title: Polynomial inequality


问:
Show that the sequence $x_{n+1}=5x_n-4x_{n-1}$ diverge
Show that the sequence $(x_n)$ defined as $x_{n+1}=5x_n-4x_{n-1}$ with $x_0=0,x_1=1$ diverge.

It is clear that the sequence diverge and this sequence is not really interesting, but still i'd like to know if my proof is correct (can't complete).

Suppose by absurd that the sequence $(x_n)$ converge and is a Cauchy sequence. So by définition we must have:

$\forall \varepsilon>0 \ \exists N \ \forall m,n\ge N$ $(m>n)$: $|x_m-x_n|<\varepsilon$

Moreover, $\forall p>1$: $|x_p-x_{p-1}|=|4(x_{p-1}-x_{p-2})|=...=|4^{p-1}|$

So we have that $|x_m-x_n|\le|x_m-x_{m-1}|+|x_{m-1}+x_{m-2}|+...+|x_{n+1}-x_n|\le 4^{m-1}+4^{m-2}+...+4^{n}=\sum_{k=n}^{m-1}4^k$

and so the sequence $(x_n)$ is not Cauchy.

The last line i'm not sure how to conclude correctly (if everything that i've done till now is correct) so need help and correction please! Thanks in advance

ContentLicense: CC BY-SA 4.0

Body:
$$X_{n+1}=5X_n-4X_{n-1}$$ Let $X_n=t^n$, then

$$t^{n+1}-5t^{n-1}+4t^n=0 \implies t^2-5t+4=0 \implies t=2,3$$
So $$X_n=p 2^n + q 3^{n} \implies X_n=-2^n+3^n \implies \lim_{n \to \infty} X_n=\infty.$$


问:
Given that $\cos\left(\dfrac{2\pi m}{n}\right) \in \mathbb{Q}$ prove $\cos\left(\dfrac{2\pi}{n}\right) \in \mathbb{Q}$
Given that $\cos\left(\dfrac{2\pi m}{n}\right) \in \mathbb{Q}$, $\gcd(m,n) = 1$, $m \in \mathbb{Z}, \, n \in \mathbb{N}$ prove that $\cos\left(\dfrac{2\pi}{n}\right) \in \mathbb{Q}.$

I know nothing about how to attack the problem. I believe I need to suppose that $\cos\left(\dfrac{2\pi}{n}\right) \not \in \mathbb{Q}$ and somehow show that $\cos\left(\dfrac{2\pi m}{n}\right) \not \in \mathbb{Q}$ what would have been a contradiction. Could you give me some hints?

答:
HINT:

For all $\theta\in\mathbb R, m\in\mathbb N$, $\cos(m\theta)$ can be expressed as a polynomial of $\cos(\theta)$ (with rational coefficients), so if $\cos(\theta)$ is rational, so is $\cos(m\theta)$.

If $m,n\in\mathbb Z$ and $\gcd(m,n)=1$, there exists $k\in\mathbb N$ such that $mk\equiv 1 \bmod  n$.

The cosine function is periodic modulo $2\pi$, that is, $\cos(\theta + 2\pi) = \cos(\theta)$ for all $\theta\in\mathbb R$.

Can you combine these facts to come up with a proof of the desired result?


问:
Derivative of complex functions
Prove that if $f'(z) = 1$ everywhere in a domain $D$ then $f(z) = z + c$
for some complex constant $c.$
And $z$ is a complex no.

答: Hint. Try showing that $g(z) = f(z) - z$ is holomorphic and constant. (I'm guessing your definition of a domain requires it to be connected.)

Title: Derivative of complex functions


问:
Proof $gcd(2^{2x},2x)=2gcd(2^{x},x)$
How do I proof $\operatorname{gcd}(2^{2x},2x)=2\operatorname{gcd}(2^{x},x)$ , I have managed to proof $\operatorname{gcd}(2^{2x},2x)=2\operatorname{gcd}(2^{2x-1},x)$ but haven't managed to get anywhere form there.

答:
Firstly,lets denote with $v_2(x)$ a positive integer k, such that $2^k$|x and $2^{k+1}\nmid x$

Lets denote with d=gcd($2^{2x-1},x$).We have that d|$2^{2x-1}$ and d|x

Because d|$2^{2x-1}$,we can say that d=$2^a$,$a \leq 2x-1$ where a is a non negative integer.

d|x so $a \leq v_2(x)$.(Because if otherwise,$d \nmid x$)

Because d is as big as possible,$a=min(2x-1,v_2(x))$

$v_2(x) < 2^{v_2(x)}\leq x \leq2x-1$ so $a=v_2(x)$ so $d=2^{v_2(x)}$.

Now if we will denote with $m=gcd(2^x,x)$ and do the exact same thing as before, we will find out that $m=2^{v_2(x)}$ so $gcd(2^x,x)=gcd(2^{2x-1},x)$

Now we can finish the proof with what you have proved by yourself.

I hope this helps.If you have any questions please ask and I will answer you.

Title: Proof $gcd(2^{2x},2x)=2gcd(2^{x},x)$


问:
Proving $R/I \otimes_R M \cong_R M/IM$ using UMP and a biadditive map not with exact sequences.
I want to Understand a paragraph of the proof of $R/I \otimes_R M \cong_R M/IM.$ in example$(8)$ on pg. 370 of Dummit and Foote (third edition)

Here is the example from Dummit & Foote:

My questions are:

1- I do not understand exactly why we need the previous observation (as stated in the book) to say that the map $N \rightarrow (R/I) \otimes_R N$ defined by $n \mapsto 1 \otimes n$ is surjective?

2-why we need $IN$ to be in the kernel?

3- why $1 \otimes a_i n_i = a_i \otimes n_i$?

4- Why the author defined $f$ by that definition?

Could anyone help me answer those questions please?

答:
It could be a useful exercise for you to prove this for yourself rather than to try to follow the proof (and consequently bear the burden of filling in all of the missing details) from a textbook. Toward this end, allow me to provide you with some suggestions that you can use in your proof.

Observe that by definition, the tensor product gives an $R$-bilinear map $\tau : (R / I) \times M \to (R / I) \otimes_R M$ defined by $\tau(r + I, m) = (r + I) \otimes_R m.$ We will need also an $R$-bilinear map $\varphi : (R / I) \times M \to M/IM.$ Hint: for a module $N$ over a commutative ring $S,$ the multiplication map $S \times N \to N$ that sends $(s, n) \mapsto s \cdot n$ is always bilinear.

Once we have two $R$-bilinear maps $\tau : (R / I) \times M \to (R / I) \otimes_R M$ and $\varphi : (R / I) \times M \to M / IM,$ the universal property of the tensor product gives a unique $R$-bilinear map $\psi : (R / I) \otimes_R M \to M / IM$ such that $\psi \circ \tau = \varphi,$ i.e., we have that $\psi[(r + I) \otimes_R m] = \varphi(r + I, m)$ for all $r + I \in R / I$ and $m \in M.$

Can you find a well-defined $R$-bilinear map $\gamma : M / IM \to (R / I) \otimes_R M$ such that $\psi \circ \gamma$ is the identity map on $M / IM$? Hint: there is a "standard" way to give a map $M \to R \otimes_R M.$ If you mimic this idea, you will find exactly what you are looking for.


问:
Linear Regression - Proof Sum Adds to Zero
In linear regression, why is $\sum(X_{i} - \mu_{x})$ = $0$?  I understand that for ($\sum$ $Y_{i}$ minus the fitted value of Y) = $\sum$ $e_{i}$ this is true but why is this other fact true?</p>

答: Let $\mu_x$ denote the mean of the observations $X_i$. Then $X_i-\mu_x$ is the deviation of $X_i$ from the mean value or "error". Positive value for $X_i-\mu_x$ shows that $X_i$ is above the mean value while negative value of $X_i-\mu_x$ shows that $X_i$ is below the mean. By a little consideration it can be seen that the sum of errors must be zero since mean is the value where errors above equal the errors below. Hence $\sum(X_{i} - \mu_{x})=0$.</p>


问:
Integral of a function in the 1st/3rd quadrant
If $f(x)\cdot x>0$ and $f(0)=0$ it is right that $\int_0^x f(x)dx>0$?

答:
No. Consider $x = 0$.

Assuming $x \neq 0$, consider the two cases of $x > 0$ and $x < 0$. Use the fact that the integral of a Riemann integrable function which is strictly positive (apart from a point) has a strictly positive integral. (See this for example.)

(The answer is "Yes" then.)


问:
Does this differential equation have a unique solution?
Consider the Initial Value Problem

$\dfrac{dy}{dx} = \sqrt{y+a}$ where $a > 0$ and $y(0) = 0$

Does this have a unique solution ?

If I try to solve it :

$\dfrac{dy}{\sqrt{y+a}} = dx$ gives,

$2\sqrt{y+a} = x + c  \ldots (1)$

Now, here using Initial Condition

$2\sqrt{y+a} = x + 2\sqrt{a}$ or, $y+a = (x+ 2\sqrt{a})^2$

But. in $(1)$ if I simplify the equation

$y+a= (\dfrac{x}{2} + \dfrac{c}{2})^2$

and here via initial condition we get two values  $c = 2\sqrt{a}$ and $c = -2\sqrt{a}$
and thus this has two different solutions.

So, can anyone explain does this Differential Equation has unique solution or multiple solutions ?

Thank you.

答:
The so-called "solution"

$$y=(x/2-\sqrt{a})^2-a$$

does not actually satisfy the differential equation $y'=\sqrt{y+a}$. Note that, according to this diff. eq., the derivative $y'$ should always be positive, since the square root of a real number, if it exists, is always positive. However, for the "solution" above,

$$y'(0)=-\sqrt{a}$$

which is negative. Thus, the only true solution is

$$\color{green}{y=(x/2+\sqrt{a})^2-a}$$


答:
Answer based on uniqueness theorem

As $a \gt 0$, the map $y \mapsto \sqrt{y+a}$ is locally Lipschitz around zero and therefore the IVP has a unique local solution according to Picard-Lindelöf theorem.

Answer based on integration

See the other answer.

Title: Does this differential equation have a unique solution?

Body:
Consider the Initial Value Problem

$\dfrac{dy}{dx} = \sqrt{y+a}$ where $a > 0$ and $y(0) = 0$

Does this have a unique solution ?

If I try to solve it :

$\dfrac{dy}{\sqrt{y+a}} = dx$ gives,

$2\sqrt{y+a} = x + c  \ldots (1)$

Now, here using Initial Condition

$2\sqrt{y+a} = x + 2\sqrt{a}$ or, $y+a = (x+ 2\sqrt{a})^2$

But. in $(1)$ if I simplify the equation

$y+a= (\dfrac{x}{2} + \dfrac{c}{2})^2$

and here via initial condition we get two values  $c = 2\sqrt{a}$ and $c = -2\sqrt{a}$
and thus this has two different solutions.

So, can anyone explain does this Differential Equation has unique solution or multiple solutions ?

Thank you.


问:
Convergence of $\sum_{k=1}^{\infty} \sum_{u=0}^{k-1}\frac{r^{u}}{(k-u)^2}$
Assume $r \in (0,1)$, I'm looking at the following sum, conjecture that it converges.

$$  \sum_{k=1}^{\infty} \sum_{u=0}^{k-1}\frac{r^{u}}{(k-u)^2} $$

This is very similar to the result of one earlier question, but I'm not sure how to cleanly deal with the $(k-u)^2$ denominator.

Edit:
Through playing around with Wolfram Mathematica, I'm getting

$$
\begin{align*}
 \sum_{k=1}^{\infty} \sum_{u=0}^{k-1}\frac{r^{u}}{(k-u)^2} &=  \sum_{k=1}^{\infty}\sum_{u=1}^{k}\frac{r^{k-u}}{u^2}\\ &=
\sum_{k=1}^{\infty}  \left(\sum_{u=1}^{k} \int_{0}^{r} \left(\frac{k}{u^2} - \frac{1}{u}\right)x^{k-u-1} dx   \right) \\
   &=   \left(\sum_{u=1}^{\infty}   \int_{0}^{r} \sum_{k=1}^{\infty} \left(\frac{k}{u^2} - \frac{1}{u}\right)x^{k-u-1} dx   \right) \\
   &=   \left(\sum_{u=1}^{\infty}   \int_{0}^{r} \frac{x^{-u} (u x-u+1)}{u^2 (x-1)^2}  dx  \right) \\
   &=   \left( \int_{0}^{r} \frac{\text{Li}_2\left(\frac{1}{x}\right)-x \log
    \left(\frac{x-1}{x}\right)+\log
    \left(\frac{x-1}{x}\right)}{(x-1)^2} dx   \right) \\
   &=   \left( -\frac{\text{Li}_2\left(\frac{1}{r}\right)}{r-1}+\text{Li}_2\left(\frac{r-1}{r}\right)+\log \left(\frac{1}{r}\right) \log
   \left(\frac{r-1}{r}\right)-\frac{\pi ^2}{6}   \right) \\
\end{align*}
$$
However, I can't validate whether all of the steps are valid and not misspecified by the software, especially the complex logarithms..

答: $$\sum_{k=1}^\infty\sum_{u=0}^{k-1}\frac{r^{u}}{(k-u)^2}=\sum_{k=1}^\infty\sum_{n=1}^k\frac{r^{k-n}}{n^2}=\sum_{n=1}^\infty\sum_{k=n}^\infty\frac{r^{k-n}}{n^2}=\sum_{n=1}^\infty\frac{1}{n^2}\frac{1}{1-r}=\frac{\pi^2}{6(1-r)}$$ (similarly to your prior question, rearranging is valid because of nonnegativity of the terms).


问:
A couple of questions about limits
I'm doing some homework for a calculus course I am taking and, though I feel silly because I've done problems like this before, I'm having some issues. I feel like I am getting the wrong answers for a couple of problems, and I'd like a second (or third, or tenth) opinion.

1) Find $\lim_{x\to 1}f(x)$ where: $$f(x) = \left\{
 \begin{array}{cc}
     x^2+2, & x\neq 1\\
     1, & x=1 \end{array} \right.$$

I'm thinking that the limit does not exist, since you have to plug 1 in to the equation, making x=1, meaning f(x) =1. Please, correct me if I'm wrong.

2) $\lim_{x\to -1} \frac{x^3-1}{x+1}$

I said it is undefined, but I'm not sure...

答:
Limits are not about "plugging in" a value of $x$.  The closest explanation of how to think of a limit would be to "plug in" values of $x$ that are close to the limiting value of $x$.  So in the first exercise, you should consider $x$ very close to $1$, and notice that values very close to $3$ come out.  Do something similar for the second limit after you factor the numerator as $(x+1)(x^2-x+1)$.

If you simply let $x$ be the value $1$ or $-1$ in these problems, then you do not yet really understand what it means to take a limit.

Title: A couple of questions about limits


问:
Dominating function of $f_n(x)=\frac{nx\sin(x)}{1+(nx)^{\frac{3}{2}}}$
I'd like to use the dominated Convergence theorem to solve $\lim_{n\to\infty}\displaystyle\int_0^1\frac{nx\sin(x)}{1+(nx)^{\frac{3}{2}}} \, dx$ but am a bit stuck on actually proving that $f_n(x)$ is dominated (meaning there is an integrable function $g(x)$ such that $|f_n(x)| < g(x) \forall n \in \mathbb{N}$). So far I have $|f_n(x)| \leq \biggl|\frac{nx}{1+(nx)^{3/2}}\biggr| \leq \biggl|\frac{nx}{(nx)^{3/2}}\biggr| = \Bigl|\frac{1}{\sqrt{nx}}\Bigr|$ but I don't really know how to proceed from here.

答:
You were almost done as

$$\int_0^1 \frac{1}{\sqrt x} \ dx$$ converges.


问:
Derivative of a particular logarithmic function
I have a vector $\mathbf{x} = [x_1 \,\, x_2 \,\, x_3] \in \mathbb{R}^3$, and a function $f\left(\mathbf{x} \right) = \log\left(x_1 - 3x_2^2 + 5x_3^2 \right)$.

What is the derivative of $f\left(\mathbf{x} \right)$ w.r.t. $\mathbf{x}$? Do I have to perform the derivative w.r.t. $x_1$, $x_2$ and $x_3$ at the same time? For example, is it correct to calculate it as below?

\begin{equation}
\frac{\partial \, f\left(\mathbf{x} \right)}{\partial \, \mathbf{x}} = \frac{1 - 6x_2 + 10x_3}{x_1 - 3x_2^2 + 5x_3^2}\,.
\end{equation}

Any help will be very appreciated!

答:
In your case you have

$$f(x_1, x_2, x_3) = \ln(x_1 - 3x_2^2 + 5x_3^2)$$

Which is a function in three variables. When it comes to take the derivative of this function wrt $x_1, x_2, x_3$ the gradient comes into play

Gradient

$$\nabla f(x_1, x_2, x_3) = \frac{\partial f}{\partial x_!} \hat{i} + \frac{\partial f}{\partial x_2} \hat{j} + \frac{\partial f}{\partial x_3} \hat{k}$$

$$\frac{\partial f}{\partial x_1} = \frac{1}{x_1 - 3x_2^2 + 5x_3^2}$$

$$\frac{\partial f}{\partial x_2} = -\frac{6x_2}{x_1 - 3x_2^2 + 5x_3^2}$$

$$\frac{\partial f}{\partial x_3} = \frac{10x_3}{x_1 - 3x_2^2 + 5x_3^2}$$


问:
Proving $\sum_{cyc} \frac{a(a^2+2bc)}{b+c}\ge \frac{{(a+b+c)}^2}{2}$

Prove that (where $a,b,c>0$) $$\sum_{cyc} \frac{a(a^2+2bc)}{b+c}\ge \frac{{(a+b+c)}^2}{2}$$

I have found a proof for this problem but it is very lengthy and is not nice.(Its not using computer though).I shall post it later.

Background:
Here however is a very similar problem

If $a,b,c>0$ then prove that $$\sum_{cyc} \frac{a(a^2+bc)}{b+c}\ge a^2+b^2+c^2$$

There is a beautiful proof of  this by Michael Rozenberg (arqady)

WLOG $a\ge b\ge c$ rewrite the inequality as $$\sum_{cyc}\frac{a(a-b)(a-c)}{b+c}\ge 0$$ but  $$\sum_{cyc}\frac{a(a-b)(a-c)}{b+c}\geq\frac{a(a-b)(a-c)}{b+c}+\frac{b(b-a)(b-c)}{a+c}=$$
$$=(a-b)(\frac{a(a-c)}{b+c}-\frac{b(b-c)}{a+c})\geq0$$

However this method doesnt work...

I am looking for a clean and smooth proof (without using BW,uvw or complete expanding)

I am not planning to disclose how I proved the inequality as it may spoil the fun!

答:
We want to show
$$
\sum_{cyc} \frac{a(a^2+2bc)}{b+c}\ge \frac{{(a+b+c)}^2}{2}
$$

The following steps mainly show how this inequality can be maximally simplified.
The final steps then have been proved by the OP Albus Dumbledore himself, and others.

Due to homogeneity, let $a+b+c =1$, then equivalently

$$
\sum_{cyc}  \frac{(a-1+1)(a^2+2bc)}{1-a} \ge \frac{1}{2} $$
$$-\sum_{cyc} (a^2+2bc) + \sum_{cyc} \frac{a^2+2bc}{1-a} \ge \frac{1}{2}$$
$$-(a+b+c)^2+ \sum_{cyc} \frac{a^2+2bc}{1-a} \ge \frac{1}{2}$$
$$\sum_{cyc} \frac{a^2-1+1+2bc}{1-a} \ge \frac{3}{2}
$$
$$-\sum_{cyc} (1+a)+\sum_{cyc} \frac{1+2bc}{1-a} \ge \frac{3}{2}
$$
$$\sum_{cyc} \frac{1+2bc}{b+c} \ge \frac{11}{2}
$$
Now we have
$$\frac{2bc}{1-a} =\frac{2bc}{b+c} = b+c -a-1  + \frac{1 -(a^2+b^2+c^2)}{b+c} $$
which leads to
$$(2 -(a^2+b^2+c^2))\sum_{cyc} \frac{1}{b+c} \ge \frac{15}{2}
$$
Now we have isolated a single sum, which is the main benefit of this answer.

This sum can be evaluated:
$$\sum_{cyc} \frac{1}{1-a} = \frac{3 -  (a^2+b^2+c^2)}{1 - 2abc - (a^2+b^2+c^2)}
$$
which leaves to show
$$(2 -(a^2+b^2+c^2))(3 -(a^2+b^2+c^2)) \ge \frac{15}{2} (1 - 2abc - (a^2+b^2+c^2))
$$
Let $x = a^2+b^2+c^2$ then we have to show
$$2x^2 + 5x + 30 abc \ge 3
$$

Note that by now, no single change has been made to the original question, since all transformations are equivalences.

The last inequality  can be proved (amongst other methods)  by Schur,  which has been done by the OP Albus Dumbledore himself, and others, see here. $\qquad \Box$


问:
$A \cup C \sim B \cup C$ with $A \cap C = \emptyset$ and $B \cap C = \emptyset$ such that $A \sim B?$
Let $A,B$ and $C$ be sets. Suppose that $A \cup C \sim B \cup C$ and that $A \cap C = \emptyset$ and $B \cap C = \emptyset$. Is it necessarily the case that $A \sim B?$ Give a proof or counterexample.
Where $\sim$ means bijection.

Actually the reciprocal is true and I already tried it, but my intuition tells me that this is not true. I would like to know if I am wrong or the counterexample is very exotic. Thanks

答:
No. Let $A = \{1, 2\}, B = \{2\}, C = \{3, 4, \ldots\}$.

This satisfies $A \cap C = \emptyset$ and $B \cap C = \emptyset$. Then $A \cup C = \{1, 2, 3, \ldots\}$ and $B \cup C = \{2, 3, \ldots\}$. We have the bijection $f: A \cup C \to B \cup C$ where $f(n) = n+1$. But $A$ and $B$ have different cardinalities and clearly cannot have a bijection.


问:
Prove that three points are aligned in the complex plane
So here is my problem: Let $A$, $B$ and $C$ be three distinct points of affixes $z_A$, $z_B$ and $z_C$. We define $Z=\frac{z_C-z_A}{z_B-z_A}$, show that $Z$ is real if and only if the points $A$, $B$ and $C$ are aligned.

This is what I've done so far: If $Z$ is real then $Z=\overline{Z}\longrightarrow$ $\frac{z_C-z_A}{z_B-z_A}=\frac{\overline{z_C}-\overline{z_A}}{\overline{z_B}-\overline{z_A}}\longrightarrow$  $(z_C-z_A)(\overline{z_B}-\overline{z_A})=(\overline{z_C}-\overline{z_A})(z_B-z_A )$

After expand: $z_C(\overline{z_B}-\overline{z_A})-z_A\overline{z_B}=\overline{z_C}(z_B-z_A)-\overline{z_A}z_B$

And now I don't know what to do with that, I am dumb, stupid or dumb?

Thanks for your help!

答:
Given is, $Z=(z_C-z_A)/(z_B-z_A)$ is real. Therefore $\arg Z =0, \pi$. Now

$$\arg Z= \arg (z_C-z_A) -\arg (z_B-z_A) \in \{ 0,\pi \}$$
$$\Rightarrow \arg (z_C-z_A) = \arg (z_B-z_A) $$
$$ \text{or,} \quad \arg (z_C-z_A) = \pi + \arg (z_B-z_A)$$

So either  $(z_C-z_A)$ has same slope as $(z_B-z_A)$ or $(z_C-z_A)$ has same slope as $-(z_B-z_A)$.

In both cases $A,B,C$ lie on same line. In first case $B,C$ lie on same side of $A$. In second case $B,C$ lie on opposite sides of $A$.


答:
A counterexample is the following:

$A=\{-2,-1\}$

$B=\{-5,-4,-3\}$

$C=\mathbb{N}$

where $\mathbb{N}$ is the set of all positive integers.

Your property is also true for infinite sets such that the cardinality of $C$ is less or equal to both the cardinalities of $A$ and $B$.

Indeed, if the cardinality of $C$ is less or equal to both the cardinalities of $A$ and $B$, then $A\sim A\cup C\sim B\cup C\sim B$.

Title: $A \cup C \sim B \cup C$ with $A \cap C = \emptyset$ and $B \cap C = \emptyset$ such that $A \sim B?$

Body:
Let $A,B$ and $C$ be sets. Suppose that $A \cup C \sim B \cup C$ and that $A \cap C = \emptyset$ and $B \cap C = \emptyset$. Is it necessarily the case that $A \sim B?$ Give a proof or counterexample.
Where $\sim$ means bijection.

Actually the reciprocal is true and I already tried it, but my intuition tells me that this is not true. I would like to know if I am wrong or the counterexample is very exotic. Thanks


问:
Confusion in a Simple probability problem
Suppose if i roll a fair die 6 times, what is the probability of getting (1, 1, 1, 1, 1, 1)?

The answer is $(\frac{1}{6})^6$, and the reasoning is as follows: Assuming independence,
$$P(1, 1, 1, 1, 1, 1)= P(1)\cdots P(1)$$
Since the die is fair, $P(1)=\frac{1}{6}$, and hence the result.

My question is, What does independence mean here?  Is it independent of the event or something else?

In short, I want to know why $P(1, 1, 1, 1, 1, 1)= P(1)\cdots P(1)$?. (I have no problem with assuming independence). Please help.

答:
"Two events are independent... if the occurrence of one does not affect the probability of occurrence of the other." Source

The action of rolling any die, in this question, doesn't magically make any other die 'unfair', thus, the event of each die landing on a certain number is independent, and the probability of all dice landing on $1$, is equal to the multiplication of probabilities of each die landing on $1$.


问:
Generating Pythagorean Triples Using a New Method?
Using a right triangle with side lengths $(a,b,c)$ where $a , b < c$,
I was thinking about how the area of a Pythagorean triple can be found using the Pythagorean triple right before it and I came across something that worked for a large number of Pythagorean triples,  $12r^2 + a_{k- 1}b_{k - 1} = a_kb_k$, a recursive formula where $k$ represents the $k$th term in a sequence. This seemingly generates a sequence of Pythagorean triples that I could not find used in any other formula. Its important to note that $12r^2$ is  twice the area of Pythagorean triples that stem from side lengths $(3,4,5)$. Using this formula we can find the $1st$ term of sets where the inradius of each Pythagorean triple is $r + r^2k$ and the relationship between the side lengths are still defined by our recursive formula.

These $1st$ terms are triplets with an even value of $a$ where $r$ increases by $2$:

$(8,15,17),(12,35,37),(16,63,65)...$

Note: We find this using $(8,15,17)$ as we have a recursive formula as well as the knowledge that $r = \frac{a + b - \sqrt{a^2 + b^2}}2$,  which lets us find the side lengths of each Pythagorean triple.

Here is a sample of what they generate:
$$\begin{array}{c|c|c|c}
set_1&15,8,17&33,56,65&51,140,149&69,260,269 \\ \hline
set_2&35,12,37&85,132,157&135,352,377&185,672,697 \\ \hline
set_3 &63,16,65&161,240,289&259,660,709&357,1276,1325& \\ \hline
set_4&99,20,101&261,380,461&423,1064,1145&585,2072,2153  \\ \hline
\end{array}$$

I couldn't seem to find any similar formulas to this one or any method of generating Pythagorean triples that follow this sequence, I am looking for a proof.

答:
Your formula does generate Pythagorean triples but misses most of them and appears to require seeds to work.

I'm not sure what you are generating. You do generate triples where $C-A=2$ in the first column but that can be generated more easily by
$\quad A=4n^2-1\quad B=4n\quad C=4n^2+1.\quad $ The rest of the table shows no pattern that I can see, like a consistent side difference within a set or consistent increment of side values within a set. The following formula generates all primitives and a few that are not but there is a consistent
$C-B=(2n-1)^2\quad$ and $\quad A_{n+1}-A_{n}=2(2n-1).\quad$ It is the formula derive when
$A=(2n-1+k)^2-k^2,\space
B=2(2n-1+k)k,\space
C=(2n-1+k)^2+k^2$

\begin{align*}
A=(2n-1)^2+ \quad &2(2n-1)k\\
B=\hspace{55pt} &2(2n-1)k\quad+2k^2\\
C=(2n-1)^2+ \quad &2(2n-1)k\quad +2k^2
\end{align*}
Here is a sample of what it generates
$$\begin{array}{c|c|c|c|c|}
n & k=1 & k=2 & k=3 & k=4   \\ \hline
Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41  \\ \hline
Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65 \\ \hline
Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 \\ \hline
Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137 \\ \hline
Set_{5} &99,20,101 &117,44,125 &135,72,153 &153,104,185 \\ \hline
\end{array}$$
Your formula generates the first column but nothing with a pattern I can see in the other cells. If you do want to work with areas, there is a list of them here.  If you can figure out how to generate this sequence, I can show you how to find all of the
$1,\space 2, \text{ or } 3\space $ triples
that correspond to each area.


答: First and foremost, the answer is definitely not $1/36$, as $P(1,1,1,1,1,1) = P(1) \cdots P(1) = (1/6)^6$. In probability theory, two events $E$ and $F$ are independent if $P(E \cap F) = P(E) P(F)$, which in the context of the problem means that $P(1)$ always occurs with probability $1/6$. You used the condition that events are independent in your assertion that $P(1, 1, 1, 1, 1, 1) = P(1) \cdots P(1)$.

Title: Confusion in a Simple probability problem

Body:
Suppose if i roll a fair die 6 times, what is the probability of getting (1, 1, 1, 1, 1, 1)?

The answer is $(\frac{1}{6})^6$, and the reasoning is as follows: Assuming independence,
$$P(1, 1, 1, 1, 1, 1)= P(1)\cdots P(1)$$
Since the die is fair, $P(1)=\frac{1}{6}$, and hence the result.

My question is, What does independence mean here?  Is it independent of the event or something else?

In short, I want to know why $P(1, 1, 1, 1, 1, 1)= P(1)\cdots P(1)$?. (I have no problem with assuming independence). Please help.


答: The chain rule is true of complex-valued functions of a real variable. If $s(t)$ moves smoothly within the domain as $r\in\mathbb R$ changes, then $\dfrac d {dt} \big(f(s(t)) -s(t)\big) = f'(s(t))s'(t)-s'(t).$ If $f'(s(t))=1$ for all values of $t,$ then this derivative is $0.$ A theorem from real variables says that if the derivative of a real-valued function of a real variable is $0$ on an interval, then the function is constant on the interval. That that is also true of complex-valued functions of a real variable follows just by considering the real and imaginary parts.

Title: Derivative of complex functions

Body:
Prove that if $f'(z) = 1$ everywhere in a domain $D$ then $f(z) = z + c$
for some complex constant $c.$
And $z$ is a complex no.


问:
BMO$2$ $2021$ Q$3$: Prove AI and KD meet on the circumcircle.
I’m stuck on question $3$ from BMO$2$ $2021$ (https://bmos.ukmt.org.uk/home/bmo2-2021.pdf)

So the first observation I’ve made is that K is the intersection of the circle with diameter AI with $\Gamma$ (converse of Thales)

From there the only idea that came to me was to somehow try and leverage symmetry along the line AIT (where T is the midpoint of the arc BC), since the incircle and the circle with diameter AI are symmetric along here and AC and AB are each others reflections in this way.

My diagram (r denotes reflection in AT): https://i.stack.imgur.com/lAjwd.jpg

And I thought it would be easier to prove the equivalent result that K D and T are colinear (here I define T as the midpoint of the arc BC or equivalently the intersection of AI and $\Gamma$) But I don’t really see any way to proceed.

答:
Draw $AO$ and extend it to meet the circumcircle at $E$. Next, draw $IE$ and extend it to meet the circle again at $K$. $\angle AKI$ will be $90^{\circ}$.$AI$, when extended, will meet the circumcircle at $M$, the midpoint of arc $BC$. It is enough to prove that $\angle IMD=\angle IAE$ because if the points are colinear, then $\angle IMD=\angle IMK=\angle IAE$. Draw the line $IO$ and let it intersect the circle at points $P$ and $Q$.

If the circumradius is $R$ and the distance between the circumcentre and incentre is $d$, then, notice that, $AI\cdot IM=PI\cdot IQ=R^{2}-d^{2}$. Now, if the inradius is $r$, then, by Euler's Theorem, $AI\cdot IM=2Rr$. Rearranging this a bit gives, $\frac{AI}{AE}=\frac{ID}{IM}$. Also, observe that, in $\triangle AIE$ and $\triangle IDM$,$\angle DIM=\angle IAE$. Equipped with the ratio derived earlier, $\triangle AIE\sim \triangle IDM$ and thereafter $\angle IMD=\angle AEI$.


问:
Given that $\cos\left(\dfrac{2\pi m}{n}\right) \in \mathbb{Q}$ prove $\cos\left(\dfrac{2\pi}{n}\right) \in \mathbb{Q}$
Given that $\cos\left(\dfrac{2\pi m}{n}\right) \in \mathbb{Q}$, $\gcd(m,n) = 1$, $m \in \mathbb{Z}, \, n \in \mathbb{N}$ prove that $\cos\left(\dfrac{2\pi}{n}\right) \in \mathbb{Q}.$

I know nothing about how to attack the problem. I believe I need to suppose that $\cos\left(\dfrac{2\pi}{n}\right) \not \in \mathbb{Q}$ and somehow show that $\cos\left(\dfrac{2\pi m}{n}\right) \not \in \mathbb{Q}$ what would have been a contradiction. Could you give me some hints?

答:
Let $\zeta=\exp(2\pi i/n)$ be a primitive $n$-th root of unity. You know that $\Re(\zeta^m)\in\mathbb Q$ and you want to show that $\Re(\zeta)\in\mathbb Q$. Since $m$ and $n$ are coprime $\zeta^m$ is a primitive $n$-th root of unity as well. This implies that we have $(\zeta^m)^k=\zeta$ for a $k\in\mathbb N$.

So it is enough (even better) to prove that for a number $\zeta \in \mathbb C$ with $|\zeta|=1$ and $\Re(\zeta) \in \mathbb Q$ you have $\Re(\zeta^n) \in \mathbb Q$ for all $n \in \mathbb N$.

We have
$$\zeta^n=(x+iy)^n  = \sum_{k=0}^n  {{n}\choose{k}} x^k (iy)^k $$
The real part of this is given by only giving a shit on the even $k$. But then we have an even power of $y$ and we have $y^2=1-x^2 \in \mathbb Q$, so everything is in $\mathbb Q$.


问:
Can the image of a continuous, but nowhere-differentiable, function be a borel set?
I apologize if this question is short, but in essence, I wonder:

Does there exist any function, $f(t): [0,1] \to \mathbb{R}^2$, which is continuous but nowhere differentiable, such that the set $$\{f(t):t\in[0,1]\} \in \mathcal{B}(\mathbb{R}^2)?$$

答:
That's a compact set, so it's always Borel. Therefore the question reduces to whether or not there is a continuous nowhere differentiable function $f:[0,1]\to\Bbb R^2$, which I don't believe I need to persuade you of. If you substitute $t\in[0,1]$ with, say, $t\in\Bbb R$, then you can write the set as $$\{f(t)\,:\,t\in\Bbb R\}=\bigcup_{n\in\Bbb N}\{f(t)\,:\, t\in[-n,n]\}$$

which is a $F_\sigma$ set.


问:
Integral with modified Bessel functions of the first and second kind
Consider the following integral
$$I\left(a, b\right)=\int\limits _0^{+\infty}\cos\left(x\right)I_1\left(ax\right)K_1\left(bx\right)dx,$$
where $a$ and $b$ are parameters such that $b\geq a \geq 0$, and $I_\nu\left(x\right)$ and $K_\nu\left(x\right)$ are the modified Bessel functions of the first and second kind, respectively.

Is it possible to express this integral in terms of elementary and basic special functions?

Similar issue has already been addressed; the result involves complete elliptic integrals.

答:
$$\int_0^{\infty } \cos (x) I_1(a x) K_1(b x) \, dx=\frac{Q_{\frac{1}{2}}\left(\frac{a^2+b^2+1}{2 a b}\right)}{2 \sqrt{a b}}$$

where:$Q_{\frac{1}{2}}\left(\frac{a^2+b^2+1}{2 a b}\right)$ is  Legendre Q Function.

from Book: Table of Integrals, Series, and Products 8th Edition on page: 726 in: 6.672.4


问:
An Einstein summation problem
Is the expression $(T_{abc}+T_{cab}+T_{bca})X^aX^bX^c$ equal to $3T_{abc}X^aX^bX^c$, using Einstein summation notation, even if we allow a,b,c to range over different values? I understand that they are probably equal if $a,b,c$ range over the same values, but I do not think that $\sum_a\sum_b\sum_c T_{abc}X^aX^bX^c$ is always equal to $\sum_a\sum_b\sum_c T_{cab}X^aX^bX^c$ if $a,b,c$ range over different values. Thanks in advance!

答:
As you thought, if they range over different values, this is not true. For instance, take the sum
$$\sum_{a=1}^1 \sum_{b=1}^1 \sum_{c=1}^2 =3T_{111}X^{1}X^{1}X^{1} + (T_{112} + T_{121} + T_{211})X^{1}X^{1}X^{2}$$
and it is not necessary that the last term is equal to $3T_{111}X^{1}X^{1}X^{2}$ since it is necessary that the $T_{ijk}$ are equal.

Title: An Einstein summation problem


问:
Finding Laurent of $ \frac{e^{1/z}}{1-z} $
The question is: Determine the Laurent expansion in the region $0<|z|<1$?
$$ \frac{e^{1/z}}{1-z} $$
Here is how i have tried to solve it but my answer is not correct and i don't know why?
$$e^{1/z}=\sum_{n=1}^{\infty}\frac{1}{z^kk!}=\frac{1}{z}+\frac{1}{z^22!}+\frac{1}{z^33!}+... $$
$$\frac{1}{1-z}=\sum_{n=0}^{\infty}z^n=1+z+z^2+...$$
$$\left(\frac{1}{z}+\frac{1}{z^22!}+\frac{1}{z^33!}+...\right)\left(1+z+z^2+...\right)  $$
Any suggestion would be great, thanks

答:
Hint

The Laurent series centered on zero has the form

$$\sum_{n=-\infty}^\infty a_n z^n.$$

Hence you need to use Cauchy product to write

$$\left(1 +\frac{1}{z}+\frac{1}{z^22!}+\frac{1}{z^33!}+...\right)\left(1+z+z^2+...\right)  $$

with the appropriate form.

Alternatively, use the fact that

$$a_n =\frac{1}{2i \pi} \oint_\gamma \frac{f(z) dz}{z^{n+1}}$$ where $\gamma$ is the counterclockwise circle centered on $0$ with radius $1/2$.


问:
NFA to DFA Conversion Help
Consider the language of all words over {a, b} that either begins and ends with aa or bb (but both are
the same within each word). That is, {ccycc | c ∈ {a, b}, y ∈ Σ
∗}

I have drawn the resulting NFA for the language described above, what would the corresponding DFA look like? What would the steps look like to get from an NFA to a DFA?

答: In this case you are very close to a DFA already.  First add a 'dead state' (non-final) with additional transitions to it from q1, q2, q5, and q6 on symbols b, a, b, and a, respectively, plus an arrow from the state to itself on a and b.  Second, split q7 into two final states q8, q9, with q5->q8 on a and q6->q9 on b, and add loops q8->q8 on a, q9->q9 on b, and cross-transitions q8->q6 on b, q9->q5 on a.</p>

Title: NFA to DFA Conversion Help


问:
How to compute $\sum^{n}_{i=0}(2t)^i$
How can I work out this easy sum?

$$\sum^{n}_{i=0}(2t)^i$$

答: Hint: This is the sum of the first $n+1$ terms of a geometric series. Use the formula.

Title: How to compute $\sum^{n}_{i=0}(2t)^i$


问:
Having deposited $\$5,000$ at $9\%$ annual interest, for how many years can you withdraw $\$500$? What's a good way to iterate this?
NOTES:
I hope the tags are appropriate, this is a homework problem EDIT (another note): if it makes a difference, this question may involve calculus since it's in my calculus homework... I doubt it would make sense to use calculus in any way here

If you deposit $5,000$ dollars at $9\%$ annual interest, for how many years can you withdraw $500$ dollars (starting after the first year) without running out of money?

I've tried several equations, such as: $f(x)=5000(1.09)^x-500x$ but this never reaches $0$

The most recent thing I've tried is setting up an iteration equation: $x_{n+1}=x_n(1.09)-500$ where $x_0=5000$ which I will iterate until $x_{n+1}\le0$.

Question: Can I easily make Desmos (and the like) repeat iteration equations to infinity? AND Is there a way I can solve this problem without having to iterate (an equation that doesn't use the previous output as the next input)?

Bonus: If there's a way to turn my iteration into a "regular"<--(lack of technical terms) equation, can all iteration equations be made that way as well? Is there a general way to take an iteration equation and turn it into a simple $f(x)$ function?

答:
Let $A=5000$ be the Amount you start with, and let $W=500$ be the amount you Withdraw at the end of each year. Let $r=.09$ be the annual interest rate, and, for convenience, let $R=1+r$. The amount that's left after your first withdrawal is $AR-W$. On compounding this, the amount that's left after your second withdrawal is $(AR-W)R-W=AR^2-W(R+1)$. Compounding this remainder, the amount left after the third withdrawal is

$$(AR^2-W(R+1))R-W=AR^3-W(R^2+R+1)$$

and so forth. I.e., the amount left after the $n$th withdrawal is

$$AR^n-W(R^{n-1}+R^{n-2}+\cdots+R+1)=AR^n-W{R^n-1\over R-1}$$

Our only constraint is that this amount must be non-negative, that is

$$AR^n\ge W{R^n-1\over R-1}$$

A little algebra rewrites this as

$$(1+r)^n=R^n\le{W\over W-(R-1)A}={W\over W-rA}$$

which implies

$$n\le{\log(W/(W-rA))\over\log(1+r)}$$

Plugging in our values for $A$, $W$, and $r$, we find that

$$n\le{\log(10)\over\log(1.09)}\approx26.719$$

which implies the number of times you can withdraw $\$500$ without running out of money is $26$.

Remark: It's important here that $W\gt rA$. That is, if your account earns more each year than you withdraw, you'll never run out of money!


问:
Let (X, d) be a metric space in which a finite subset A of X is dense. Determine all open sets in (X, d).
This is the question given to us. I know the conditions for an open set but I don't know how to find it when we have been given that A of X is dense. Can someone help me out with this</p>

答: Hint: Every finite subset $F$ of a metric space $(X,d)$ is a closed subset of $X$, and therefore $\overline F=F$.


问:
How should I go about this approaching-infinite limit problem?
I'm doing some exercises of limits approaching infinite, most are simple polynomials where only the highest degree term will matter in the end but for this one I couldn't find a solution (not correct at least).

$$\lim_{x\to-\infty}\frac{x^2+x+1}{(x+1)^3-x^3}$$

How should I proceed to get the correct answer? ($\frac{1}{3}$)

Also, while simplifying some questions this question took my mind: is it correct to say that $\sqrt{a+b}$ is equal to $\sqrt{a}+\sqrt{b}$?

Ps.: Without using l'Hôpital's rule.

答:
In general $\sqrt{a+b} \neq \sqrt{a} + \sqrt{b}$ (e.g. $\sqrt{2} \neq \sqrt{1} + \sqrt{1} = 2$).

The first problem can be solved by writing down what $(x+1)^3$ means in terms of a polynomial.


问:
Convergence of $\sum_{k=1}^{\infty} \sum_{u=0}^{k-1}\frac{r^{u}}{(k-u)^2}$
Assume $r \in (0,1)$, I'm looking at the following sum, conjecture that it converges.

$$  \sum_{k=1}^{\infty} \sum_{u=0}^{k-1}\frac{r^{u}}{(k-u)^2} $$

This is very similar to the result of one earlier question, but I'm not sure how to cleanly deal with the $(k-u)^2$ denominator.

Edit:
Through playing around with Wolfram Mathematica, I'm getting

$$
\begin{align*}
 \sum_{k=1}^{\infty} \sum_{u=0}^{k-1}\frac{r^{u}}{(k-u)^2} &=  \sum_{k=1}^{\infty}\sum_{u=1}^{k}\frac{r^{k-u}}{u^2}\\ &=
\sum_{k=1}^{\infty}  \left(\sum_{u=1}^{k} \int_{0}^{r} \left(\frac{k}{u^2} - \frac{1}{u}\right)x^{k-u-1} dx   \right) \\
   &=   \left(\sum_{u=1}^{\infty}   \int_{0}^{r} \sum_{k=1}^{\infty} \left(\frac{k}{u^2} - \frac{1}{u}\right)x^{k-u-1} dx   \right) \\
   &=   \left(\sum_{u=1}^{\infty}   \int_{0}^{r} \frac{x^{-u} (u x-u+1)}{u^2 (x-1)^2}  dx  \right) \\
   &=   \left( \int_{0}^{r} \frac{\text{Li}_2\left(\frac{1}{x}\right)-x \log
    \left(\frac{x-1}{x}\right)+\log
    \left(\frac{x-1}{x}\right)}{(x-1)^2} dx   \right) \\
   &=   \left( -\frac{\text{Li}_2\left(\frac{1}{r}\right)}{r-1}+\text{Li}_2\left(\frac{r-1}{r}\right)+\log \left(\frac{1}{r}\right) \log
   \left(\frac{r-1}{r}\right)-\frac{\pi ^2}{6}   \right) \\
\end{align*}
$$
However, I can't validate whether all of the steps are valid and not misspecified by the software, especially the complex logarithms..

Body:
Using the Iverson bracket:

\begin{align}
\sum_{k=1}^\infty\sum_{u=0}^{k-1}\frac{r^u}{(k-u)^2}
& = \sum_{k=1}^\infty\sum_{u=0}^{\infty}\frac{r^u[u<k]}{(k-u)^2}\\
& = \sum_{u=0}^\infty r^u\sum_{k=1}^{\infty}\frac{[u<k]}{(k-u)^2}\\
& = \sum_{u=0}^\infty r^u\sum_{k=u+1}^{\infty}\frac{1}{(k-u)^2}\\
& = \sum_{u=0}^\infty r^u\sum_{k=1}^{\infty}\frac{1}{k^2}
= \frac{1}{1-r}\frac{\pi^2}{6}
\end{align}


问:
Solve system of non-linear equations $\frac{1}{x^2}-\frac{1}{(y-x)^2}-\frac{1}{(1-x)^2}= \frac{1}{y^2}+\frac{1}{(y-x)^2}-\frac{1}{(1-y)^2}=0$
Whilst solving a problem, the following system of non-linear equations arose:

$$0= \frac{1}{x^2}-\frac{1}{(y-x)^2}-\frac{1}{(1-x)^2}$$

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

I have tried to solve it via brute force and by trying to add them both in a way that could give me the answer, but I still haven't got it.

I can't manage to find the intersection of the two parametric equations that these equations give. I know that I can find approximate solutions with the Wolfram Alpha tool, but I would like to know, if any, an analytical or strong numerical way to proceed.

Note: Out of all the solutions ($8$ it seems, $4$ real and $4$ complex), I only need the one that takes real values for $x$ and $y$, with $0<x<y<1$.

答:
Provided that $x\ne0;\;x\ne 1;\;y\ne0;\;y\ne 1;\;x\ne y$ we simplify the denominators and get
$$
\begin{cases}
-x^4+4 x^2 y-2 x y^2-2 x y+y^2=0\\
-2 x^2 y+x^2+4 x y^2-2 x y+y^4-4 y^3+2 y^2=0\\
\end{cases}\tag{1}
$$

add the two equation in $(1)$
$$-x^4+4 x^2 y-2 x^2 y+x^2+4 x y^2-2 x y^2-2 x y-2 x y+y^4-4 y^3+2 y^2+y^2=0$$
which can be factored
$$(y-x) (x+y-1) \left(x^2+x+y^2-3 y\right)$$

First factor $y-x=0$ is discarded for the conditions above

$x+y-1=0\to y=1-x$ give
$$x^4+6 x^3-11 x^2+6 x-1=0$$
two real solutions
$$(x=-7.5619;\;y=8.5619);\;(x=0.31926;\;y=0.68074)$$

$x^2+y^2+x-3 y=0$

I can't see elementary methods to get the other solutions. A graph can help, though.

$$...$$


问:
Prove that $x^3 -2$ is irreducible over $\mathbb{Q}(\sqrt{5})$.
I'm trying to prove what the tittle says. So far i know that $\sqrt[3]{2} \not \in \mathbb{Q}(\sqrt{5})$ and that $\{1,\sqrt{5} \}$ is a basis of $\mathbb{Q}(\sqrt{5})$ over the field $\mathbb{Q}$. I have tried to write $$x^3 -2=m(x)q(x)$$  With $m,q \in \mathbb{Q}(\sqrt{5})[x]$ and express the coefficients of $m $ and $q$ as a linear combination of the basis in an attempt to get a contradiction but in the last step it gets quite messy and so i think tha way is not going to have result.  Any help would be appreciated.

答:
Hint

If
$$x^3 -2=m(x)q(x)
$$
with non-constant $m,q$,
then one of $m,q$ must be linear and hence must have a root in $\mathbb{Q}(\sqrt{5})$. This root is a root of $x^3-2$.


问:
Confusion about simple probability
This Question stems from Ronald Fishers Tea problem in how to design experiments which I have reworded.

Anyway, suppose you have 2 different items in front of you, 5 of each which we can just call red and blue. So you have 10 items total,
and you want to randomly select the 5 red ones. The probability of getting this correct is $\frac{1}{10 \choose 5} $ but I can't shake the intuition that it would be $ \frac{5}{10} $ and I can't reason with myself as to why it wouldn't be. Can someone explain intuitively and not mathematically why it wouldn't be because I am able to conclude mathematically that it's not.

答:
The probability of randomly drawing one red item is $\frac{5}{10}$. But, after drawing that first one, the probability of drawing another red item is now $\frac{4}{9}$, since 9 items remain in total, and of them, 4 are still red. The next one is $\frac{3}{8}$, then $\frac{2}{7}$, and the final one has probability $\frac{1}{6}$.

The probability of all five of these events occurring is their product:

$$ p = \frac{5}{10}*\frac{4}{9}*\frac{3}{8}*\frac{2}{7}*\frac{1}{6} $$

$$ = \frac{5*4*3*2*1}{10*9*8*7*6} $$

$$ = \frac{5!}{\frac{10!}{5!}} $$

$$ = \frac{1}{\frac{10!}{5!5!}} $$

$$ = \frac{1}{10 \choose 5} $$

Title: Confusion about simple probability


问:
Proof $gcd(2^{2x},2x)=2gcd(2^{x},x)$
How do I proof $\operatorname{gcd}(2^{2x},2x)=2\operatorname{gcd}(2^{x},x)$ , I have managed to proof $\operatorname{gcd}(2^{2x},2x)=2\operatorname{gcd}(2^{2x-1},x)$ but haven't managed to get anywhere form there.

答:
All positive integers my be written in the form $n= 2^kw$ where $w$ is odd.(If $n$ is odd then $k =0$.  If $n$ is even than there is a highest power of $2$ that divides $n$ and the remaining quotient will be odd).

Now the only prime factors $n=2^kw$ and $2^m$, some power of $2$, will in common will be $2$ so $\gcd(2^m,n)= 2^j$ for some $j$.  And it's easy to convince ourselves $\gcd(2^m, 2^kw) = \gcd(2^m, 2^k) = 2^{\min(m,k)}$.

So let $x = 2^kw$

Then $\gcd(2^{2x}, 2x)= \gcd(2^{2x}, 2^{k+1}w) = 2^{\min(2x,k+1)}=2^{\min(2^{k+1}w, k+1} = 2^{k+1}$.  (Obviously $k+1 < 2^{k+1} \le 2^{k+1}w$)

Likewise $\gcd(2^x, x)= \gcd(2^x, 2^kw) = 2^{\min(x, k)}=2^{\min(2^kw,k)} = 2^k$ and the result follows.

.....

It may seem tricky but the gist is $2^{something}$ is $2$ to a higher power than the factor of two that divides $something$ itself so $\gcd(2^x, x) = 2^{the\ highest\ power\ that\ divides\ the\ something}$

So $\gcd(2^{huge\ therefore\ ignorable}, 2x)= 2^{one\ more\ power\ than\ the\ highest\ power\ of\ 2\ that\ divides\ x}= 2\cdot 2^{the\ highest\ power\ of\ 2\ that\ divides\ x}=2\gcd(2^{not\ as\ huge\ but\ still\ ignorable}, x)$

We could just as easily asked you to prove $\gcd(2^{x^2 + 7x+5}, 6x)=2\gcd(2^{9x-1},7x)$ for all the obfuscation involved.

Title: Proof $gcd(2^{2x},2x)=2gcd(2^{x},x)$


答: If the probability of getting $5$ reds were $\frac{5}{10}=\frac{1}{2}$, then by symmetry, we would expect the probability of getting $5$ blues to also be $\frac{1}{2}$. From this, we would conclude that the probability of getting a mix of reds and blues would be $1-\frac{1}{2}-\frac{1}{2}=0$. However, we know that the probability of getting a mix of reds and blues is non-zero, leading to a contradiction. Therefore, the probability of getting $5$ reds cannot by $\frac{5}{10}$.

Title: Confusion about simple probability

Body:
This Question stems from Ronald Fishers Tea problem in how to design experiments which I have reworded.

Anyway, suppose you have 2 different items in front of you, 5 of each which we can just call red and blue. So you have 10 items total,
and you want to randomly select the 5 red ones. The probability of getting this correct is $\frac{1}{10 \choose 5} $ but I can't shake the intuition that it would be $ \frac{5}{10} $ and I can't reason with myself as to why it wouldn't be. Can someone explain intuitively and not mathematically why it wouldn't be because I am able to conclude mathematically that it's not.


问:
Finding the Laurent series of $\frac{1+cos(z)}{(z-\pi)^3}$
How would one go about determining the Laurent series of such a series as

$$\frac{1+\cos(z)}{(z-\pi)^3}$$

I would like to expand this about the point $z= \pi$ in order to determine the residue however, I tried substitution $x = z-\pi$
so that we can expand;

$$\frac{1+\cos(x+\pi)}{x^3}$$
however the cos term is giving me trouble when it comes to expanding as it always provides so many x terms of the same order and I do not know how to account for this.

答:
Hint.

$\cos(x+\pi)=-\cos(x)$.

Body:
Hint.

$\cos(x+\pi)=-\cos(x)$.


问:
How do I examine the behavior of the Reciprocal Multifactorial Series $\Sigma \frac{1}{n!^{(k)}}$ as k approaches infinity.
This question is a direct continuation of my previous question.

How to prove the formula for the Reciprocal Multifactorial constant?

User metamorphy provided a lucid proof for the closed form equation of the Reciprocal Multifactorial Constant. And also a statement on its asymptote as k approaches infinity. Which is as follows:

$m(k)=1+e^{1/k}(H_k-\Delta_k)\ \text{and,}\\
> \Delta_k=\frac1k\underbrace{\int_0^1\frac{1-t}{1-t^{1/k}}t^{1/k}\frac{1-e^{-t/k}}{t}\,dt}_{\text{has
> a finite $k\to\infty$ limit}}\underset{k\to\infty}{\longrightarrow}0.$

I was aware that the Reciprocal Multifactorial would eventually approach a harmonic series due to the definition of a multifactorial, which is as follows:

$n!^{(k)} =   \begin{cases}
1 & \text{if $n=0$} \\
n & \text{if $0<n\leq k$} \\   n\left((n-k)!^{(k)}\right) & \text{if $n>k$}   \end{cases}$

As k approaches infinity the $k^{th}$ multifactorial will tend towards the natural numbers. And the reciprocal series would approach $1+H_k$

However I am asking this question to know how to prove this using the closed form formula as metamorphy did in the previous question. I wasn't able to solve it myself.

Unrelated: Do you think there are any other interesting properties about this series worth examining, since I've found it quite interesting?

答:
I think I have to answer it myself. Let the point of departure be $$m(k)=1+\frac{e^{1/k}}{k}\sum_{r=1}^k\int_0^1 t^{r/k-1}e^{-t/k}\,dt,$$ appearing in my post immediately before the substitution $t=kx$.

The statement being discussed just comes from $e^{-t/k}=1-(1-e^{-t/k})$: $$\int_0^1 t^{r/k-1}e^{-t/k}\,dt=\frac{k}{r}-\int_0^1 t^{r/k-1}(1-e^{-t/k})\,dt,$$ giving $m(k)=1+e^{1/k}(H_k-\Delta_k)$ as written: $$\Delta_k=\frac1k\int_0^1\left(\sum_{r=1}^k t^{r/k}\right)\frac{1-e^{-t/k}}{t}\,dt=\frac1k\int_0^1\frac{(1-t)(1-e^{-t/k})}{t(t^{-1/k}-1)}\,dt.$$

For curiosity, here's the asymptotics of $\Delta_k$ in more detail:

\begin{align*}
\Delta_k&=\frac1k f\Big(\frac1k\Big),\\
f(\lambda)&=\int_0^1\frac{(1-t)(1-e^{-\lambda t})}{t(e^{-\lambda\log t}-1)}\,dt\asymp\sum_{n=0}^{(\infty)}a_n\lambda^n,\\
a_0&=\int_0^1\frac{1-t}{-\log t}\,dt=\color{blue}{\log 2},\\
a_1&=-\frac12\int_0^1\frac{1-t}{-\log t}(t-\log t)\,dt=\color{blue}{-\frac14-\frac12\log\frac32},\\
a_2&=\frac{1}{12}\int_0^1\frac{1-t}{-\log t}(2t^2-3t\log t+\log^2 t)\,dt=\color{blue}{\frac{5}{48}+\frac16\log\frac43},\\
a_3&=-\frac{1}{24}\int_0^1\frac{1-t}{-\log t}(t^3-2t^2\log t+t\log^2 t)\,dt=\color{blue}{-\frac{11}{864}-\frac{1}{24}\log\frac54},
\end{align*}
and so on. For $m(k)$, use the power series of $e^{1/k}$, and known asymptotics of $H_k$...


Title: Finding the Laurent series of $\frac{1+cos(z)}{(z-\pi)^3}$

Body:
How would one go about determining the Laurent series of such a series as

$$\frac{1+\cos(z)}{(z-\pi)^3}$$

I would like to expand this about the point $z= \pi$ in order to determine the residue however, I tried substitution $x = z-\pi$
so that we can expand;

$$\frac{1+\cos(x+\pi)}{x^3}$$
however the cos term is giving me trouble when it comes to expanding as it always provides so many x terms of the same order and I do not know how to account for this.


问:
Row-wise derivative of a vector with respect to a matrix
Consider a column $n\times 1$ vector $\overline
{z}$ and an $n\times m$ matrix $W$. What would one call and denote an $n\times m$ matrix of derivatives defined by $M_{ij}=\frac{\partial z_i}{\partial W_{ij}}$?

To provide some context, this matrix appears when deriving the matrix form of backpropagation updates for neural networks, whereby $\overline
{z}$ is the vector of inputs to a particular layer (which are linear combinations of actions received from the previous layer) and a particular training instance, and $W$ is the matrix of weights at that layer.

答:
$\def\p#1#2{\frac{\partial #1}{\partial #2}}$The
network layer equation in matrix and index notation is
$$\eqalign{
z &= Wa + b \\ z_i &= W_{i\ell}a_\ell +b_i \\
}$$
When taking the derivative in index notation, take care not to repeat an index, because that implies summation over that index (like the $\ell$-index in the above equation).

With that in mind, the derivative of $z$ with respect to $W$ should be calculated as
$$\eqalign{
\p{z_i}{W_{kj}}
  &= \left(\p{W_{i\ell}}{W_{kj}}\right)a_\ell \\
  &= \left(\delta_{ik}\delta_{j\ell}\right)a_\ell \\
  &= \delta_{ik}\,a_j \\
}$$
This third-order tensor is the correct derivative to use for backpropagation problems.

If you set the index $k=i$, that usually implies summation (aka the Einstein convention)
$$\eqalign{
\p{z_i}{W_{ij}}
  &= \delta_{ii}\,a_j \\
  &= \sum_{i=1}^N \delta_{ii}\,a_j \\
  &= Na_j \\
}$$
If instead, you wanted to equate the indices without summing then you'd obtain
$$\eqalign{
M_{ij} = \delta_{ii}\,a_j = {\tt1}_i\,a_j \\
}$$
but this matrix is not the right quantity to use for backprop.


答:
Here is an intuitive answer: by the same reasoning, if there are $100 \, 000$ balls, $50 \, 000$ red and $50 \, 000$ blue, then the chance of picking $50 \, 000$ reds would be $50 \, 000/100 \, 000 = 1/2$. But this is absurd! If you make $50 \, 000$ selections, then the chance of all of them being red would be tiny.

You might be confusing this with a different scenario. If there are $10$ balls and $5$ reds, then the chance of picking a red is equal to $5/10 = 1/2$. However, this is different from making $5$ selections, and considering the probability that all $5$ of them are red.

Title: Confusion about simple probability


问:
How do you respond to "I was always bad at math"?
Here in the U.S., it is my experience that over 75% of adults I meet socially will volunteer that phrase or a variation upon learning that I am a mathematician.  I find this frustrating, since almost nobody would  brag about being bad at history or English or really any other subject.   Normally I change the subject, or perhaps say they must have had a bad classroom experience, but that feels unsatisfying and turns the conversation in an ugly direction.

Surely some of the bright minds here on Math.SE have found a good response. Note that I'm not looking to  alienate a new acquaintance, so let's keep it positive.

答:
Math is hard for everyone. It takes long hours of practice to become competent. I think that's what people want to hear; and fortunately, it's true.

This reply has the advantage of steering the conversation in a somewhat more positive direction---in my experience, usually towards whatever it is that the person does like to do and has practiced long hours at. This is usually more interesting than having a conversation about poorly taught math courses.


问:
These functions are very similar. Is there an explanation?
Blue: $x \ln x-x$

Brown: $\ln \frac{\Gamma(x+1/2)}{\sqrt{\pi}}$

答:
Using Sterling’s approximation for the Gamma function, $$\Gamma(z)\sim \sqrt{\frac{2\pi}{z}}\left(\frac{z}{e}\right)^z$$
In this case \begin{align*}
\Gamma\left(x+\frac{1}{2}\right)&\sim \sqrt{\frac{2\pi}{x+\frac{1}{2}}}\left(\frac{x+\frac{1}{2}}{e}\right)^{x+\frac{1}{2}}\\
\ln\Gamma\left(x+\frac{1}{2}\right)&\sim \frac{1}{2}\ln 2+\ln \sqrt \pi -\frac{1}{2}\ln \left(x+\frac{1}{2}\right)+\left(x+\frac{1}{2}\right)\ln\left(x+\frac{1}{2}\right)-x-\frac{1}{2}\\
&=\frac{1}{2}\ln 2+\ln \sqrt \pi +x\ln \left(x+\frac{1}{2}\right)-x-\frac{1}{2}\\
&\sim \ln \sqrt\pi +x\ln x-x\\
\ln \left(\frac{\Gamma\left(x+\frac{1}{2}\right)}{\sqrt\pi}\right)&\sim x\ln x-x
\end{align*}


问:
Linear Approximation help
I need some help with the following problem. I know how to do a linear approximation, but in here they defined a function as another which throws me off. Can anyone point me in the right direction

I'm thinking the following. A linear approximation takes the form of $y-y_1=f'(x)(x-x_1)$

so after I derive $g(x)$ I get $-4/(f(x)^3)$ times $f'(x)$ and if I replace $x$ for $-1$ I get

$\dfrac{-7}{16}$ which should be the slope but I'm lost here as I don't see my second point $(x_1,y_1)$. Please lmk

Title: Linear Approximation help

Body:
The linear approximation you seek can take the form
$$
g(x) \approx g(-1) + g'(-1)(x+1)
$$

Now, you can get both $g(-1)$ and $g(-1)$ from the table:
$$
g(-1)=\frac{2}{f(-1)^2}= \frac{2}{(-4)^2} = \frac 18
$$

$$
g'(-1) = -4 f'(-1) f(-1)^{-3}=\frac{-4 \times (-7)}{(-4)^3}=-\frac{7}{16}
$$

and finally get
$$
g(x)\approx \frac 18 -\frac{7}{16}(x+1)
$$


问:
How can this property of definite integrals be true?
In this question, people are saying that the definite integral of $f(x)$ from $0$ to $a$ is equal to the integral of $f(a-x)$ from $0$ to $a$. How can that be true? Simple examples don't work. </p>

答:
Consider the function
$$g(x):=f(a-x)\qquad(0\leq x\leq a)\ .$$
It follows that
$$g\left({a\over2}+y\right)=f\left({a\over2}-y\right) \qquad\left(-{a\over2}\leq y\leq{a\over2}\right)\ .$$
Therefore the graph of $g$ is the graph of $f$, reflected at the vertical $x={a\over2}$. Assuming for simplicity that $f$, and therefore $g$, are $\geq0$, it is obvious that the areas $A_f$ and $A_g$ in question are the same.</p>


问:
Odd primes more than a certain number
I am tasked to write a simple logical statement on this statement "All primes $n$, $n$ greater or equals to $3$, are odd.".

For a start, I have

$\forall n \in \mathbb{Z}^+ (n \geq 3 ...)$

I apologise for the formatting, quite new though I followed the MathJax guide.

Can anyone please point me in the right direction?

答:
$$\forall n > 2 \wedge n \in \mathbb{P}, \exists k \in \mathbb{Z}^+ {\rm s.t.}\  n = 2k + 1$$

Where ${\rm s.t.}$ is read "such that."

Title: Odd primes more than a certain number


问:
How to read this number: 2 - 3e-4
e in this case is not the euler number.
Would I read this number, $$2 - 3e-4$$, as $$2 - 3 * 10^{-4}?$$

thanks

答:
Question: $$ 2-3e-4 $$

Using pure mathematical notation, this is two subtractions, so the order of operations (and left to right) suggests the answer is: $$-2-3e$$ where $e$ is Euler's number.

However, in many programming languages, and in calculators, $E$ or $e$ is used to denote orders of magnitude, so: $$ 2-3e-4 = 2-3(10)^{-4}$$

I would argue that the later is intended, as otherwise a simplification would have been made to the former. But,if this is part of a calculation, it may be laid out like this for ease of seeing the steps taken. e.g. for $x=2$
$$x-3e-x^2$$
$$= 2-3e-4$$
Therefore, more context is required, but it is fairly safe to guess the order of magnitude option is intended.

Title: How to read this number: 2 - 3e-4


答: $\forall p\in P: p \ge 3 \Rightarrow \exists n \in \mathbb N \; \& \; p=2n+1$, where $P$ denotes the set of all primes.

Title: Odd primes more than a certain number

Body:
I am tasked to write a simple logical statement on this statement "All primes $n$, $n$ greater or equals to $3$, are odd.".

For a start, I have

$\forall n \in \mathbb{Z}^+ (n \geq 3 ...)$

I apologise for the formatting, quite new though I followed the MathJax guide.

Can anyone please point me in the right direction?


问:
Given $n+1$ points, bound the product of the distances from one of them
We have $n+1$ real numbers $x_1,\cdots,x_{n+1}$ such that $-1\leq x_i\leq 1$ for all $1\leq i\leq n+1$.

I am wondering whether the following fact is true:

There exists some $j$ such that $\prod_{\substack{i=1\\i\neq j}}^{n+1}\left |x_j-x_i\right |\leq \frac{n+1}{2^{n-1}}$.

Some base steps are easy. Assume that this holds for $n-1$.

Applying this inductive hypothesis to $\left\{x_i:i\in\mathbb{N}_{\leq n+1}\right\}\setminus\left\{x_j\right\}$ for each $j\in\mathbb{N}_{\leq n+1}$, we conclude that for each $j\in\mathbb{N}_{\leq n+1}$ there exists $h\left(j\right)\in\mathbb{N}_{\leq n+1}\setminus\left\{j\right\}$ such that $\prod_{\substack{i=1\\i\neq j\\i\neq h\left(j\right)}}^{n+1}\left |x_{h\left(j\right)}-x_i\right |\leq\frac{n}{2^{n-2}}$.

Now, if for some $j\in\mathbb{N}_{\leq n+1}$ we have $\left |x_{h\left(j\right)}-x_j\right |\leq \frac{1}{2}$, then $\prod_{\substack{i=1\\i\neq h\left(j\right)}}^{n+1}\left |x_{h\left(j\right)}-x_i\right |\leq \frac{1}{2}\cdot\frac{n}{2^{n-2}}\leq \frac{1+\frac{1}{n}}{2}\frac{n}{2^{n-2}}=\frac{n+1}{2^{n-1}}$, which gives what we want. But what about the case in which $\left |x_{h\left(j\right)}-x_j\right |>\frac{1}{2}$ for all $j\in\mathbb{N}_{\leq n+1}$?

Is there any other way?

答:
Yes, it is true. Consider the polynomial $p(x)=\prod_{i=1}^{n+1}(x-x_i)$ and the monic Chebyshev polynomial $T_n(x)=x^n+\dots$ of degree $n$, so $|T_n|\le 2^{-(n-1)}$ on $[-1,1]$. Now apply the residue theorem to the integral of the rational function $Q(z)=\frac{T_n(z)}{p(z)}$ in a huge disk centered at $0$. On the one hand
$$
\oint_{|z|=R}Q(z)\,dz\approx \oint_{|z|=R}\frac{dz}{z}=2\pi i
$$
as $R\to\infty$ because $Q(z)=\frac 1z+O(|z|^{-2})$ as $z\to\infty$.

On the other hand, it is $$2\pi i\sum_j{\rm Res}_{x_j}Q=2\pi i\sum_j\frac{T_n(x_j)}{p'(x_j)}\,.$$ Since $|T_n(x_j)|\le 2^{-(n-1)}$, we conclude that $$\sum_j\frac{1}{|p'(x_j)|}\ge 2^{n-1}\,,$$ so there must exist $j$ with $|p'(x_j)|\le \frac{n+1}{2^{n-1}}$. But $|p'(x_j)|$ are exactly the products you are interested in.


问:
Prove compact SVD using normal SVD
Please read the EDIT first - it's the bit that will be helpful to anyone coming across this post.

Question is in the title (SVD means SVD decomposition). The answer from my textbook says that $U$ and $V$ both contain bases for the column and row space respectively (which is true - I can understand this myself with a little bit of effort). It then goes on to say that because $C$ and $B$ also contain the same respectively, then you can write $U=CF$ and $V=BG$ for two invertible $r \times r$ matrices $F$ and $G$ (where $r=rank(A)$). But here is where the issue arises for me: SVD decomposition splits an $m \times n$ matrix $A$ into the following matrices: an $m \times m$ $U$, an $m \times n$ $\Sigma$ and an $n \times n$ $V^T$. But both $C$ and $B$ have at max $r$ columns because they are limited by rank (they will be $m \times r$ and $n \times r$ respectively). This means that the products $CF$ and $BG$ will come out as being the same dimensions as $C$ and $B$ originally and respectively - not as square matrices. What am I missing here? (for this textbook proof) Or is there another method for proving the title question?

EDIT: After some searching $A=CMB^T$ is actually compact SVD ($A=U_1\Sigma_1 V_1^T$, with $U \in \mathbb{R}^{m \times r}$, $\Sigma \in \mathbb{R}^{r \times r}$ and $V \in \mathbb{R}^{n \times r}$) instead of "normal" SVD. A proof will be below for any weary travellers who stumble open this post.

答:
Let $A \in \mathbb{R}^{m \times n}$, with $m$, $n \mathbb{R}$
Assume $m < n$.
We will use the usual SVD decomposition $A=U\Sigma V^T$ with $U \in \mathbb{R}^{m \times m}$, $\Sigma \in \mathbb{R}^{m \times n}$ and $V \in \mathbb{R}^{n \times n}$. Let $u_i$ denote the $i$th column vector of $U$.
$$A=U\Sigma V^T$$
$$\implies{A=\sum^m_{i=1}\sigma_iu_iv_i^T}$$
We know that $\sigma_i=0$ $\forall i$ s.t. $r < i \leq m$ (number of non-zero eigenvalues is the same as the rank for diagonalizable matrices - good proof here: Proving number of non zero eigen values.)
$$\implies{A=\sum^r_{i=1}\sigma_iu_iv_i^T}$$
$$\therefore A=U_1\Sigma_1V_1^T \text{ with } U \in \mathbb{R}^{m \times r}\text{, } \Sigma \in \mathbb{R}^{r \times r}\text{, (with }E^{-1}\text{ existing) and }V \in \mathbb{R}^{n \times r}$$
To prove the compact SVD for $m > n$, it's trivial to prove it from the first proof given here - with some minor alterations to where the dimension variables are used.

Title: Prove compact SVD using normal SVD


问:
How should I go about this approaching-infinite limit problem?
I'm doing some exercises of limits approaching infinite, most are simple polynomials where only the highest degree term will matter in the end but for this one I couldn't find a solution (not correct at least).

$$\lim_{x\to-\infty}\frac{x^2+x+1}{(x+1)^3-x^3}$$

How should I proceed to get the correct answer? ($\frac{1}{3}$)

Also, while simplifying some questions this question took my mind: is it correct to say that $\sqrt{a+b}$ is equal to $\sqrt{a}+\sqrt{b}$?

Ps.: Without using l'Hôpital's rule.

答:
Start by expanding:

$$\lim_{x\to-\infty}\frac{x^2+x+1}{(x+1)^3-x^3}= \lim_{x\to-\infty}\frac{x^2+x+1}{x^{3} + 3x^{2} + 3x + 1 -x^3}$$

Simplify to obtain:

$$\lim_{x\to-\infty}\frac{x^2+x+1}{3x^{2}+3x +1}$$

Now divide every term by $x^{2}$:

$$\lim_{x\to-\infty}\frac{1+\frac{1}{x}+\frac{1}{x^{2}}}{3+\frac{3}{x}+\frac{3}{x^{2}}}$$

Now take the limit to obtain $\frac{1}{3}$.


问:
How to prove the variance of this function of a normal random variable is decreasing?
Partial answers found here and here.

Suppose $\phi(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$ and $\Phi(x)=\int_{-\infty}^x\phi(t)dt=\frac{1}2 \text{Erfc}[-\frac{x}{\sqrt{2}}]$ where Erfc is the complementary error function. These are the density and distribution function of a standard normal random variable.

Define the function $g(\mu)=\int_{-\infty}^{\infty}\Phi(x) \phi(x-\mu)dx$.

$g(\mu)$ is the expected value of $\Phi(X)$ when $X$ has a normal distribution with mean $\mu$ and variance 1.

Differentiate under the integral to observe that $$g'(\mu)=\int_{-\infty}^{\infty}\Phi(x)(x-\mu) \phi(x-\mu)dx=\frac{\phi\left(\frac{\mu}{\sqrt{2}}\right)}{\sqrt{2}}$$

Since $g'(\mu)$ is always positive, $g(\mu)$ is an increasing function.

Furthermore, define the function $h(\mu)=\int_{-\infty}^{\infty}(\Phi(x)-g(\mu))^2 \phi(x-\mu)dx$.

$h(\mu)$ is the variance of $\Phi(X)$ when $X$ has a normal distribution with mean $\mu$ and variance 1.

I can shown that:

$h(0)=\frac{1}{12}$

$h'(0)=0$

$h''(0)=-\frac{1}{2 \pi}+\frac{1}{2 \sqrt{3}\pi} \approx -0.06727$

$\lim_{\mu\rightarrow \infty}h(\mu)=0$

$h(\mu)=h(-\mu)$ so that $h$ is an even function.

I can draw $h(x)$ for $x \in [0,4]$

I can also find all the terms of the Taylor series for $h(\mu)$ at $\mu=0$. But, I cannot show the derivative is never $0$ for $\mu \ne 0$.
Is it possible to prove that $h(\mu)$ is decreasing for $\mu>0$?

答:
Computing the derivative of $h$ we can readily see that

$$h'(\mu)=2\int_{-\infty}^{\infty}dx~\Phi(x+\mu)\phi(x+\mu)\phi(x)-2g(\mu)g'(\mu)$$

However it is true that

$$\phi(x)\phi(x+\mu)=\phi(\frac{\mu}{\sqrt{2}})\phi(x\sqrt{2}+\frac{\mu}{\sqrt{2}})$$

whereupon we can simplify the above integral to the form

$$h'(\mu)=2g'(\mu)\left(\int_{-\infty}^{\infty}dt~\Phi(\frac{t}{\sqrt{2}}+\mu)\phi(\frac{\mu}{\sqrt{2}}+t)-g(\mu)\right)\equiv2g'(\mu)\Lambda(\mu)$$

It suffices to show that $\Lambda(\mu)\leq 0$ for $\mu\geq 0$. We know that $\Lambda(0)=0, \Lambda(\infty)=0$ and we compute (the derivation can be done similarly as above)

$$\Lambda'(\mu)=\frac{\phi(\mu/\sqrt{6})}{\sqrt{6}}-\frac{\phi(\mu/\sqrt{2})}{\sqrt{2}}$$

It is a trivial exercise in algebra to find the sign of this function and in fact we see that $\Lambda$ possesses a single minimum at $\mu=\sqrt{3\ln3}$ and a maximum at $\mu=-\sqrt{3\ln3}$. We are only interested in positive values however, so from the above data we conclude that $\sup_{x\geq 0}\Lambda(x)=0$ and thus for all $x\geq 0$ the result follows.

We conclude that $h$ is decreasing on the positive x-axis.


答:
You can expand the denominator and use polynomial division:

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

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

Title: How should I go about this approaching-infinite limit problem?

Body:
I'm doing some exercises of limits approaching infinite, most are simple polynomials where only the highest degree term will matter in the end but for this one I couldn't find a solution (not correct at least).

$$\lim_{x\to-\infty}\frac{x^2+x+1}{(x+1)^3-x^3}$$

How should I proceed to get the correct answer? ($\frac{1}{3}$)

Also, while simplifying some questions this question took my mind: is it correct to say that $\sqrt{a+b}$ is equal to $\sqrt{a}+\sqrt{b}$?

Ps.: Without using l'Hôpital's rule.


问:
Calculate $\int_\gamma\frac{|z|e^z}{z^2}dz$, where $\gamma$ is the circle of radius 2 centered at the origin.
Calculate $$\int_\gamma\frac{|z|e^z}{z^2}dz,$$ where $\gamma$ is the circle of radius 2 centered at $0\in\mathbb{C}$.

I wanted to apply the Residue Theorem, but I think it is not possible because $|z|$ is not analytic at the origin. Also, I tried to calculate the integral using the parametrization $z=2e^{i\theta}$, $0\leq\theta\leq 2\pi$. Then $$\int_\gamma\frac{|z|e^z}{z^2}dz=\int_0^{2\pi}\frac{|2e^{i\theta}|e^{2e^{i\theta}}}{2e^{2i\theta}}2ie^{i\theta}d\theta= 2i\int_0^{2\pi}\frac{e^{2e^{i\theta}}}{e^{i\theta}}d\theta.$$ But I have not been able to solve that integral.

I thank any suggestion.

答: As @leoli1 and @Gary hint,$$\int_{|z=2|}\frac{|z|e^zdz}{z^2}=\int_{|z=2|}\frac{2e^zdz}{z^2}=2\pi i\operatorname{Res}_{z=0}\frac{2e^z}{z^2}=4\pi i.$$Alternatively,$$2i\int_0^{2\pi}\frac{e^{2e^{i\theta}}}{e^{i\theta}}d\theta=2i\int_0^{2\pi}2d\theta=4\pi i$$because $\int_0^{2\pi}e^{ik\theta}d\theta=2\pi\delta_{k0}$ for $k\in\Bbb Z$.


问:
NFA to DFA Conversion Help
Consider the language of all words over {a, b} that either begins and ends with aa or bb (but both are
the same within each word). That is, {ccycc | c ∈ {a, b}, y ∈ Σ
∗}

I have drawn the resulting NFA for the language described above, what would the corresponding DFA look like? What would the steps look like to get from an NFA to a DFA?

答: I believe this DFA would work.

Title: NFA to DFA Conversion Help

Body: I believe this DFA would work.


问:
Why is the Mapping Cylinder a fibering?
If $p:E \rightarrow B$ is a fibration with fiber $F$, in other words we have the fiber space $(E,B,F,p)$. Let $M_p$ be the mapping cylinder of $p$, $M_p = ((E\times[0,1]) \sqcup B)/\sim  $, in which $(x,0)\sim p(x)$. Then we can define the map $$\overline{p}:M_p \rightarrow B$$ given by $\overline{p}([x,t])=p(x)$ for $(x,t)\in E\times [0,1]$, and $\overline{p}(b)=b$ for $b\in B$.

My question is , how to show that $\overline{p}$ is a fibration with fiber $C(F)$ (the cone of $F$, i.e., $C(F)=F\times[0,1]/\sim$ with $(x,0)\sim(y,0)$ for all $x,y\in F$).

I tried to show that $\overline{p}$ satisfies the homotopy lifting property, but had no luck in defining the lifting homotopy in a general case. I think that in some point I have to use the fact that $p$ itself is a fibration, but I don't see how to use this fact. Also I have tried to use the fact that $M_f$ strong deformation retracts onto $B$, but again had no luck in showing that $\overline{p}$ is a fibration.

Edit: I think I understand why the fiber would be $C(F)$, if we simply annalyse $\overline{p}^{-1}(b)$ for $b\in B$, we get that it must be equal to $((p^{-1}(B)\times[0,1])\sqcup \{b\})/\sim$ with $(x,0)\sim b = p(x)$ for all $x$, thus the fiber is $C(p^{-1}(B)) = C(F)$. It remains to show that $\overline{p}$ is in fact a fibration.

答:
Here is a partial answer which makes use of some additional assumptions. We'll assume

$p:E\rightarrow B$ is a locally trivial map with typical fibre $F$.

$B$ is locally compact Hausdorff.

Under these assumptions we'll prove that the projection $q:M_p\rightarrow B$ is locally trivial.

The key point to notice is that if $X,Y$ are spaces and $X$ is locally compact, then
\begin{array}{ccc}
X\times Y &\xrightarrow{in_0} & X\times Y\times I \\
pr_X\downarrow & &\ \downarrow \\
X &\rightarrow& X\times CY
\end{array}
is a pushout diagram, where $CY=(Y\times I)/(Y\times\{0\})$. This is easy to see because the square is obtained by applying the functor $X\times(-)$ to the pushout diagram defining $CY$. The functor $X\times(-)$, as a left adjoint, preserves all colimits, and in particular preserves the pushout.

Now take the map $p$ and form the mapping cylinder $M_p$ as the pushout in the square
\begin{array}{ccc}
E &\xrightarrow{in_0} & E\times I \\
p\downarrow & &\ \downarrow \\
B &\rightarrow& M_p.
\end{array}
Our assumptions that $p$ is locally trivial and that $B$ is locally compact means that any $x\in B$ has a compact neighbourhood $K\subseteq B$ over which $p$ is trivial. Because $B$ is Hausdorff, $K$ is compact Hausdorff, and hence locally compact. Thus we have a pushout
\begin{array}{ccc}
K\times F &\xrightarrow{in_0} & K\times F\times I \\
pr_K\downarrow & &\ \downarrow \\
K &\rightarrow& K\times CF.
\end{array}
which induces a map $\varphi:K\times CF\rightarrow M_p$ factoring the projection $K\times CF\rightarrow K\subseteq B$.

Let $\overline p:M_p\rightarrow B$ be the canonical map. The inverse image $\overline p^{-1}(K)$ is the pushout of $p^{-1}(K)\cap B\cong K$ against $p^{-1}(K)\times I\cong K\times F\times I$. This is exactly $K\times CF$, and we conclude that $\varphi$ is a homeomorphism onto its image in $M_p$.

Thus we have produced a chart for $\overline p$ in a neighbourhood of any given point $x\in B$. Our conclusion follows: the map $\overline p:M_p\rightarrow B$ is locally trivial.

Discussion: For our statements $p$ need not be a fibation. Even if $p$ is a fibration, then we cannot conclude that $\overline p$ is a fibration. If we have the additional assumption that $B$ is paracompact, then both $p$ and $\overline p$ are necessarily fibrations. This will be the case, for instance, if $B$ is a manifold or CW complex.

I do not know if the result you have asked for holds in the generality that $p$ is a fibration. What is true is that if $f:X\rightarrow Y$ is any map, then the projection $M_f\rightarrow Y$ is a Dold fibration. That is, $f$ is homotopy equivalent to a fibration in the category $Top_Y$ of spaces over $Y$. This means that $f$ will satisfy the weak homomotopy lifting property. The map I suggest in the comments is an example of a Dold fibration. It is not always possible to lift homotopies exactly, but for any $\epsilon>0$ a solution can be found for the cylinder $[-\epsilon,1]$ which lifts the required homotopy over $[0,1]$.

Thus $\overline p$ will aways be a Dold fibration, but there has been nothing gained by the assumption that $p$ is a fibration. If you are interested in producing a counterexample to see that $\overline p$ may fail to be an actual fibration, then here is a strategy. The map $\overline p$ is a homotopy equivalence. Thus for it to be a fibration it is necessary and sufficient that a diagonal filler $X\rightarrow M_p$ can be found for any strictly commuting diagram
\begin{array}{ccc}
A &\xrightarrow{f} & M_p \\
\downarrow & &\ \downarrow \overline p \\
X &\xrightarrow{g}& B
\end{array}
in which $A\subseteq X$ is a closed cofibration. I would suggest to take $B=I$ and $A=\{1\}$, with $f$ mapping to a point in $E\times \{1\}\subseteq M_p$, and $g$ being a path in $B$. The idea is that any diagonal map would need to 'jump' over the whole interval in $M_p$ and would be discontinuous. I'm afraid I don't have time right now to try and make this rigourous. I'll remark again that I don't know the full answer to your query.

Title: Why is the Mapping Cylinder a fibering?


问:
Let $ \mathbb{N}$, $a \in \mathbb{N} \to a+1 \in \mathbb{N}$
I need to prove the following:

"Let $ \mathbb{N}$, if $a \in  \mathbb{N} \to a+1 \in  \mathbb{N}$"

thanks in advance!!

答:
There is actually  something to prove here.  $a+1$ is defined in the Peano axioms as $a+S(0)=S(a+0)=S(a)$.  Hence, if $a\in\mathbb{N}$, $a+1=S(a)\in \mathbb{N}$.

We can go further; $a+2=a+S(1)=S(a+1)=S(S(a))$.  Hence, if $a\in \mathbb{N}$, $a+2=S(S(a))\in \mathbb{N}$.  In fact, for any specific $k$, we can equally prove that if $a\in \mathbb{N}$ then $a+k\in \mathbb{N}$.


问:
Linear Regression - Proof Sum Adds to Zero
In linear regression, why is $\sum(X_{i} - \mu_{x})$ = $0$?  I understand that for ($\sum$ $Y_{i}$ minus the fitted value of Y) = $\sum$ $e_{i}$ this is true but why is this other fact true?</p>

答:
Just break up the sum into two sums, and substitute the definition of
$\mu_x$. More explicitly, $$\sum_{i=1}^n (X_i - \mu_X) = \sum_{i=1}^n X_i - \sum_{i=1}^n \mu_x
= \left[\sum_{i=1}^n X_i\right] - n\mu_x
= \left[\sum_{i=1}^n X_i\right] - n\times \left[\frac{1}{n}\sum_{i=1}^n X_i\right]
= 0.$$</p>


问:
How to prove the representation theory of permutation groups (S3) also satisfies the group definition?
Take S3 group for example, if I associate the permutation with the permutation of the three unit vectors. Then I can write:

$${p((1,2)) = \left[\begin{array}{ccc}
0 & 1 & 0\\
1 & 0 & 0\\
0 & 0 & 1
\end{array}\right]},{p((1,3)) = \left[\begin{array}{ccc}
0 & 0 & 1\\
0 & 1 & 0\\
1 & 0 & 0
\end{array}\right]}$$

Then, the product of p((1,2)) and p((1,3)) should be p((1,3,2)) because (1,2)(1,3) = (1,3,2) but instead I got:

$${p((1,2))p((1,3)) = \left[\begin{array}{ccc}
0 & 1 & 0\\
0 & 0 & 1\\
1 & 0 & 0
\end{array}\right]}$$

which is p((1,2,3)). Could someone please tell me where went wrong?

答:
You never state your definition of $p(\sigma)$, and the examples you give are insufficient to deduce the one you are using (since they both have $\sigma^{-1}=\sigma$). There is only one proper way to define $p(\sigma)$ so that it becomes a representation of $S_n$, namely such that $p(\sigma)\,e_j=e_{\sigma(j)}$ for all$~j$ (where $e_j$ is the $j$-th standard basis vector), in other words where $p(\sigma)_{i,j}=\delta_{i,\sigma(j)}$: there is an entry $1$ in row $\sigma(j)$ of column $j$. According to this definition your product matrix is indeed $p((1,3,2))$, as you would like.

Specifically since $(1,3,2)$ sends $1\mapsto3$, one has $p((1,3,2)(e_1)=e_3$ so the first column of $p((1,3,2))$ is $e_3=(0,0,1)^\top$; similarly $(1,3,2)$ sends $2\mapsto1$ and $3\mapsto2$, so the second and third columns are $e_3=(1,0,0)^\top$ respectively $e_3=(0,1,0)^\top$, as in your final matrix. The fact that $p(\sigma)\,e_j=e_{\sigma(j)}$ is the proper rule in general follows from the simple computation
$$
p(\sigma)p(\rho)\,e_j=p(\sigma)\,e_{\rho(j)} =e_{\sigma(\rho(j))} =p(\sigma\circ\rho)\,e_j\qquad\text{for all $j$}.
$$


问:
Give syntax proof of FOL Formla $\exists x.P(x) \lor \exists x.Q(x)$ $\vdash$ $\exists x.(P(x) \lor Q(x))$
So the question is pretty much described in the title already. I have to show the following result. I have tried it but am failing to do so. Anyone who can please help me in understanding its proof. I have attached the picture of formulas that may prove to be helpful.

$\exists x.P(x) \lor \exists x.Q(x)$ $\vdash$ $\exists x.(P(x) \lor Q(x))$

答:
Hint: the proof has 3 instances of an elimination rule.

Consider that the premise has as the principal operator a Vel, whereas on the right we have an existential quantifier.

Try to work backward to see what you need: clearly the last rule will be an application of $\exists$-introduction after a derivation of the formula $Pa \lor Qa$.

If you're still stuck let me know and I'll give you another hint.


问:
Reference request: the terminology "has products" in category theory
I read the accepted answer to the question

Example of a category where some pair of objects lacks a product?

in which the poster gives a definition for a category which "has products", but does so by referencing when a pair of objects in that category "has a product". It seems like this means "the product of the objects is an object in the category", but I would like to see a formal definition in a text if possible, or if not, it would be great if someone could give one here. I would also like to know how you'd prove a pair or tuple of objects has a product. The products live in the product category, and I can't seem to reconcile this conceptually with the products also living in the original category.

答:
I think the core of your confusion is captured in "The products live in the product category". Objects in a product category are ordered pairs of objects in the original categories, not products of them.

If $a$ and $b$ are objects in a category $C$, the ordered pair $(a,b)$ lives in the product category $C\times C$. But the product of $a$ and $b$ (if it exists) is another object $p$ in the category $C$, given together with arrows $\pi_1\colon p\to a$ and $\pi_2\colon p\to b$, satisfying the universal property of the product: for any object $c$ in $C$ and arrows $f_1\colon c\to a$ and $f_2\colon c\to b$, there is a unique arrow $g\colon c\to p$ such that $\pi_1\circ g = f_1$ and $\pi_2\circ g = f_2$.

If such an object $p$ exists, we usually denote it by $a\times b$. If a product $a\times b$ exists for every pair of objects $a$ and $b$, we say that $C$ has products. Note that it makes no sense at all to talk about a product of $a$ and $b$ living in some category other than $C$!


问:
Solve system of non-linear equations $\frac{1}{x^2}-\frac{1}{(y-x)^2}-\frac{1}{(1-x)^2}= \frac{1}{y^2}+\frac{1}{(y-x)^2}-\frac{1}{(1-y)^2}=0$
Whilst solving a problem, the following system of non-linear equations arose:

$$0= \frac{1}{x^2}-\frac{1}{(y-x)^2}-\frac{1}{(1-x)^2}$$

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

I have tried to solve it via brute force and by trying to add them both in a way that could give me the answer, but I still haven't got it.

I can't manage to find the intersection of the two parametric equations that these equations give. I know that I can find approximate solutions with the Wolfram Alpha tool, but I would like to know, if any, an analytical or strong numerical way to proceed.

Note: Out of all the solutions ($8$ it seems, $4$ real and $4$ complex), I only need the one that takes real values for $x$ and $y$, with $0<x<y<1$.

Body:
Using Groebner basis,

Eliminating $x$ we get

$$
y^8-12 y^7+34 y^6-52 y^5+131 y^4-188 y^3+126 y^2-40 y=-5
$$

and simplifying

$$
\left(y \left((y-1)^2 y-10\right)+5\right) (y (y ((y-10) y+13)-6)+1)=0
$$

NOTE

The roots in the range $0<x<y<1$ are

$$
\cases{
x = 0.3192601457911361\\
y = 0.6807398542088641
}
$$


问:
Calculating eigenvectors for eigenvalues of $3 \times 3$ symmetric matrix?
I've calculated the 3 eigenvalues of a $3\times3$ real symmetric matrix (actually its an inertia tensor) and confirmed that $\det(M-{\lambda}I) = 0$ for each one, how do I now calculate the eigenvectors that correspond to each eigenvalue?

答: For each eigenvalue $\lambda_i$, compute $M-\lambda_iI$, and find a vector $\mathbf{v}_i$ such that $(M-\lambda_iI)\mathbf{v}_i = \mathbf{0}$.


问:
How do you respond to "I was always bad at math"?
Here in the U.S., it is my experience that over 75% of adults I meet socially will volunteer that phrase or a variation upon learning that I am a mathematician.  I find this frustrating, since almost nobody would  brag about being bad at history or English or really any other subject.   Normally I change the subject, or perhaps say they must have had a bad classroom experience, but that feels unsatisfying and turns the conversation in an ugly direction.

Surely some of the bright minds here on Math.SE have found a good response. Note that I'm not looking to  alienate a new acquaintance, so let's keep it positive.

答: To put a slightly different spin on this (before @Zev votes to close :) ). When folks ask me what I teach, I ask them to name their least favorite subject. Occasionally I do get English or history or science. But 90% of the time I get math ... to which I reply "bingo." :) I then engage them in a discussion about how math (generically) and history (for me and many others) are ruined for the masses by secondary teachers who just don't get it and who teach relying almost totally on rote memorization/manipulation. They concur 99% of the time on this. At some point, I reassure them that there are some really good teachers out there (and I claim to be one) :P</p>


Title: Solve system of non-linear equations $\frac{1}{x^2}-\frac{1}{(y-x)^2}-\frac{1}{(1-x)^2}= \frac{1}{y^2}+\frac{1}{(y-x)^2}-\frac{1}{(1-y)^2}=0$

Body:
Whilst solving a problem, the following system of non-linear equations arose:

$$0= \frac{1}{x^2}-\frac{1}{(y-x)^2}-\frac{1}{(1-x)^2}$$

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

I have tried to solve it via brute force and by trying to add them both in a way that could give me the answer, but I still haven't got it.

I can't manage to find the intersection of the two parametric equations that these equations give. I know that I can find approximate solutions with the Wolfram Alpha tool, but I would like to know, if any, an analytical or strong numerical way to proceed.

Note: Out of all the solutions ($8$ it seems, $4$ real and $4$ complex), I only need the one that takes real values for $x$ and $y$, with $0<x<y<1$.

Body:
Take the cue of $0<x<y<1$ and observe that $x+y=1$ satisfies both equations
$$\frac{1}{x^2}-\frac{1}{(y-x)^2}-\frac{1}{(1-x)^2}=
\frac{1}{y^2}+\frac{1}{(y-x)^2}-\frac{1}{(1-y)^2}=0$$
which leads to $\frac{1}{x^2}-\frac{1}{(1-2x)^2}-\frac{1}{(1-x)^2}=0$, or
$$x^4+6x^3-11x^2+6x -1=0$$
Factorize
$$[x^2+3(1+\sqrt2)x-(1+\sqrt2)][x^2+3(1-\sqrt2)x-(1-\sqrt2)]=0$$
to obtain the solution in the domain $(0,1)$

$$x= \frac12\left( \sqrt{31+22\sqrt2}-3\sqrt2-3\right)
$$
$$ y= \frac12\left(- \sqrt{31+22\sqrt2}+3\sqrt2+5\right)$$


问:
$m+n+p = 2017$ Solve for $(x,y,z,t)$ $x+my+nz+pt =2018$
$m, n,p \in \mathbb{N}^*$ such that $m+n+p = 2017$.

Find the values of $x,y,z,t$ Such that  $x+my+nz+pt =2018$
With $x,y,z,t \in \mathbb N^*$

My Attempt:

Im not sure if this solution is correct because it looks easy to me.

If we add a $1$ to the original equation:
$$1+1(m)+(1)n+(1)p=2018$$

And fron here, It’s easy to see that the only solution to the equation is $(x,y,z,t)=(1,1,1,1)$,
Given the fact that $(x,y,z,t) \in \mathbb N^*$  so they cannot equal to $0$.

答:
Yes. You can formalize this by considering $x=1$ and $x\geq 2$ by observing that
$$x+m(y-1)+n(z-1)+p(t-1) + m + n+ p =2018 \\
\implies x+m(y-1)+n(z-1)+p(t-1) =1$$

$x=1$:
$$ m(y-1)+n(z-1)+p(t-1)=0 \\
\implies (y-1)=(z-1)=(t-1) = 0 \\
(y,z,t) = (1,1,1)
$$

$x\geq 2$:
$$ m(y-1)+n(z-1)+p(t-1)\lt 0
$$
so at least one of $(y{-}1),(z{-}1),(t{-}1)$ is less than zero and the corresponding in  $y,z,t$ is not $\in \mathbb N^*$

For style you could perhaps define $y',z',t'\in \mathbb N_0$ to be the decremented versions of $y,z,t$

@zwim in comments points out that we can extend the initial reformulation:
\begin{align}
x+m(y-1)+n(z-1)+p(t-1) + m + n+ p &=2018 \\
\implies x+m(y-1)+n(z-1)+p(t-1) &=1 \\
\implies (x-1)+m(y-1)+n(z-1)+p(t-1) &=0 \\
\implies (x-1)=(y-1)=(z-1)=(t-1) &=0
\end{align}

giving us $(x,y,z,t) = (1,1,1,1)$ as stated.


问:
Determine splitting field $K$ over $\mathbb{Q}$ of the polynomial $x^3 - 2$
Determine the splitting field $K$ over $\mathbb{Q}$ of the polynomial $x^3 - 2$

Also determine the basis over $\mathbb{Q}$ and its degree. Can I do this using only first principles?

答:
The roots of $x^3-2=0$ are $x_1=\sqrt[3]{2},$ $x_2= \sqrt[3]{2}\omega$, $x_3=\sqrt[3]{2}\omega^2$, where $\omega=\frac{-1+i\sqrt{3}}{2}$. It follows that the splitting field $K$ is
$$K=\mathbb{Q}(x_1,x_2,x_3)=\mathbb{Q}(\sqrt[3]{2},i\sqrt{3})$$

Since $1, \sqrt[3]{2}, \sqrt[3]{2}^2$ form a basis for $\mathbb{Q}(\sqrt[3]{2})$ over $\mathbb{Q}$ and $1, i\sqrt{3}$ form a basis for $\mathbb{Q}(\sqrt[3]{2}, i\sqrt{3})$ over $\mathbb{Q}(\sqrt[3]{2})$, a basis for $\mathbb{Q}(\sqrt[3]{2},i\sqrt{3})$ over $\mathbb{Q}$ must be $1, \sqrt[3]{2}, \sqrt[3]{2}^2,i\sqrt{3}, \sqrt[3]{2}i\sqrt{3},\sqrt[3]{2}^2i\sqrt{3} $.

Edit: I have used the following facts, taken from Robert Ashs excellent book Abstract Algebra: The Basic Graduate Year:

3.1.7

Let $E/F$ be a field extension and suppose $\alpha\in E$ is algebraic over $F$. Then
$$1, \alpha, \alpha^2, ..., \alpha^{n-1}$$
form a basis of $F(\alpha)$ over $F$, where $n=$deg Irr$(\alpha, F)$.

3.1.8

If $E/K/F$ are field extensions such that $\{\alpha_i\}_{i\in I}$ is a basis for $E$ over $K$ and $\{\beta_j\}_{j\in J}$ is a basis for $K$ over $F$, then $\{\alpha_i\beta_j\}_{i\in I, j\in J}$ form a basis for $E$ over $F$.


问:
How many 6-digit numbers are there whose digits alternate between even and odd numbers?
Just hoping to solve this question that asks how many 6-digit numbers there are, where each digit alternates between even and odd numbers? I know that you simply need to find the number of options each digit can hold but was unsure if I am supposed to add them up or multiply - given this is a mutually exclusive question. Thank you in advance!</p>

答: There are nine choices for the first digit, and five choices for each of the other five digits. The answer is therefore $9 \cdot 5^5=28125$.


答:
There are $3$ even digits and $5$ choices they each can be.  There are $3$ odd digits and $5$ choices they each can be.

You can start with either an even digit or an odd digit.

If you start with an odd digit it can not be $0$ so there are $4$ choices.

That's enough.

Do not read the next paragraph:

  I told you not to read this!  Don't read the next one either.

 $5^3\cdot 5^3 + 4\cdot 5^2\cdot 5^3$

Title: How many 6-digit numbers are there whose digits alternate between even and odd numbers?

Body: Just hoping to solve this question that asks how many 6-digit numbers there are, where each digit alternates between even and odd numbers? I know that you simply need to find the number of options each digit can hold but was unsure if I am supposed to add them up or multiply - given this is a mutually exclusive question. Thank you in advance!</p>


问:
Derivative of complex functions
Prove that if $f'(z) = 1$ everywhere in a domain $D$ then $f(z) = z + c$
for some complex constant $c.$
And $z$ is a complex no.

答:
$D$ is a domain and connected.  Assuming each point can be connected by a piecewise smooth curve $\gamma$ to every other point in the domain, and since $f^\prime$ is holomorphic and has a primitive $f$:

Without loss of generality, assume the origin is in the domain and $f(0)=z_0.$

$$\int_\gamma f^\prime(z) dz = \int_0^z f^\prime(\zeta) \,d\zeta=f(z)-f(0).$$

But $$\int_0^z d\zeta = z,$$

So $$f(z) =z + z_0.$$

Title: Derivative of complex functions