问:
Prove $p^2=p$ and $qp=0$
I am not really aware what's going on in this question. I appreciate your help.

Let $U$ be a vector space over a field $F$ and $p, q: U \rightarrow U$ linear maps. Assume $p+q = \text{id}_U$ and $pq=0$. Let $K=\ker(p)$ and $L=\ker(q)$.

Prove $p^2=p$ and $qp=0$.

$$p(p+q) = p(\text{id}_U) \Rightarrow p^2+pq=p \Rightarrow p^2 =p.$$ I actually first found this by letting $(\text{id}_U) =1$, although probably a wrong to do it.

Since $p^2 =p$ and $q$ is also a linear map defined by $q: U \rightarrow U$ then $q^2=q$. Again, I'm not sure if this actually is the right way to do it.

Then we have $$q(p+q) = q(\text{id}_U) \Rightarrow q^2+qp=q \Rightarrow q^2+ qp= q^2 \Rightarrow qp=0.$$

Prove $K=\text{im}(q)$.

For this question, I honestly do not know how to tackle it from the following definitions.

$K = \ker(p)=\left\{u \in U \ |\ p(u)=0_U\right\}$ and $\text{im}(q)=\left\{q(u) \ | \ u \in U \right\}$.

What's the correct way of doing these questions? Thank you for your time.

答:
Since $\mbox{id}_U$ is the unit of the ring $L(U)$, it is standard to denote it by $1$.

So you know that $p$ is idempotent ($p^2=p$). It follows automatically that $q=1-p$ is idempotent as $q^2=(1-p)^2=1^2-p-p+p^2=1-2p+p=1-p=q$. And then $qp=(1-p)p=p-p^2=0$ as well.

Such a pair $(p,q)$ is the $n=2$ case of what is called an orthogonal decomposition of the unit. When such a thing happens, we have the Peirce decomposition which is very handy for many purposes. In operator theory/algebras in particular.

For any given idempotent $p$, the range and the nullspace yield a direct sum decomposition of the vector space $V=\mbox{im} p\oplus \ker p$. Conversely, a direct sum decomposition of $V=I\oplus K$ defines a unique idempotent such that $\mbox{im} p=I$ and $\ker p=K$. Cf. remark below. Now let us answer your question.

The relation between the idempotents $p$ and $q=1-p$ is
  $$
\mbox{im }(1-p)=\ker p\qquad \ker (1-p)=\mbox{im} \,p.
$$

Proof: by symmetry, it suffices to prove one of these two properties.
Let us do the lhs.
Since $0=p-p^2=p(1-p)$, we get $p(1-p)x=0$ for every $x$ whence $\mbox{im}(1-p)\subseteq \ker p$.
Now take $x\in \ker p$. Then $x=x-px=(1-p)x$ belongs to $\mbox{im} (1-p)$. QED.

Remark: it is also useful to remember that the idempotents are exactly the diagonalizable operators with spectrum in $\{0,1\}$. If $p$ is nontrivial ($p\ne 0,1$), we have $\ker p$  the eigenspace of $0$, and $\mbox{im} p=\ker(1-p)$  the eigenspace of $1$. That is $\mbox{im} p=\{x\in V\,;\,px=x\}$, as opposed to $\ker p=\{x\in V\,;\,px=0$.

Title: Prove $p^2=p$ and $qp=0$


问:
$F(x)$ versus $F[x]$ where $F$ is a field
What is the difference, if there is any, between $F(x)$ and $F[x]$, where $F$ is a field and $x$ is an element of a field extension $E/F$ that is not in $F$? I have some guesses, but I was never sure about this. For example, for one specific usage in a textbook, I see $\mathbb{Q}[i]$ and $\mathbb{Q}(\pi)$. And maybe a more general question: what happens when $F$ is just a ring? For example, for the ring of Gaussian integers, the notation I usually see is $\mathbb{Z}[i]$, with brackets instead of open parentheses. I guess I never properly learned the notation.

答:
Short answer: given a ring $A$ and an element $x$, the notation $A[x]$ denotes the smallest ring that contains both $A$ and $x$. If $A$ happens to be a field, then $A(x)$ denotes the smallest field that contains both $A$ and $x$. That's all.

Longer answer: given a ring $A$, we denote by $A[x]$ the ring of polynomials in the variable $x$ with coefficients in $A$. Starting from the description in the above paragraph, if a ring contains $x$, it must also contain $x^2,x^3$, and so on. And sums of these. And the results of multiplying these with one another and with the elements of $A$. Without any more context, these are just polynomials the way you're used to them.

If $A\subseteq B$ and $x\in B$, then $x$ inherits any relations it may satisfy in $B$ when used in this polynomial ring. One example is the ring of Gaussian integers $\Bbb Z[i]$, where $i^2=-1$. If $x$ satisfies no relations that can be described using only elements of $A$, then it acts like in the previous paragraph, and you get regular polynomials with no special rules. Your $\Bbb Q[\pi]$ is an example.

If $F$ is a field, then $F(x)$ is the smallest field that contains both $F$ and $x$. Starting from the description in the first paragraph, this field must for the same reason contain all polynomials in the variable $x$ with coefficients in $F$. But it must also contain the reciprocal of any non-zero polynomial. Same things as in the above paragraph apply: If $x$ satisfies some algebraic relation that can be phrased using only coefficients from $F$ (we call this "$x$ is algebraic over $F$"), then that's baked into $F(x)$. Your $\Bbb Q(i)$ is an example of that, where $i^2=-1$ is the algebraic relation in question.

When $x$ is not algebraic over $F$, then $F(x)$ becomes the field of rational functions in the variable $x$ with coefficients in $F$. Which is to say, the elements of $F(x)$ are fractions where the numerators and the denominators are polynomials in $F[x]$. One concrete example is $\Bbb Q(\pi)$. But the fact that $\pi$ lies inside a well-known extension of $\Bbb Q$ is unnecessary. $\Bbb Q(x)$ works just as well, where $x$ is just an abstract variable.

If $x$ is algebraic over $F$, then $F[x]$ and $F(x)$ turn out to be isomorphic: fractions in $F(x)$ can always be expanded so that the denominators have no $x$ in them. Thus you will see either one used, which can be a point of confusion. For instance, $\Bbb Q(i)$ and $\Bbb Q[i]$ are, for all practical intents and purposes, the same thing.

ContentLicense: CC BY-SA 4.0


问:
Remodelling random points on a circle: Arc length between points distributed on circle is uniform?
An update to this: Modelling random points on a circle , but I hope my posts are self-contained.

Suppose 3 (distinct) points are uniformly and independently distributed on a circle of unit length (smaller than a unit circle!). This is really circle and not disc. Call these points $A,B,C$.

There is a claim here:

Let $X$ be the anticlockwise distance along the circle's circumference from $B$ to $A$.
Let $Y$ be the anticlockwise distance from $B$ to $C$.

Since the three points are independently uniformly distributed along the circumference of the circle, $X$ and $Y$ are iid variables with uniform distributions on $[0,1).$

Question: What exactly are $X$ and $Y$ (like the specific formulas in terms of the random variables $A,B,C$, and why are they iid Unif(0,1)? Please consider modifying what I did below to answer this question.

What I tried for identical distributed Unif(0,1): I tried thinking of something related to my wrong model.

For each $\omega \in \Omega$, we'll have either a line up of Case 1: $A(\omega),B(\omega),C(\omega)$ or Case 2: $C(\omega),B(\omega),A(\omega)$.

This model involves some point on the arc between A and C not containing B is selected as like the point 0/1 in viewing the circle as the unit interval with end points glued together. (Unfortunately, I think this choice of point kind of depends on a choice of $\omega$. Perhaps a choice of bijection $\phi$ from circle to unit interval doesn't depend on $\omega$ directly, but I think it depends on the choice of point.)

So, if we somehow are able to well-definedly imagine $A,B,C$ as members of the unit interval, then we have $A<B<C$ in Case 1 and then $C<B<A$ in Case 2.

Here, I might model $X$ as is something like $X=B-A$ in Case 1 and $X=1+(B-A)$ in Case 2. Similarly, $Y=B-C$ in Case 2 and $Y=1+(B-C)$ in Case 1.

To compute cdf of $X$: (btw, I'm assuming inequalities as strict or loose here are not really relevant)

5.1. $F_X(x) = P(X \le x)$ is $1$ for $x \ge 1$ and $0$ for $x \le 0$

5.2. For $x \in (0,1), F_X(x) = P(X \le x)$

$= P(X \le x, 1 > B-A > 0) + P(X \le x, -1 < B-A < 0)$

$= P(0 < B-A \le x < 1) + P(0 < 1+(B-A) \le x < 1)$

5.3. If $P(0 < B-A \le x < 1) = x - \frac12 x^2$ and $P(0 < 1+(B-A) \le x < 1) = \frac12 x^2$, then I have proven $X \sim Unif(0,1)$. Similarly, $Y \sim Unif(0,1)$.

5.4. I was able to make those computations in (5.3), but I'm not exactly sure what those assumptions were. Assuming the integrand is just '$1 \ da \ db$', the regions of integration have areas of, resp, $x - \frac12 x^2$ and $\frac12 x^2$ and so the areas are the values of the integrals. I think assuming $A,B$ are iid Unif(0,1), then the computation is correct, but I'm not sure that we can view $A,B$ as iid Unif(0,1) based on that my model was wrong. Well perhaps when converting $A,B,C$ from uniform on circle to uniform on $(0,1)$, we get that at least $A$ and $B$ are independent and that $B$ and $C$ are independent even if all 3 together are not independent (and even if they are not pairwise independent, i.e. we have pairwise independence except for independence of $A$ and $C$). So, perhaps the converted $A,B,C$ are not iid Unif(0,1), but maybe there is some partial independence here, and I believe the converted $A,B,C$ have $(0,1)$ (or $[0,1]$, whatever) as images.

What I tried for independence:

(Btw, since there's no measure theory here, I'm assuming the independence definition here is that the joint pdf splits up.)

I think this depends on the formulas for $X$ and $Y$, so I'm thinking to just wait on resolution on what the formulas are. Currently, there's a $B$ in each of the formulas for $X$ and $Y$, but I'm hoping this information with $B$ cancels out somehow. Something like how in the Gauss-Bonnet Theorem/Formula all the geometry information is cancelled out to get a topological quantity. Or maybe the $B$ isn't relevant after you compute the pdf's because...the joint pdf's are just constant whenever they are nonzero. Idk.

These questions are all related, but I hope I made each self-contained

Random points on a circle

More random points on a circle

Most random points on a circle

Modelling random points on a circle

Remodelling random points on a circle: Arc length between points distributed on circle is uniform?

答:
As $A,B,C$ are independent, so are $X$ and $Y$ and by symmetry they are identically distributed.

Now for any $b$ (the position of a point being measured as the counterclockwise distance from a common origin), $a$ is uniform in $[0,1)$, $a-b$ is uniform in $[-b,1-b)$ and $x=(a-b)\bmod1$ is uniform in $[0,1)$.


问:
Asymptotic matching of a first order ODE
Consider the following ODE:
\begin{equation}
\frac{\mathrm{d}y(t)}{\mathrm{d}t} = \varepsilon y(1-y)+y^2-\varepsilon \frac{y^2}{1-y} \quad \quad \text{for} \quad \quad t>0, \tag{1}
\end{equation}
where $y(0)=c$ and $\varepsilon\ll1.$ I would like to find an asymptotic solution that is valid for all $t>0.$ I have attached a figure of the numerical solution for $\varepsilon = 0.01$ and $c=0.5.$

For the outer solution (the solution that satisfies the initial conditions), I take $y(t) \sim y_0(t) +\varepsilon y_1(t)$ and substitute this into equation (1), which at leading order, gives
\begin{equation}
\frac{\mathrm{d}y_0(t)}{\mathrm{d}t} = y_0^2, \quad \quad \text{where} \quad \quad y_0(0)=c. \tag{2}
\end{equation}
Solving yields
\begin{equation}
y_0(t) = \frac{c}{1-ct}. \tag{3}
\end{equation}
Im relatively comfortable that this is correct outer solution.

Now, I need to find the inner solution which matches with this outer solution. Taking $y_{\infty}$ to be the steady state solution to (1), I make the variable transform
\begin{equation}
\tau = \frac{t-t_0}{\varepsilon^{\alpha}} \quad \quad \text{which provides} \quad \quad \frac{\mathrm{d}}{\mathrm{d}t}= \varepsilon^{-\alpha}\frac{\mathrm{d}}{\mathrm{d}\tau},\tag{4}
\end{equation}
for $\alpha>0.$ Here $t_0$ is the overlap between the outer and inner solutions; this can be estimated by seeing when $y_0(t)$ first meets $y_{\infty}$, meaning we just solve $y_0(t_0)=y_{\infty}.$ Doing so yields $t_0 = (y_{\infty}-c)/y_{\infty}c.$ For the inner region, the numerical simulations suggest that we take $Y(\tau) \sim y_{\infty}-\varepsilon Y_1(\tau). $ Substituting this into (1) yields
\begin{equation}
-\varepsilon^{1-\alpha}\frac{\mathrm{d}Y_1}{\mathrm{d}\tau} = -2\varepsilon y_{\infty} Y_1+\mathcal{O}(\varepsilon^2). \tag{5}
\end{equation}

This is where I'm stuck. There is no possible balance except from taking $\alpha = 0$, which is not allowed. Furthermore, say if $\alpha=0$, equation (5) solves to give $Y_1 = k \exp (2y_{\infty}\tau),$ which is clearly secular.

Am I using the wrong variable transform? I'm clearly doing something wrong, since $Y_1\rightarrow \infty$  as $\tau \rightarrow \infty$.

Thanks for any help in advanved.

答:
As you approach $y=1$, or better $y=1-ε$, the next term to become "active" is the third one. So try $y=\frac1{1+εu}$ with $u>1+O(ε)$ to get
$$
-εu'=ε^2u+1-\frac{1+εu}{u}=\frac{ε^2u^2+(1-ε)u-1}{u}\tag1
$$
Thus approximately
$$
-εu'=1-\frac1u\iff u'+\frac{u'}{u-1}=-\frac1ε
\\
u(t)+\ln(u(t)-1)=-\frac{t}ε+d
\\
(u(t)-1)e^{u(t)-1}=De^{-\frac{t}ε}\iff u(t)=1+W(De^{-\frac{t}ε})
,~~~ y(t)=\frac1{1+ε+εW(De^{-\frac{t}ε})}\tag2
$$
using the Lambert-W function

This is a rather short segment of width $ε$ around the $t_0$ given in the question.

Then even closer to the asymptote, set $u=1+δv$ and insert in (1)
$$
-εδv'=ε^2(1+δv)+1-\frac{1}{1+δv}-ε=ε^2(1+δv)+\frac{δv}{1+δv}-ε.
$$
This balances on the right for $δ=ε$ to give in the leading order terms
$$
-εv'=v-1\implies v(t)=1+Fe^{-tε},~~~ y(t)=\frac1{1+ε+ε^2+ε^2Fe^{-tε}}
$$

Now select the constants so that all fits together...


问:
How to go about finding whether this limit exists: $\lim_{x \to 0} x [[\frac{1}{x}]]$?
I am trying to solve few challenge questions on Real Analysis from Kaczor and Nowak's Problems in Mathematical Analysis, to become more proficient and stimulate thinking. I'd like someone to (a) verify if my proof is correct (b) is there a way to rigorously show that the limit does not exist?

Note. $[[x]]$ is the greatest integer less than or equal to $x$ for all $x \in \mathbf{R}$.

Find the limits or state that they do not exist.

Problem 1.1.1 (b) $\lim_{x \to 0}x\cdot[[\frac{1}{x}]]$

Proof.

Consider the sequences $a_n:=\frac{1}{n}$ and $b_n=-\frac{1}{n}$. Both these sequences converge to zero.

The corresponding image sequences are,

\begin{align*}
(f(a_n)) &= \frac{1}{1}\cdot 1, \frac{1}{2} \cdot 2, \frac{1}{3}\cdot 3, \ldots \\
&= 1,1,1,1,1,\ldots
\end{align*}

\begin{align*}
(f(b_n)) &= -\frac{1}{1}\cdot (-1), -\frac{1}{2} \cdot (-2), -\frac{1}{3}\cdot (-3), \ldots \\
&= 1,1,1,1,1,\ldots
\end{align*}

Also, consider the sequence $c_n:=\frac{1}{\sqrt{n}}$. The sequence $(c_n)$ also converges to zero.

\begin{align*}
(f(c_n)) &= \frac{1}{\sqrt{1}}\cdot [[\sqrt{1}]], \frac{1}{\sqrt{2}} \cdot [[\sqrt{2}]], \frac{1}{\sqrt{3}}\cdot [[\sqrt{3}]], \frac{1}{\sqrt{4}} \cdot [[\sqrt{4}]], \ldots \\
&= 1,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{3}},1,\frac{2}{\sqrt{5}},\frac{2}{\sqrt{6}},\frac{2}{\sqrt{7}},\frac{2}{\sqrt{8}},1,\ldots
\end{align*}

Thus, the image sequence oscillates between $0$ and $1$ and is not convergent. This violates the definition of functional limits. We require that, for all sequences $(x_n)$ in the domain of the function $f$, such that $(x_n) \to a$, $x_n \ne a$ for all $n \in \mathbf{N}$, the image sequence $f(x_n)$ converges to $L$. Then, $\lim_{x \to a}f(x) = L$. So, the above limit does not exist.

Body: I'll make a try to show that $\lim\limits_{x\to0}x[[\frac{1}{x}]]=1$. For $x\not=0$ it is $[[\frac{1}{x}]]\leq \frac{1}{x}<[[\frac{1}{x}]]+1$ hence $\frac{1}{x}-1<[[\frac{1}{x}]]\leq \frac{1}{x}$ which means $1-x<x[[\frac{1}{x}]]\leq 1$ if $x>0$ and $1-x>x[[\frac{1}{x}]]\ge 1$ if $x<0$. So $\lim\limits_{x\to0}x[[\frac{1}{x}]]=\lim\limits_{x\to0^+}x[[\frac{1}{x}]]=\lim\limits_{x\to0^-}x[[\frac{1}{x}]]=1$


问:
Remainder Term for a Taylor Series and Polynomials
If $f(x)=e^x$ and $f_n(x)$ is the $n^{th}$ order Taylor Polynomial of this function, then we know that the Taylor Theorem states there exists $t\in [x_0, x]$ such that the remainder term of is given by,

$$r_n(x)=\frac{f^{(n+1)}(t)}{(n+1)!}(x-x_0)^{n+1}=\frac{e^t}{(n+1)!}(x-x_0)^{n+1}$$

as $f^{(n+1)}(t)=f(t)=e^t$. Now we also know,

$$f=f_n+r_n$$

$$\Rightarrow e^x = f_n(x) + \frac{e^t}{(n+1)!}(x-x_0)^{n+1}$$

On the right hand side, $f_n$ is a polynomial and so is the expression for the remainder term. However $e^x$ is not a polynomial. How is it possible to write a non polynomial function in terms of a polynomial? We can use a similar procedure and arrive at this contradiction for other non polynomial functions as well. So what is the gap in my understanding?

答:
$t$ is just a value between $x_0$ and $x$ and $e^t$ is how to evaluate this constant.
It does depend on $x$, but nobody said that Taylor polynomial is expressing the function in full. It has an error in its approximation.

The term with $t$ is that error term, so it talks about an error in Taylor polynomial approximation. It is not part of polynomial and of course that it depends on $x$, because typically more you are away from $x_0$ worse the error.

Just because it has $(x-x_0)^{n+1}$ in it does not make it a part of the polynomial, there are other expressions for the error term that look less like an additional polynomial term.

So, yes, besides $(x-x_0)^{n+1}$ the error term may have one additional dependency on $x$, or it may not, we do not know. All we know is that $t$ is between the two. In general it is movable regardless if this move is expressible as a function on $x$ or not.

Typically this $t$ is used to say that max error cannot be worse than, in your case, the maximum value of $e^t$ where $t$ is between $x$ and $x_0$ together with other values in that term. Some functions will have this max value fixed for quite some possible range between $x$ and $x_0$, some will not. But anyway $t$ is not really useful beside deciding this max error, because we know nothing else about $t$ except that it is between $x$ and $x_0$.

If $t$ would be a fixed value then the error term would be in form of $k(x-x_0)^{n+1}$, meaning every function would be expressible as a polynomial of $n+1$ degree, and unless the function is such a polynomial that is not the case.


答:
Hint: $\frac 1x-1\lt [[\frac 1x]]\le 1/x $ for all $x\ne 0$

By Squeeze theorem, $\lim_{x\to 0^+} x[[\frac 1x]]=1=\lim_{x\to 0^-} x[[\frac 1x]]$

Title: How to go about finding whether this limit exists: $\lim_{x \to 0} x [[\frac{1}{x}]]$?

Body:
I am trying to solve few challenge questions on Real Analysis from Kaczor and Nowak's Problems in Mathematical Analysis, to become more proficient and stimulate thinking. I'd like someone to (a) verify if my proof is correct (b) is there a way to rigorously show that the limit does not exist?

Note. $[[x]]$ is the greatest integer less than or equal to $x$ for all $x \in \mathbf{R}$.

Find the limits or state that they do not exist.

Problem 1.1.1 (b) $\lim_{x \to 0}x\cdot[[\frac{1}{x}]]$

Proof.

Consider the sequences $a_n:=\frac{1}{n}$ and $b_n=-\frac{1}{n}$. Both these sequences converge to zero.

The corresponding image sequences are,

\begin{align*}
(f(a_n)) &= \frac{1}{1}\cdot 1, \frac{1}{2} \cdot 2, \frac{1}{3}\cdot 3, \ldots \\
&= 1,1,1,1,1,\ldots
\end{align*}

\begin{align*}
(f(b_n)) &= -\frac{1}{1}\cdot (-1), -\frac{1}{2} \cdot (-2), -\frac{1}{3}\cdot (-3), \ldots \\
&= 1,1,1,1,1,\ldots
\end{align*}

Also, consider the sequence $c_n:=\frac{1}{\sqrt{n}}$. The sequence $(c_n)$ also converges to zero.

\begin{align*}
(f(c_n)) &= \frac{1}{\sqrt{1}}\cdot [[\sqrt{1}]], \frac{1}{\sqrt{2}} \cdot [[\sqrt{2}]], \frac{1}{\sqrt{3}}\cdot [[\sqrt{3}]], \frac{1}{\sqrt{4}} \cdot [[\sqrt{4}]], \ldots \\
&= 1,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{3}},1,\frac{2}{\sqrt{5}},\frac{2}{\sqrt{6}},\frac{2}{\sqrt{7}},\frac{2}{\sqrt{8}},1,\ldots
\end{align*}

Thus, the image sequence oscillates between $0$ and $1$ and is not convergent. This violates the definition of functional limits. We require that, for all sequences $(x_n)$ in the domain of the function $f$, such that $(x_n) \to a$, $x_n \ne a$ for all $n \in \mathbf{N}$, the image sequence $f(x_n)$ converges to $L$. Then, $\lim_{x \to a}f(x) = L$. So, the above limit does not exist.


问:
If Banach space is countable union of closed sets, Can I conclude space should be one of the closed set?
Let $X$ be Banach space. $F_n$ be closed subsets of X. If $X = \cup F_n$, Can I conclude $X = F_{n_0}$ for some $n_0$? If right, suggest a hint.

答: If $(X,d)$ is unbounded, select $y\in X$ and consider $F_n=\{x\in X\,:\, d(x,y)\le n\}$.


问:
Decomposition of the difference of cubes
I have to study the existence condition of the following algebraic fraction:
$$p=\frac{1}{a^3-b^3}$$
Since I can decompose the difference of the cube of $a$ and $b$ in this way: $a^3-b^3=(a-b)(a^2+ab+b^2)$, I can rewrite $p$ as:
$$p=\frac{1}{(a-b)(a^2+ab+b^2)}$$
So the existence conditions will correspond to the denominator different from $0$ and so:
$\begin{cases} a\neq b\\ a^2+ab+b^2\neq 0
\end{cases}$

I have a doubt: my Professor says that the terms "$a^2+ab+b^2$" is what we can call with "false square" and it is always $>0$.

But I think it is $\geq 0$, since if $a=b=0$ then it is null. I have found out that the false square is always $\neq 0$ also on the web, so I can't understand why. Can you hlep me?

答: $a^2+ab+b^2=\Big(a+\frac{b}{2}\Big)^2+\frac{3b^2}{4}$ and the sum of two squares is always positive, unless $a=b=0$.


问:
Proof of lemma on parameters in natural deduction
I could not understand the proof of the Lemma on parameters on page 29 of this book (Natural Deduction) by Dag Prawitz. Why if all proper parameters in a derivation $\pi$ are pure, $\pi$ will satisfy 3 clauses as stated in the Lemma?
(On page 28 and 29, a pure parameter $a$ has the properties of (a) being the only proper parameter of an application $\alpha$ of $\forall I$ or $\exists E$, and  (b) all formula occurrences $B$s where $a$ occurs in $B$s are linked by $a$ to the premiss or discharged premiss of $\alpha$.)

Specifically,

Why in a deduction $\pi$ whose proper parameters are pure, every proper parameter used by a quantifier rule $\alpha$ ($\forall I$ or $\exists E$) is a proper parameter of exactly one quantifier rule in the derivation? (Or put it differently, how the second property of a pure parameter $a$ is violated if it occurs twice in 2 different $\alpha$ ?)

Why if a pure parameter $a$ is used by an application $\alpha$ of $\forall I$ or $\exists E$, then it occurs only in the part of the derivation above $\alpha$?

I tried to cook up a counter example as follows. Let $\pi$ be the derivation $\forall xPx$ $\rightarrow$ $Pa$ $\rightarrow$ $\forall xPx$ $\rightarrow$ $Pa\supset\forall xPx$. The bottom formula ($Pa\supset\forall xPx$) occurs below $Pa$ and is linked to $Pa$ by $a$ (by clause 4 of the definition of a connection between 2 formulas in page 28). So $a$ is still a pure parameter in $\pi$, but it occurs below the consequence of $\forall I$.

答:
The issue of the Lemma [page 29 of Dag Prawitz's Natural Deduction (1965)] regards the management of parameters in the quantifier rules.

A couple of rules has restrictions on the parameter used: the simple example is the restriction of the $(∀ \text I)$ rule where the parameter $a$ must not occur free in some open assumption [reason: we cannot "generalize" on the assumption $x=0$ to derive the incorrect $∀x(x=0)$].

The simplest way to fulfill the restrictions is that:

every parameter used by a quantifier rule is used only once in the derivation, and

if a parameter $a$ is used by an application $\alpha$ of a rule, then it occurs only in the part of the derivation above $\alpha$.

The Lemma describe a procedure that applied to a derivation $\Pi$ transforms it in a new derivation where all parameters are pure: in a nutshell, start from the uppermost application $\alpha$ of a quantifier rule with restriction and if the parameter $a$ of $\alpha$ is not pure, replace it with a fresh parameter $b$ that is not used in $\Pi$.

The gist of the "purification" procedure is that, in order to avoid potential clash between variables, we can always transform a derivation $\Pi$ in an "equivalent" one (same assumptions, same conclusion) $\Pi'$ such that:

no variable occurs in $\Pi'$ both free and bound: and this is achieved in Prawitz's system using parameter (free) and variables (bound), and

all parameters are distinct and different from the remaining parameters occurring in $\Pi'$: they are used only once.


问:
Quadratic variation of independent processes
when reading some lecture notes of stochastic calculus, I have notice that often the professor seems to use the fact that:

given $B^1$ and $B^2$ as two independent Brownian motion, then we have:
$$ \langle B^1, B^2 \rangle_t = 0$$
Is this true? Can anybody explain why?

TBN: angle brackets mean "quadratic variation of the processes".

答:
The process $B = (B^1, B^2)$ is a 2-dimensional Brownian motion, hence a standard Wiener-process (on the natural filtration), so the Levy-theorem states that the covariation matrix is diagonal (we might need that they are continuous). I believe there should be an easier way though.

Edit: Looking back into my lecture, we showed once by direct calculation that the approximation of the quadratic covariation $ \tau_n ( \langle B^1 , B^2 \rangle )_t $ tends to 0 for a Riemann decomposition, which should also yield the claim.
I think another option would be to prove that $ B^1 B^2 $ is already a martingale (wrt some natural filtration), as the quadratic covariation is the only process in $ \mathcal{V} $ (the class of continuous processes starting in 0 and with bounded variation) s.t. $ B^1 B^2 - \langle B^1 , B^2 \rangle$ is a local martingale. I'm not sure, but this might also require continuity of $B^1$ and $B^2$ .

I hope this helps


问:
Motivation for Ramanujan's mysterious $\pi$ formula
The following formula for $\pi$ was discovered by Ramanujan:
$$\frac1{\pi} = \frac{2\sqrt{2}}{9801} \sum_{k=0}^\infty \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}\!$$

Does anyone know how it works, or what the motivation for it is?

答:
Here is a nice article Entitled: "Ramanujan's Series for $\displaystyle\frac{1}{\pi}$ : A Survey", by Bruce C.Berndt. This article appeared in the American Mathematical Monthly *August/September* 2009. You can see it here.

http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.158.2533&rep=rep1&type=pdf

ContentLicense: CC BY-SA 2.5

ContentLicense: CC BY-SA 3.0


问:
How to show that $|z| \geq 0$ and $|z|=0$ if and only if $z=0$?
Teaching myself Linear Algebra and got stuck on the following question for complex numbers:

Show that $|z| \geq 0$ and $|z|=0$ if and only if $z=0.$

Now, the question itself seems pretty obvious where $z=x+iy$ and $|z|=\sqrt{x^2+y^2}$ but I am a bit confused regarding how to tie everything together. Thanks in advance.

答: Now, $|Z|= 0$ implies that $\sqrt{x^2+y^2}=0$ which implies that $x^2+y^2=0$. Under what circumstances will the last equality hold?</p>


问:
Chromatic number of Simple Connected Planar Graph
The chromatic number of a Simple Connected Planar Graph is $\le 6$.

How do I proceed to prove this?

答:
I'm giving you some hints.

Theorem: For a connected, simple planar graph $G=(V,E)$, $
\chi(G) \leq 6$.

Proof Sketch: We prove via induction on $|V(G)|=n$. The base case $1 \leq n \leq 6$ is trivial, since a graph on $6$ or less vertices can always be coloured with $6$ colours. Suppose that $G$ has $k \geq 6$ vertices and $G$  is $6$-colourable. Consider any simple, connected planar graph $G'$ on $k+1$ vertices. $G'$ contains a $v' \in V(G')$ with $deg(v') \leq 5$ (This is the "tricky" part. Why is this true? Hint: Consider the average degree and use Eulers Formula). What happens if we remove $v'$ from $G'$?


问:
How to solve : $ \frac{\log(5x-6)}{\log(x)} = \log (x)$
I'm struggling with this equation :

$$ \frac{\log(5x-6)}{\log(x)} = \log (x)$$

I've tried to do this:

$$ \log(5x-6) = \log^2(x)$$

But really I can' do more. . I just need a hint

答:
I'll try to get escape of the logarithm to see what's going on.

Let $5x-6=e^t$,

$$t=\ln^2\left(\frac{e^t+6}{5}\right)$$

$$\sqrt t=\ln \left(\frac{e^t+6}{5}\right)$$

$$\frac{e^t+6}{5}=e^{\sqrt t}$$

$$e^t+6=5e^{\sqrt t}$$

Let $\sqrt t=u$,

$$5e^u-e^{u^2}-6=0$$

$$ \color {gold}{\boxed {\color{black} {e^{u^2}-5e^u+6=0.}}}$$

There is no standard mathematical function that can solve the last equation.


问:
How to solve this step to get this
$\sqrt{2A^2(1+\cos \theta)} $

to

$\sqrt{4A^2\cos^2 \dfrac {\theta} {2}}$  (Here, divided by $2$ is only under $\theta$)

I have solved till : $\dfrac {2A\sqrt{1+\cos\theta}}{2}$ (divided by $2$ is whole under $1 + \cos \theta$)

答: Hint: Using the trigonometric identity (a variation of the half-angle formula) $$\cos^2 \dfrac {\theta}{2} = \dfrac {1 + \cos \theta}{2}$$ rewrite this expression in terms of $1 + \cos \theta$ only, and the answer will be immediate.

Title: How to solve this step to get this


问:
Questions on ring fixed by automorphisms
Suppose $A$ is an integral domain and $F$ is its field of fractions. Let $G \leq Aut(F)$ be a group of automorphisms of $F$. Assume $g(a) \subseteq A$ for all $g \in G$ and let $A^G$ be the fixed ring
$$A^G=\{a \in A\ \mid g(a)=a, \forall g\in G\}.$$
I have managed to show that $A$ is integral over $A^G$ by considering the polynomial $\prod_{g\in G}(x-g(a))$ and showing it has coefficients in $A^G[x]$. Clearly it is monic, so the result follows. I have some more questions about this setup. I would also like to show that the fraction field of $A^G$ is $F^G$ and I am not sure how to do it. I think I might have to use the norm or trace maps $N,T:A \rightarrow A^G$
$$N(a)=\prod_{g\in G} g(a),$$
$$T(a)=\sum_{g\in G} g(a).$$
Also, I would like to show that $A$ Noetherian $\implies A^G$ Noetherian and $A$ Dedekind $\implies A^G$ Dedekind.

答:
There is the following lemma: If $A \subseteq B$ is a ring extension with $x_1,..,x_k \in B$ with $B\cong A[x_1,..,x_k]$ and such that each $x_i$ is integral over $A$ then $B$ is a finitely generated $A$-module.

It seems the following holds:

Proposition. Assume $G$ is a finite group. If there is a finite set of elements $x_1,..,x_k \in A$ with $A\cong A^G[x_1,..,x_k]$ (as a ring) then $A$ is a finitely generated $A^G$-module. Hence if $A$ is Noetherian and $A$ is a finitely generated $A^G$-algebra it follows $A^G$ is Noetherian.


问:
Proving that a function of the form f(x+y)=f(x)*f(y) is continuous
I have the to solve the following problem:

Let $f$ be a function from the real numbers to the real numbers. The function satisfies $f(x+y) = f(x)f(y)$ for all real $x,y$. Prove that if $f$ is continuous in $0$ then $f$ is continuous in every point.

I think I have a solution but I would like to know if it's correct:

By letting $y=0$ we get that $f(x)=f(x)f(0)$ meaning that $f(0)=1$. Thus we know that for every $\epsilon > 0$ and for every $x$ there exists a $\delta$ such that $|f(x)-1| < \frac{\epsilon}{|f(a)|} $ when $|x| < \delta$. Since this holds for all $x$ we can replace $x$ with $x-a$. Rewriting this we get that whenever $|x-a| < \delta$ we have that $\,|f(a)f(x-a)-f(a)|<\epsilon \implies |f(x)-f(a)| < \epsilon$.

答:
In your proof,doing $\epsilon/|f(a)|$ may be problematic as $f(a)$ could be $0$ also! You may proceed as below also.

$f(x+y)=f(x)f(y)\implies f(0)=f^2(0)\implies f(0)=0 $ or $f(0)=1$

If $f(0)=0$, then $\forall x\in \mathbb R, f(x)=f(x+0)=f(x)f(0)=0$ and we are done.

If $f(0)=1$, then for any $x_0\in \mathbb R: x_0\ne 0$:

$|f(x_0+h)-f(x_0)|=|f(x_0)(f(h)-1)|\le (|f(x_0)|+1)|f(h)-1|$. It'll be clear later why $1$ was added here.

$\forall \epsilon\gt 0, \exists \delta\gt 0: |h|\lt \delta\implies |f(h)-1|\lt \frac{\epsilon}{(|f(x_0)|+1)}$. It follows that:

$|f(x_0+h)-f(x_0)|\lt (|f(x_0)|+1)\times \frac{\epsilon}{(|f(x_0)|+1)}=\epsilon $

We have proved that: $\lim_{h\to 0} (f(x_0+h)-f(x_0))=0$

Now write $f(x_0+h)=(f(x_0+h)-f(x_0))+f(x_0)$. By limit rules, we have

$\lim_{h\to 0}f(x_0+h)=0+f(x_0)=f(x_0)$

$x_0\ne 0$ was any arbitrary point and hence $f$ is countinuous everywhere.


问:
Generalizing sequence of fractions
I need to construct a formula for the $n$th term of the sequence $$\{2, 2/3, 3/4, 4/5,\cdots \}$$

It's easy to come up with someting for every term after $2$, something like $\cfrac{n+1}{n+2}$ with lower limit $n=1$.

But after countless attempts I have not been able to come up with a formula that includes the very first term.

答:
If you really need to, you can come up with some artificial way to express this. For instance:

$$u_n=\frac{n+1}{n+2}+\frac32\left\lfloor\frac{1}{n+1}\right\rfloor$$
where $\lfloor\cdot\rfloor$ is the floor function (so that $\lfloor\frac{1}{n+1}\rfloor$ is zero unless $n=0$).

But that is pointless, and just serves to obfuscate the fact that $n=0$ is a special case. Better is simply
$$
  u_n =
\begin{cases}
2  & \text{if $n=0$} \\
\frac{n+1}{n+2} & \text{if $n\ne 0$}
\end{cases}
$$

Title: Generalizing sequence of fractions


问:
What property says that $\int_{x_1}^{x_2} g'w' dx = 0$ for $w \in H_0^1(x_1,x_2)$, since $g$ is linear?
What property says that $$\int_{x_1}^{x_2} g'w' dx = 0$$ for $w \in H_0^1(x_1,x_2)$, since $g(x)$ is linear?

$w$ is in $H_0^1(0,1)$.

Body: $g'$ is constant, so $$\int_{x_1}^{x_2} g'w'~dx = g'\int_{x_1}^{x_2}w'~dx = g' (w(x_2)-w(x_1)) = 0.$$


问:
What is the formula for the derived sequence?
what the sequence is derived from

Once the sequence is derived how is that turned into a formula and what is it?
Ive been working on this for hours im not getting anywhere. Very thankful to anyone answering my maths problem.

Body:
Hint
$$b_{n+1}-b_{n}=n-2\tag 1$$
$$b_{n+2}-b_{n+1}=n+1-2\tag 2$$

$$(2)-(1)= ???$$


问:
$\left\|\int_a^b x(t) dt \right\| \leq \int_a^b \left\| x(t) \right\| dt$
I have seen this inequality:

$$\left\|\int_a^b x(t) dt \right\| \leq \int_a^b \left\| x(t) \right\| dt$$

How could I prove that it holds for any $x:[a,b]\longrightarrow \mathbb{R}^n$, assuming that $x$ is integrable.

I would like to see a prove of this for a general norm, but if it is easier to do it for a particular norm, it would also help me to get a sense of why it is true.

Thanks in advance.

答:
Let $\zeta := \int_a^b x(t)\, dt$. If $\zeta = 0$ the claim is trivial.
Assume $\zeta \neq 0$. It holds that
$$
|\zeta|^2 = \zeta \cdot \int_a^b x(t)\, dt = \int_a^b \zeta\cdot x(t)\, dt
\leq \int_a^b |\zeta\cdot x(t)|\, dt
\leq \int_a^b |\zeta|\, |x(t)|\, dt
= |\zeta| \int_a^b |x(t)|\, dt
$$
and the claim follows.

Edit: here $|\cdot|$ denotes the standard Euclidean norm.
For a general norm, you can prove first the inequality for simple functions.
Assume that
$$
x(t) = \sum_{i=1}^N \chi_{I_i}(t) \, x_i\,.
$$
Then
$$
\left\| \int_a^b x(t)\, dt\right\| = \left\|\sum_{i=1}^N m(I_i)\, x_i\right\|
\leq \sum_{i=1}^N m(I_i)\, \|x_i\| = \int_a^b \|x(t)\|\, dt.
$$


问:
Heat equation with periodic initial conditions
Let $f:\mathbb{R}^n \to \mathbb{R}, f \in C(\mathbb{R})$ be $[0,1]^n-$periodic.

Consider then the problem:
$$\left\{\begin{matrix}\partial_t u -\Delta_x u = 0\quad t>0\\u(x,0) = f(x)\quad t=0\end{matrix}\right.$$

Show that $$\lim_{t \to \infty}u(x,t) = \int_{[0,1]^n}f(y)dy$$

We can find a solution of the PDE by taking
$$u(x,t) = \int_\mathbb{R^n} \Phi(x-y,t) f(y)dy$$

with $\Phi(x-y,t) = \frac{e^{-\frac{|x-y|^2}{4t}}}{(4\pi t)^{\frac{n}{2}}}$ the heat kernel. I know that $\int_\mathbb{R^n} \Phi(x-y,t)dy=1$ for all $t>0$.

Consider the set of points in $\mathbb{z}^n$ and take the partition of $\mathbb{R}^n = \bigcup_{z \in \mathbb{Z}^n} \left([0,1]^n+z\right)$

We have then that
$$u(x,t) = \sum_{z \in \mathbb{Z}^d}\int_{[0,1]^n+z}\Phi(x-y,t)f(y)dy = \sum_{z \in \mathbb{Z}^d}\int_{[0,1]^n}\Phi(x-y+z,t)f(y+z)dy$$
$$=\sum_{z \in \mathbb{Z}^d}\int_{[0,1]^n}\Phi(x-y+z,t)f(y)dy$$

Now I'd like to find some way to justify something like
$$\int_{[0,1]^n}\Phi(x-y+z,t)f(y)dy \to \left[\int_{[0,1]^n}f(y)dy\right]\left[\int_{[0,1]^n}\Phi(x-y+z,t)dy\right]\mbox{ as }t \to \infty$$

which would solve the question. How would I do that?

Attempt to show it:

$$\left|\int_{[0,1]^n}\Phi(x-y+z,t)f(y)dy - \left[\int_{[0,1]^n}f(w)dw\right]\left[\int_{[0,1]^n}\Phi(x-y+z,t)dy\right]\right| = $$
$$=\left|\int_{[0,1]^n}\Phi(x-y+z,t)\left[f(y) - \int_{[0,1]^n}f(w)dw\right]dy\right|$$
$$\leq \sup_{y \in [0,1]^n}\left[f(y) - \int_{[0,1]^n}f(w)dw\right] \int_{[0,1]^n}\Phi(x-y+z,t)dy$$

We can guarantee that $sup$ is finite because $f$ is periodic on $[0,1]^n$ and continuous, therefore bounded.

Now, as $t \to \infty$ we have that the function $\Phi(x-y+z,t)$ goes to $0$ by the definition. As the integral of $\Phi$ is $1$ for all $t>0$ we must have that, in the bounded square $[0,1]^n$ the integral must go to $0$.

and so we have proved the convergence we needed.

Is this correct? Thanks in advance.

Edit: $\color{red}{\mbox{My attempt is incorrect}}$. As I colleague pointed out to me, my argument would be valid for any value instead of $\int f(y)dy$. I only saw that every term in the summation goes to $0$, which isn't enough. I'd need some form of uniform convergence to pass the limit inside the summation in order to use any results concerning the limit of the terms on summation.

Our professor said to use the fact that periodic functions can be approximated uniformly by trigonometric polynomials, so we'd solve the problem for a trigonometric function and then by the uniform convergence we would have solved for any periodic function. Can anyone help on that? Thanks.

答:
I think it goes more or less as follows or at least I hope it helps,

$$
\begin{split}
u(x,t)
 & =\lim_{t\to 0}\sum_{z \in \mathbb{Z}^d}\int\limits_{[0,1]^n+z}\Phi(x-y,t)f(y)dy\\
& =\lim_{t\to 0}\sum_{z \in \mathbb{Z}^d}\int\limits_{[0,1]^n+z}\Phi(y-x,t)f(y)dy\\
& =\lim_{t\to 0}\sum_{z \in \mathbb{Z}^d}\int\limits_{[0,1]^n}\Phi(y+z-x,t)f(y+z)dy\\
& =\lim_{t\to 0}\sum_{z \in \mathbb{Z}^d}\int\limits_{[0,1]^n}\Phi\big((z-x)+y,t\big)f(y)dy\\
& =\lim_{t\to 0}\int\limits_{[0,1]^n}f(y)\bigg(\sum_{z \in \mathbb{Z}^d}\Phi\big((z-x)+y,t\big)\bigg)dy\\
& =\lim_{t\to 0}\int\limits_{[0,1]^n}f(y)\bigg(\sum_{z \in \mathbb{Z}^d}\hat{\Phi}(z-x,t)e^{2\pi i zy}\bigg)dy\qquad(\text{Poisson summation formula (PSF))}\\
& =\int\limits_{[0,1]^n}\sum_{z \in \mathbb{Z}^d}\Big(\lim_{t\to 0}\hat{\Phi}(z-x,t)\Big)f(y)e^{2\pi i zy}dy\\
& =\int\limits_{[0,1]^n}\sum_{z \in \mathbb{Z}^d}\Big(\lim_{t\to 0}e^{-t(z-x)^2}\Big)f(y)e^{2\pi i zy}dy\\
& =\int\limits_{[0,1]^n}\sum_{z \in \mathbb{Z}^d}f(y)e^{2\pi i zy}dy\\
& =\int\limits_{[0,1]^n}f(y)\bigg(\sum_{z \in \mathbb{Z}^d}e^{2\pi i zy}\bigg)dy\\
& =\int\limits_{[0,1]^n}f(y)\bigg(\sum_{z \in \mathbb{Z}^d}\delta(z-y)\bigg)dy\qquad\qquad\;\;\qquad(\text{PSF)}\\
&=\int\limits_{[0,1]^n}f(y)dy
\end{split}
$$


答:
The difference of two successive $b$'s is a linear polynomial so you can expect that a $b$ follows a quadratic polynomial, because $(n+1)^2-n^2=2n+1$ (quadratic yields linear).

We set

$$b_n=pn^2+qn+r$$

and solve

$$b_{n+1}-b_n=p(2n+1)+q=n-2$$ and by identification

$$p=\frac12,\\q=-\frac{5}2.$$

Hence

$$b_n=\frac{n^2-5n}2+r$$ where $r$ is free.

Title: What is the formula for the derived sequence?

Body:
what the sequence is derived from

Once the sequence is derived how is that turned into a formula and what is it?
Ive been working on this for hours im not getting anywhere. Very thankful to anyone answering my maths problem.


问:
weak convergence in L2
Let $f_n\rightarrow f$ strongly in $L^2(\Omega)$ for $\Omega\subset \mathbb{R}^n$ open an bounded.
And $g_n \rightarrow g$ weakly in $L^2(\Omega)$. It is said that

$$
\int_{\Omega} f_n g_n \rightarrow \int_{\Omega}fg
$$

Does anyone have a proof for this convergence?

答:
$\newcommand\abs[1]{\left|#1\right|}$

First, a standard application of Uniform Boundedness shows that $||g_n||_2$ is bounded; say $||g_n||\le A$ for all $n$.

Let $\epsilon>0$. Choose $N$ so $$||f_n-f||\le\frac{\epsilon}{2A}\quad(n\ge N).$$If $n\ge N$ then $$\abs{\int (f_ng_n-fg)}\le\abs{\int(f_n-f)g_n}+\abs{\int f(g_n-g)}=I+II.$$Cauchy-Schwarz shows that $I\le\epsilon/2$ for all $n\ge N$, while weak convergence implies that $II\le\epsilon/2$ for large enough $n$.

Title: weak convergence in L2


问:
Show that the equation of the form $x^2$$-$$y^3$$=$$1$ has solutions in $\mathbb{Z}$
Show that the equation of the form $x^2$$-$$y^3$$=$$1$ has solutions $x$ and $y$ $\in$  $\mathbb{  Z }$</p>

Body:
$$0^2-(-1)^3=(-1)^2-0^3=1^2-0^3=3^2-2^3=(-3)^2-2^3=1$$

It turns out these are all possible solutions.


问:
Probability of sampling linearly independent vectors
Let $q$ be a prime so that $\mathbb{Z}_q$  is a field. I would like to sample $n$ vectors independently and uniformly from $\mathbb{Z}^n_q$, to get a set $V = \{ v_i \}_{i=1}^n$.
What is the probability that $V$ is a basis for $\mathbb{Z}^n_q$? I had the following proof idea, by induction. The problem reduces to what is the probability of sampling $n$ linearly independent vectors, and as such, we proceed by induction on the number of sampled vectors $k$. For $k = 1$, we only require that we are not sampling the zero vector, and as such we have that $\Pr[\{v_1\} \text{is LI} ] = 1 - q^{-n}$. Now let us suppose that we have $k$ linearly independent vectors $\{ v_i \}_{i=1}^k$. We sample a new vector $v_{k+1}$. In order for it to be linearly independent, it must be that it is not a linear combination of the others, and as such there are at most $q^k$ possible 'forbidden' values. Namely these correspond to the possible coefficients $\alpha_i$ of the   linear combinations $\sum_{i=1}^k \alpha_i v_i$. So we have that $\Pr[\{v_i\}_{i=1}^{k+1} \text{is LI} \, | \, \{v_i\}_{i=1}^{k} \text{is LI}] \geq 1 - q^{k - n}$. As such we should have that

$$\Pr[\{v_i\}_{i=1}^n \text{is LI}] = \Pr[\{v_i\}_{i=1}^n \text{is LI} | \{v_i\}_{i=1}^{n-1} \text{is LI}] \Pr[\{v_i\}_{i=1}^{n-1} \text{is LI}] \geq \prod_{i=1}^n (1 - q^{i - n})$$

Do you think this analysis is correct? What I am also interested in is the following generalizations, on which I, unfortunately, had less success.

Suppose we set $n=3k$ and that instead of sampling the vectors independently we sample three matrices $M, M_0, M_1 \in \mathbb{Z}^{3k\times k}_q$ such that each matrix is of full rank. What would be an upper bound on the probability of the columns of said matrices spanning $\mathbb{Z}_q^{3k}$? I have found in the literature (Lemma 8) a claim that it should be at least $1 - \frac{2k}{q}$ but have not found any good argument

Let us relax the very first statement, and admit that $q$ could be composite. Does this change the argument significantly? I am interested in this result both in the vector case and in the three matrices sample case.

Thank you, I hope the formatting is clear enough :)
For context, I am investigating whether the construction in the linked paper would hold in nonprime order groups.

答:
Your analysis is correct. There is possibly a typo: $i$ should be $k$. Moreover, I think it is equality. More precisely, $\{v_i\}_{i=1}^{k+1} \text{is LI} \, | \, \{v_i\}_{i=1}^{k} \text{is LI}$ if and only if $v_{k+1}$ is not a linear combination of $v_1,\ldots,v_k$. The only thing that you need to make sure is that two different linear combinations of $\{v_i\}_{i=1}^{k}$ give rise to two different vectors, which would be true since $\{v_i\}_{i=1}^{k}$ is LI. Thus $\Pr[\{v_i\}_{i=1}^{k+1} \text{is LI} \, | \, \{v_i\}_{i=1}^{k} \text{is LI}] = \frac{q^n-q^k}{q^n}$.

If it is not a field (that is $q$ is composite) then the equality may not hold as two different linear combinations can give you the same vector. To see this take $q=6$, then $v_1=(1,0,0,0,0,0)', v_2=(5,2,0,0,0,0)$, note that here $v_1+v_2 = 4v_1+4v_2=(0,2,0,0,0,0)$.

The inequality that you wrote still holds (after the typo correction:)).

Matrix Case: write $M_0|M_1|M_2$ for the augmented $3k\times 3k$ matrix.
Now $M_0|M_1|M_2 \text{ is LD}$ if one of the columns in $M_1$ is linear combination of columns in $M_0$ or one of the column in $M_2$ is linear combination of columns in $M_0$ and $M_1$. Now probability that a fixed column in $M_1$ is linear combination of columns in $M_0$ is at most $\frac{q^k}{q^{3k}}\leq \frac{1}{q}$, and probability that a fixed column in $M_2$ is linear combination of columns in $M_0$ and $M_1$ is at most $\frac{q^{2k}}{q^{3k}}\leq \frac{1}{q}$. Therefore using union bound
$$
\mathbb{P}(M_0|M_1|M_2 \text{ is LD}) \leq  \frac{2k}{q}.
$$
You can see that this bound holds even if $q$ is not a prime.


答:
You are looking for the zero of function $$f(x)=\log_{10}(5x-6) - \log_{10}^2(x)$$

This is smaller than
$$\log_{10}(5x) - \log_{10}^2(x)=\log_{10}(5)+\log_{10}(x) - \log_{10}^2(x)$$ To have an estimate, solve the quadratic equation in $\log_{10}(x)$. This gives
$$\log_{10}(x)\sim 1.47415\implies x\sim 10^{1.47415}\sim 29.7955$$ Now, use Newton method with this overestimate
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 29.795516 \\
 1 & 29.153010 \\
 2 & 29.156608
\end{array}
\right)$$

Title: How to solve : $ \frac{\log(5x-6)}{\log(x)} = \log (x)$

Body:
I'm struggling with this equation :

$$ \frac{\log(5x-6)}{\log(x)} = \log (x)$$

I've tried to do this:

$$ \log(5x-6) = \log^2(x)$$

But really I can' do more. . I just need a hint


问:
Integrate $f(z)=\frac{\pi\cot\pi z}{(u+z)^2}$

Suppose $N$ integral, $N\geq |u|$, prove that
$$
\lim_{N\rightarrow\infty}\int_{|z|=N+\frac{1}{2}}\frac{\pi\cot\pi z}{(u+z)^2}dz =0.
$$

I have got
$$
\left|\int_{|z|=N+\frac{1}{2}}f(z)dz\right|\leq\frac{2\pi^2(N+\frac{1}{2})}{{(N+\frac{1}{2}-|u|})^2}\sup_{|z|=N+\frac{1}{2}}|\cot z|.
$$

If we can prove that $\sup_{|z|=N+\frac{1}{2}}|\cot z|=o(N) (N\rightarrow \infty)$, then we complete the proof. But I stuck here. And I want to use the result and the residue formula to prove an indentity, so the formula is of no use here. Appreciate any help!

答:
Hint: Consider the square closed contour
$$
Q:\quad (\nu,i\nu)\to(-\nu,i\nu)\to(-\nu,-i\nu)\to(\nu,-i\nu)\to (\nu,i\nu)
$$
where $\nu=N+1/2$.

Since the function $f(z)$ is analytic inside the area enclosed by $Q$ and $C:\ |z|=(N+1/2)$ the integrals along both contours are equal.

Now recall that
$$
\cot(x+iy)=\frac{\sin 2x-i\sinh2y}{\cosh2y-\cos2x},
$$
so that
$$
|\cot(x+iy)|^2=\frac{\sin^22x+\sinh^22y}{(\cosh2y-\cos2x)^2}
=\frac{\cosh^22y-\cos^22x}{(\cosh2y-\cos2x)^2}=
\frac{\cosh2y+\cos2x}{\cosh2y-\cos2x}.
$$


问:
Rotation matrix - rotate a ball around a rotating box
I've a 3D box: center point = (a,b,c), width = w, height = h, depth = d.

the center point isn't the origin.

I have a ball on the box(touch each other), its center and radius.

I can rotate the box(around the X axis but its center STAYS the same..) and I want the ball to stay stuck to the box. so the ball needs to be rotated WITH the box.

the angle of the rotation is 45 degrees.

I tried to do this:

I defined the Rotation Matrix around the X axis:

```<code>mat[3][3]
1,    0   ,    0
0, cos(45), -sin(45)
0, sin(45), cos(45)
</code>```

and multiply it by the ball center vector:

```<code>(ball.Center().m_x , ball.Center().m_y, ball.Center().m_z) * mat
</code>```

so I got:

```<code>Point3D new_center(ball.Center().m_x,
                   ball.Center().m_y*cos(45) + ball.Center().m_z*sin(45),
                   -(ball.Center().m_y)*sin(45) + ball.Center().m_z*cos(45));
ball.Center() = new_center;
</code>```

the ball is really rotated when the box is rotated but too far. How can I fix it?

答: You are rotating the ball center's vector with respect to the origin. If you just want to rotate the box with respect to the ball center. Then instead of computing ball center vector times the rotation matrix, compute (box.Corner() - ball.Center())*mat + ball.Center(), and you will get the new corner coordinates for the box.


问:
LRT tests and sufficiency
If $T$ is sufficient for $\theta$, let $\lambda$ be an LRT statistic for $\mathbf{X}$ and $\lambda^*$ and LRT statistic for $T$. Is it true that $\lambda=\lambda^*$?

According to theorem 8.2.4 of the book "Statistical Inference" by Casella and Berger, seems to be the case, but the proof relies on an artifact of the factorization theorem that I think is only true with discrete distributions. The proof is as follows
\begin{equation}
\begin{aligned}
\lambda(\mathbf{x})&:=\frac{\sup_{\Theta_0}L_{\mathbf{x}}(\theta)}{\sup_{\Theta}L_{\mathbf{x}}(\theta)}=\frac{\sup_{\Theta_0}f_{\theta}(\mathbf{x})}{\sup_{\Theta}f_{\theta}(\mathbf{x})}\\&=\frac{\sup_{\Theta_0}g_{\theta}(T(\mathbf{x}))h(\mathbf{x})}{\sup_{\Theta}g_{\theta}(T(\mathbf{x}))h(\mathbf{x})}=\frac{\sup_{\Theta_0}g_{\theta}(T(\mathbf{x}))}{\sup_{\Theta}g_{\theta}(T(\mathbf{x}))}=\lambda^*(T(\mathbf{x}))\text{, }
\end{aligned}
\end{equation}
according to the book, the last equality is true as long as $g_{\theta}$ is the density function of $T$, but I think this only holds if the population is discrete (watch this).

答:
The test you showed is the Generalized one...but just to understand the problem take as an example the more suitable Neyman Pearson Lemma for simple hypothesis.

You have that the test is

$$\frac{L(\theta_0|\mathbf{x})}{L(\theta_1|\mathbf{x})}\leq k$$

now observe that "IF" a sufficient estimator $t(\mathbf{x})$ esists NP lemma becomes

$$\frac{h(\mathbf{x})\cdot g(t(\mathbf{x}),\theta_0)}{h(\mathbf{x})\cdot g(t(\mathbf{x}),\theta_1)}\leq k$$

$$\frac{ g(t(\mathbf{x}),\theta_0)}{ g(t(\mathbf{x}),\theta_1)}\leq k$$

thus the rejection region depends  by the data ONLY through a Statistic, that is $t(\mathbf{x})$, sufficient for the model.

Example 1:

$X\sim Exp(\theta)$

You perform the test using $T=\Sigma_iX_i$, sufficient statistic (in this case, Minimal, Sufficient and Complete) and you can use chi-squared tables,

Example 2

$X\sim N(\mu;1)$

you perform your test again using $T=\Sigma_iX_i$ or, what is the same, the sample mean...again $T$ is CSS (complete and sufficient)

Example 3

$X\sim U(0;\theta)$

and you perform your test basing it on $T=X_{(n)}$, the maximum of your observations, Complete and sufficient statistics

Other UMP test generalizing NP Lemma or Generalized LR Test (which is not anymore UMP) work in the same way: if a Sufficient estimator exists, the test is based on this important estimator... and it is very intuitive because the Sufficient estimator is so named as it include all the information you need to estimate the parameter...

Title: LRT tests and sufficiency


问:
Substructures in first order logic

Let $\sigma = \{f \}$ be a signature with a unary function symbol and let $\mathcal{B} = (\mathbb{Z}, f^\mathcal{B})$ be a signature with $f^\mathcal{B} = x-1$ for all $x \in \mathbb{Z}$.

$(i)$ Find a substructure $\mathcal{A}$ of $\mathcal{B}$ with  $\mathcal{A} \neq \mathcal{B}$.

$(ii)$ Is there a substructure of $\mathcal{B}$ with a finite universe?

$(iii)$ Prove or refute, $\mathcal{B}$ has an infinite number of pair-wise different substructures.

My attempt:

$(i)$ Since $\mathbb{N} \subset \mathbb{Z}$, we take  $\mathcal{A} = (\mathbb{N}, f^\mathcal{A})$ with $f^\mathcal{A} = x - 1$. Since we don't have any relation symbols, we only have to worry that $f$ "acts the same" in the substructure  $\mathcal{A}$.

$(ii)$ Take the set which only has the element $1$ and define the same $f$.

I haven't been able to solve $iii$.

I'm highly unsure of my answers and would really appreciate some feedback.

答:
Point (i)

$\mathbb N$ can't be taken as the domain for a substructure $\mathcal A$. The issue is that the image of $1$ under $f$, namely $0$ doesn't belong to $\mathbb N$. But following a similar idea, you can take $A = \{0, -1, - 2, \dots\}$ as a domain of a proper substructure $\mathcal A \subsetneq \mathcal B$.

Point (ii)

A substructure of $\mathcal B$ can't have a finite universe as the image under $f$ of its minimum element won't be defined.

Point (iii)

This is true. For any $n \in \mathbb Z$, $\mathcal A_n$ with domain $A_n = \{n - m \mid m \in \mathbb N\}$ are pair-wised different substructures.

Title: Substructures in first order logic


问:
Newton's root finding method that copes with stationary points
I am using Newton's method to solve an equation of the form $y=f(x)$ for increasing values of $y$. The value of $x$ found for some $y$ is used for the next $y$ as a starting approximation and this works fine.

$f$ is increasing but may have stationary points, and I do have situations where the range of interest of $y$ starts or ends on an extremum or very close. This causes the algorithm to experience a division by zero or yield very inaccurate roots.

Is there a simple way to cope ? Is there a known modified Newton's method able to deal with this situation ?

I know that double roots can be handled by the modified iterations $$x\leftarrow x-2\frac{f(x)}{f'(x)}$$ but this still does not work at the root, and one needs to know when to activate the factor $2$.

For information, my function $f$ is a cubic polynomial, but I want to avoid the explicit computation of the roots by Cardano's formulas.

Update:

If we restate Newton's method at a stationary point, we get to the next term of Taylor's development,

$$f(x+h)-y\approx f(x)-y+\frac{h^2}2f''(x),$$

so that $$h\approx\sqrt{-\frac{f(x)-y}{f''(x)}}.$$

In fact, the approximation

$$f(x+h)-y\approx f(x)-y+h f'(x)+\frac{h^2}2f''(x)$$ always holds, giving the general formula

$$h\approx \frac{f'(x)\pm\sqrt{f'^2(x)-2f''(x)(f(x)-y)}}{f''(x)}.$$

Due to the increased complexity, I don't want to use that formula all the time.

答:
There is an easy modification called modified Newton's method. The idea is easy: every time the derivative is zero, one uses a value of the derivative from the last iteration, when it was nonzero (couple of times if needed). This changes the method to the secant kind of method for these couple of iterations. When it becomes nonzero again, one switches back to Newton's method. In more dimensions it works as well.

Additional thing, which you can always try in case of convergence problems, is the damped version of Newton's method. It means, that you reduce each increment by some fixed value smaller than $1$, for example by $0.9$.

Can you post the function you solve and your starting point?


问:
uniqueness of complete space of a given metric space X
A given metric space X, consider its smallest complete metric sapce Y which contains X(" smallest" means if Z is an arbitary complete metric space which contains X, then Z contains Y) ("contain"means " there is an injuctive map f:Y->Z, d(f(a),f(b))=d(a,b) )
Question : Is Y is unique up to isomorphism of metric space</p>

答: $Y$ is called the completion of $X$. And it is indeed unique up to isometry. See Wikipedia entry to get references.


问:
Working out eigenvalues of this matrix without any calculations
$$
    \begin{pmatrix}
    110 & 55 & -164 \\
    42 & 21 & -62 \\
    88 & 44 & -131 \\
    \end{pmatrix}
$$

This is the matrix. I have worked out so far that because the columns are linearly dependent, then one of the eigenvalues is $0$. The sum of the diagonal values is $110+21+(-131)=0$ and this is the sum of the eigenvalues. So I know the other two eigenvalues are $a$ and $-a$. How would I work out $a$ without writing down any calculations? Any hints/tips are greatly appreciated!

I also notice that the 3rd column is very close to being the negative sum of the first two columns.

Thanks for your time :)

答: What do you count as "calculation"? Note that the row sums are all equal to $1$. Hence, $(1,1,1)^T$ is an eigenvector with eigenvalue $1$. Since the trace is $0$ the remaining eigenvalue must be $-1$.


问:
How to define the amount of additions. E.g. $1+2+3+4+5+6+7+8+9$
How to define the amount of additions. E.g. $1+2+3+4+5+6+7+8+9$

Are there $9$ additions, because of the nine numbers that are added together.
Or can you also say that there are $8$ additions, because there are only $8$ '$+$' signs.

答: Addition is an operation and hence we are performing 8 additions as we repeat the operation of 'adding' 8 times. In contrast, I would say that we are adding 9 numbers.</p>


问:
Confusion in a multivariable cubic polynomial.
WARNING:-

This question may have wrong tags.

HISTORY:-

After watching an old numberphile video(https://youtu.be/wymmCdLdPvM)(this video has nothing to do with my question) ,I got interested in the equation: $x^3+y^3+z^3=k$

Now being a high school student ,it is not likely that I am gonna prove some hundred years old uncracked problem surrounding the equation.

But you know, we all have false hopes, and so I started playing around with the beautiful equation.

After writing down the equation in a page, I remembered "Oo, I have to be smart to solve problems... ooooo." But I still kept playing with the equation and decided to write the sum of three cubes as a sum of 2 cubes

Here's what I did:

Assume that $x\neq -y$

$x^3+y^3+z^3=k$

$\implies x^3+y^3=k-z^3$

-eq(1)

$\implies (x+y)(x^2-xy+y^2)=k-z^3$

$\implies x^2-xy+y^2=\frac{k-z^3}{x+y}$

$\implies (x+y)^2=\frac{k-z^3}{x+y}+3xy$

$\implies (x+y)^3=k-z^3+3xy(x+y)$

$\implies (x+y)^3+z^3=k-3x^2y-3xy^2$

$\implies (x+y)^3+z^3-x^3-y^3=k-(x+y)^3$

$\implies 2(x+y)^3+z^3=k+x^3+y^3$

Combining this with eq(1) will give:

$2(x+y)^3+z^3=k+k-z^3$

$\implies (x+y)^3+z^3=k$

Confusion and my question:-

I just found that $x^3+y^3+z^3=k=(x+y)^3+z^3$

$\implies x^3+y^3+z^3=(x+y)^3+z^3$

What does that supposed to mean? I gave absolutely no restrictions to $x,y,z$ except the assumption in my first step, but that assumption shouldn't really matter.

The equation $x^3+y^3+z^3=(x+y)^3+z^3$

$\implies$ that either $x$ or $y$ is $0$.

Which means that I cannot add three cubes. What does that supposed to mean?!!

Please tell me what I did wrong in the steps.

答:
You made some sign errors.

From the line
$$(x+y)^3=k-z^3+3xy(x+y)$$
the next line should be
$$
(x+y)^3+z^3=k+3x^2y+3xy^2
$$
Even allowing that error, from the line
$$
(x+y)^3+z^3=k-3x^2y-3xy^2
$$
the next line should be
$$
(x+y)^3+z^3+x^3+y^3=k-(x+y)^3
$$
An easy way to have found the error on your own is to choose $x,y,z,k$ such that
$$
x^3+y^3+z^3=k
$$
e.g., $x=1,y=2,z=3,k=36$, and then substitute line-by-line until equality fails on some line.


问:
How to find the exponential decay function
I am studying bioinformatics. I am trying to solve a problem. So we have a gene, whose initial value
$x_0$ at time $t = 0$ is $x_0 = 1$. A perturbation of factor  $-0.9789812$ is applied to it, such that, at time $t = 10$, its value is  $0.0210359$. The gene is measured at time-points $t = [0,1,2...,10]$. How can I know the calculate the values at time $t = 1,2,...,10$, given the only information I have is the value at $x_0$ and and the perturbation applied ?
I am giving the data here

$x = [ 1.0000000 ,0.3482754, 0.1304151, 0.0575881, 0.0332433, 0.0251052, 0.0223848, 0.0214755, 0.0211715, 0.0210698, 0.0210359]$

But I want to know how would I generate this data. I tried the exponential decay function and the values do not correspond to the ones I have in my data set.

So if the gene has an initial value of 1 and a perturbation of −0.9789812 is applied to it, at the final reading of the gene at time t = 10 is 1−0.9789812 = 0.0210188. The decay is not linear, it seems like it has an exponential curve but I dont know how to fit it. Note that the only information I have is the initial value of the gene, the perturbation applied. I want to be able to calculate the values of gene at any time given this information.

答:
Even if your model is correct and exponential decay is expected, real work data rarely fits perfectly.

I expect that you are familiar with cases of linear relationships e.g. $y = ax$ or $y = ax + b$.  In these cases, you could plot the points on a graph and find the line of best fit.

If the expected relationship is $y = e^{-ax}$ then this does not work but note that $\log(y) = -ax$ so the relationship between $\log(y)$ and $x$ is linear.  So, plot $\log(y)$ rather than $y$ and try to find the line of best fit.


问:
Can rankT = rank$T^2$ implies $T = T^2$?
*Note: this is a very elementary question on linear algebra, and there has been indeed some solutions on similar questions on this site, but most of them are too concise that cannot solve my question.

Suppose we have a linear transformation T : $V \to V$ and rank(T) = rank($T^2$)

Then we have dim({$T(v): v \in V$} = dim({$T^2(v): v \in V$} (definition)

Since {$T^2(v): v \in V$} is a subspace of {$T(v): v \in V$}, and their dimension equal, we  have

$\{T^2(v): v \in V\}$ = $\{T(v): v \in V\}$, i.e. R(T) = R($T^2$)

Then I can proceed no further since the range of these transformations equaling does not tell me about the relations of T and $T^2$

答:
No!

As the linear transformation could be represented by a matrix.

We could take the $T: \mathbb{R}^2 \to \mathbb{R}^2$ such that the matrix form of $T$ is
$$
A_T =
\left(
\begin{matrix}
1 & 1 \\
0 & 1
\end{matrix}
\right).
$$
Then, the $T^2$ has matrix form as the following
$$
A_{T^2} = A_T^2 =
\left(
\begin{matrix}
1 & 2 \\
0 & 1
\end{matrix}
\right),
$$
which demonstrates that $T \neq T^2$.

If you want to write out the formula for $T$. We can define $T$ as $T(x,y) = (x,x+y)$ then $\mathrm{rank}(T) = \mathrm{rank}(T^2)$, however, $T$ is totally different from $T^2$.


问:
about harmonic polynomial on R^n
here is an exercise from axler's book: linear algebra done right 3ed

Suppose Q is a polynomial on $\mathbb{R^n}$. Prove that there exists a harmonic
polynomial P on $\mathbb{R^n}$ such that P(x)=Q(x) for every x $\in$ $\mathbb{R^n}$ with
$ \parallel x\parallel=1$.
[The only fact about harmonic functions that you need for this exercise
is that if P is a harmonic function on $\mathbb{R^n}$ and  $P(x)=0$ for every x $\in$ $\mathbb{R^n}$ with $ \parallel x\parallel=1$ then $P=0$

Axler gives a hint that the desired harmonic polynomial P is of
the form Q+$ (1-\parallel x\parallel^2)$R for some polynomial R. and he ask to prove the existence of such
polynomial R on $\mathbb{R^n}$ that verifies $Q+ (1-\parallel x\parallel^2)R$ is harmonic by defining
an operator T on a suitable vector space by Tr=$\Delta((1-\parallel x\parallel^2)r$

and then showing that T is injective and hence surjective.
My attempt:
first I prove that T is injective (T is defined on the set of polynomial of $\mathbb{R^n}$ to itself) :

assume $\Delta((1-\parallel x\parallel^2)r=0$ so $(1-\parallel x\parallel^2)r$ is harmonic polynomial and since this polynomial equal zero for every $\parallel x\parallel=1$ I conclude using the hint that $(1-\parallel x\parallel^2)r=0$ therefore $r=0$ so T is injectif.The problem is that I can't prove that T is surjective (I can't deduce from injectivity because it's not a finite dimentional vector space).

if I succeed at proving that T is surjective then I can conclude that T is a isomorphisim and since $-\Delta Q \in $ (the image of T) I deduced the existence of a polynomial r st $\Delta((1-\parallel x\parallel^2)r)=-\Delta Q$ which is equivalent to say $\Delta(Q+(1-\parallel x\parallel^2)r)=0$. is my  logic right? or did I need to change the vector space to a finite one?
note: this is an exerice from inner product chapter

答:
Restrict the map to polynomials of degree less than equal to degree of q.
Note that here degree means highest sum of the indices of the monomials with nonzero coefficients.</p>

Title: about harmonic polynomial on R^n


问:
Quick Integration Clarification
$ \int_ {-\pi}^{\pi}\ f(x) \ dx  $ is equivalent to integrating $f(x)$ over $-\pi < x < \pi$ (strictly less than) and also $-\pi \le x \le \pi$ (less than or equal). So I just need to reassure myself that the bounds of integration don't change when considering either less than or less than or equal. A quick confirmation will do.

Cheers.

UPDATE:Just applying this to a question to do with fourier series: $\
f(x) = \begin{cases}x+π & -π<x\le 0 \\ π & 0 < x < π\\  \end{cases} $ with 2π periodicity.
Can I proceed to compute:

$a_0= {1\over 2L}[\displaystyle \int \limits_{-π}^{0} \ x+π \ dx + \int \limits_{0}^{π} π\ dx]$

答: No, the bounds are not included which is crucial when evaluating improper integrals. For example, if $ f(x) = \sqrt{\left|\tan\left(\frac{x}{2}\right)\right|} $, then the integral converges despite the integrand not being defined at the bounds. </p>

Title: Quick Integration Clarification


问:
If I define one or more binary operations that form group(s) over a set, is it correct to call it an algebra?
More specifically, I defined the operations multiplication (non-Abelian, composition of permutations) and addition (Abelian, modulo addition of bounded naturals) over the same set, that can be seen as a bounded set of naturals or permutations equivalently.
Is it formally correct to call it an algebra?

Is it acceptable to call it an algebra in any informal sense?

The two operations are not (at least doesn't seem to be) related to each other in any way like the usual ones (+ and *). Although, it would be interesting to know about an incidental relation.

Some entries like this seem to relate the concept of algebra to vectors, which is not my case (I suppose).

A use that seems non-related to vectors is in relational algebra
which borrows the word algebra from the fact it uses algebraic structures.

答:
Any set of (finitary) operations, obeying any list of axioms, can be called an algebra according to the terminology of the field of universal algebra.  These are special cases of algebras for monads.

But generally the more interesting question is whether that class of operations and axioms is of a particular form, has particular behavior, etc.  For example, my first question about your two operations is whether they obey a compatibility condition and what form it takes.


问:
Proving that a function of the form f(x+y)=f(x)*f(y) is continuous
I have the to solve the following problem:

Let $f$ be a function from the real numbers to the real numbers. The function satisfies $f(x+y) = f(x)f(y)$ for all real $x,y$. Prove that if $f$ is continuous in $0$ then $f$ is continuous in every point.

I think I have a solution but I would like to know if it's correct:

By letting $y=0$ we get that $f(x)=f(x)f(0)$ meaning that $f(0)=1$. Thus we know that for every $\epsilon > 0$ and for every $x$ there exists a $\delta$ such that $|f(x)-1| < \frac{\epsilon}{|f(a)|} $ when $|x| < \delta$. Since this holds for all $x$ we can replace $x$ with $x-a$. Rewriting this we get that whenever $|x-a| < \delta$ we have that $\,|f(a)f(x-a)-f(a)|<\epsilon \implies |f(x)-f(a)| < \epsilon$.

答:
Putting $x=y=0$, $f(0)=(f(0))^2$,   if $f(0)=0$ then for any $x \in \mathbb{R}$, $f(x)=f(x)f(0)=0$ and we are done.

So take $f(0)=1$. Then note that $1=f(0)=f(x)f(-x)$ holds for any $x \in \mathbb{R}$ which implies $f(x)\neq 0$ for all $x \in \mathbb{R}$ in this case. Hence your proof is perfectly right.

Sequence can also be used to prove this: $\lim_{n \rightarrow \infty} x_n=a\Rightarrow \lim_{n \rightarrow \infty} (x_n-a)=0\Rightarrow \lim_{n \rightarrow \infty} f(x_n-a)=f(0)=1 \Rightarrow \lim_{n \rightarrow \infty}f(x_n)f(-a)=1 \Rightarrow \lim_{n \rightarrow \infty} f(x_n)=\frac{1}{f(-a)}=f(a)$.


问:
Using dimensional analysis to ‘get’ planck’s ${\lambda}$ equation with avogadro’s constant
This is for when you calculate bond energy in: ${j\cdot mol(n)^{-1}}$ and want to use planck’s equation for photon energy to calculate ${\lambda}$

Planck’s equation for photon energy:
$${\lambda=\frac{h\cdot c}{E}}$$

Planck’s equation for photon energy with avogadro’s constant:

$${\lambda=\frac{h\cdot c\cdot N_A}{E}}$$

I don’t understand how ${N_A}$ (${\frac{1}{n}}$) is at the top as:

${n=\frac{g}{g\cdot n^{-1}}\Rightarrow{n=\frac{1}{n}}}$

Energy in this case:

${E=\frac{j}{mol\ or\ n \ (1/n)}}$

I can’t rearrange ${1/n}$ to get it to the to the top in order to get rid of the ${n}$ at the bottom so that:

${E=j}$

Can someone show me verbosely with dimensional analysis how to do this?

Units:

${\lambda=m}$

${E=j}$

${h=js}$

${c=ms^{-1}}$

${N_A=n^{-1}}$

Dimensional analysis:

${\lambda=js\cdot ms^{-1}\cdot j^{-1}(1^{-1}n^{-2})}$

${\lambda=\frac{j\cdot s\cdot m}{j\cdot s(\frac{1}{n})}}\Rightarrow{\lambda=\frac{j\cdot s\cdot m}{\dfrac {j\cdot s}{n}}}\Rightarrow{\lambda=\frac{m}{\dfrac{1}{n}}}$

Then I can’t get ${\lambda=m}$

Is this correct, and if not then why?:

${\frac{j}{mol(1/n)}=\frac{j}{\dfrac{1}{n}}}$

Bonus question:

I have been using dimensional analysis to ‘get’ equations before I even knew that word, as my memory is bad and I couldn’t remember equations, I used to think it was a trick

Are there more/better ways to ‘get’ equations and cases where dimensional analysis won’t work?

I really need help here and would appreciate it very much

答:
$${E=j\cdot \frac{1}{n}\Rightarrow{E=j}}$$

This does not ever happen. In the same framework of $E$ this is impossible (unless of course if $1/n$ equals $1$)

The two lambdas described in

${\lambda=\frac{h\cdot c}{E}}$

and

${\lambda=\frac{h\cdot c\cdot N_A}{E}}$

are NOT the same equation. Using the same variable($\lambda$) for two different quantities without context is negligent.


问:
If $\lim_{x \to x_0} f(x) = f(x_0)$, as $x \to x_0$, can we always say $f(x) \to f(x_0)$
Assuming $$\lim_{x \to x_0} f(x) = f(x_0)$$

As $x$ approaches $x_0\ $i.e. $(x\to x_0)$, can we always say $f(x)$ approaches $f(x_0)\ $ i.e. $(f(x)\to f(x_0))$? From what I understand, this is the motivation for the $\epsilon$-$\delta$ definition of limit.

The $\epsilon-\delta$ definition states that- $\forall \epsilon$, $\exists \delta$, such that:

$0<|x-x_0|<\delta\Rightarrow|f(x)-L|<\epsilon$. For any value of $\epsilon$ if I can provide a $\delta$, then $L$ is the limit of $f(x)$ at $x_0$.

Approach is apparently an informal idea, and not what the limit defines. So the idea of $f(x)$ getting closer to $f(x_0)$, as $x$ gets closer to $x_0$ is inaccurate and not always true.

Update: The statement apparently is true, when $f(x)$ is continuous at $x_0$. Though I have been unable to find a general proof for it yet.

答:
Yes. If $\lim_{h\to 0}f(x_0+h)=L$ then for every $\varepsilon>0$ there exists $\delta>0$ such that whenever $|h|<\delta$ then $|f(x_0+h)-L|<\varepsilon$.

This means, for the same value of $\varepsilon$ and $\delta$, if $|x-x_0|<\delta$ then, setting $h=x-x_0$ we have $|f(x)-L|=|f(x_0+h)-L|<\varepsilon$, so the definition of $f(x)\to L$ as $x\to x_0$ is satisfied.

Note that this is actually slightly stronger than saying $f(x)\to L$ as $x\to x_0$, as $\lim_{h\to 0}f(x_0+h)=L$ also implies that $f(x_0)=L$, whereas in the definition of "$f(x)\to L$ as $x\to x_0$" it doesn't matter whether $f(x_0)$ even exists (or, if it does, what value it takes).


问:
Can anyone please tell me why it is called Parametric equation?
Parametric representation of the parabola $y^2 = x$  is $(t^2 , t)$.

Can anyone please tell me why it is called Parametric representation ?

Can I say $(x , \pm \sqrt x)$  is a parametric representation of $y^2 = x$ ?

I know for a family of equations for every value of parameter we get a particular member  of that family.

$x^2+y^2 = a^2$  is the family of circles with center at origin. If we put $a= 1$ , we get a particular circle. $a$ is also varying but for a particular value of $a$ we are getting a member. That is why , this is little different from variable.

答:
As with any language, mathematics is allowed to use the same word for different notions.

One sense of "parameter" (the second in the question) is that of "characteristic" or perhaps "ingredient"; a value that helps describe or define an object. Like how the parameters of a debate define the scope of topics that can be discussed. So, parameters can include the radius and/or center of a circle, the aspect ratio of a rectangle, the $x$- and $y$-intercepts of a line. The "Earliest Known Uses of Some of the Words of Mathematics" pages mention that "parameter" was first used in 1631 specifically for the latus rectum of a conic.

The "parametric equation" sense of "parameter" is that of "other measurement", in particular with regard to graphing curves and surfaces. (See this ancient answer of mine to the question "What is a the intuition behind a parametric equation?".) Sticking with curves ... While we often graph by expressing $y$ in terms of $x$, or $x$ in terms of $y$, there are significant advantages (beyond the scope of discussion here) to expressing both $x$ and $y$ in terms of something else ---an "other measurement", or parameter--- that helps convey a graph as a path of travel. Whereas the cartesian equation $x^2+y^2=1$ for the unit circle is perfectly effective at defining a static set of points, the parametric forms $(\cos t,\sin t)$ and $\left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\right)$ describe (very) different dynamic journeys along the curve.

So, the $a$ in $x^2+y^2=a^2$ and the $t$ in $(t,t^2)$ are both parameters, but in different senses of the word. (Attempts to fashion a single meaning to the term that covers both use cases are a bit strained.)

Consequently, we have the situation that $(a\cos t, b\sin t)$ uses the parameter $t$ to traverse an elliptical path defined by parameters (namely, semi-axes) $a$ and $b$. Language is funny like that sometimes. (See "Buffalo buffalo Buffalo buffalo buffalo buffalo Buffalo buffalo".)

As to whether $(x,\pm \sqrt{x})$ can be considered a parametric representation of parabola $x=y^2$ ...

It's a not-unreasonable representation of the parabola, in the same way as one writes $y=\pm\sqrt{x}$. It tells you something about the points on the curve. So, fine.

However, $(x,\pm\sqrt{x})$ doesn't describe a path of travel along the curve; rather, at each $x$, an ostensible traveler either chooses between candidate $y$-values, or somehow visits both of them simultaneously. That defeats the purpose of describing a path. A proper "parametric representation" doesn't give a choice; the value of the parameter should tell you exactly where you are. Formally: the $x$ and $y$ coordinates should be functions of the parameter.


问:
Are all solutions to this coupled PDE constant?
This is a self-answered question. I posted it here since it wasn't trivial for me. I would be happy to see alternative solutions.

Suppose $\theta(x,y),\tilde \theta(x,y)$ are smooth functions on a connected domain, and define
$$
c=\cos(\theta),s=\sin(\theta),\tilde c=\cos(\tilde \theta),\tilde s=\sin(\tilde \theta).
$$
Suppose that the following coupled system of PDE's is satisfied:
$$
(1) \,\,\,\,\partial_x(\tilde sc)=-\partial_y(c\tilde c),
$$
$$
(2) \,\,\,\, \partial_y(\tilde s c)=\partial_x(c\tilde c), \,\,\,\,\,
$$
$$
(3) \,\,\,\,\partial_x(s\tilde c)=\partial_y(s\tilde s),\,\,\,\,\,\,
$$
$$
(4) \,\,\,\,\partial_y(s\tilde c)=-\partial_x(s\tilde s).\,\,\,\,
$$

How to prove that $\theta,\tilde \theta$ are constant functions?

答:
The system $(1)-(2)$ implies that $f=-\tilde sc+i(c\tilde c)=i\cos(\theta)e^{i\tilde \theta}$ is holomorphic, and $(3)-(4)$ imply that $g=s\tilde c+i(s\tilde s)=\sin(\theta)e^{i\tilde \theta}$ is holomorphic.
Thus $(if)^2+g^2=e^{2i\tilde \theta}$ is holomorphic. Since its image lies inside a circle, the open mapping theorem implies that it is constant. Thus $\tilde \theta$ is constant.

Together with the holomorphicity of $f$, we deduce that $\cos(\theta)$ is holomorphic, so $\theta$ is also constant.


答:
$$
\sum_{i=1}^n a_i = a_1 + \cdots + a_n
$$
There are $n$ terms in the addition but what if there are minuses? Hence counting terms are kinda less ambiguous. Also we can do:
$$
\sum_{i=1}^n a_i = 0+ a_1 + \cdots + a_n
$$
and we have $n$ additions now.</p>

Body:
How to define the amount of additions. E.g. $1+2+3+4+5+6+7+8+9$

Are there $9$ additions, because of the nine numbers that are added together.
Or can you also say that there are $8$ additions, because there are only $8$ '$+$' signs.


问:
To calculate an improper integral
Is $$\int_1^\infty \frac{\log x}{x^2}dx$$
finite? How to solve this?

答:
Use integration by parts. Keep trying different parts until it works.

This doesn't work:

$u= \frac1x$, $dv=\frac{\ln(x)dx}{x}$

$du= -\frac{dx}{x^2}$, $v=\frac{\ln^2(x)}{2}$

This doesn't work:

$u= \frac1{x^2}$, $dv=\frac{\ln(x)}{1}$

$du= (...)$, $v=(...)$

This works

$u=\frac{\ln(x)}{1}$, $dv= \frac1{x^2}$

$du=\frac{1}{x}$, $v= -\frac1{x}$

Title: To calculate an improper integral


问:
Is maximum of sum exists if each function has a maximum?
Let $f,g:D \rightarrow \mathbb R$ be multivariable functions over the open set $D$. Assume that $f,g$ both have a global maximum point inside $D$. Can we be sure that $f+g$ also has a global maximum point inside $D$? I know that the maximum point if exists must satisfying $\max(f+g) \leq \max f + \max g$. But will the maximum point exist in general? If not, is there any condition that is needed for the existence of the global maximum?

答:
Let $D$ be a bounded simply connected open set, and $\vec{v}, \vec{w}$ be any two points in $D$. Then define $h$ as $$h:D\rightarrow \mathbb{R}:\vec{x}\mapsto \|\vec{x}-\vec{v}\|\cdot\|\vec{x}-\vec{w}\|$$
This function reaches a maximum on $\overline{D}$. If the set $D$ is sufficiently large, this maximum will be achieved on the boundary, rather than in the local extremum $\frac{\vec{v}+\vec{w}}{2}$. Let this maximal value be $M$, then define $f$ and $g$ in the following way.

$$f(\vec{x}) =
\begin{cases}
h(\vec{x}) &\text{if } \vec{x}\neq\vec{v}\\
M &\text{if } \vec{x} = \vec{v}
\end{cases}$$

And similarly for $g$ and $\vec{w}$.
Then the functions $f$ and $g$ will achieve their maximal values in $\vec{v}$ respectively $\vec{w}$, but the sum $f+g$ has a limit point in $\overline{D}$ where the limit is equal to $2M$, which is not achieved in any of the interior points of $D$.

This answer the first question negatively.

To figure out a criterion in which the maximum is achieved, first the open set needs to be simply connected. Otherwise any $f$ and $g$ that satisfy the condition on $D$ where $f+g$ reaches its maximum in $\vec{x}$, will fail to achieve a maximum on $D\setminus \{\vec{x}\}$.

Any kind of differentiability criterion also doesn't work. The example is discontinuous, but by replacing the discontinuity with a sufficiently small peak of the form $\frac{M}{c\|\vec{x}-\vec{v}\|^2+1}$ for some large positive value of $c$ you can make even an analytic counterexample.
$$
\begin{align}
f &= h + \frac{M}{c\|\vec{x}-\vec{v}\|^2+1}\\
g &= h + \frac{M}{c\|\vec{x}-\vec{w}\|^2+1}
\end{align}
$$

You can require both $f$ and $g$ to be concave and the set $D$ to be convex, which implies $f+g$ is concave as well, and the function $f+g$ reaches its maximum on an interior point of $D$. This seems to be the simplest criterion, albeit a very restrictive one.


答:
Applying lim to both sides as y to 0 $$f(x+y) =f(x)f(y)$$

We get $$\lim_{y\to 0}f(x+y)=\lim_{y\to 0}f(x)f(y)=f(x)\lim_{y\to 0}f(y)=f(x)f(0)=f(x)$$
So by definition $f(x)$ is defined everywhere and the limit exists and is equal $f(x) $. Those 3 facts imply continuity.

Title: Proving that a function of the form f(x+y)=f(x)*f(y) is continuous

Body:
I have the to solve the following problem:

Let $f$ be a function from the real numbers to the real numbers. The function satisfies $f(x+y) = f(x)f(y)$ for all real $x,y$. Prove that if $f$ is continuous in $0$ then $f$ is continuous in every point.

I think I have a solution but I would like to know if it's correct:

By letting $y=0$ we get that $f(x)=f(x)f(0)$ meaning that $f(0)=1$. Thus we know that for every $\epsilon > 0$ and for every $x$ there exists a $\delta$ such that $|f(x)-1| < \frac{\epsilon}{|f(a)|} $ when $|x| < \delta$. Since this holds for all $x$ we can replace $x$ with $x-a$. Rewriting this we get that whenever $|x-a| < \delta$ we have that $\,|f(a)f(x-a)-f(a)|<\epsilon \implies |f(x)-f(a)| < \epsilon$.


问:
Number theory problem book
Can anyone suggest me some good book that has problems on classical elementary number theory with solutions?</p>

答:
The book by Kuldeep Singh Number theory step by step is ideally for self-study as it provides complete solutions to all problems and video topics with subtitles at the following URL:
https://maths-for-all.co.uk/number-theory/</p>

Title: Number theory problem book


问:
Showing that the roots of a polynomial with descending positive coefficients lie in the unit disc.
Let $P(z)=a_0+a_1z+\cdots+a_nz^n$ be a polynomial whose coefficients satisfy $$0<a_0<a_1<\cdots<a_n.$$

I want to show that the roots of $P$ live in unit disc. The obvious idea is to use Rouche's theorem, but that doesn't quite work here, at least with the choice $f(z)=a_nz^n, g(z)=$ (the rest).

Any ideas?

答:
The previous answer establishes that roots lie inside or on the unit circle. By a slight modification to
$$
Q(z)=(1-z)P(Rz)
$$
this argument remains true as long as $0<a_0\le Ra_1 \le R^2a_2\le ... \le R^na_n$. Let $R$ be minimal with this property, then $R<1$. Now if $z$ is any root of $P$, then $z/R$ is a root of $Q$ and thus in the unit disk. Thus $|z|\le R<1$.


问:
How to correctly reformulate the universal property of the direct sum of abelian groups?
In this post I learned that if $G$ is an abelian group we may state (without diagrams) the universal property of the direct sum of finitely many abelian groups by saying that that map from there is a group isomoprhism between $\prod \mathrm{Hom}(G_i,G)$ and $\mathrm{Hom}(G_1\times\cdots\times G_n,G)$. I would like to know if it is essential to say that the map is a group isomorphism. I was looking at this post and I think that we just need that map between $\prod \mathrm{Hom}(G_i,G)$ and $\mathrm{Hom}(G_1\times\cdots\times G_n,G)$ to be bijective. Why would we also need it to be a group homomorphism? I know it is, but is it truly necessary to say that when writing down the universal property?

答: It is not necessary. When an object satisfies some universal property, that means that it is an initial or terminal object of some category. But the group structure on $\text{Hom}(G,H)$ is not part of the structure of the category of abelian groups. This additional group structure is the result of all abelian groups being group objects in the category of groups, so the fact that the map you described is an isomorphism is an additional property, and not really part of the definition.


答: Yes that is true (assuming a proper riemann integral, otherwise you must take a limit). Think of it in terms of the area under a curve, ie. you want to calculate the area under the curve f(x) where the end points are $-\pi$, and $\pi$. Clearly the end point don't contribute a nonzero amount to the area.</p>

Title: Quick Integration Clarification

Body:
$ \int_ {-\pi}^{\pi}\ f(x) \ dx  $ is equivalent to integrating $f(x)$ over $-\pi < x < \pi$ (strictly less than) and also $-\pi \le x \le \pi$ (less than or equal). So I just need to reassure myself that the bounds of integration don't change when considering either less than or less than or equal. A quick confirmation will do.

Cheers.

UPDATE:Just applying this to a question to do with fourier series: $\
f(x) = \begin{cases}x+π & -π<x\le 0 \\ π & 0 < x < π\\  \end{cases} $ with 2π periodicity.
Can I proceed to compute:

$a_0= {1\over 2L}[\displaystyle \int \limits_{-π}^{0} \ x+π \ dx + \int \limits_{0}^{π} π\ dx]$


问:
Why is $]-\infty,0]$ a closed interval.
Maybe i'm confused, but why is $]-\infty,0]$ a closed interval? I mean, we know that $-\infty$ is a limit point of $]-\infty,0]$, but isn't contained in the interval.

答:
$-\infty$ is not a limit point of $(-\infty, 0]$, because it is not a point. In the context of real numbers, $\infty$ and $-\infty$ are not numbers, they are simply symbols used to shorten some notations. For example, the set $$A = \{x\in\mathbb R| x < a\}$$ can be written more compactly as $A=(-\infty, a]$. In writing this, $-\infty$ is just a symbol used to write down an expression. Saying that the symbol $\infty$ is a "point" makes, in this case, just as much sense as saying that the symbol "$($" is a point. It is not. Out of all the symbols in the expression $(-\infty, 0]$, only one of them represents an actual number.

The set $(-\infty, 0]$ is a closed interval for a simple reason: because it is:

An interval.

A closed set.

Title: Why is $]-\infty,0]$ a closed interval.


问:
Does asymptotic normally imply uniform integrability?
If a set of random variables $\{X_n\}_{n \ge 1}$ is asymptotically normal is it also uniformly integrable?

Intuitively it seems like this should be true as $\{X_n\}_{n \ge 1}$ becomes 'nicer' as $n$ gets larger with the density becoming more symmetrical about the mean.

So is it true? If not, are there additional conditions we can impose on the set of asymptotically normal random variables to make them uniformly integrable?

答:
Let $Z$ be a standard normal random variable, and $B_n$'s are i.i.d. Bernoulli$(\frac{n-1}{n})$ independent of $Z$, then define
$$X_n = ZB_n +n^2 (1-B_n).$$

Note that
$$
\mathbb{P}(X_n\leq x) = \left(1-\frac{1}{n}\right)\mathbb{P}(Z\leq x) + \frac{1}{n}\mathbb{P}(n^2 \leq x),
$$

The last probability is either $0$ or $1$, thus $X_n$ converges to standard normal as $n \to \infty$. Moreover $\mathbb{E}(|X_n|) = n$. Now from here, conclude $X_n$'s are not UI. Also the link gives you necessery and sufficient conditions for UI.


问:
$\left\|\int_a^b x(t) dt \right\| \leq \int_a^b \left\| x(t) \right\| dt$
I have seen this inequality:

$$\left\|\int_a^b x(t) dt \right\| \leq \int_a^b \left\| x(t) \right\| dt$$

How could I prove that it holds for any $x:[a,b]\longrightarrow \mathbb{R}^n$, assuming that $x$ is integrable.

I would like to see a prove of this for a general norm, but if it is easier to do it for a particular norm, it would also help me to get a sense of why it is true.

Thanks in advance.

答: I am assuming that the integral is defined component wise. Hint: There exist $\mathbb R^{n}$ valued simple functions $x_k$ converging pointwise to $x$ with $\|x_k(t)\|' \leq \|x(t)\|'$ for all $t$ where $\|.\|'$ is the Euclidean norm. Now use the fact any norm on $\mathbb R^{n}$ is equivalent to the Euclidean norm and use the triangle inequality for the given norm.


问:
Energy norm. Why is it called that way?
Let $\Omega$ be an open subset of $\mathbb{R}^n$. The following

$$\lVert u \rVert_{1, 2}^2=\int_{\Omega} \lvert u(x)\rvert^2\, dx + \int_{\Omega} \lvert \nabla u(x)\rvert^2\, dx$$

defines a norm on $H^1(\Omega)$ space, that is sometimes called energy norm.

I don't feel easy with the physical meaning this name suggests. In particular, I see two non-homogeneous quantities, $\lvert u(x)\rvert^2$ and $\lvert \nabla u(x)\rvert^2$, being summed together. How can this be physically consistent?

Maybe some example could help me here. Thank you.

答:
When u is the velocity field of an viscous, incompressible homogeneous fluid with density set to 1, then the first integral is proportional to the kinetic energy of the fluid flow. The second integral is called enstrophy (see Wikipedia).

Enstrophy is in this situation something different than energy, therefore the term "energy norm" is slightly misleading. For a viscuous flow described by the Navier-Stokes equations, the enstrophy measures how fast the fluid flow dissipates energy due to friction, it is possible to show under some mild assumptions that
$$
\frac{d}{dt} \text{energy} = - \text{viscosity} * \text{enstrophy}
$$
So, physically speaking, fluid flows with finite energy and finite enstrophy are precisely those whose velocity fields are elements of $H^1$. And physically interesting fluid flows should have finite enstropy so that the process of energy loss due to friction is described by the model.

Title: Energy norm. Why is it called that way?


问:
Why does $w(z)=z^2$ represent a conformal map from the $z$ plane to the $w$ plane?
Doesn't every straight line $y=kx$ in the $z$-plane become a straight line $u=v\cdot2m/(1-m^2)$ in the $w$-plane such that each point having a phase angle $\theta$ in the $z$-plane has a phase angle $2\theta$ in the $w$-plane? This would mean that if the angle between two straight lines in the $z$-plane is $\alpha-\beta$, the corresponding angle between the corresponding lines in the $w$-plane should be $2\alpha-2\beta=2(\alpha-\beta)$. Then, how come is the angle preserved?

答: The angle is not preserved. And that's not a problem since $w'(0)=0$. The angles must be preserved at those points $z$ at which $w'(z)\ne0$.


问:
Prove that $0<11 - 6 \sqrt{3} <1$
This point was used as part of a solution I was seeing. A way to do it with approximation is using taylor's theorem:

$$ \sqrt{3} \approx \sqrt{4} - \frac{1}{2\sqrt{4} } = 1.75$$ and hence we find that $$ 11-6 \sqrt{3}\approx.5$$

But, are there ways to prove the above without appealing to taylor's theorem?

P.s: I would prefer a proof which is not based on contradiction methods.

答: $\sqrt3\ge0$, therefore \begin{align}&11-6\sqrt3>0\Leftrightarrow 11>6\sqrt3\Leftrightarrow 121>36\times 3=108\\ &11-6\sqrt3<1\Leftrightarrow 10<6\sqrt3\Leftrightarrow 100<108\end{align}

Title: Prove that $0<11 - 6 \sqrt{3} <1$


问:
Size of units of a quotient of a discrete valuation ring
Let $R$ be a discrete valuation ring and let $\mathfrak m$ be its maximal ideal. Suppose that $R/\mathfrak m$ is a finite field of characteristic $p$. Let $\# R/\mathfrak m:=p^f$ for some $f \in \mathbb N_{\geq 1}$. If we take some $r \in \mathbb N_{\geq 1}$, how do we calculate  $\#(R/\mathfrak m^r)^{\times}$?

答:
$\mathfrak{m}= (\pi)$ is principal.

Take $S$ a set of representatives of $R/(\pi)$ in $R$.

We have the disjoint union $$R/(\pi^r)=\bigcup_{s\in S} s+(\pi)/(\pi^r)=\bigcup_{s\in S} s+\pi \ (R/(\pi^{r-1}))$$

Whence $$|R/(\pi^r)| = |S| \ |R/(\pi^{r-1})|=|R/(\pi)|^r$$

The units are simply

$$|R/(\pi^r)^\times| = |R/(\pi^r)\ -\ (\pi)/(\pi^r)| = |R/(\pi)|^r-|R/(\pi)|^{r-1}$$


答:
Using the difference of two squares identity:

$$0<11 - 6 \sqrt{3} < 1 \Longleftrightarrow 0 < 11^2-6^2\cdot3<11+6 \sqrt3 \Longleftrightarrow 0 < 13 < 11 + 6 \sqrt3$$

and this is true since $0<13 < 11 + 6 \cdot 1 < 11 + 6 \sqrt{3}$.

Title: Prove that $0<11 - 6 \sqrt{3} <1$

Body:
This point was used as part of a solution I was seeing. A way to do it with approximation is using taylor's theorem:

$$ \sqrt{3} \approx \sqrt{4} - \frac{1}{2\sqrt{4} } = 1.75$$ and hence we find that $$ 11-6 \sqrt{3}\approx.5$$

But, are there ways to prove the above without appealing to taylor's theorem?

P.s: I would prefer a proof which is not based on contradiction methods.


问:
Why we take closed interval in defining curve in complex plane?
A curve in the complex plane is defined as a continuous function from a closed interval of the real line to the complex plane:$ z(t) : [a, b] → \mathbb{C}$, $a\leq t\leq b$.

Doubt: Why we choose closed interval to define curve in the complex plane? Can't we take an open interval to define curve?

Thanks for help

答:
Edit: A continuous function $(a,b)\to X$ is equivalent to a continuous function $\mathbb R\to X$, because $\mathbb R$ is homeomorphic (and diffeomorphic and probably more) to an open interval. With a curve $[a,b]\to X$, we actually get something different.

Basically because we want a curve to have endpoints. Of course you could consider at open-ended curve $(a,b)\to X$, but that's not as useful.

Probably the most important reason is that it forces the (image of the) curve to be a closed and bounded set. Otherwise $z(t)=\frac 1t$ for $t\in(0,1)$ would be a curve that goes off to infinity. Even worse, you could have very complicated curves such as infinite spirals and whatnot.

As a bonus, it makes it easy to talk about the endpoints by just writing $z(a)$ and $z(b)$.


答:
@Henry's comment should be the answer!  {I wrote this answer before seeing Henry's comment}

For $x>0$ we have
$$
x-6\sqrt3 > 0 \quad\Longleftrightarrow\quad x>6\sqrt{3}
 \quad\Longleftrightarrow\quad x^2 > 108
$$
Investigate when $x$ is a perfect square, $10^2 = 100 < 108 < 121 = 11^2$.  Therefore
$$
10-6\sqrt3 < 0,\quad 11-6\sqrt3 > 0
$$

Title: Prove that $0<11 - 6 \sqrt{3} <1$


Title: Prove that $0<11 - 6 \sqrt{3} <1$

Body:
Originally a comment and similar to Gae. S.:

You could say $$\begin{matrix} & 121 &>& 108 &>& 100 \\ \implies& \sqrt{121} &>& \sqrt{108} &>& \sqrt{100}\\ \implies& 11 &>& 6 \sqrt{3} &>& 10\\ \implies& -11 &<& - 6 \sqrt{3} &<& -10\\ \implies &0 &<& 11- 6 \sqrt{3} &<& 1\end{matrix}$$


答:
Let the roots of a quadratic be $11 - 6\sqrt3$ and $11 + 6\sqrt3$. Then their sum is $22$ and their product is $(11 - 6 \sqrt3)(11 + 6 \sqrt3) = 11^2 - 6^2 \cdot 3 = 13$, leading to the quadratic $f(x) = x^2-22x+13$.

Then $f(0) = 13$ but $f(1) = -8$, and since $f(0)$ and $f(1)$ are of opposite signs,  there must be a root in between $0$ and $1$ (IVT in disguise), and it cannot be $11 + 6 \sqrt3$. Thus $0 < 11 - 6 \sqrt3 < 1$.

Title: Prove that $0<11 - 6 \sqrt{3} <1$


问:
Computing a partial sum and finding its sequence
Let $S=\sum_{n=1}^{\infty}a_n$ be an infinite series such that its partial sum is given by $S_n=8-\frac{2}{n^2}$.

Find $\sum_{n=1}^{10}a_n$ and $\sum_{n=4}^{16}a_n$

Find a general formula for $a_n$

I found the first part of (1). I computed $S_{10}$ to get $8-1/50$. I'm struggling with the second part of (1). I figured it was $S_{16}-S_4$, but that didn't work.

For problem (2) I found a couple terms of the sequence, $$6, \frac32, \frac{5}{18}, \frac{7}{72}, \cdots$$I know that $S=\lim_{n\rightarrow \infty}S_n=8$. Assuming this is correct, I see no pattern here to get the sequence.

答:
Remember this well-known formula:

$S_n-S_{n-1}=a_n$

Then use: $S_n=\sum_{i=1}^{n}a_i$

If you ask, where does this formula come from?

It comes from here:

$$ \begin{align}S_n=\underbrace {a_1+a_2+a_3+\cdots +a_{n-1}}_{S_{n-1}} +a_n =S_{n-1}+a_n \end{align}$$

$$ \implies a_n= S_n-S_{n-1}$$


答:
By definition of partial sums, $$S_n=\sum_{i=1}^n a_n$$ therefore by definition $$\sum_{n=1}^{10}a_n=S_{10}=8-\frac{1}{50} \text{ and } \sum_{n=4}^{16}a_n=S_{16}-S_3.$$

To find $a_n$, note that $a_n=S_{n}-S_{n-1}$.

Title: Computing a partial sum and finding its sequence

Body:
Let $S=\sum_{n=1}^{\infty}a_n$ be an infinite series such that its partial sum is given by $S_n=8-\frac{2}{n^2}$.

Find $\sum_{n=1}^{10}a_n$ and $\sum_{n=4}^{16}a_n$

Find a general formula for $a_n$

I found the first part of (1). I computed $S_{10}$ to get $8-1/50$. I'm struggling with the second part of (1). I figured it was $S_{16}-S_4$, but that didn't work.

For problem (2) I found a couple terms of the sequence, $$6, \frac32, \frac{5}{18}, \frac{7}{72}, \cdots$$I know that $S=\lim_{n\rightarrow \infty}S_n=8$. Assuming this is correct, I see no pattern here to get the sequence.


问:
Equivalent continuation of a metric
Hello fellow mathematicians,

I am confronted with the following, supposedly not too difficult, problem:

Let $(E,f_1)$ be a normed space and $F \subset E$ a linear subspace. Let $f_2$ be a norm on E which is equivalent with the norm $f_1$ on $F$.
Prove that there is a norm $g$ on $E$ which is equivalent to $f_1$ on $E$ and whose restriction on $F$ is $f_2$.

I tried with all the lemmata and corollaries connected with Hahn-Banach theorem, but, even though I still feel the proof should be easy, I have found no way to tackle the problem.

I would be very thankful for pointers!

答:
Idea: fix a number $c>0$ such that $f_2(x)\le cf_1(x)$ for each $x\in F$ and put $g(x)=\inf \{f_2(y)+cf_1(z)|y\in F, z\in E, y+z=x\}$ for each $x\in E$. At least, I used a similar construction  for topological groups (see Lemma 2 of my paper [1]).

[1] Ravsky O.V. On Extension of (Pseudo-)Metrics from Subgroup of Topological Group onto the Group // Matematychni Studii. – 1999. – 11, #1 – P.31-39.

Title: Equivalent continuation of a metric


问:
Let relation $R=\{(1,1), (2,2), (3,3), (1,2), (2,1), (2,3) (3,2)\}$ be defined on $A=\{n:3^n>4^{n-1}\}$. Find type of relation.
Clearly R is a symmetric relation.

Now for finding the set A, I don’t know the proper method to do it, but I managed to find $A=\{1,2,3,4\}$ (please let me know how to solve without trial and error)

Now, $(4,4)$ isn’t in the relation, so does that prohibit it from reflexive? I am just having a problem in the technicality of the question

答:
Note that the function $f(x) = 3^x - 4^{x-1}$ is eventually decreasing after some value of $x$. Since we are interested in only positive integer solutions and since $f(5) < 0$, your set $A$ is complete.

The relation $R$ is indeed symmetric but not reflexive and transitive. I am sure you can find out why.


问:
Wrong defined/boundless $g \rightarrow\sum_{n = 1}^\infty\frac{g(\frac1n)}{2^n}$
Let's consider:

$$f\colon (C[0,1], \Vert\cdot \Vert_1) \ni g\rightarrow \sum_{n=1}^\infty\frac{g(\frac1n)}{2^n}$$

I'm trying to check if this object is well defined.

where $\Vert f \Vert_1 = \int_0^1|f(t)|dt$

Let's take function $g(x) = e^{\frac1x}$

Then this sum above will be turned into:

$$f(g) = \sum_{n=1}^\infty (\frac{e}{2})^n$$

But this sequence cannot converge, becuase $(\frac e2)^n \rightarrow \infty, n \rightarrow \infty$.

So - I just found a function that $f(\text{that_function}) = \infty$.

Am I proved that funcional $f$ is wrongly defined or I just proved that $f$ it's boundless ?

My queries are derived from the fact that in my way of thinking I haven't used $\Vert \cdot \Vert_1$ anywhere.

答:
If $g$ belongs to $V=(\mathcal C([0,1], \mathbb R), \Vert \cdot \Vert_1)$, $g$ is continuous and therefore bounded on the compact segment $[0,1]$. The series

$$f(g) = \sum_{i=1}^\infty\frac{g(\frac1n)}{2^n}$$ is absolutely convergent and convergent. Conclusion: $f$ is defined on $V$.

Note: the norm $\Vert \cdot \Vert_1$ has nothing to do regarding the question of $f$ being well defined or not. The norm would obviously play a role on questions like the continuity of the functional $f$.


问:
Showing that $1200x_1^2-400x_2+202>0$ given that $x_1^2-x_2>-0.005$
I am trying to construct a simple argument to verify that the inequality $$1200x_1^2-400x_2+202>0$$ is true, given that $x_1^2-x_2>-0.005$. My argument is as follows:

\begin{align}
1200x_1^2-400x_2+202&\geq400x_1^2-400x_2+202 \\
&=400(x_1^2-x_2)+202 \\
&>400(-0.005)+202 \\
&=200 \\
&>0.
\end{align}
The first inequality follows from the fact that $\alpha x^2\geq \beta x^2$ $\forall x\in\mathbb{R}$ where $\alpha>\beta$. Is this simple argument valid?

Body:
you answer is already neat, just to add you can also do it like this to avoid confusion
$$1200x_{1}^2-400x_{2}+202 > 0$$
$$1200x_{1}^2-400x_{2}+202 + 400x_{1}^2 > 0 +400x_{1}^2$$
but $x_{1}^2-x_{2} > -0.005$, say $\alpha$ is a positive real number
$$x_{1}^2-x_{2} = -0.005 + \alpha $$
$$1200x_{1}^2+202 + 400 \cdot (x_{1}^2 - x_{2} ) > 0 +400x_{1}^2$$
$$1200x_{1}^2+202 + 400 \cdot ( -0.005 + \alpha ) > 0 +400x_{1}^2$$
$$1200x_{1}^2+202 - 2 +400 \alpha >  0 +400x_{1}^2$$
$$1200x_{1}^2 - 400x_{1}^2 + 200 + 400 \alpha > 0  $$
$$800 x_{1}^2 + 200 + 400 \alpha > 0$$
since $\alpha$ is a positive real number and " $x_{1}^2$ " is also a positive number, then the identity is true