问:
Even weighted codewords and puncturing
My question is below:

Prove that if a binary $(n,M,d)$-code exists for which $d$ is even, then a binary $(n,M,d)$-code exists for which each codeword has even weight.

(Hint: Do some puncturing and extending.)

答:
Breaking this down to individual steps. Assume that $d=2t$ is an even integer.
Assume that an $(n,M,d)$ code $C$ exists.

Show that puncturing the last bit from the words of $C$, you get a code $C'$ with parameters $(n-1,M,d')$, where $d'\ge d-1$. Actually we could puncture any bit, but let's be specific. Also we fully expect to have $d'=d-1$, but can't tell for sure, and don't care.

Let us append each word of $C'$ by adding an extra bit chosen in such a way that the weight of the word is even. Call the resulting code $C^+$. Show that $C^*$ has parameters $(n,M,d^*)$, where $d^*\ge d'\ge 2t-1.$ Observe that all the words of $C^*$ have an even weight.

Show that the minimum distance of $C^*$ must be an even number. Conclude that $d^*\ge d$.

Observe that we did not assume linearity at any step.

Title: Even weighted codewords and puncturing


问:
If $X_n \stackrel{d}{\to} X$, does $c_n X_n$ “converge” to $c_n X$?
Suppose $X_n \stackrel{d}{\to} X$. Let $c_n$ denote a bounded, deterministic sequence that does not converge. Is there a sense in which $c_n X_n$ “converges” to $c_n X$?

答:
No.

For example, if $X \sim Bern(0.5)$ and $c_n = (-1)^n$, then $c_n X$ itself doesn't have limit, so we need smth. like $ c_n X_n - c_n X \stackrel{d}{\to} 0$.

Thus  we have $X_n  \stackrel{d}{\to} X$ and want to check if the condition $ c_n X_n - c_n X  = c_n (X_n - X)\stackrel{d}{\to} 0$ holds.

Let us show than even if we have $c_n \to c$ and $X_n  \stackrel{d}{\to} X$ then it doesn't follow that $ c_n (X_n - X)\stackrel{d}{\to} 0$.

Example 1: suppose that all $X_n$ are defined on different probability spaces.

And even if all $X_n$ are defined on one different probability space then still nothing works.

Example 2: Suppose that $X = 1$ and $X = -1$ with probability $\frac12$, $c_n = 1 =c$, $X_{2n} = X$, $X_{2n-1} = -X$ for all $n\ge 1$. Hence $c_n X_n \stackrel{d}{\to} c_n X$ but it's not true that $ c_n (X_n - X)\stackrel{d}{\to} 0$.

If $c_n \to c$ and $X_n \stackrel{d}{\to} X$ then $c_n X_n \stackrel{d}{\to} c X$ by Slutsky's theorem, but it's not what you want.

ContentLicense: CC BY-SA 4.0


问:
$\int \frac{\cos x}{x} dx $ using contour integration
I saw a solution of how to use complex analysis to integrate
$\int_{-\infty}^{\infty} \frac{\cos x}{x} dx$ which uses the contour that traverses an upper semi circle of radius $R$ in the positive direction followed by the line segment from $-R$ to $-\epsilon$ then the upper semi circle of radius $\epsilon$ in the negative direction followed by the line segment from $\epsilon$ to $R$. Everything makes sense to me except computing $\lim_{\epsilon \rightarrow 0} \int_{C_\epsilon} \frac{e^{iz}}{z}dz$. Everywhere I have looked, the limit is computed by taking the limit inside the integral. Why can this be done? I think it has to do with dominated convergence but what is the dominating integrable function?

Thanks in advance.

答:
To be completely rigorous and apply the dominated convergence theorem, we need to convert the function family $\{e^{ire^{i\theta}}\}_{r>0}$ to a sequence $\{e^{\frac{i}{n}e^{i\theta}}\}_{n=1}^{\infty}$ via the change of variable $r=\frac{1}{n}$. Keeping in mind that we are on the measure space $[-\pi,0]$, it is not too hard to see that $|e^{-\frac{1}{n}\sin\theta}| \leq e^{-\sin\left(\frac{-\pi}{2}\right)} = e$ on our measure space. So now we are ready to show the crucial inequality which will allow us to apply the dominated convergence theorem:

$$|e^{\frac{i}{n}e^{i\theta}}| = |e^{-\frac{1}{n}\sin\theta}| < e \text{ for all } n \in \mathbb{N} \text{ and } \theta \in [-\pi,0]$$

Thus, we can use the constant function $g \equiv e$ as our dominating function. This is integrable because we are on a finite measure space.


问:
Integrate $x^3\sqrt{x^2-9}\,dx$ with trig substitution
$\newcommand{\arcsec}{\operatorname{arcsec}}$This expression can be integrated without trig substitution, but I wanted to try using it anyways and got a different answer.

Starting with finding the indefinite integral $$\int x^3\sqrt{x^2-9} \, dx$$
I then substituted $x = 3\sec\theta$, $\theta = \arcsec(\frac{x}{3})$, and $d\theta = \frac{3}{\sqrt{x^2-9} |x|} \, dx$.
Plugging these values into the original integral,
$$\int 27\sec^3\theta\sqrt{9\sec^2\theta - 9}\frac{3}{\sqrt{9\sec^2\theta-9} |3\sec\theta|} \, d\theta$$
$$\int \frac{81\sec^3\theta3\tan\theta}{3\tan\theta |3\sec\theta|} \, d\theta$$
$$\int \frac{81\sec^3\theta}{|3\sec\theta|} \, d\theta$$
$$\int 27\sec^2\theta \, d\theta$$
$$27\tan\theta + C$$
Now substituting $x$ back in and simplifying:
$$27\tan(\arcsec(\frac{x}{3})) + C$$
$$9\operatorname{sgn}(x)\sqrt{x^2-9} + C$$

This does not seem close at all to the solution I found by $u$-substitution,
$$\frac{(x^2-9)^\frac{3}{2}(x^2+6)}{5} + C$$

I am relatively new to integration so I think I made a mistake. What did I do wrong? Any help appreciated.

答:
With a u-substitution:

$\int x^3\sqrt{x^2-9} \ dx\\
\int \frac 12 x^2\sqrt{x^2-9} (2x\ dx)\\
u = x^2 - 9\\
x^2 = u+9\\
du = 2x\ dx\\
\frac 12\int (u+9)\sqrt{u}\ du\\
\frac 12(\frac 23 (9u^\frac 32) + \frac 25 u^\frac 52)+ C\\
3(x^2-9)^\frac 32 + \frac 15(x^2-9)^\frac 52 + C$

With a trig substitution:

$\int x^3\sqrt{x^2-9} \ dx\\
x = 3\sec\theta\\
dx = 3\sec\theta\tan\theta\\
3^5\int \sec^4\theta\tan^2\theta\ d\theta\\
3^5\int (\tan^2\theta\ + \tan^4\theta)\sec^2\theta d\theta\\
3^5 (\frac 13\tan^3\theta\ + \frac 15\tan^5\theta) + C\\
3^5 (\frac 13 (\frac {x^2}{9}-1)^\frac 32 + \frac 15(\frac {x^2}{9}-1)^\frac 52) + C\\
3^5 (\frac 13 \frac {(x^2-9)^\frac 32}{3^3} + \frac 15\frac {(x^2-9)^\frac 52}{3^5}) + C\\
3(x^2-9)^\frac 32 + \frac 15(x^2-9)^\frac 52 + C$

If you want to use the substitution

$\theta = \sec^{-1} \frac {x}{3}\\ dx = \frac {3}{x\sqrt{x^2-9}}$

$\int x^3\sqrt{x^2-9} \ dx\\
\int \frac 13 x^4(x^2-9)\left(\frac {3}{x\sqrt{x^2-9}}\ dx\right)\\
\frac 13 \int  (3^4\sec^4 \theta) (3^2\tan^2\theta)\ d\theta\\
$

And continue as above.


问:
Write Multiplication Using Sigma?
Let's suppose I have the following two numbers:

$$ a= a_{N/2} a_{N/2-1} \cdots a_1 $$

and

$$ b = b_N b_{N-1} b_{N-2} \cdots b_1 $$

Where a_1 is the first digit of a, a_2 the second digit and so on.

Then how can I write their multiplication using sigmas?

For example something like:

答:
Hint: Multiply both numbers by corresponding powers of 10 to make them integers. Then, any $n$ digit integer can be written as a sum:
$$a=\sum_{i=0}^{n-1} a_i\cdot 10^i$$
To multiply two such sums together, use the distributive property of real numbers.

Title: Write Multiplication Using Sigma?


问:
Is this Solution Correct. Simple Analysis Question Bounded Function From Below.
For constants $a_1,a_2>0$, $r\in(0,1)$ the function $f:\mathbb{R}_+\to \mathbb{R}$

$$ f(x) =a_1 x-a_2(1-x)^r $$ is bounded from below.

Is the following solution correct? a simple yes is enough
$\underline{Solution :}$

Since $f$ is continuous its bounded below on any interval $[0,c],~c<\infty$. $f'(x)=a_1-ra_2(1-x)^{r-1}$ which is positive for some $x$ large enough (depending on $a_1,a_2,r$).

答:
I don't really understand why you calculate $f'$ since you only want to minorate $f$.

As you said $f$ is minorated on any interval $[0, c]$ (the image of a segment by a continuous function is a segment).

Since $a_1 >0$, we have $f(x)\rightarrow +\infty$ when $x\rightarrow +\infty$.
Let's take $c$ such that $f(x) > 0$ for all $x>c$.

Then $f$ is minorated on $[0, c]$, and minorated by $0$ on $[c,+\infty]$.

So $f$ is minorated.


问:
Question about proof in problem with a sequence
I have a sequence which is defined as $\sqrt{a_n}$ if $a_n$ is a square number and $a_n+2$ otherwise. I am told that $a_1>1$ and I have to prove that the sequence ends with $3,5,7,9$ if the first term is odd and with $2,4$ if it is odd.

Anyways I did try to prove it but I think it only applies if $a_1=2$ in the case of even and $a_1=3$ and okay in a couple of other cases but I don't think it applies for any random $a_1>1$.

So my question is, is there any rule or theorem that makes me have  $a_1=2$ if I know it is even and $a_1=3$ in the case of odd?

Maybe it's trivial or maybe I am missing something but I would be really thankful for some help.

答:
For $a_1=2$ sequence is $$2 \to 4 \to 2 \to 4 \to \ldots$$

For $a_1=3$ sequence is $$3 \to 5 \to 7 \to 9 \to 3 \to \ldots$$

The sequence increases in steps of $2$ until it hits a perfect square. Thus the terms always maintain their parity.

For $a_1=2k+1$ where $m^2 < a_1 < n^2$, the terms increase in steps of $2$ until an odd square is hit. Then the sequence drops to an odd square root. This goes on and finally $3 \to 5 \to 7 \to 9 \to 3 \to \ldots$ is obtained.

Same for $a_1=2l$.


问:
What is the inverse of the function $f(x)=x^2-9$? $f(x)= \sqrt{x+9}$ or $f(x)= -\sqrt{x+9}$ or both?
What is the inverse of the function $f(x)=x^2-9$? $f(x)= \sqrt{x+9}$ or $f(x)= -\sqrt{x+9}$ or both?

I have to find the inverse of the function $f(x)=x^2-9$ for $x \in \mathbb{R}$.

I can see that the range of this function is $y \in [-9, ∞)$.

Inverse relation:

$f(x)= \sqrt{x+9}$ for  $x  \in [-9, ∞)$

$f(x)= -\sqrt{x+9}$ for  $x  \in [-9, ∞)$

So range becomes $y \in [0, ∞)$ for positive part inverse function and $y \in [0, -∞)$ for negative part inverse function. But since its a function It has to be either 1 to 1 or many to 1 mapped, hence it can't be both, so which part to keep as the answer and why?

Also I have read that range and domain of inverse functions are swapped, if so, we must include both parts of the inverse relation, but that would result in violation of the {(1 or many) to 1} mapping. I am a bit confused how to go about it. Thanks for the help.

答:
Whether you choose $f^{-1}(x)=\sqrt{x+9}$ or $f^{-1}(x)=-\sqrt{x+9}$ will depend on the restriction you put on the domain of $f(x)$. A restriction is necessary here because, as you correctly point out, $\pm\sqrt{x+9}$ is not a function. This is due to the fact that $f(x)$ is not injective ("one-to-one"), so its inverse will not pass the vertical line test (i.e. not be a function).

Let's say you restrict the domain of $f(x)$ to be $[0,\infty)$. This means the range of the function will be $[-9,\infty)$. The domain of $f(x)$ will be the range of $f^{-1}(x)$ and the range of $f(x)$ will be the domain of $f^{-1}(x)$, as you also correctly point out. Notice that the domain of $f(x)$ is nonnegative; this means the range of $f^{-1}(x)$ will be nonnegative, so $f^{-1}(x)=\sqrt{x+9}$ in that case.

In the case that you choose the domain restriction of $(-\infty,0]$, the range of the function will be $[-9,\infty)$. Notice that the domain of $f(x)$ is now nonpositive; this means the range of $f^{-1}(x)$ will be nonpositive, so $f^{-1}(x)=-\sqrt{x+9}$ in that case.


问:
Is there a Ramsey's theorem equivalent for vertex colorings?
Ramsey's theorem states that any 2-color edge coloring of a sufficiently large graph contains a monochromatic complete subgraph. Does there exist something like Ramsey's theorem but for vertex colorings? Like, for any 2-color vertex coloring of a sufficiently large graph contains a monochromatic complete subgraph (but this time with the vertices colored and not the edges). Or is there someway to translate the vertex problem in an edge coloring problem such that we can use Ramsey's theorem?</p>

答:
For vertex colouring the theorem is called "The Pigeon Hole Principle". It says that no matter how you colour the vertices of the complete graph with $(n-1)r+1$ vertices with $r$ colours you can always find a mono-chromatic copy of $K_n$.

Moreover, there exists a colouring of the complete graph with $(n-1)r$ vertices with $r$ colours with no monochromatic $K_n$.

Nobody cares about this because the answer is trivial.


问:
Morphism between integral domain and field is injective?
I just read in my linear algebra's notes the following statement :
Let A be an integral domain and K a field. Any nonzero ring morphism $\phi : A \to K$ is injective.

I think this statement is false by considering the morphism $$\phi : \mathbb Z \to \mathbb Z /2 \mathbb Z$$
$$n \to [n]$$
This is a morphism between an integral domain and a field but clearly not injective.

So is the statement wrong ? I am quite sure of the counterexample but each time I disagreed with my teacher's notes, I was wrong.

答:
You're correct. Here are two possibilities for what the statement should have been:

Any morphism from $K \to A$ is injective (because the kernel is an ideal of $K$ and the only ideals are $(0)$ and $(1) = K$). It doesn't matter so much that $A$ is an integral domain here other than to know that $A \neq 0$. If $A$ were $0$ then $K \to 0$ is non-injective.

The map $A \to \operatorname{Frac}(A)$ is injective.


问:
Determine if the sequence $\{a_n\}:a_n=(-3)^n$ is an increasing or a decreasing sequence
Determine if the sequence $\{a_n\}:a_n=(-3)^n$ is an increasing or a decreasing sequence.

The $(n+1)^{st}$ term is $a_{n+1}=(-3)^{n+1}.$ So $\dfrac{a_{n+1}}{a_n}=\dfrac{(-3)^{n+1}}{(-3)^n}=\dfrac{(-3)^n(-3)^1}{(-3)^n}=-3<1.$ Doesn't this mean that $a_{n+1}<a_n$ for every $n$ which obviously isn't true? The sequence looks like $-3,9,-27,81...,$ so it isn't monotonic, but why the usual method tells us else? We can use it if $a_n\ne0$ for every $n$ which holds here.

答:
It is because $\frac{a_{n+1}}{a_n} < 1$ does not tell you the sequence is decreasing.

What we would all like to do is to multiply both side by $a_n$, but WE CAN'T, since $a_n$ is not necessarily positive !

So $\frac{a_{n+1}}{a_n} < 1$ does not imply $a_{n+1} < a_n$.


问:
$ E\left( \left|\frac{1}{n}\sum_{j=1}^n X_j\right|^p \right) \le \left( \frac{1}{n}\sum_{j=1}^n E(|X_j|^p)^{1/p} \right)^p$
The following is problem 14 of section 3.2 from Chung's "A Course in Probability Theory".

If $p>1$, we have

$$\left| \frac{1}{n}\sum_{j=1}^n X_j \right|^{p} \le \frac{1}{n}\sum_{j=1}^n |X_j|^p$$

and so

$$ E\left( \left|\frac{1}{n}\sum_{j=1}^n X_j\right|^p \right) \le \frac{1}{n}\sum_{j=1}^n E(|X_j|^p)$$

we have also

$$ E\left( \left|\frac{1}{n}\sum_{j=1}^n X_j\right|^p \right) \le \left( \frac{1}{n}\sum_{j=1}^n E(|X_j|^p)^{1/p} \right)^p$$

I understand how the the first two inequalities follow from just using the well known power mean inequality, but how to show the last inequality?

答:
The first is applying Jensen's Inequality to the convex function $x\mapsto|x|^p$:
$$
\color{#C00000}{\left|\color{#000000}{\frac1n\sum_{j=1}^n}\color{#00A000}{X_j}\right|^p}
\le\frac1n\sum_{j=1}^n\color{#C00000}{\left|\color{#00A000}{X_j}\right|^p}\tag{1}
$$
Next, simply take the expectation of both sides of $(1)$, using the linearity of expectation
$$
\mathrm{E}\left(\left|\frac1n\sum_{j=1}^nX_j\right|^p\right)
\le\frac1n\sum_{j=1}^n\mathrm{E}\left(|X_j|^p\right)\tag{2}
$$
The last is Minkowski's Inequality in a probabilistic setting, as shown below. First, let's show that $\displaystyle\mathrm{E}(X^p)^{1/p}=\sup_{\mathrm{E}(Y^q)=1}\mathrm{E}(XY)$.

Assume $X,Y\ge0$. Suppose that $\mathrm{E}(X^p)=1$ and $\mathrm{E}(Y^q)=1$ where $\frac1p+\frac1q=1$. Then, because $XY\le\frac{X^p}{p}+\frac{Y^q}{q}$, we get that
  $$
\begin{align}
\mathrm{E}(XY)
&\le\frac1p\mathrm{E}(X^p)+\frac1q\mathrm{E}(Y^q)\\
&=1\\
&=\mathrm{E}(X^p)^{1/p}\mathrm{E}(Y^q)^{1/q}\tag{3}
\end{align}
$$
  Equation $(3)$ scales properly with $X$ and $Y$, so we can remove the restrictions that $\mathrm{E}(X^p)=1$ and $\mathrm{E}(Y^q)=1$.

Suppose that $Y=X^{p-1}$, then $Y^q=X^p$ and
  $$
\begin{align}
\mathrm{E}(XY)
&=\mathrm{E}(X^p)\\
&=\mathrm{E}(X^p)^{1/p}\mathrm{E}(Y^q)^{1/q}\tag{4}
\end{align}
$$
  Combining $(3)$ and $(4)$, we see that
  $$
\mathrm{E}(X^p)^{1/p}=\sup_{\mathrm{E}(Y^q)=1}\mathrm{E}(XY)\tag{5}
$$

Lift the assumption that $X\ge0$. Applying $(5)$ twice, we get
$$
\begin{align}
\left(\mathrm{E}\left(\left|\frac1n\sum_{j=1}^nX_j\right|^p\right)\right)^{1/p}
&\le\left(\mathrm{E}\left(\left(\frac1n\sum_{j=1}^n\left|X_j\right|\right)^p\right)\right)^{1/p}\\
&=\sup_{\mathrm{E}(Y^q)=1}\mathrm{E}\left(\frac1n\sum_{j=1}^n\left|X_j\right|Y\right)\\
&\le\frac1n\sum_{j=1}^n\sup_{\mathrm{E}(Y^q)=1}\mathrm{E}\left(\left|X_j\right|Y\right)\\
&=\frac1n\sum_{j=1}^n\mathrm{E}\left(|X_j|^p\right)^{1/p}\tag{6}
\end{align}
$$
The last inequality in $(6)$ follows because we can choose a different $Y$ for each $j$ on the right, while we can only choose one on the left. Equation $(6)$ shows, as desired,
$$
\mathrm{E}\left(\left|\frac1n\sum_{j=1}^nX_j\right|^p\right)
\le\left(\frac1n\sum_{j=1}^n\mathrm{E}\left(|X_j|^p\right)^{1/p}\right)^p\tag{7}
$$

ContentLicense: CC BY-SA 3.0


问:
Find the Closest Vector to a Given Vector from a Given Set Defined by Set of Multiplications
Let $x=\left(x_t\right)_{t=1}^n$ be a vector such that
$$
x_t = \prod_{i=1}^t u_i, \tag{1}
$$
where each parameters $u_i$ can take any of two value
$$
u_i \in \left\{a,b \right\} = \left\{ 1.3, 0.8 \right\}.
$$
Each vector $x$ represents a path in a binomial tree.

Given an arbitrary vector $y=\left(y_t\right)_{t=1}^n$,
where each $y_t$ is in $\mathbb R^+$, how to solve the problem
$$
\min_{x \in X} \lVert y - x \rVert^2 = \sum_{t=1}^n \left(y_t - x_t \right)^2
$$
where $X$ is the set of all possible vector of the form (1).

So, given a realization of the stock, find the closest binomial path.

The only way I see to solve this problem is to do Gram-Schmidt on $X$,
and then project $y$ on the orthonormal basis found.

But, my $n$ is large, say $100$.

This induces
$$\mathrm{cardinality} X = 2^{100},$$
so I don't think Gram-Schmidt is viable.

I would use least-square or Newton-Raphson, but the variable are not continuous.

答:
I would got with one of the approaches (They are not closed form solutions but might be good enough):

Try estimating $ {u}_{i} $ by looking at the vector $ \left[ {y}_{1}, {y}_{2} / {y}_{1}, {y}_{3} / {y}_{2}, \ldots, {y}_{n} / {y}_{n - 1} \right] $ then taking the nearest value out of $ \left\{ 0.8, 1.3 \right\} $.

Apply the $ \ln \left( \cdot \right) $ on each element of $ \boldsymbol{y} $. Then each element becomes sum of $ {u}_{i} $ which means you can solve a problem of the form $ A u = \ln \left( y \right) $ (Pay attention to $ A $ as it is only 1 and 0 in a convolution like form). From the estimated $ \boldsymbol{u} $ you may build $ \boldsymbol{x} $.


问:
The Sobolev embedding inequality on manifolds
Let $(M,g)$ be a (smooth) compact Riemanian manifold of dimension $n$. I expect that the following inequality is true for any smooth function $f$:

$$(\int_{M} |f|^{\beta})^{1/\beta} \leq C \;(\int_{M} |\nabla f|_{g}^{2}+\int_{M} f^{2})^{1/2},$$
for $\beta=2n/(n-2)$. Since this is true for $M=\mathbb{R}^{n}$.

Could anyone point out a clear reference for this?

In the book: Aubin T. Nonlinear analysis on manifolds: Monge Ampere equations, page 50, the author stated this result with the extra assumption: $M$ has a constant curvature and positive injectivity radius! Are these conditions necessary?

答:
I found the result you referenced in page 50 of the following text:

Aubin, Thierry, Some nonlinear problems in Riemannian geometry, Springer Monographs in Mathematics. Berlin: Springer. xviii, 396 p. (1998). ZBL0896.53003.

I assume is the book you are looking at. The result of compact manifolds without boundary is proved in Theorem 2.20 the same text, and a slightly more general result is proved on page 45 which asserts the following.

2.21 Theorem. The Sobolev imbedding theorem holds for $M_n$ a complete manifold with bounded curvature and injectivity radius $\delta>0.$ Moroever, for any $\varepsilon>0,$ there exists a constant $A_q(\varepsilon)$ such that every $\varphi \in H_q^q(M_n)$ satisfies
$$ \lVert \varphi \rVert_p \leq \left[K(n,q)+\varepsilon\right] \lVert \nabla \varphi \rVert_q + A_q(\varepsilon) \lVert \varphi \rVert_q,$$
with $1/p=1/q-1/n>0,$ and $K(n,q)$ is the optimal constant for Sobolev embedding in $\Bbb R^n.$

Note that if $M$ is smooth and compact, the injectivity radius is always positive and the curvature is always bounded so this applies. The result in page 50 (Theorem 2.28) shows that if $M$ is compact and either $n=2$ or $M$ is flat, one can additionally take $\varepsilon=0$ in the above.

In general the main obstruction to the Sobolev embedding result is the degeneration of the metric; you can think of having $M = \Bbb R^n$ with a metric $g(x) = w(x)^{\frac12} g_0$ with $g_0$ the usual Euclidean metric, and $w>0$ a suitable weight function, the inequality reduces to
$$ \left( \int_M |f|^{\beta} \,\mathrm{d} x \right)^{\frac1\beta} \leq C \left( \int_M w(x)|\nabla f|^2 \,\mathrm{d} x + \int_{\Bbb R^n} |f|^2 \,\mathrm{d}x\right)^{\frac12}.$$
From here we observe the validity of this inequality depends on choice of weight function $w.$ If $M$ is compact this cannot occur however, and a straightforward partition of unity argument given in Theorem 2.20 will do the trick.

In the case of the results in Aubin, one is interested in obtaining sharp bounds for the associated constants, which has implications on the geometry of the space $M.$ For this one needs the more refined arguments given in the book and the papers cited therein, involving the exponential map and using curvature bounds in addition.


问:
How does Ramsey Theory relate to Godel Incompleteness theorem?
My knowledge in math is not too advanced but I am super interested in this topic and although I think I know part of the answer, I do not know for sure how these two theories are related.

Also, if there is some proof done in Peano Arithmetic, that would be helpful to see . Thanks

答:
There is a two-step connection between Ramsey theory and unprovability:

Theorems in Ramsey theory often correspond to the totality of particular fast-growing computable functions - namely, functions of the form $F(k)=$ the least $n$ such that every "A-configuration" of size $n$ has a "B-configuration" of size $k$ for some notions of "A-configuration" and "B-configuration."

In general, we can show that a sentence is unprovable in a theory $T$ by analyzing the provably total functions of $T$ and then showing that that sentence implies the totality of a computable function not in that class.

For example, the provably total functions of $I\Sigma_1$ are exactly the primitive recursive functions, so any sentence implying for example that the Ackermann function is always defined is automatically unprovable in $I\Sigma_1$. The provably total functions of $\mathsf{PA}$ are much harder to describe, but we can still get similar results. The particular role Ramsey theory plays is simply as a vehicle for producing fast-growing functions.


问:
Proof that $\text{Hom}(E_1,F_1)\otimes\text{Hom}(E_2,F_2)$ and $\text{Hom}(E_1\otimes E_2,F_1\otimes F_2)$ are isomorphic
Let $C$ be a commutative ring and $E_1,E_2,F_1,F_2$ $C$-modules. Suppose $E_1$ and $F_1$ are projective and finitely generated. I want to show that the canonical mapping
$$\lambda:\text{Hom}(E1,F1)\otimes\text{Hom}(E_2,F_2)\rightarrow\text{Hom}(E_1\otimes E_2,F_1\otimes F_2)$$
is an isomorphism. Is there a short categorical proof of this? The proof I have has 4 lemmas and is almost two pages; it implicitly uses functors and natural transformation. It treats $\lambda$ as a natural transformation between the functor $\text{Hom}(-,F_1)\otimes\text{Hom}(E_2,F_2)$ and $\text{Hom}(-\otimes E_2,F_1\otimes F_2)$ and also as natural transformation between $\text{Hom}(C,-)\otimes\text{Hom}(E_2,F_2)$ and $\text{Hom}(C\otimes E_2,-\otimes F_2)$.

Any way to clean this proof up?

答:
First prove it for $E_1 = F_1 = C$ where it's very easy. Then prove that if it's true for two values of $E_1$ then it's true for their direct sum, and similarly for $F_1$; this proves it for $E_1, F_1$ finitely generated free. Then prove that if it's true for some value of $E_1$ then it's true for any retract of it, and similarly for $F_1$; this proves it for $E_1, F_1$ finitely generated projective.

The general pattern here is that "retracts are absolute colimits," which means they are preserved by any functors whatsoever. See the section titled "The facts of life about idempotents and retracts" in my blog post Tiny objects for details.


问:
Is it possible that Taylor Remainder be a negative number?
To bound the Error of the approximation $\sin(x)\approx x$ for $-\frac{\pi}4\le x\le \frac{\pi}{4}$ I used Taylor Remainder formula and I get $R_2(1)=\frac{-\sqrt{2}}{12}$. I want to make sure that is it possible that this value be a negative number?

答:
Yes it's possible.

Let's take a Taylor development around $0$ :

$$f(x) = a_0 + a_1 x + a_2 x^2 + ...{}{}{}$$

You can have a function $f$ with whatever coefficient you want.

Then if you take one such, for example, all $a_k<0$, then when you evaluate with a $x>0$, you always get a negative remainder.

A simple example is to take a polynomial with only negative coefficients : then its Taylor development is the polynomial (if you're at a rank high enough).


问:
Find the highest natural number which is divisible by $30$ and has exactly $30$ different positive divisors.

Find the highest natural number which is divisible by $30$ and have exactly $30$ different positive divisors.

What I Tried: I am not sure about any specific approach to this problem. Of course as the number is divisible by $30$ , some of the factors of the number would be $1,2,3,5,6,10,20,30$ , but that only makes $8$ divisors and there will be $12$ more which I have no idea of .

Can anyone help me?

答:
If $n$ has $k$ prime divisors then $$\sigma (n) \geq \underbrace{(1+1)(1+1)\cdots (1+1)}_{k\; \rm times} = 2^k$$

So $2^k\leq 30$ and thus $k\leq 4$. So $n$ has $3$ or $4$ prime divisors.

If $\color{red}{k=4}$ then $n= 2^a3^b5^cp^d$, where $p$ is prime greater than $5$. Now we have $$(a+1)(b+1)(c+1)(d+1)= 30$$ which is impossible.

If $\color{green}{k=3}$ then $n= 2^a3^b5^c$. Now we have $$(a+1)(b+1)(c+1)= 30$$ which is possible if $a=1$, $b=2$ and $c=4$ or any permutation of them.


问:
convergence of series with absolute value
prove or show false:

if $\sum_{n=1}^{\infty}\left |a_{n}  \right |$ converges, then $\sum_{n=1}^{\infty}\frac{n+1}{n}a_{n}$ converges as well.

Thank you very much in advance,

Yaron.

答:
Hints:

Show that, in fact, $\sum\limits_{n=1}^\infty {n+1\over n} a_n$ converges absolutely.  This will imply the series converges.

To show absolute convergence, you may use the Comparision test.  Note that for each $n$, one has ${n+1\over n}=1+{1\over n}\le 2$.


问:
Extension of analytic function to the boundary of its domain using non-tangential limits
Let $\mathbb{D}$ denote the open unit disk of the complex plane and let $f : \mathbb{D} \rightarrow \mathbb{C}$ be an analytic function. Assume that $f$ has non-tangential limits on a dense set on $\partial\mathbb{D}$ such that the values can be extended to a continuous function on $\partial\mathbb{D}$. Is it enough to deduce that $f$ can be extended to a continuous function on $\overline{\mathbb{D}}$ that is analytic on its interior?

Thank you very much!

答:
Here's a counterexample: Let $E\subset \partial\mathbb D$ be a dense subset of measure $0.$ Then there exists a positive harmonic function $u$ on $\mathbb D$ such that

$$\tag 1 \lim_{z\to \zeta,z\in \mathbb D}u(z)=\infty$$

for each $\zeta\in E.$*

Using $(1),$ let $v$ be a harmonic conjugate of  $u.$ Set $f=e^{-(u+iv)}.$ Then

$$\lim_{z\to \zeta,z\in \mathbb D}f(z)=0$$

for each $\zeta\in E.$ Note that these limits allow full approach within the disc, not just nontangential approach.

I'll leave the verification that this $f$ is indeed a counterexample. But ask if you have questions.

*See Exercise 18, Chapter 6, Harmonic Function Theory by Axler, Bourdon, and Ramey  https://www.axler.net/HFT.pdf


问:
Integrate $x^3\sqrt{x^2-9}\,dx$ with trig substitution
$\newcommand{\arcsec}{\operatorname{arcsec}}$This expression can be integrated without trig substitution, but I wanted to try using it anyways and got a different answer.

Starting with finding the indefinite integral $$\int x^3\sqrt{x^2-9} \, dx$$
I then substituted $x = 3\sec\theta$, $\theta = \arcsec(\frac{x}{3})$, and $d\theta = \frac{3}{\sqrt{x^2-9} |x|} \, dx$.
Plugging these values into the original integral,
$$\int 27\sec^3\theta\sqrt{9\sec^2\theta - 9}\frac{3}{\sqrt{9\sec^2\theta-9} |3\sec\theta|} \, d\theta$$
$$\int \frac{81\sec^3\theta3\tan\theta}{3\tan\theta |3\sec\theta|} \, d\theta$$
$$\int \frac{81\sec^3\theta}{|3\sec\theta|} \, d\theta$$
$$\int 27\sec^2\theta \, d\theta$$
$$27\tan\theta + C$$
Now substituting $x$ back in and simplifying:
$$27\tan(\arcsec(\frac{x}{3})) + C$$
$$9\operatorname{sgn}(x)\sqrt{x^2-9} + C$$

This does not seem close at all to the solution I found by $u$-substitution,
$$\frac{(x^2-9)^\frac{3}{2}(x^2+6)}{5} + C$$

I am relatively new to integration so I think I made a mistake. What did I do wrong? Any help appreciated.

答:
sometimes its easier to break it up into other expressions first:
$$I=\int x^3\sqrt{x^2-9}$$
$x=3u\Rightarrow dx=3du$
$$I=\int(3u)^3\sqrt{3^2(u^2-1)}(3du)$$
$$I=3^5\int u^3\sqrt{u^2-1}du$$
now we want to look for an identity that matches this, we know:

$\cosh^2\theta-\sinh^2\theta=1$

$\cos^2\theta+\sin^2\theta=1$

so we want to choose one of these that can be rearranged to our form, the nicest would be: $\cosh^2\theta-1=\sinh^2\theta$ so:

$u=\cosh(v)\Rightarrow du=\sinh(v)dv$ now replace everything:
$$I=3^5\int\cosh^3v\sqrt{\cosh^2v-1}\sinh v dv$$
$$I=3^5\int\cosh^3v\sinh^2vdv$$


问:
Why do we say some infinite products "diverge" when the limit is zero, a finite value?
Infinite series (sums) are discussed a lot, but infinite products less so. Despite having tried hard to find reading material on infinite products, I have only found lecture notes and not proper texts.

Having read these, eg "Background Notes Theory of Infinite Products", I am puzzled as to why infinite products that do have a limit of 0 are said to "diverge".

Here is one example:

$$ \prod_{n=2}^{\infty}(1-\frac{1}{n}) $$

As usual we take the partial sum and consider the limit.

$$ \require{cancel} \begin{align} \prod_{n=2}^{N}(1-\frac{1}{n}) &=  \prod_{n=2}^{N}(\frac{n-1}{n}) \\ \\ &= \frac{1}{\cancel{2}} \times \frac{\cancel{2}}{\cancel{3}} \times \frac{\cancel{3}}{4} \times  \ldots \times \frac{N-1}{N} \\ \\ &= \frac{1}{N} \end{align} $$

As $N \rightarrow \infty$, this partial sum tends to 0.

Because this limit is finite, albeit 0, I feel this should be considered convergent. What key insight am I missing?

Update: according to lecture notes, "diverge to zero" means there are an infinite number of factors that are zero, so this specific example doesn't "diverge to zero".

答:
The simplest way to explain this convention is that it ensures that an infinite product $\prod r_i$ of positive reals converges iff its logarithm $\sum \log r_i$ converges in the usual sense, which reduces the theory of infinite products (of positive reals) to the theory of infinite sums. If an infinite product diverges to $0$ then its logarithm diverges to $-\infty$. The divergence of the infinite product you describe turns out to be equivalent to the divergence of the harmonic series, for example (Example 5 in the notes you've linked).

This convention also has the desirable property that it makes convergence closed under taking inverses: $\prod r_i$ converges to $\ell$ iff $\prod r_i^{-1}$ converges to $\ell^{-1}$.

More abstractly, you might say that when we deal with infinite products we restrict our attention to the positive reals $\mathbb{R}_{+}$, because taking products involving $0$ is not very interesting. So $0$ is excluded and takes on the role of a "point at infinity" (and again this can be motivated by thinking about the logarithm $\log : \mathbb{R}_{+} \to \mathbb{R}$).

(Considering the nonzero reals is not much of an upgrade over the positive reals, because for convergence only finitely many terms can be negative anyway. There are reasons to consider the nonzero complex numbers though, e.g. when considering the Euler product of the Riemann zeta function for complex values of $s$.)


问:
How to simplify $x^{x^{-5}(5)}=5^{-\frac{25}{\sqrt[5]{5^{16}}}}$ and find a value from it?
The problem is as follows:

First simplify:

$$x^{x^{-5}(5)}=5^{-\frac{25}{\sqrt[5]{5^{16}}}}$$

Then using $x$ find: $x^{-50}$

The alternatives in my book are as follows:

$\begin{array}{ll}
1.&5\\
2.&\frac{1}{5}\\
3.&25\\
4.&125\\
\end{array}$

I'm confused about solving this problem, how should it be approached?

$x^{x^{-5}(5)}=5^{-\frac{25}{\sqrt[5]{5^{16}}}}$

$5^{-\frac{25}{\sqrt[5]{5^{16}}}}=5^{-\frac{1}{5\sqrt[5]{5}}}$

$5^{-\frac{1}{5\sqrt[5]{5}}}=5^{-\frac{1}{5^{\frac{6}{5}}}}$

But that's where I'm stuck. Where to go from there?. I can't seem to find a way to get that problematic $x$. Can someone help me with the radical manipulation?. Does this requires any additional multiplication?

答:
Im slightly confused about your notation, do you mean:
$$x^{5x^{-5}}$$ or:
$$\left(x^{x^{-5}}\right)^5$$

Anyways, we can start with the other side, we have:
$$5^{-\frac{25}{\sqrt[5]{5^{16}}}}\tag{1}$$
if we look at the power first:
$$-\frac{25}{\sqrt[5]{5^{16}}}=-\frac{5^2}{(5^{16})^{1/5}}=-\frac{5^2}{5^{16/5}}=-5^{2-16/5}=-5^{-6/5}$$
so we can simplify $(1)$ to:
$$5^{-5^{-6/5}}$$
now equate the LHS to this


问:
Sharp map on differential
I do understand the definitions of $g^{\flat}\colon TM \to T^*M$ and $g^{\sharp}\colon T^*M \to TM$ [1] but I can not make the connection to obtain
$$\mathrm{grad} \, f=(df)^\#$$

Is it possible to obtain analytically the definition of gradient using the definitions of $g^{\flat}$ and $g^{\sharp}$?

答:
It seems like maybe your confusion is that you have a preconceived notion of what the gradient is, and are wondering why this definition agrees with what you have previously thought of as the "gradient"?

Maybe you are thinking (from elementary calculus) of the gradient of $f$ as the vector with components $\nabla f = \left( \frac{\partial f}{\partial x_1}, \; \dots, \; \frac{\partial f}{\partial x_n} \right)$. This in general does not agree with the $df^\sharp$ definition. It only agrees when you take the standard Euclidean metric on $\Bbb{R}^n$ (the ordinary "dot product"):
$$ g(v,w) = \left<v,w\right> = v_1w_1 + v_2w_2 + \cdots + v_nw_n $$

But of course you do not have to take this as your metric. A general metric looks like
$$ g(v,w) = \sum_{ij} g_{ij} v_iw_j $$
where $g_{ij}$ are some collection of smooth functions (obeying certain relations from the definition of a metric, like symmetry : $g_{ij} = g_{ji}$, etc.). Then the gradient (i.e. $df^\sharp$) is the vector field so that $df(v) = g(df^\sharp,v)$ for all vectors $v$. Let's use $c_i$ for the components of $\mathrm{grad} f = df^\sharp =(c_1,c_2,\dots,c_n)$. By plugging in $v=e_i=(0,0,\dots,1,\dots,0)$, you see that
$$ \begin{align*}
df(e_i) &= g(df^\sharp,e_i) \\
\frac{\partial f}{\partial x_i} &= \sum_j g_{ji} c_j
\end {align*}$$
So you see, in general $c_i \neq \frac{\partial f}{\partial x_i}$. By solving these equations for $c_i$, you get
$$c_i = \sum_j g^{ji} \frac{\partial f}{\partial x_j}$$
where $g^{ij}$ are the components of the inverse of the matrix $g_{ij}$.

Title: Sharp map on differential


问:
Show that $[l_1 \cdot l_2 \cdot l_3 ] = [l_1 + l_2 + l_3] \in H_1(X)$ The first Homology group of X
Let $l_1$ , $l_2$ and $l_3$ be three paths in X with $l_1 (0) =
l_3 (1)$, $l_1 (1) = l_2 (0)$ and $l_2 (1) = l_3 (0)$. Define the loop $l = l_1 \cdot l_2 \cdot l_3 $ (based at $l_1 (0)$).

Show that $l$ and $l_1 + l_2 + l_3$ are 1-cycles. (This can just be shown by applying the boundary map, $\partial$, on the paths are to get $0$. Hence they are in $\ker(\partial)$, which defines them as 1-cycles).

Furthermore, show that $[l] = [l_1 + l_2 + l_3] \in H_1(X)$

My confusion is regarding treating this as an algebraic problem, or proving it through the definition of homotopy equivalence, i.e using the definition of the class $[l]$ as everything homotopy equivalent to the path $l$

答: Consider a cylinder $[0,1]\times S^1$ based an a circle $S^1$, where the "left" boundary $\{0\}\times S^1$ is partitioned into a single $0$-cell and a single $1$-cell, while the "right" boundary component $\{1\}\times S^1$ is partitioned into three $1$-cells by three $0$-cells.  Clearly this can be extended to a triangulation of the cylinder.  Now map this to the manifold $X$ by a function $f: [0,1]\times S^1 \to X$, where $f(s,t)$ is independent of $s\in [0,1]$ and only depends on $t\in S^1$.  This induces a 2-chain showing that the two cycles you defined are homologous.  Hence they define the same homology class in $H_1(X)$.</p>


问:
Positive integers expressable as sums of powers of 2
I need to prove that any positive integer is expressable as
$$x=2^{j_0}+2^{j_1}+2^{j_2}+...+2^{j_m}$$
where $m\ge 0$ and $0\le j_0\lt j_1\lt j_2\lt ... \lt j_m. $

I think I get the gist of the proof; what I mean is I think I intuitively understand what is happening here, but looking for verification, either a more simplistic argument or proper rigor.

I start by saying a positive integer is expressable as an even or odd integer.  So, a positive integer, $x=2k$ or $x=2k+1$ for some positive integer $k$.  If $x$ is odd, then we can write $1=2^0$.  Now, an integer $k$ is also expressable as an even or odd integer.  Thus we now have $2^2=4$ possibilities; if $x$ is even;
$$x=2k=2(2k_1)=2^2k_1$$
$$x=2k=2(2k_1+1)=2^2k_1+2^1$$
if $x$ is odd;
$$x=2k+1=2(2k_1)+1=2^2k_1+2^0$$
$$x=2k+1=2(2k_1+1)+1=2^2k_1+2^1+2^0$$
it is clear that $k>k_1$ since $k=2k_1$.  Now we repeat the process with $k_1=2k_2$ or $k_1=2k_2+1$ and eventually, $k_n=2k_{n+1}$ or $k_n=2k_{n+1}+1$.  Since $x$ is a positive integer, $k_i$ is positive for all $i=1,2,...,n$ and thus the larger the $i$, the smaller the $k_i$.  This process eventually terminates since $k_i>0$ and thus $k_n=1$

Now how do I simplify this argument, assuming it's correct.   If it is not correct,, how do I repair it or make it more rigorous?

答:
Theorem: any non-negative integer is a sum of 0 or more distinct powers of 2.

Proof: By the well-odering principle. Assume there is a set, $C$, of positive integers that can not be expressed as a sum of 0 or more powers of 2. Since $C$ is a subset of the positive integers, by the well-ordering principle, there must be a smallest element, $m$, in C.

It's clear that 0 is the sum of 0 powers of 2, so $m>0$.

Now, let's subtract from $m$ the largest power of two, $2^n$, that is still smaller than $m$ itself. Let's call it $j$:

$j = m-2^n$

Now let's use cases analysis. There are 2 cases:0

$j$ is a sum of 0 or more distinct powers of 2.;

$j$ is not a sum of 0 or more distinct powers of 2.

In the first case, by manipulation:

$m = j + 2^n$

In this case, $m$ is expressed as the sum of 2 raised to distinct powers. Since $2^n$ is the largest power of 2 in $m$ so $2^n>j$. It makes it a contradiction and, therefore, $C$ must be empty.

In the second case, $j$ is smaller than $m$. It's a contradiction since $m$ should be the smallest counterexample in $C$.

QED


问:
When is it equivalent to do a joint optimization over two variables in sequence over each variable?
Say I have a function $f(x,y)$. When is
$$\min_{x,y}f(x,y)=\min_x\min_y f(x,y)=\min_y\min_x f(x,y)$$ and what happens when there are constraints?

答:
The below equation is always true,
$$
\min_{x,y}f(x,y)=\min_x\min_y f(x,y)=\min_y\min_x f(x,y)
$$

Let's take any point $x_0, y_0$. The following is always true by definition of $\min$,
$$
f(x_0, y_0) \geq \ \min_y f(x_0, y) \geq \ \min_x\min_y f(x, y)
$$
Note that the above holds for any point $x_0, y_0$. What if $x_0, y_0 = \arg\min_{x, y} f(x, y)$?

$$
\implies \min_{x, y} f(x, y) \geq \ \min_x\min_y f(x, y)
$$

Let's prove the reverse direction for any point $x_0, y_0$,

$$
\min_{x, y} f(x, y) \leq f(x_0, y_0)
$$

The above holds for any $y_0$, What if $y_0 = \arg\min_{y} f(x_0, y)$?

$$
\min_{x, y} f(x, y) \leq \min_{y} f(x_0, y)
$$

The above holds for any $x_0$, What if $x_0 = \arg\min_{x} \arg\min_{y}f(x, y)$?

$$
\min_{x, y} f(x, y) \leq \min_x \min_{y} f(x, y)
$$

Combining both eqns. We get the required result.


问:
Let be $A=\left\{\frac{n}{n+1}: n \in \mathbb{N}\right\} \subset \mathbb{R}$ prove there exists $\sup A$ and find said supreme.
$A=\left\{\frac{n}{n+1}: n \in \mathbb{N}\right\} \subset \mathbb{R}$

I did:
How $1\geq n$ because $n\in\mathbb{N}$

then $n+1\geq n \rightarrow \frac{n+1}{n}\geq1 \rightarrow \frac{n}{n+1}\leq1$

then $\forall n\in\mathbb{N}  \frac{n}{n+1}\leq1 $ then $\sup A = 1$
is right?

答: Yes it is. Let’s prove it. We already know that $1$ is an upper bound of $A$. We need to show it’s the least upper bound. Thus for any $\epsilon > 0$ we need to find an $n$ such that : $\dfrac{n}{n+1} > 1 - \epsilon \implies n > \dfrac{1}{\epsilon} - 1$. Those natural $n$ satisfies this inequality does it. This proves that $1$ is the least upper  bound for $A$.


问:
How to find the cost of a tv set when it is given as an algebraic expression?
The problem is as follows:

Gabriel wants to purchase a new TV for her mother as a Christmas gift. The cost is $2P$ thousand dollars. In order to fulfill this wish he begins saving his earnings as a pharmacist clerk for six months. Assume that the money he earns is the same each month and $P$ to be the sum of the coefficients of the rational integer algebraic expression and not null of the form

$$M(x,y)=\left(\frac{n+4}{3}\right)x^{5-n}y^{4-m}-2x^{n-3}y^{m-4}+(n-3)y^{n+2}$$

If Gabriel decides to buy the gift after $4$ months instead. By how much additional dollars should Gabriel have to save each month with respect of his original plan so that he can buy the tv?

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{300 dollars}\\
2.&\textrm{400 dollars}\\
3.&\textrm{500 dollars}\\
4.&\textrm{200 dollars}\\
\end{array}$

How exactly should be the relationship between the given condition of the integer rational integer algebraic expression should be used?. Can someone help me here?.

So far what I've been able to spot was this:

$P=\frac{n+4}{3}-2+(n-3)$

$P=\frac{4n-11}{3}$

But that's the part where I'm stuck, where exactly should I go from here?. I don't know what else to do with the information related with those $4$ months.

答:
Note that rational integer algebraic expression means that the coefficients are integers, the exponents are rational and the expression is rational of the form $\frac{P(x,y)}{Q(x,y)}$. Also note that $n$ is in the exponent of the variables $x$ and $y$.

Thus $n$ is rational, more so $n$ must be an integer because we have the coefficient $n-3$, furthermore $\frac{n+4}{3}$ must be an integer, and  $3$ divides $n+4$.

Because $\frac{n+4}{3}$ must be an integer, so must $P$, and since Gabriel saves an equal amount each month he must save either $M$ or $M’$ dollars every month respectively:

$$6M=2000P$$

$$4M’=2000P$$

Since $P$ is an integer so are $M$ and $M’$.

$$M’-M=2000P \cdot \left(\frac{1}{4} - \frac{1}{6}\right)=\frac{500P}{3}$$

Since $M’-M$ is an integer because $M’$ and $M$ are both integers $\frac{P}{3}$ must be an integer.

Given $P=\frac{4n-11}{3}$, $9$ divides $4n-11$ thus $n=9n'+5$ for arbitrary $n' \in \mathbb{Z}$. Thus $\frac{P}{3}=4n’+1$

$$M’-M=500\cdot (4n’+1)$$

Could we restrict $n’$ further given the fact that the algebraic expression is rational and $n$ appears in the exponent? Every polynomial is a rational function, thus there are no further restrictions on $n$, and therefore $n’$ as well.

From there, the only available answer that is a multiple of $500$ is $(3).$


问:
Is the linear approximation of the product of two functions the same as the product of the linear approximations of the two functions?
I have just seen a lecture about linear approximation, in which it was established that when $x$ is near $0$:

The linear approximation of $e^x$ is $1+x$

The linear approximation of $(1+x)^r$ is $1+rx$

Afterwards, the lecturer explained how to find—using 1 and 2—the linear approximation of:

$e^{-3x} \times (1+x)^{-1/2}$.

His next line is:

$(1-3x)(1-0.5x)$

So it seems that he was acting under the assumption that the linear approximation of the product of two functions is the product of the linear approximations of the functions.

Is that true in general for any order of approximations? Maybe just for linear ones? Maybe just when x is near zero? Maybe only when the approximation is linear and x near 0?

Here is a link to the lecture. The comments were made at 26:30.

答:
The linear approximation of a product equals the product of the linear approximations modulo terms of degree $2$.

Define the operator $M_1$ that takes a differentiable function and giving the linear approximation of it:
$$
M_1 f(x) := f(0) + f'(0) x
$$

Then,
$$\begin{align}
M_1 f(x) \, M_1 g(x)
&= (f(0)+f'(0)x)(g(0)+g'(0)x) \\
&= f(0)g(0) + f(0)g'(0)x + f'(0)g(0)x + f'(0)g'(0)x^2 \\
&= (fg)(0) + (fg)'(0)x + f'(0)g'(0)x^2 \\
&= M_1(fg)(x) + O(x^2)
.
\end{align}$$


问:
Calculating the degree of a splitting field of $f(x)=x^4-4x^3-4x^2+16x-8$ over $\mathbb{Q}$.
I have shown that the above polynomial is irreducible in $\mathbb{Q}[x]$ and that $\alpha=1+\sqrt{2}+\sqrt{3}$ is a root.  As $f(x+1)$ is an even function, $f$ is symmetric in the line $x=1$, so $2-\alpha$ is also a root. Graphing $f$, one sees that $f$ has two real roots larger than $1$. One of these is $\alpha$ so if we call the other one $\beta$, by the symmetry we've already noted $2-\beta$ would also a root. We could then factor $f$ into linear factors in the following way:
\begin{equation*}
f(x)=(x-\alpha)(x+\alpha-2)(x-\beta)(x+\beta-2)
\end{equation*}
Therefore a splitting field of $f$ over $\mathbb{Q}$ is given by $\mathbb{Q}(\alpha,\beta)$. By the tower law,
\begin{equation*}
[\mathbb{Q}(\alpha,\beta):\mathbb{Q}]=[\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)]\cdot [\mathbb{Q}(\alpha):\mathbb{Q}]
\end{equation*}
and we know $[\mathbb{Q}(\alpha):\mathbb{Q}]=4$ ($f$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$). It remains to calculate $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)]$. I know that $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)]\leq [\mathbb{Q}(\beta):\mathbb{Q}]=4$ so $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}] \leq 16$. This is a small improvement on the general bound that the splitting field of a degree $4$ polynomial is less than or equal to $4!=24$. I'd appreciate any advice on how to continue, or feedback on my approach so far.

答:
In this case, I think the best thing to do is to actually compute the roots! Your observation that $f(x+1)$ is even is very useful: it means that $f(x+1)$ has no odd-degree terms, so that $f(x+1)$ will be a polynomial in $x^2$. Indeed, we find $f(x+1) = x^4 - 10x^2 + 1 = (x^2)^2 -10(x^2) + 1$. Thus, $f(x+1) = 0$ if and only if
$$x = \pm \sqrt{5 \pm 2 \sqrt{6}}.$$
Equivalently, the roots of $f$ are
$$1 \pm \sqrt{5 \pm 2 \sqrt{6}}$$
$\alpha$ is the root $1 + \sqrt{5 + 2\sqrt{6}} = 1 + \sqrt{2} + \sqrt{3}$. Then we must have $\beta = 1 + \sqrt{5 - 2 \sqrt{6}} = 1 - \sqrt{2} + \sqrt{3}$.

Now we notice that $\mathbb{Q}(\alpha, \beta) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3})$, and also that
$$\sqrt{2} = \frac{\alpha - \beta}{2} \in \mathbb{Q}(\alpha, \beta)$$
$$\sqrt{3} = \frac{\alpha + \beta - 2}{2} \in \mathbb{Q}(\alpha, \beta)$$
Thus, $\mathbb{Q}(\alpha,\beta) = \mathbb{Q}(\sqrt{2},\sqrt{3})$. Now we have
$$[\mathbb{Q}(\alpha,\beta) : \mathbb{Q}] = [\mathbb{Q}(\sqrt{2},\sqrt{3}) : \mathbb{Q}] = [\mathbb{Q}(\sqrt{2},\sqrt{3}) : \mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}) : \mathbb{Q}] \\ = 2 [\mathbb{Q}(\sqrt{2},\sqrt{3}) : \mathbb{Q}(\sqrt{2})],$$
so we only need to compute $[\mathbb{Q}(\sqrt{2},\sqrt{3}) : \mathbb{Q}(\sqrt{2})]$. This is just the degree of the minimal polynomial of $\sqrt{3}$ over $\mathbb{Q}(\sqrt{2})$. Since $\sqrt{3} \notin \mathbb{Q}(\sqrt{2})$ (exercise!), $x^2 - 3$ does not have any roots in $\mathbb{Q}(\sqrt{2})$, so $x^2 - 3$ is the minimal polynomial of $\sqrt{3}$ over $\mathbb{Q}(\sqrt{2})$. Thus, $[\mathbb{Q}(\sqrt{2},\sqrt{3}) : \mathbb{Q}(\sqrt{2})] = 2$, and we get
$$[\mathbb{Q}(\alpha,\beta) : \mathbb{Q}] = 4.$$


问:
Intuitive explanation of the Fundamental Theorem of Linear Algebra
Can someone explain intuitively what the Fundamental Theorem of Linear Algebra states? and why specifically it is called the above? Specifically, what makes it 'Fundamental' in the broad scope of the theory.</p>

答:
Imagine a projection, for example, from the whole $\mathbb{R}^3$ to the $x$-$y$ plane. It compresses each line that parallels to $z$-axis to a point on the plane. So there is a one-to-one relationship between the lines and the points. Notice that the lines are the translations of the $z$-axis -- that is also what quotient means. And the $z$-axis, is just the kernel of the projection, so we can see that $\operatorname{im} A\simeq V/\ker A$.

As for the dimension, the dimension of $\ker A$ measures how much we compress, while the dimension of $\operatorname{im} A$ measures how much we leave -- the amount we compress, plus the amount we leave, equals the whole thing, intuitively.

ContentLicense: CC BY-SA 2.5


答:
With your justification, you get that sup $A$ exists and is $\leqslant 1$, but you don't get the equality.

To get it, you just need to remark that

$$\frac{n}{n+1} \rightarrow 1 \text{ when } n\rightarrow +\infty $$

And since as you said $\frac{n}{n+1}\leqslant 1$ for all $n$, you have that :

$$\sup A = 1$$

Title: Let be $A=\left\{\frac{n}{n+1}: n \in \mathbb{N}\right\} \subset \mathbb{R}$ prove there exists $\sup A$ and find said supreme.

Body:
$A=\left\{\frac{n}{n+1}: n \in \mathbb{N}\right\} \subset \mathbb{R}$

I did:
How $1\geq n$ because $n\in\mathbb{N}$

then $n+1\geq n \rightarrow \frac{n+1}{n}\geq1 \rightarrow \frac{n}{n+1}\leq1$

then $\forall n\in\mathbb{N}  \frac{n}{n+1}\leq1 $ then $\sup A = 1$
is right?


问:
A subspace $Y$ of a topological space $X$ is discrete iff $\exists$ open cover of $X$ in which every open subset contatins exactly one element of $Y$
I have proved the one side. Suppose that there is an open cover $\{U_\alpha\}_{\alpha\in I}$ of $X$ such that $|U_{\alpha}\cap Y|=1$ for all $\alpha\in I$.

Let $y\in Y\subset X$, so there is $\alpha_y\in I$ such that $y\in U_{\alpha_y}$.

Since $|U_{\alpha_y}\cap Y|=1$, we have $\{y\}=U_{\alpha_y}\cap Y$, this implies $\{y\}$ is open in $Y$. Hence, any subset of $Y$ is open in $Y$.

$\therefore Y$ is discrete.

But I can't argue the converse part i.e. if subspace topology of $Y$ is discrete then there is an open cover of $X$ in which every open subset contains exactly one element from $Y$.

I am not sure about the construction of open cover. Can anyone help me in this regard? Thanks for help in advance.

答:
The converse need not hold.

A Counter-Example:

Let $X = \{ a, b, c \}$, $Y = \{ a, b \}$, and let
$$
\mathscr{T}_X = \big\{ \emptyset, \{ a \}, \{ b \}, \{ a, b \}, X \big\}.
$$
Then the subspace topology on $Y$ is clearly the discrete topology. However, we cannot construct an open cover of $X$ in which every set intersects $Y$ in exactly one point.

As a general pattern, let $Y$ be any discrete space, and let $X$ be the set obtained by adjoining an extra element to set $Y$, that is, if $p$ is that element, then
$$
X = Y \cup \{ p \},
$$
and of course $p \not\in Y$.

Now let the topology of $X$ be given by
$$
\mathscr{T}_X = \mathscr{P}(Y) \bigcup \{ X \}.
$$


答: No, it is not correct. First of all, if $n\in\Bbb N$, you have $n\geqslant1$, instead of $1\geqslant n$. And, from the fact that $(\forall n\in\Bbb N):\frac n{n+1}\leqslant1$, you get that $A$ is bounded above and that $\sup A\leqslant1$. But you still have to prove that $\sup A=1$. That's not hard, in fact: if $a<1$, then, by the Archimedian property, there is some $n\in\Bbb N$ such that $n+1>\frac1{1-a}$. But\begin{align}n+1>\frac1{1-a}&\iff 1-a>\frac1{n+1}\\&\iff a<1-\frac1{n+1}=\frac n{n+1}.\end{align}So, $a$ is not an upper bound of $A$ and therefore $1$ is indeed the least upper bound. In other words, $\sup A=1$.


问:
All vectors $(a, b, c, d)\in K^4$ are described by a 2×2-matrix
I am currently taking a course in linear algebra and for the first time in my life I will probably fail a maths test. Since there are about 100 theorems in my linear algebra book and each theorem has an average of 10 rules and some theorems use each other, it gets pretty tough. I decided to try to understand the material instead of memorizing every rule.

When watching 3blue1brown's video entitled "matrix multiplication  as composition | Essence of linear algebra, chapter 4" He describes the following matrix multiplication

$$
\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}e&f\\g&h\end{bmatrix} = \begin{bmatrix}ae+bg&af+bh\\ce+dg&cf+dh\end{bmatrix}
$$

I understand the rule and can memorize it... however it makes no sense to me. How my professor and my book describes matrices is, not exactly but basically, as follows:

Matrices (plural of matrix) are made up of equations. For example:
$$
2x+3y+6z\\
4x+1y+5z
$$

which is the same as:

$$
\begin{bmatrix}2&3&6\\4&1&5\end{bmatrix}
$$

Meaning the first column is the x value for each equation, the second column contains all the y values, and so on.

I would think that only the x values of a matrix could affect the x values of another matrix. However in the first example the first element in the matrix $ae+bg$ is a contradiction to my assumption in that $bg$ is the y value of one equation times the x value of another equation.

I can memorize the rule but I have no way of visualizing it (however 3blue1brown's video did a great job of visualizing what he described it makes not sense. Is the first elements of the first column, going doing the rows, x and y? I thought that the first row was an equation and that x and y is the two elements of the row.) nor a logical explanation as to why we derived this method. Can anyone guide me further?

P.S
English isn't my first language so if I drifted of in a nonsensical way somewhere, please comment or propose an edit and I'll change it as soon as I can!

Edit:
I am getting down votes, how can I improve my question? The ways I can improve it by myself is limited because I have limited knowledge.

答:
Instead of thinking of the matrix as just the expressions $2x+3y+6z$ and $4x+y+5z$, think of it as the map/function
$$ \begin{pmatrix}x\\y\\z\end{pmatrix} \mapsto \binom{2x+3y+6z}{4x+y+5z} $$
Then matrix multiplication is composition of these maps.
If you first do $\begin{pmatrix} e&f\\g&h\end{pmatrix}$, you get
$$ \binom{x}{y} \mapsto \binom{ex+fy}{gx+hy} = \binom{x'}{y'} $$
Now apply the second matrix to $x'$ and $y'$:
$$ \binom{x'}{y'} \mapsto \binom{ax'+by'}{cx'+dy'} $$
Substitute in the expressions $x'=ex+fy$ and $y'=gx+hy$ to get $$\binom{a(ex+fy)+b(gx+hy)}{c(ex+fy)+d(gx+hy)} = \binom{(ae+bg)x+(af+bh)y}{(ce+dg)x+(cf+dh)y} $$
Notice that this is the same as applying the product matrix
$$ \begin{pmatrix}ae+bg & af+bh \\ ce+dg & cf+dh \end{pmatrix} $$
directly to $\binom{x}{y}$.


问:
Convergence of Bernoulli numbers infinite sum
In https://en.wikipedia.org/wiki/Harmonic_number I found that harmonic numbers admit asymptotic expansion as:

$$H_n \approx \ln n + \gamma_0 + \frac{1}{2n} - \sum_{k=1}^{\infty}{\frac{B_{2k}}{2kn^{2k}}}$$

but for a certain n, when trying to get some values on infinite sum on RHS it appears to diverge always, even if I try Cesaro sums pairing consecutive terms.

What is the real meaning of that infinite sum and how can I evaluate it?

Additionaly I am interested in some other infinite sums (in same depicted expression) that could formally converge without any exotic sumation argued.

答:
This is an example of an asymptotic power series. Its meaning is that for each $N\geq 1$, there exist $A_N>0$ and $C_N>0$ such that
$$
H_n  = \log n + \gamma  + \frac{1}{{2n}} - \sum\limits_{k = 1}^{N - 1} {\frac{{B_{2k} }}{{2k}}\frac{1}{{n^{2k} }}}  + R_N (n),
$$
where
$$
\left| {R_N (n)} \right| \le \frac{{C_N }}{{n^{2N} }},
$$
whenever $n\geq A_N$. In particular, $R_N (n)=\mathcal{O}\!\left( {\frac{1}{{n^{2N} }}} \right)$ as $n\to +\infty$. In this specific example, more can be said: for any $n\geq 1$ and $N\geq 1$,
$$
\left| {R_N (n)} \right| \le \frac{{|B_{2N}| }}{{2N}}\frac{1}{{n^{2N} }}.
$$
Note that
$$
\left| {B_{2k} } \right| = \frac{{2(2k)!}}{{(2\pi )^{2k} }}\zeta (2k) > \frac{{2(2k)!}}{{(2\pi )^{2k} }},
$$
so the terms of the series do not tend to $0$ for any $n\geq 1$ as $k\to +\infty$, i.e., the series cannot converge. Note however, that for each fixed $n$, the terms of the series initially decrease in absolute value and reach a minimum around $k \approx \pi n$. If the series is truncated at this optimal point, i.e., $N = \left\lfloor {\pi n} \right\rfloor$, then the error term satisfies, using Stirling's approximation for the factorial,
$$
\left| {R_N (n)} \right| \le \frac{{\left| {B_{2N} } \right|}}{{2N}}\frac{1}{{n^{2N} }} = \frac{{(2N)!}}{{(2\pi n)^{2N} N}}\zeta (2N) = \mathcal{O}\!\left( {\frac{{e^{ - 2\pi n} }}{{\sqrt n }}} \right).
$$
This is called superasymptotics: the error is exponentially small. It is possible to re-expand the remainder into a new series which will improve the accuracy beyond superasymptotics. These are called hyperasymptotic expansions. In this particular example, the re-expansion will be a convergent series in terms of incomplete gamma functions which "terminates" the divergent asymptotic series. For a similar example of an asymptotic series like this, consider the Stirling series for $\log n!$.


问:
Prove that $a ( n ) = b( n + 2)$

let $a(n)$ denotes the number of ways of expressing the positive integer $n$ as an ordered sum of 1's and 2's. Let $b(n)$ denote the number of ways of expressing n as an ordered sum of integers greater than 1. prove that  $a(n) = b(n+2)$. for $n=1,2,3...$

My approach:
$a(1) = 1$ (only) , $b(3) = 3$

$a(2) = 2$, because $2=1+1=2$ ,and  $b(4)=2$, because $4=2+2=4$

$a(3) = 3$, because $3=1+1+1=1+2=2+1$ and, $b(5) = 3$ , because $5=2+3=3+2=5$

$a(4)= 5$, because $4=1+1+1+1=2+2=1+1+2=1+2+1=2+1+1$ and, $b(6)=5$, because $6=3+3=2+4=4+2=2+2+2=6$

$a(5)=8$, because $5=1+1+1+1+1=2+1+1+1=1+2+1+1=1+1+2+1=1+1+1+2=2+2+1=2+1+2=1+2+2$ and, $b(7)=8$, because
$7=3+2+2=2+3+2=2+2+3=3+4=4+3=2+5=5+2=7$

By this way i am able to show that  $a(n)=b(n+2)$. but is there any general method for this problem. I mean any recursion relation which  i can  understand.

Background:-This problem is from pathfinder for Olympiad mathematics.

答:
It is also possible to exhibit a bijection between compositions of $n$ using only $1$ and $2$ and compositions of $n+2$ that do not use $1$.

Suppose that $n=c_1+c_2+\ldots+c_k$, where $c_i\in\{1,2\}$ for $i=1,\ldots,k$. Let $c_0=2$; then $n+2=c_0+c_1+\ldots+c_k$. Let $i_1<\ldots<i_\ell$ be the set of indices $i$ such that $c_i=2$, and let $i_{\ell+1}=k+1$. For $j=1,\ldots,\ell$ let

$$d_j=\sum_{i=i_j}^{i_{j+1}-1}c_i\;;$$

then $n+2=d_1+\ldots+d_\ell$ is a composition of $n+2$ that does not use $1$.

Example: The composition $1+2+2+1+1+2$ of $9$ is first extended to the sum $2+1+\color{blue}2+\color{red}{2+1+1}+\color{green}2$, which is then reduced to the composition $3+\color{blue}2+\color{red}4+\color{green}2$ of $11$.

This mapping is clearly injective, and it’s not hard to see that it is surjective as well. Let $n+2=d_1+\ldots+d_\ell$ be a composition of $n+2$. Replace each $d_i$ by its composition

$$2+\underbrace{1+\ldots+1}_{d_i-2}$$

and drop the leading $2$ to get the composition

$$n=\underbrace{1+\ldots+1}_{d_1-2}+2+\underbrace{1+\ldots+1}_{d_2-2}+\ldots+2+\underbrace{1+\ldots+1}_{d_\ell-2}\;;$$

applying the original procedure to this composition of $n$ recovers the composition $n+2=d_1+\ldots+d_\ell$ of $n+2$.

Title: Prove that $a ( n ) = b( n + 2)$


问:
can y1 and y1+(t,t) be solutions for ODE
the question is:

a. can $ y_1$ and$y_1 + (t,t)^{T}$ be solutions for ODE of the form $ \dot {\overrightarrow y}=A(t)\overrightarrow y + \overrightarrow b(t) $
where A(t) is a continous matrix and $\overrightarrow b(t)$ is a continious vector.

b. same question about $\overrightarrow y_1, 2\overrightarrow y_1$

I have more questions of the same but I think understanding the basic concept will be enough for me to solve it.

I thought about showing that the wronskian is equal to zero but I'm not sure that's the way to do it...

Help would be appreciated.

答:
Denote dimension of $\vec{y}$ by $d$.

a) If $y_1$ is  solution then it is a vector. But if we have sum $y_1 + (t,t)^{T}$ then $y_1$ in 1-dimensional since $(t,t)^{T}$ is 1-dimensional.

Hence for all $d \ge 2$ the answer is negative as $y_1 + (t,t)^{T}$ doesn't have sence.

Now consider the case $d=1$ and suppose that $y_1$ and $y_1 + t^2$ are solutions. Thus
${y'}_1 = A(t) y_1 + b(t)$ and ${y'}_1 + ( t^2)' = A(t) (y_1 + t^2) + b(t)$. It follows that $2t = A(t) t^2$. Thus $A(t) = \frac2{t}$.

For example, if $t \in [1,2]$ we may take $b = 0$, $A(t) = \frac2{t}$, then $y(t) = t^2 + const$, so the answer is positive.

If $t \in \mathbb{R}$ then the answer is negative since $A(t) = \frac2{t}$.

Maybe by $y_1$ you meant the first coordinate of $\vec{y}$, which is $d$-dimensional.

In this case the answer is: "sometimes yes, but in general no", as it was shown in the previous example in case $d=1$.

b) The answer is: "sometimes yes, but in general no".

Suppose that $y_1$ and $2 y_1$ are solutions. Put $d=1$ for simplicity. Thus
$y'_1 = Ay_1 + b$ and $2y'_1 = 2Ay_1 + b$. Thus $b = y'_1 - Ay_1 = \frac{b}2$ and $b=0$.

So in case $d=1$ your statement is true iff $b = 0$ for all $t$. Thus the answer to b) is: "sometimes yes, but in general no".


问:
Find the maximum and minimum value of $2x^2 + y^2 + z^2$ subject to $x + y + z = 10$.
I am working on inequality problems for mathematical olympiads. I have come across this problem:

Find the maximum and minimum value of $2x^2 + y^2 + z^2$ subject to $x + y + z = 10$, with $x,y,z \ge 0 $

I found the minimum using Cauchy-Schwarz: $$100 = \left(\frac 1 {\sqrt 2} \left(\sqrt 2 x\right) + y + z\right)^2 \le \left(\left(1/\sqrt 2\right)^2 + 1^2 + 1^2\right) \left(\left(\sqrt 2 x\right)^2 + y^2 + z^2\right)=2.5\left(2x^2 + y^2 + z^2\right)$$
so $2x^2 + y^2 + z^2 \ge 40$ with equality if $x=2$ and $y=z=4$.

I am not sure how to find the maximum though. Could I have a hint? Ideally I would like to do it without using calculus, though if it is easy to do with calculus feel free to share the method.

答:
Substitute the value of $z$ in the expression: $$2x^2 +y^2 +(x+y-10)^2 $$ As increasing $x$ and $y$ increases the value of this expression, the maximum occurs when $x+y$ is maximum, i.e. $x+y=10$: $$2x^2 +(10-x)^2 = 3x^2 -20x +100$$

This is a parabola which attains its maximum value on $[0,10]$ at $x=10$, which is $200$.


问:
Question about proof in problem with a sequence
I have a sequence which is defined as $\sqrt{a_n}$ if $a_n$ is a square number and $a_n+2$ otherwise. I am told that $a_1>1$ and I have to prove that the sequence ends with $3,5,7,9$ if the first term is odd and with $2,4$ if it is odd.

Anyways I did try to prove it but I think it only applies if $a_1=2$ in the case of even and $a_1=3$ and okay in a couple of other cases but I don't think it applies for any random $a_1>1$.

So my question is, is there any rule or theorem that makes me have  $a_1=2$ if I know it is even and $a_1=3$ in the case of odd?

Maybe it's trivial or maybe I am missing something but I would be really thankful for some help.

答:
We take out all perfect squares from the sequence and form a new sequence $\{b_n\}$ and prove the following:

If $a_1$ is odd, thus all $b_n$'s are odd, then $b_n \ge 25 \implies b_{n+1} < b_n$.

Assume $b_n=(2m+1)^2, m \ge 2$. If $2m+1$ is a perfect square then $b_{n+1}=2m+1 < b_n$ we are done. If not, assume $(2k-1)^2 < 2m+1 < (2k+1)^2 \implies b_{n+1} = (2k+1)^2$, we now prove that $b_{n+1} = (2k+1) < (2m+1)^2=b_n$, or equivalently, $k<m$. If not, then $k\ge m$,

$$2m+1 > (2k-1)^2 \ge (2m-1)^2 \\
\implies 0 > (2m-1)^2-(2m+1) = 2m(2m-3) > 0$$ a contradiction.

If $a_1$ is even, thus all $b_n$'s are even, then $b_n \ge 16 \implies b_{n+1} < b_n$.

Assume $b_n = (2m)^2, m\ge 2$. If $2m$ is a perfect square then $b_{n+1} = 2m < b_n$ we are done. If not, assume $(2k-2)^2 < 2m < (2k)^2 \implies b_{n+1} = (2k)^2$, we now prove that $b_{n+1} = (2k)^2 < (2m)^2=b_n$, or equivalently, $k<m$. If not, then $k \ge m$,

$$2m > (2k-2)^2 \ge (2m-2)^2 \implies 0 > (2m-2)^2-2m \\
\implies 0 \ge (2m-2)^2-2m+2 = 2(m-1)(2m-3) > 0$$ another contradiction (note that $0> 2A \implies 0\ge 2A+2$, where $2A=(2m-2)^2-2m$ is even).

Therefore if $b_n$ is greater than $9$ the next perfect square is always smaller. Only when $b_n=4$ or $9$ do we get a loop, which is either $4\to 2 \to 4$ or $9 \to 3 \to 5 \to 7 \to 9$.


问:
Show that $\lim \frac1{n^2 +1} =0$
I am not very familiar to epsilon... But this question belongs to my Sallybus and I want to understand it with the help of epsilon definition.. Kindly help me to solve it</p>

答:
In the first place, it is worth to emphasize the definition of convergence of sequences.

Let $x_{n}\in X$, where $(X,d)$ is a metric space. We say that $x_{n}$ converges to $x_{0}\in X$ iff
\begin{align*}
(\forall\varepsilon > 0)(\exists n_{\varepsilon}\in\mathbb{N})\,\,\text{s.t.}\,\,(n\geq n_{\varepsilon} \Rightarrow d(x_{n},x_{0}) < \varepsilon)
\end{align*}

At the present case, $(X,d) = (\mathbb{R},d)$, $x_{n} = (n^{2}+1)^{-1}$ and $x_{0} = 0$, where $d(x,y) = |x - y|$.

Having said that, we can proceed.

To begin with, I would start with noticing that

\begin{align*}
\frac{1}{n^{2}+1} < \frac{1}{n^{2}} < \varepsilon \Rightarrow n > \frac{1}{\sqrt{\varepsilon}}
\end{align*}

Thus, for every $\varepsilon > 0$, there corresponds a natural number $n_{\varepsilon} = \Bigl\lceil\dfrac{1}{\sqrt{\varepsilon}}\Bigr\rceil$ s.t.
\begin{align*}
n\geq n_{\varepsilon} \Rightarrow \frac{1}{n^{2}+1} < \varepsilon
\end{align*}

Hopefully this helps!

Title: Show that $\lim \frac1{n^2 +1} =0$


答: Let $\varepsilon > 0$ be the goal to approach. We want to find an integer $N>0$ such that whenever $n > N$, we have $\frac{1}{n^2+1} < \varepsilon$. As we know $\frac{1}{n^2+1} < \frac{1}{n}$, let's pick $N$ be the smallest integer greater than $1/\varepsilon$. Then whenever $n>N$, we have $n>N\geq1/\varepsilon$ and $\frac{1}{n^2+1} < \frac{1}{n}<\varepsilon$. That's exactly what we want to prove.

Title: Show that $\lim \frac1{n^2 +1} =0$

Body: I am not very familiar to epsilon... But this question belongs to my Sallybus and I want to understand it with the help of epsilon definition.. Kindly help me to solve it</p>


问:
For any smooth manifold there exists a smooth atlas whose coordinate charts are homeomorphisms to open balls.
I'm trying to show that on any smooth n-manifold $M$, there exists a smooth atlas $ \{ ( U_{\alpha},\varphi_{\alpha})\} $ such that $ \varphi_{\alpha}(U_{\alpha}) $ is an open ball in $\mathbb{R^{n}}$ for all $\alpha$.

In the definition we used for a topological n-manifold $(M,T)$ one condition was that each $x \in M $ has a neighborhood $U \in T$ such that $U$ is homeomorphic to some open subset $V$ of $\mathbb{R^{n}}$. I did exercise 1.1 in Lee as discussed here which asks to show that this condition is equivalent to each point having a neighborhood homeomorphic to either $\mathbb{R^{n}}$ or an open ball in $\mathbb{R^{n}}$. I was thinking maybe this could be used to show that every $U \in T$ is homeomorphic to $\mathbb{R^{n}}$ and then every atlas would have the required property, but that feels wrong. Is this fact useful in obtaining my desired result?
Or is it necessary to make some use of the smoothness criteria?

答:
Since every smooth manifold is a topological manifold, then I would say that it is always possible to choose such an atlas.

For every $p \in M$, where $M$ is a differentiable real manifold, up to restricting the open set on the manifold, it is always possible to make the charts homeomorphic to an open ball in $\mathbb{R}^n$ and the $\mathcal{C}^{\infty}$-compatibility still holds on the restriction of the transition functions.


答:
For a calculus-based method that delivers both solutions, you want a Lagrange multiplier. Let: \begin{align*}f\left(x,y,z\right) &= 2x^2 + y^2 + z^2 \\ g\left(x,y,z\right) &= x + y + z - 10\end{align*} The condition for an extremum is $$\vec{\nabla} f\left(x,y,z\right) = \lambda \vec{\nabla} g\left(x,y,z\right)$$ Calculating this out, we have a system with four equations and four unknowns: \begin{align*}4x &= \lambda \\ 2y &= \lambda \\ 2z &= \lambda \\ 0 &= x+y+z-10\end{align*}

Solving for $\lambda$ first is easiest: $$0 = \frac{\lambda}{4} + \frac{\lambda}{2} + \frac{\lambda}{2} - 10 \hspace{2.00cm} \rightarrow \hspace{2.00cm} \lambda = 8$$ With $\lambda=8$ figured out, we know one extreme point to be: $$x = 2  \hspace{2.54cm} y = 4 \hspace{2.54cm} z = 4$$

So what about the maximum? The best hint, which undoubtedly drives the other solutions to this, is the factor of $2$ in front of $x^2$, meaning this term grows faster than its counterparts. It only makes sense to try $y=z=0$, so that $x=10$ is evident from inspection. This makes the whole $\lambda$-apparatus somewhat useless, but at least the form is there in case the problem was trickier than it looked.

Title: Find the maximum and minimum value of $2x^2 + y^2 + z^2$ subject to $x + y + z = 10$.

Body:
I am working on inequality problems for mathematical olympiads. I have come across this problem:

Find the maximum and minimum value of $2x^2 + y^2 + z^2$ subject to $x + y + z = 10$, with $x,y,z \ge 0 $

I found the minimum using Cauchy-Schwarz: $$100 = \left(\frac 1 {\sqrt 2} \left(\sqrt 2 x\right) + y + z\right)^2 \le \left(\left(1/\sqrt 2\right)^2 + 1^2 + 1^2\right) \left(\left(\sqrt 2 x\right)^2 + y^2 + z^2\right)=2.5\left(2x^2 + y^2 + z^2\right)$$
so $2x^2 + y^2 + z^2 \ge 40$ with equality if $x=2$ and $y=z=4$.

I am not sure how to find the maximum though. Could I have a hint? Ideally I would like to do it without using calculus, though if it is easy to do with calculus feel free to share the method.


问:
Computing $ \int_{-\infty}^{\infty} e^{-ax^4+bx^2} dx$
I was wondering if it was possible to compute $ \int_{-\infty}^{\infty} e^{-ax^4+bx^2} dx$  explicitly.

Completing the square doesn't seem to work, since there is no factor of $x$ in order to use the Gaussian integral.

Any hint would be greatly appreciated.

Thanks!

答:
With the change of integration variable $x = (2a)^{ - 1/4} \sqrt t$ and the representation http://dlmf.nist.gov/12.5.E1, we find
\begin{align*}
\int_{ - \infty }^{ + \infty } {e^{ - ax^4  + bx^2 } dx} & = 2\int_0^{ + \infty } {e^{ - ax^4  + bx^2 } dx}  = (2a)^{ - 1/4} \int_0^{ + \infty } {t^{ - 1/2} e^{ - \frac{1}{2}t^2  + \frac{b}{{\sqrt {2a} }}t} dt}
\\ & = (2a)^{ - 1/4} \sqrt \pi  e^{\frac{{b^2 }}{{8a}}} U\!\left( {0, - \tfrac{b}{{\sqrt {2a} }}} \right).
\end{align*}
Here $U$ is the parabolic cylinder function. By analytic continuation, this formula is valid whenever $\Re a>0$ and $b$ is any complex number. By http://dlmf.nist.gov/12.7.E10, this is also equal to
$$
\frac{1}{2}\sqrt { - \tfrac{b}{a}} e^{\frac{{b^2 }}{{8a}}} K_{1/4}\! \left( {\tfrac{{(-b)^2 }}{{8a}}} \right),
$$
where $K_\nu$ is the modified Bessel function. One has to choose the phase of $-b$ consistently in this form to make the function single-valued (and real valued for real $b$).


问:
Is there a Ramsey's theorem equivalent for vertex colorings?
Ramsey's theorem states that any 2-color edge coloring of a sufficiently large graph contains a monochromatic complete subgraph. Does there exist something like Ramsey's theorem but for vertex colorings? Like, for any 2-color vertex coloring of a sufficiently large graph contains a monochromatic complete subgraph (but this time with the vertices colored and not the edges). Or is there someway to translate the vertex problem in an edge coloring problem such that we can use Ramsey's theorem?</p>

答:
Questions of Ramsey theory often look something like this:

We have a set with $n$ elements

We assign colors to its $k$-sets (that is, its $\binom n k$ subsets of size $k$).

Must there be a $w$-set whose $k$-sets are all one color?

Phrased in this way, the typical example, coloring the edges of $K_6$, looks like this:

We have a set with $6$ elements

We assign one of two colors to each of its $2$-sets, (there are $\binom 62=15$ of these)

Is there a $3$-set whose $2$-sets are all the same color?

The answer is that there must be, but if we replace $n=6$ in the first line with $n=5$, there need not be.  (The essentially unique counterexample is
$\{
\color{darkred}{\{1, 2\}},
\color{darkblue}{\{1, 3\}},
\color{darkblue}{\{1, 4\}},
\color{darkred}{\{1, 5\}},
\color{darkred}{\{2, 3\}},
\color{darkblue}{\{2, 4\}},
\color{darkblue}{\{2, 5\}},
\color{darkred}{\{3, 4\}},
\color{darkblue}{\{3, 5\}},
\color{darkred}{\{4, 5\}}
\}$.)

Note that we've scrubbed out the graph terminology completely.  Graph edges have become $2$-sets, but nothing has really changed.  Graph edges can be completely identified with $2$-sets of vertices; there's no difference.

If you're asking about graph vertices instead of edges, you've changed the $2$-sets to $1$-sets.  But $1$-sets, that is subsets with only 1 element,  are no different from elements themselves.  So the formulation is:

We have a set with $n$ elements

We assign one of $r$ colors to each of its… elements

Is there a $k$-set whose… elements… are all the same color?

And the answer is pretty obvious.  (As N.S. and Qiaochu pointed out, it's essentially the pigeonhole principle.  Which, I should add, is usually regarded as a simple example of a Ramsey-type theorem.)

But you can go in the other direction too. Instead of looking at $1$-sets, you can look at $3$-sets:

We have a set with $n$ elements

We assign one of several colors to each of its 3-sets (there are $\binom n3$ of these)

Is there a $k$-set whose 3-sets are all the same color?

Here we're coloring the “triangles” of a complete graph, and asking if there must be a smaller complete graph whose “triangles” are all the same color.  As with the $2$-set case, the answer is yes, if $n$ is sufficiently large.  And in fact the general theorem is true:

Let $k, s, r$ be given.  Take a set of $n$ elements, and assign $r$ colors to each of its $s$-sets, then there must always be a $k$-element subset whose $s$-sets have all been assigned the same color, if $n$ is sufficiently large.

Taking the special case $s=2$ gives us the usual Ramsey graph theorem; taking $s=1$ gives us the pigeonhole principle.

The keyword to search for here is “hypergraph Ramsey numbers”.


问:
Separate result of multiplication using Sigma
Let's suppose I have two numbers $a=a_{N/2}...a_1$ and $b={b_N/2}...b_1$ where $a_1, b_1$ etc are single digits of the number (So each number has $N/2$ digits).

How can I write the result of $a*b$ using sigmas such that I want to separate each column.

What I mean?

When we do $a_3a_2a_1*b_3b_2b_1$ then first column is $b_1*a_1+0+0$ second columns is $a_2*b_1+a_1*b_2+0$ and so on.

In other words I want to seperate the parts which are multiplied by 10^0, 10^1, 10^2 ...

Body:
You can write

$$a= \sum_{k=1}^{N/2}a_k10^{k-1} \\ b= \sum_{k=1}^{N/2}b_k10^{k-1}$$

So

$$ a*b= \sum_{k=1}^{N/2}\sum_{j=1}^{N/2}a_kb_j10^{k+j-1} $$

Or

$$a *b= \sum_{i=1}^n ( \sum_{k+j=i}a_kb_j) 10^{i-1} $$

I hope it helps you :)


问:
Is $x=2,y=13$ the unique solution?
Problem:
Find all positive integers $x$ and $y$ satisfying:
$$12x^4-6x^2+1=y^2.$$

If $x=1, 12x^4-6x^2+1=12-6+1=7,$ which is not a perfect square.

If $x=2, 12x^4-6x^2+1=192-24+1=169=13^2$, which is a perfect square. Thus, $x=2,y=13$ is a solution to the given Diophantine equation.

However, after testing a few more small cases, it seems as if $12x^4-6x^2+1$ can never be a perfect square if $x>2$. I have tried to prove this, but to no avail. Here is the gist of what I considered:

$12x^4-6x^2+1=y^2 \iff 12x^4-6x^2 = (y+1)(y-1)$. Since the L.H.S. is a multiple of $2$, it follows that $2 \mid (y+1)(y-1) \Rightarrow y$ is odd $\Rightarrow y+1$ and $y-1$ are both even.

Hence, $4 \mid (y+1)(y-1) \Rightarrow 4 \mid  12x^4-6x^2 \Rightarrow 4 \mid 6x^2 \Rightarrow x $ is even. Let $x=2m$ and $y+1=2k$, so $12x^4-6x^2=192m^4-24m^2=4k(k-1) \Rightarrow 48m^4-6m^2=k(k-1)$. At this point, I am at a loss of how to continue. We could continue with divisibility arguments, but it seems to be a never-ending process?

Another method I tried was to let $y=x+k$, thus $12x^4-6x^2+1 = x^2 + 2xk+k^2 \iff k^2+2xk-(12x^4-7x^2+1)=0$, which is a quadratic in terms of $k$. However, stuff like the discriminant or sum and product of roots did not seem to yield any important information.

Any hints provided to point me in the right direction will be much appreciated.

答:
We try to get a perfect square term on the LHS.

Note that multiplying both sides by 4 keeps the right side as a square:
$$48x^4 - 24x^2 + 4 = (2y)^2$$
Now, completing the square yields: $$3(4x^2 -1)^2 +1 = (2y)^2$$
Try to now find the solutions to $3(2x-1)^2(2x+1)^2 = (2y-1)(2y+1).$

Title: Is $x=2,y=13$ the unique solution?


问:
Is learning linear algebra fundamentally different from other mathematical fields and how do you actually learn it?
I am currently taking a linear algebra course and I have an issue I have never ever experienced before. I understand nothing. Nada. I have never been a genius, but I learn fast. I have never struggled with mathematics before and I just scored near perfect on the mathematical analysis test last week. However next week we have a linear algebra test and my professor has posted the old linear algebra tests, along with solutions, for us students to test ourselves. I got a clean 9 points out of 70 points total.

There are 800+ pages in our linear algebra book and we've been given 10 weeks to study, along with the professor doing classes on Zoom. My issue is with the fact that there are over 150+ theorems and each theorem have an average of 10 rules + exceptions. This means that in 10 weeks not only do we have to memorize about 150*(10 rules + exceptions) which is approximately 1500+ unique things but we also have to understand them to actually get a grade I hope one day achieve.

So my three questions are:

(1) Why does everything have multiple names?

(2) What might be a good way to not confuse these names together?

(3) What makes linear algebra so fundamentally different from other mathematical courses and where do I go to learn these new attributes?

In the book they talk about unit vectors, but the question is about standard basis vectors and the professor is talking about the natural basis (and they mix around unit, basis and natural all the time). I always have to have a table with synonyms or words that at times is something that I already know but is presented in such a way that I get thrown off. This just makes things more tedious because then I learn something only to later learn that one word only means "basis vector" if such and such criteria are met, otherwise they refer to this and that.

I can solve most question in the book easily, but I don't understand why I calculate, I only go after the book and my professors examples. I have gone through things over and over and I just keep getting worse at linear algebra.

答:
First let me say that I had a somewhat similar experience at one point. I had always done great in math, but at one point I was learning linear programming, in which there's a fundamental algorithm called the simplex method. And I didn't get it. Not even close. I had no idea what I was supposed to be doing nor how the calculation gave me a solution to a problem. Fortunately a good friend was also learning the subject and spent a lot of patient time taking me through it multiple times until I slowly but surely got there.

The point I want to make is: I know what it's like to suddenly experience, really for the first time, absolutely not getting something mathematical, after a lifetime-thus-far of always getting mathematics quickly. It freaked me out, it was unsettling and frightening. But today I am a professional mathematician, so clearly I managed to get through it! I couldn't perform the simplex method today, but I remember the gist of it and what it accomplishes, and I remember that it really is a mechanical procedure, so I'm confident that I could learn it again in a day or a week if I needed to.

So I hope you will have faith that in the future, you too will look back at how you feel now and say "Wow, that was really unsettling and scary, but I got through it and actually understood a reasonable amount of it, and the new math I'm encountering is making sense, so things are okay in the end."

As for how to maybe get from here to there: I would absolutely recommend coming at the subject seeking understanding rather than memorization. Of course those things can't be 100% separated and reinforce each other, but to me what you're describing is a desire for greater understanding more than a desire for better memorization. Yes, linear algebra and abstract algebra feel, and are, much different sorts of subject than calculus and analysis, and there absolutely is a precision to definitions of individual words that we have to come to grips with. But to me, the best way to come to grips with all these details is to see the overall picture.

For example, I'm not sure what you mean when you say that each theorem has 10 rules and exceptions. A theorem, by definition, doesn't have exceptions; it might have conditions under which a conclusion is valid ("if the matrix has nonzero determinant", for example), but the implication itself that is the content of a theorem always holds true. It is, however, true that the assumptions of linear algebra theorems do tend to be more exacting than the assumptions of calculus theorems: instead of things like "as long as the function is differentiable" (which makes sense for statements about derivatives and is true for most functions anyway, to the point where we forget it's a necessary assumption), we get things like "if a square matrix is nonsingular/is invertible/has nonzero determinant" (yep, equivalent ways of saying the same thing—and that fact is itself a theorem!), or "if the set of vectors is linearly independent" or "is a basis" or "spans the vector space", or "if the rank of the matrix is $r$"—and all of these assumptions are necessary and don't hold as generically as "differentiable".

So, is linear algebra fundamentally different? I would say, "yes, there are enough differences to notice a different feel", but also "no, linear algebra still consists of mechanical procedures, and ways to use them to solve certain kinds of problems, and statements that situation A always implies conclusion B, and leveraging old facts and relationships among mathematical objects to prove new facts and relationships".

As for concrete advice: I would suggest going back to the beginning (I know that can be hard when things are time-sensitive) and getting a more comfortable mastery with smaller pieces of the course, maybe one chapter or lecture-week at a time. Go over the material again in those smaller chunks, and ask yourself dispassionately which parts of it you get pretty well, which parts of it you feel sort of close to getting, and which parts you feel really unsettled by; then concentrate on the more confusing parts and really find a more solid understanding before trying to move on to the next chunk. When so much of a statement's meaning is tied to the exact meaning of the individual mathematical words, we don't want to build higher on a shaky foundation ... so okay, let's be willing to go back to the foundation itself. It's worth the investment of time and energy and should hopefully lead to a more positive experience in the later material.

After a quick search, I can suggest Chapter 1 of this book as a reasonable attempt to organize the main concepts of linear algebra, together with a lot of its terminology. It might be that you find this source (or some other resource) makes more sense to you.

But finally let me re-emphasize my first point: this might well be the first time you've ever felt defeated by mathematics. It's okay to feel that way!—not pleasant, sure, but it doesn't mean that you're incapable of getting through it. Feel the unpleasant feeling, take a deep breath, shrug and say oh well, I'll just be methodical about this particular subject. And just like everything else in mathematics (and life), the more we practice, the better we get; your challenge at this moment is to find ways of practicing linear algebra that work for you.


答:
Option:

Let $n \ge1;$

$\dfrac{1}{n^2+1} \le\dfrac{1}{n^2} \le \dfrac{1}{n};$

Let $\epsilon >0$ be given.

Archimedian principle:

There exists a $n_0 \in \mathbb{N}$ s. t.

$n_0 >1/\epsilon;$

For $n\ge n_0:$

$\dfrac{1}{n^2+1}\le \dfrac{1}{n}<\dfrac{1}{n_0} <\epsilon. $

Title: Show that $\lim \frac1{n^2 +1} =0$


问:
Can topology be studied without $\mathbb{R}$?
My friend is a programmer who works with countable math objects (like countable graphs, or groups). He often reads the professional literature where a lot of topological concepts are mentioned. That's why he's trying to study topology on his own. Alas, in those textbooks, there are a lot of examples concerning $\mathbb{R}$. Thus, since he usually works with countable objects, he starts thinking that not much of the concepts would be helpful. Can he omit the examples with $\mathbb{R}$ without loss of generality? Are there general or algebraic topology textbooks where all such examples are omitted, i.e. they are concentrated on finitistic objects?

答:
Yes, it is possible, and very useful to build an intuition of topology without $\mathbb R$. It is possible to view topology as capturing data about semi-decidability. If your friend knows Haskell (it's a very math-influenced purely functional programming language), ask them to read Martín Escardó's so-called Barbados notes, number 46 in https://www.cs.bham.ac.uk/~mhe/papers/index.html, AKA "Synthetic topology of data types and classical spaces".

The general idea is that the open sets in the topology are those properties which we can decide are true in finite time, but cannot decide if they are false in finite time.

To expand, the topology on $\mathbb R$ allows us to semi-decide if a point $p$ belongs to some interval $I$ when we are able to access the point $p$ one digit at a time. For example, let's pick $I = (0, 1)$.

Now, if we see a point $p = 0.1\dots$ we know immediately that this point is in $I$, so we can answer that $p \in_? I = \texttt{TRUE}$.

If we see a point $p = 2.\dots$, we know immediately that this point $p$ is not in $I$, so we can answer $p \in_? I = \texttt{FALSE}$

Now consider the point $p = 0.9999\dots$ When do we decide that this point $p$ does not belong to $I$?

we can't decide after seeing two digits $(0.9\dots)$, for the point $p$ could be $0.98$ and it would then belong to $I$

we can't decide after seeing  three digits ($0.99\dots)$, for the point $p$ could be $0.998$ and it would then belong to $I$

in general, we would need to read the entire number $0.999\dots$ to decide that $p \not \in I$. Clearly, this takes infinite time!

So, if $p \in I$, then we will always answer $p \in_? I = \texttt{TRUE}$ in finite time. But if $p \not \in I$, then we may not answer $p \in_? I = \texttt{FALSE}$ in finite time. Hence, the relation $\in_?$ is semi-dedicable, and the topology of $\mathbb R$ precisely captures the semi-decidable nature of membership in an interval.

More broadly speaking, discrete topology is vital to computer science, since finite topology is the study of preorders and partial orders. This is precisely the notion that is necessary to assign mathematical meaning to computer programs, through a branch of mathematics called as denotational semantics

I've written some blog posts about these issues that may be of some help to get started, since I collect links as well:

Topology is really about computation

Finite topologies and DFS numbers

Body:
Yes, it is possible, and very useful to build an intuition of topology without $\mathbb R$. It is possible to view topology as capturing data about semi-decidability. If your friend knows Haskell (it's a very math-influenced purely functional programming language), ask them to read Martín Escardó's so-called Barbados notes, number 46 in https://www.cs.bham.ac.uk/~mhe/papers/index.html, AKA "<a href="https://pdf.sciencedirectassets.com/272990/1-s2.0-S1571066104X00177/1-s2.0-S1571066104051357/main.pdf?X-Amz-Security-Token=IQoJb3JpZ2luX2VjEHwaCXVzLWVhc3QtMSJHMEUCIQCUHPaMhXIxR5ntki261HQ0rfhfVOSKmrZnxQxCu8EbLAIgaUrO%2B1NOieevNSJgqIYxGmF7aXB7G5WDxBHhLuIfRtkqtAMINBADGgwwNTkwMDM1NDY4NjUiDLXEqlNcrvWx8rgxqCqRA5WHSCrlPlhL8rxOK2qpHmVb0zZkNi7EyP%2B4WI8mXy5u8khz%2BkpXVCmjBj3ZCFbrZfsvl%2FecroIhcYDVqQSl7we4AVyXwW3Zb79Go%2F8Nz1%2FtIJOK0GELEKGbI%2FM4rZwRvVtJBsUn64xjP6lHBrLZOXAkC%2ByXlTZDlTRxfAlHYZURRLcGv8h8o%2BYWXb2d0eOIsMx5yMyXxM%2FZ5aUjcxoXaj3N5hNPo5bjni1aHfLEfEDRZlxbdszxRLVuzvbmTxAnychD7O202VANV3vr8JRdDcNTCmmrxl3yUYvmYrnmiFSM0u4C1bwtAyy4cYM%2FJNDpGLyYTGdxHDwVALblPBWVZxBN2VAFcCohCxCZjNXyFQ2nLZdaLFkLdYxPLJ3ktfBCMhduf0BFvWZIkwqJM3qo2u6Wli0e9j1eo3%2FyyJg026sKigWl04CzDgwD6hQdnENdQrSC%2BkrfJUuPfkkROcrh1IBtCXARS2v5BS2uglERQoeS2SFFo2Lh1C3fv%2BwDtAtE7Dd50tKrrMTeyev11%2BXzc0aQMIm7o%2F8FOusBLyr9%2ByezwCA4qaqn4Jr2vulLMV98FrevmpbLYcsqtUGmohhFj%2FkAbiMowIUsoguogKowqjjM3fKB8luUlJMzdP%2F5v3zI0LFDmdqxpLvHhtK%2FATJOyhTL9bJm3QRSmNyaYZAP4vEj%2Bb6Ih7my0jhd6dIqeem7P8sb66PLUNJK3nJq7C7RaBRjlfukFbOLmf1zpiGnv%2FSPJNn%2Bz1aEJgTazVih5WnjozllOW9FnpOfr7qTXybBtGW3edcPD36DagRwZNx8%2Fr38fxITNWSoTOJtJ%2BchN7Ne4NvFlB%2FvaRUZLKLSM4PW16Xr4Hfthg%3D%3D&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Date=20201227T202705Z&X-Amz-SignedHeaders=host&X-Amz-Expires=300&X-Amz-Credential=ASIAQ3PHCVTY376YV3HK%2F20201227%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Signature=33198b49665d78c6443ed6053122b6f94e23223142de95353bb275224b7392ee&hash=1bd7e2be21ff0c16f5b4f835ae4d6feb278d9d3e7223273f2b7eccf56e0fa83a&host=68042c943591013ac2b2430a89b270f6af2c76d8dfd086a07176afe7c76c2c61&pii=S1571066104051357&tid=spdf-205b519b-5eaf-41e9-8e18-b6dc0536de14&sid=10069daf2e0ff3406118641-573fbfb7c5f5gxrqb&type=client" rel="noreferrer">Synthetic topology of data types and classical spaces".

The general idea is that the open sets in the topology are those properties which we can decide are true in finite time, but cannot decide if they are false in finite time.

To expand, the topology on $\mathbb R$ allows us to semi-decide if a point $p$ belongs to some interval $I$ when we are able to access the point $p$ one digit at a time. For example, let's pick $I = (0, 1)$.

Now, if we see a point $p = 0.1\dots$ we know immediately that this point is in $I$, so we can answer that $p \in_? I = \texttt{TRUE}$.

If we see a point $p = 2.\dots$, we know immediately that this point $p$ is not in $I$, so we can answer $p \in_? I = \texttt{FALSE}$

Now consider the point $p = 0.9999\dots$ When do we decide that this point $p$ does not belong to $I$?

we can't decide after seeing two digits $(0.9\dots)$, for the point $p$ could be $0.98$ and it would then belong to $I$

we can't decide after seeing  three digits ($0.99\dots)$, for the point $p$ could be $0.998$ and it would then belong to $I$

in general, we would need to read the entire number $0.999\dots$ to decide that $p \not \in I$. Clearly, this takes infinite time!

So, if $p \in I$, then we will always answer $p \in_? I = \texttt{TRUE}$ in finite time. But if $p \not \in I$, then we may not answer $p \in_? I = \texttt{FALSE}$ in finite time. Hence, the relation $\in_?$ is semi-dedicable, and the topology of $\mathbb R$ precisely captures the semi-decidable nature of membership in an interval.

More broadly speaking, discrete topology is vital to computer science, since finite topology is the study of preorders and partial orders. This is precisely the notion that is necessary to assign mathematical meaning to computer programs, through a branch of mathematics called as denotational semantics

I've written some blog posts about these issues that may be of some help to get started, since I collect links as well:

Topology is really about computation

Finite topologies and DFS numbers


问:
Combining Nesterov Acceleration with Exact Line Search for Least Squares Problem
Nesterov accelerated gradient descent is one way to accelerate the gradient descent methods.

The exact line search is also one way to find the optimal step size along the gradient direction for the least-squares problems.

However, I have not seen anything related to the combination of Nesterov acceleration and exact line search.

Is there something that impedes the combination?

答:
You may have a look at A Field Guide to Forward Backward Splitting with a FASTA Implementation (Project Website).

They have some heuristics to combine both acceleration and line search.


问:
Some questions regarding the $L^2$ space
Let $(\Xi,\boldsymbol{\Xi},\mu)$ be a measure space, and let $(\mathbb{R},\mathcal{B}_\mathbb{R})$ be the Borel space associated to $\mathbb{R}$.
Also, let
$$\mathcal{L}^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})=\bigg\{f:(\Xi,\boldsymbol{\Xi})\to(\mathbb{R},\mathcal{B}_\mathbb{R})\,\,\Big|\,\int f^2\,\mathrm{d}\mu<\infty\bigg\}$$
be the set of all (real-valued) square-integrable (measurable) functions with respect to (the measure) $\mu$.

Then can we say the following?

When the Lebesgue space $L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})$ is viewed as a set, it is defined as the quotient set:
$$L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})_\text{set}:=\mathcal{L}^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})\,\,/\sim,$$
where $\sim$ is an equivalence relation on $\mathcal{L}^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})$ such that:
$$f\sim g\quad:\Leftrightarrow\quad f-g=0\,\text{ $\mu$-almost everywhere.}$$
Is this the definition of $L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})$ as a set?

If so, what does '$\mu$-almost everywhere' actually mean in the above definition?
Some authors seem to imply that (but of course I may be wrong with this interpretation):
$$f(\xi)\neq g(\xi)\quad\Rightarrow\quad \mu(\xi)=0$$
when they say that if $f=g$ then $f$ and $g$ differ only on a set of measure zero. Here $\xi\in\Xi$.
However, for this to work, we require that $\{\xi\}$ is an element of the $\sigma$-algebra $\boldsymbol{\Xi}$ to make sure that $\{\xi\}\mapsto\mu(\xi)$ is well-defined.
This therefore imposes a hidden requirement on the definition of $\boldsymbol{\Xi}$ (which I don't like).
Or is this the correct interpretation?
$$f-g=0\,\text{ $\mu$-almost everywhere}\quad\Leftrightarrow\quad\int (f-g)^2\,\mathrm{d}\mu=0.$$

The reason why we want to define such an equivalence relation $\sim$ on $\mathcal{L}^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})$ is that if we equip $L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})$ with the bilinear map:
$$\langle\,\cdot\,,\cdot\,\rangle:L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})\times L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})\to\mathbb{R}\quad:\Leftrightarrow\quad\langle f,g\rangle=\int fg\,\mathrm{d}\mu,$$
then $\langle\,\cdot\,,\cdot\,\rangle$ qualifies as an inner product on $L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})$ because it satisfies the positive-definite condition of an inner product.
Is this the correct motivation?

$L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})$ is by definition a vector space over $\mathbb{R}$, and no further structure is assumed on $L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})_\text{set}$.

If nothing is said, all (equivalence classes of) measurable functions in $L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})$ are always assumed to be maps from $(\Xi,\boldsymbol{\Xi})$ into the Borel measurable space $(\mathbb{R},\mathcal{B}_\mathbb{R})$, aren't they?

(Disclaimer: I am not a mathematician but an engineer studying functional analysis; thus, the notation used here may be non-standard or simply incorrect. If that is the case, please let me know. I would appreciate it very much.)

Thank you,
Frederick.

答:
Concerning (2), we say that $f = g$ $\mu$-a.e. (or $f - g = 0$ $\mu$-a.e.) if $\mu(\Theta \setminus \{f - g = 0\}) = 0$.  This makes sense as long as $f$ and $g$ are measurable functions from $(\Theta,\mathbf{\Theta})$ into $(\mathbb{R},\mathscr{B}_{\mathbb{R}})$ since then $\{f - g = 0\} \in \mathbf{\Theta}$.

It turns out that this is, in fact, equivalent to the condition $\int_{\Theta} (f - g)^{2} \, d \mu = 0$.  However, this requires that we define the Lebesgue integral, whereas the previous definition I gave is a "primitive notion."

The answer to (3) is "yes."  However, it's a bit like saying "We define $\mathbb{R}$ because $\mathbb{Q}$ isn't a complete metric space."  That may technically contain some kernel of truth, but there are "philosophical" or "emotional" reasons why we would like to work with $\mathbb{R}$ in addition to $\mathbb{Q}$ (i.e. we might like to believe that a "continuum" or a "line" has some physical significance).  Similarly, in the settings in which measure theory is relevant, it tends to be natural to think of two functions that agree $\mu$-almost everywhere to be the same.  ($\mu$ sees them the same way, let's say; or we added "some dust" to one of the functions, which is irrelevant.)

Concerning (5), yes, technically, an element of $L^{2}(\Theta,\mathbf{\Theta},\mu;\mathbb{R})$ is an equivalence class of measurable functions mapping $(\Theta,\mathbf{\Theta})$ into $(\mathbb{R},\mathscr{B}_{\mathbb{R}})$.  The one caveat is nothing is lost or gained if we replace $(\Theta,\mathbf{\Theta})$ by $(\Theta,\mathbf{\Theta}_{\mu})$, where $\mathbf{\Theta}_{\mu}$ is the completion of $\mu$ with respect to $\mathbf{\Theta}$.  Hence, formally, it might be better to incorporate completeness of $(\Theta,\mathbf{\Theta},\mu)$ into the definition of $L^{2}(\Theta,\mathbf{\Theta},\mu;\mathbb{R})$ at the outset.  Someone can correct me if I'm wrong, but I'm not aware of any instances where this matters.  (It's a bit like the space $\mathcal{L}(V)$ of continuous linear operators on a normed space $V$.  $\mathcal{L}(V)$ is unchanged if $V$ is replaced by its completion.)


问:
Is $d$ a metric for $\mathbb{R}^2?$
In $\mathbb{R}^2 $  define $d(x,y)$ = smallest integer greater or equal to usual
distance between $x$ and $y$. Is $d$ a metric for $\mathbb{R}^2?$

My attempt : I think No, take $x=[0.1] , y=[0.2]$ then $d(x,y)=0$ but $x\neq  y$

Is its correct ?

答:
We have that $d(x,y)=\lceil d_2(x,y)\rceil$ where $d_2(x,y)=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}$ is a metric. Indeed, $d$ is a metric if we change $d_2$ for any metric on $\mathbb{R}^2.$

$d(x,x)\ge 0?$

Yes, since $d(x,y)=\lceil d_2(x,x) \rceil \ge \lceil 0 \rceil \ge 0.$

$d(x,x)=0 \iff x=y?$

Yes: $x=y\iff d_2(x,y)=0 \iff \lceil d_2(x,y)\rceil=0\iff d(x,y)=0.$

$d(x,y)=d(y,x)?$

Yes, because $d(x,y)=\lceil d_2(x,y) \rceil=\lceil d_2(y,x) \rceil=d(y,x).$

$d(x,y)\le d(x,z)+d(y,z)?$

Yes: $d(x,y)=\lceil d_2(x,y) \rceil\le \lceil d_2(x,z)+d_2(y,z) \rceil \le \lceil d_2(x,z)\rceil + \lceil d_2(y,z) \rceil=d(x,z)+d(y,z).$

Title: Is $d$ a metric for $\mathbb{R}^2?$