答:
Another different proof may be to use this nice and not so well-known criterion, and apply for $p=1$:

\begin{aligned}
&\text { Theorem. For any } p, 1 \leq p<\infty, \text { a bounded subset } K \text { of } \ell^{p} \text { is relatively compact if and }\\
&\text { only if } \lim _{n_{\epsilon}} \sum_{i=n_{\epsilon}}^{\infty}\left||k_{i}|\right|^{p}=0, \text { uniformly for } k \in K
\end{aligned}

If you need a proof of the theorem, you can click here.

Title: Compactness of Subset of $\ell^1$.

Body:
Let $X$ be a finite-dimensional normed space and let $\ell^1(X)$ denote the Banach space of all sequences in $X$ for which $\sum_{n=1}^{\infty} \|x_n\|_X<\infty$.  For $s\geq 0$, is the set
$
C = \left\{
(x_n)\in \ell^1(X):\, \sum_{n=1}^{\infty} n\|x_n\|\leq s
\right\}
$
compact?

What I've tried:

I tried using the Grothendieck compactness principle to show that $C$ is contained in the closed convex-hull of the sequence $(\frac{s}{n})$ but I'm not sure this is the case..


问:
Plotting graph given parametric form satisfying $x=f(t)$ and $y=-f(-t)$
I had to plot the curve given by the parametric definition as follows. In attempting to simplify doing so, here is what I see holds between $x$ and $y$.

Plot the curve given by $$x=\frac{(t+2)^2}{t+1} ,y=\frac{(t-2)^2}{t-1}$$

$$\boxed{x=f(t)\\ y=-f(-t)}$$

Is there somehow we can use that to plot the graph easily. Any hints are appreciated. Thanks.

答:
write $x$ as $\dfrac{(1+t+1)^2}{t+1} = t+1+2+\dfrac{1}{t+1}$ and $y = t-1 -2 + \dfrac{1}{t-1} $

$$\begin{align} x+y &= 2t \Big(\dfrac{t^2}{t^2-1} \Big) \\
y - x &= 6 - \dfrac{2}{t^2-1}\\
x+y &= t^3 \cdot (6+x-y) \end{align}$$

and now you can just put  value of $t$ in terms of $x,y$

to get $$(8+x-y)^3 = (x+y)^2 (6+x-y) $$

here is what geogebra gave

ContentLicense: CC BY-SA 4.0


问:
{$\frac{1}{x}$} , $0<x\leq1$ in terms of {$x$}, $x\geq 1$, {$x$} is fractional part of $x$
$f(x)$ $=$ {$\frac{1}{x}$}, $0<x\leq1$ where {$x$} denotes the fractional part of $x$.

$g(x) =$ {$x$}, $x\geq 1$

I want an expression for $f(x)$ in terms of x and $g(x)$.

My try-

If $x\in \mathbb{Z} , x\geq 1$ , g(x)={x}=0

f(x)={$\frac{1}{x}$}=$\frac{1}{x}$
$$f(x)=\frac{1}{x}+g(x)$$
What happens when x is a fraction?
Jacobi theta function $\theta(x)= \sum_{n=-\infty}^\infty e^{-n^2\pi x}$ This function satisfies the functional equation $\theta(\frac{1}{x})=\sqrt{x}\theta(x)$. I want a similar functional equation of f(x) in terms of g(x)

答:
There is no expression for $f$ solely in terms of $g(x)$.

Proof:

Take $x_1 = 2, x_2=3$. Then, $g(x_1)=g(x_2)$, but $f(x_1)\neq f(x_2)$.

Intuitive explanation: an expression "in terms of $g(x)$" is an expression that only depends on the value of $g(x)$ and does not directly depend on the value of $x$. Therefore, the expression, whatever it is, will have the same value for $x_1$ and $x_2$. But since $f(x_1)\neq f(x_2)$, the expression cannot possibly capture both values, $f(x_1)$ and $f(x_2)$.

Rigorous explanaiton: An expression of $f$ in terms of $g(x)$ is, mathematically speaking, a function $h$ such that $f(x)=h(g(x))$.

Assuming that such a function $h$ exists, however, brings us to a contradiction, since any such function means that

$$f(x_1)=h(g(x_1))=h(g(x_2))=f(x_2)$$ which is a contradiction, because we know that $f(x_1)\neq f(x_2)$.

An expression "in terms of $g(x)$ and $x$", on the other hand, trivally always exists, since $f(x)$ itself is already an expression in terms of $x$.


问:
The property of the functor "Hom"
If $ i\in I$ and $\ I $ is an infinite index set, $\ A,A_i ,B,C$ are all R-module,  is it true for$\ Hom_R(\bigoplus A_i,B) \cong Hom_R(\prod A_i,B)$?

And when is it true for $\ Hom_R( A,B\bigoplus C) \cong Hom_R(A,B)\bigoplus Hom_R(A,C)$?

All the above is true, but it seems $\ Hom_R(B,\bigoplus A_i) \ne \bigoplus Hom_R(B,A_i)$ but $\ Hom_R(B,\prod A_i)\cong \prod Hom_R(B,A_i)$?

答:
Both isomorphisms are true. For the first one, note that for any $i$ there is a natural map (by restriction)

$$Hom_R (\bigoplus A_i, B) \to Hom_R (A_i, B)$$

so this lifts to a map $ Hom_R (\bigoplus A_i, B) \to \prod Hom_R (A_i, B)$. To check this is indeed an isomorphism, notice that it has to be injective, since the morphism at the $A_i$ allows us to determine the morphism at the direct sum. It must also be surjective because you can define the morphism to be however you want on the $A_i$ and this gives a valid morphism on the direct sum. The map will send this to the product of each of the $A_i$ morphisms.

For the other isomorphism, you could actually have an arbitrary direct sum, and this factors as a direct sum as well. The idea is very similar since by inclusion we have a natural map

$$Hom_R (A, B_i) \to Hom_R (A, \bigoplus B_i)$$

This then lifts to a map on the direct sum and you can apply a similar reasoning.

Title: The property of the functor "Hom"


问:
Prove that the elements of the group satisfy following equality
Question: Let G denote a group generated by the transformation $S = -1/{\gamma}$ and $T^2$, where $T(\gamma)=\gamma +1$ . Now if matrix $ \begin{pmatrix}
a & b  \\
 c & d   \\
 \end{pmatrix}$ $\in G $ , then show that $a\equiv d \pmod 2$  and $b\equiv c \pmod 2$.

Any element of $T^{2k}$ will be given by $ \begin{pmatrix}
1 & 2k  \\
 0 & 1   \\
 \end{pmatrix}$ but I am not able to represent general element of $S^{k}$ by this as

$S= \begin{pmatrix}
0 & -1 \\
 1 & 0   \\
 \end{pmatrix}$.

$S^{2}= \begin{pmatrix}
-1 & 0  \\
  0& -1   \\
 \end{pmatrix}$

$ S^{3}= \begin{pmatrix}
0 & 1  \\
 -1 & 0   \\
 \end{pmatrix}$

$S^4=S$.

So, I am not able to find an expression similar to expression of $T^{2k}$ which will be used for multiplying both and then to find a,b,c,d.

Can you please tell how should I find them. Thank you .

答:
Since any element of $G$ has a form of $S^{a}T^{2b}S^{c}T^{2d}\cdots$, it is enough to show that the condition holds for $Sg$ and $T^{2}g$ if it holds for $g$, which shouldn't be hard. Since the identity matrix satisfies the condition, we can complete the proof by using induction.

If $g = \left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right)$ satisfies $a \equiv d$ and $b \equiv c$, then $Sg = \left(\begin{smallmatrix} -c & -d \\ a & b \end{smallmatrix}\right)$ satisfies $-c \equiv c \equiv b$ and $-d \equiv d \equiv a$. Also, we have $T^{2}g = \left(\begin{smallmatrix} a + 2c & b + 2d \\ c & d \end{smallmatrix} \right)$ and $a + 2c \equiv a \equiv d$ and $b + 2d \equiv b \equiv c$.


答: I assume you mean the following: If $I$ is an arbitrary indexed set it holds that $$\text{Hom}(\bigoplus_{i \in I} M_i,N) \cong \prod_{i \in I}\text{Hom}(M_i,N).$$ However it "commutes" with direct sums, whenever you have a finite direct sum, meaning: $$\text{Hom}_R(\bigoplus_{i=1}^n M_i,N) \cong \bigoplus_{i=1}^n \text{Hom}(M_i,N)$$ for any $n \in \mathbb{N}$. Note that this also holds for products/sums in the second variable. One can prove the first statement by looking at the map $$\text{Hom}(\bigoplus_{i \in I} M_i,N) \to \prod_{i \in I}\text{Hom}(M_i,N), \varphi \mapsto (\varphi \circ \iota_i)_{i \in I}.$$

Title: The property of the functor "Hom"

Body:
If $ i\in I$ and $\ I $ is an infinite index set, $\ A,A_i ,B,C$ are all R-module,  is it true for$\ Hom_R(\bigoplus A_i,B) \cong Hom_R(\prod A_i,B)$?

And when is it true for $\ Hom_R( A,B\bigoplus C) \cong Hom_R(A,B)\bigoplus Hom_R(A,C)$?

All the above is true, but it seems $\ Hom_R(B,\bigoplus A_i) \ne \bigoplus Hom_R(B,A_i)$ but $\ Hom_R(B,\prod A_i)\cong \prod Hom_R(B,A_i)$?


问:
Prove the convergence of the sequence.
Prove the convergence of the following sequence:

$$x_1 = \sqrt{a}$$

$$x_{n+1} = \sqrt{a + x_n}$$

答:
(Of couse we need $a\ge 0$ as premise)

Show by induction that $0\le x_n<a+1$

Show by induction that $x_{n+1}\ge x_n$

(You can facilitate things a bit by prepending a term $x_0=0$. Not that then still $x_1=\sqrt{a+x_0}$)

ContentLicense: CC BY-SA 3.0

Title: Prove the convergence of the sequence.


问:
Decompose a matrix into diagonal term and low-rank approximation
For a matrix $A$ the Singular Values Decomposition allows getting the closest low-rank approximation $$A_K=\sum_i^K\sigma_i \vec{v}_i \vec{u}_i^T$$ so that $\|A-A_k\|_F$ is minimal.

I'd like to do the same but allow for an additional diagonal term; that is, for a given square matrix $A$ and an positive integer $K$ find a diagonal matrix $D$ and low-rank approximation  $$A_K=D+\sum_i^K\sigma_i \vec{v}_i \vec{u}_i^T$$ so that like above $\|A-A_k\|_F$ is minimal.

The problem originated in the context of correlations matrices. Thus answers which further assume $A$ is symmetric, positive semi-definite are also welcome.

答:
The solution I ended up implementing is publicly available:
https://github.com/sompolinsky-lab/dnn-object-manifolds/blob/master/library/optimal_low_rank_structure2.m

The idea behind it is to denote $A\in\mathbb{R}^{N\times P}$ the data, such that $C=AA^T$, then find an orthonormal set $V\in\mathbb{R}^{N\times K}$ such that the residual $R=A-V(V^TA)$ is approximately diagonal; this is done by solving the following optimization problem:

$$V^*=\arg\min_{V\ :\ V^TV=I_K} \sum_{i,j} (RR^T)_{ij}/\sqrt{(RR^T)_{ii}(RR^T)_{jj}}$$

This can be done using existing tools for optimization under orthonormality constraints. It has no free parameters you need to fit or optimize over and for my problems it gives very solid results, i.e. finds very consistent solutions despite the fact that it converges is to a local minima, with no assurance if it's a good one relative to the global minima. To achieve this for every $k=1..MAX\_K$ it repeats for $N\_REPEATS$ time the search for a minima assuming the rank is $k$, using the solution for $k-1$ as initial conditions. Then the $k$ with the minima of optimization target is used.

Please refer to the paper if you use this method: https://www.nature.com/articles/s41467-020-14578-5


问:
How to find all $x$ of function $f(x)$ where $x,f(x)\in\mathbb{Z}$?
To start of with some background info on my problem:

I was/am trying to get all integer zeros the function $f(x,y)=\frac{xy-10x-10y}{10xy}$, so $x$, $y$ and $f(x,y)$ should be integers. So I did $\frac{xy-10x-10y}{10xy}=0\Leftrightarrow y=\frac{10x}{x-10}$.

And so it has become my goal to determine all $x$ for which $g(x)=\frac{10x}{x-10}$ and $x$ are integers because so I can get the $x$ and $y$ coordinates for the zeros of $f(x,y)=\frac{xy-10x-10y}{10xy}$.

But how do I go about finding all integer solutions for $g(x),x\in\mathbb{Z}$? Or maybe there is a generally better way to find all integer zeros of a function with two variables?\

Thanks in advance!

答: Observe that, $$f(x)=\frac{10x}{x-10}=\frac{10x-100+100}{x-10}=10+\frac{100}{x-10}$$ So, you should find all $x \in \mathbb Z$ such that $x-10$ divides $100$, which can be done easily considering the factors of $100$.


问:
Creating a network based on arrays of numbers
Intro:

I am not a mathematician, but I am a bit familiar with the graph theory. I am also comfortable with Python and NetworkX.

Example data:

Array A: [10, 20, 30, 40, 50]

Array B: [20, 30, 40, 50, 90]

Array C: [100, 200, 300, 400, 500]

I would like to create a network based on similarities of the elements in each array. By intuition, the values in array A are more closely related to array B, and in turn to array C. In some cases, I also want to weight certain elements more, where the last element could be more important than the earlier ones.

Why I think the network approach is suitable?

The number of pairwise comparison needed is the factorial of the number of arrays. For a large number, this will simply become computationally infeasible.

Question:

Am I correct in my thinking that I can use a network approach to find pairs that are closely related to each other? If yes, how do I go about creating the network using the example data above?

Thank you.

答:
I guess the amplitude of your values and their position matter, and maybe the sequences all have the same length like in your example. Then, it may make sense to see them as time series, and use one of the various approaches to measure similarity between time series to build a graph.

If order does not matter, then you may measure similarity between sequences through their value distributions, either with something like a KS test, or by comparing their parameters (mean and variance, e.g.). If amplitude does not matter, you may normalize the sequences.

You may also consider edit distances: how many modifications (like removing or adding an entry, rescaling one, swapping two entries, or shifting or rotating the array, e.g.) are required to transform one array into another. The cost of each modification may be weighted.

Clusters in such graphs, and other properties like their degree distributions, would certainly say something on your set of sequences, although it may be difficult to interpret.

The approaches above exist in the literature, and I suggest searching the web for the highlighted keywords.

Hope this helps!


答:
Hint :

Prove that the sequence $\{x_n\}$ is bounded and monotonic

Title: Prove the convergence of the sequence.

Body:
Prove the convergence of the following sequence:

$$x_1 = \sqrt{a}$$

$$x_{n+1} = \sqrt{a + x_n}$$


问:
Connectedness of covering space
Suppose we have a covering map $f:E \to B$, where $B$ is a connected and path-connected space and $E$ is a p-sheeted cover of $B$.
Are the following statements equivalent?

$E$ is a non-trivial covering, i.e. $E$ is not a disjoint union of $p$ copies of $B$.

$E$ is connected.

$E$ is path connected.

There exists a loop in $B$ whose preimage is an interval in $E$.

I think the equivlance between 1 and 4 is easy to see since a trivial covering will have the property that the preimage of any loop is a disjoint union of $p$ loops. But I don't really understand the relation between 1 and 2 or 1 and 3. Definitely when $E$ is trivial, it is neither path-connected nor connected, but what about the other way around?

Or, is there any way to see if the covering space is connected/path-connected by looking at the covering map?

Any comment/help is appreciated.

答:
1 and not 2/3: a union of two non-trivial coverings.

2/3 and not 1: $f = \text{id}_B$.

2 and not 3: take the famous counterexample, make it path-connected:

and cover it spirally, as in the universal cover for $S^1$. Or do the same with the so-called extended long line, which is imo more fun. (Obviously, you can make p finite by modifying this way other $S^1$ coverings.)

Title: Connectedness of covering space


答:
$f(x) = \dfrac{10x}{x - 10}$

Let $x - 10  = k.$   So, $x = 10+k$

Therefore,  $\dfrac{10x}{x-10} = \dfrac{10(k+10)}{k} = 10 + \dfrac{100}{k}$

So $k$ divides $100$ and therefore $k = 1,2,4,5,10,...$

You can find respective values of $x$ since $x = 10 + k$

EDIT: Solutions for $f(x) = \dfrac{10x}{3x-10}$

Let $k = 3x-10$

$\dfrac{10x}{3x-10} = \dfrac{100+10k}{3k}=t$ (Lets say it is equal to t)

$100 = k(3t -10)$

Now basically you can write 100 as a product of 2 factors like

100 = 1 x 100...

Now check for which values you get t as an integer and you will get your answer.

Title: How to find all $x$ of function $f(x)$ where $x,f(x)\in\mathbb{Z}$?

Body:
To start of with some background info on my problem:

I was/am trying to get all integer zeros the function $f(x,y)=\frac{xy-10x-10y}{10xy}$, so $x$, $y$ and $f(x,y)$ should be integers. So I did $\frac{xy-10x-10y}{10xy}=0\Leftrightarrow y=\frac{10x}{x-10}$.

And so it has become my goal to determine all $x$ for which $g(x)=\frac{10x}{x-10}$ and $x$ are integers because so I can get the $x$ and $y$ coordinates for the zeros of $f(x,y)=\frac{xy-10x-10y}{10xy}$.

But how do I go about finding all integer solutions for $g(x),x\in\mathbb{Z}$? Or maybe there is a generally better way to find all integer zeros of a function with two variables?\

Thanks in advance!


问:
Intermediate Value Theorem Prove\Disprove Question
I would appreciate your help regarding this prove/disprove:

--

Let $f:(0, \infty) \to R$ be a continuous function at $x_{0}=1$ which $f(x)\cdot f(\frac{1}{x})<0$ for every $0 < x \neq 1$.

So $f(1)=0$.

--

I have tried to prove it with no luck, than tried to find a counter-example with no luck...

Any help will be amazing!

Thank you so much!

答:
Consider $x_n = 1+\frac 1n$.

Since $f$ is continuous at $x=1$ you have

$$\lim_{n\to\infty}f(x_n) = \lim_{n\to\infty}f\left(\frac 1{x_n}\right)= f(1)$$

So, it follows

$$\lim_{n\to\infty}\left(f(x_n)\cdot f\left(\frac 1{x_n}\right)\right)=\left(f(1)\right)^2 \leq 0$$

Hence, $f(1) = 0$.


问:
How can we show that there is a zero of $f$
Let $$f(x)= x^2+2020x+1$$ a polynomial.
We define $F_n(x)$ to be $f(f(f...f(x)))$ applied $n$ times .
Show that for all positive integer $n\ge1$ there exists a real $x$ such that $F_n(x)=0$

Please help me, I've tried induction but it doesn't seem to work. $n=1$ is trivial but then it's hard. I don't know much about polynomials. Any ideas?

答:
Consider the solutions to $ f(x) = x$. These are $ x_1, x_2 = \frac{ - 2019 \pm \sqrt{ 2019^2 - 4 } } { 2 } < 0  $.

We have $ f(x_i) = x_i$, so $ F_n(x_i) = x_i$.

In particular, $ F_n(x_i) < 0$.

Observe that for $x > 0$ $ f(x) > 0$ , so $ F_n(x) > 0 $ for $ x > 0 $.

Hence by IVT, there exists a $x'$ such that $F_n(x') = 0$.

Note: $ \frac{ - 2019 + \sqrt{ 2019^2 - 4 } } { 2 } \approx -0.0005$, and so we can conclude that there is a root in $(-0.0005, 0)$.


问:
Is this $\mathbb{Z}_2^n$?
What group is formed by binary strings of a fixed length, $n$, and the XOR operation (^)?

For example, we have:

For $n=1$:

```<code>A^B = B^A = B
A^A = B^B = A
</code>```

For $n=2$:

```<code>AA ^ AB = AB = AB ^ AA
BB ^ AB = BA = AB ^ BB
</code>```

For $n=3$:

```<code>ABA ^ BAA = BBA
</code>```

etc

I reckon it is written $\mathbb{Z}_2^n$ but I would like someone to either confirm this or tell me why I am wrong.

答:
Considering bit strings of length $n$ as $n$-tuples of elements of $\mathbb Z/2\mathbb Z$, the XOR operation is simply the addition in $(\mathbb Z/2\mathbb Z)^n$.
So, yes, this "is" just the group  $(\mathbb Z/2\mathbb Z)^n$.</p>

Title: Is this $\mathbb{Z}_2^n$?


问:
Confusion Regarding Comparison Theorem
Determine if the integral converges or diverges using the Comparison Theorem (CT).

$$\int_{1}^{\infty} \frac{x+1}{\sqrt{x^4 -x}}dx$$

I saw solutions online arrive at the following inequality

$$\frac{x+1}{\sqrt{x^4 -x}} \ge \frac{1}{x}$$

But this inequality only holds for  $x > 1$  and not for $x \ge 1$ since at $x=1$ the LHS of the inequality, $\frac{x+1}{\sqrt{x^4 -x}}$, is undefined.

I thought the inequality had to be true over the limits of integration, $x \ge 1$ or $[1, \infty]$ in this case, in order to use the CT.

Furthermore, the CT requires the functions being compared to be continuous over a closed interval, but at $x = 1$ function $\frac{x+1}{\sqrt{x^4 -x}}$ is infinitely discontinuous (vetical asymptote) which I'm assuming means we can't apply the CT over $[1, \infty]$.

But I have seen solutions online show that

$$\int_{1}^{\infty} \frac{1}{x}dx$$

diverges and thus $$\int_{1}^{\infty} \frac{x+1}{\sqrt{x^4 -x}}dx$$ diverges by the CT. Are these solutions correct?

As a work around I was thinking the inequality is true for $x \ge 2$ so the integral could be "split up" into the following

$$\int_{1}^{\infty} \frac{x+1}{\sqrt{x^4 -x}}dx = \int_{1}^{2} \frac{x+1}{\sqrt{x^4 -x}}dx + \int_{2}^{\infty} \frac{x+1}{\sqrt{x^4 -x}}dx$$

The CT could then be applied to the second integral. The first integral, however, still has an infinite discontinuity at $x = 1$ and thus is an "improper" integral. This would require evaluating the integral directly which defeats the purpose of using the CT in the first place to determine convergence or divergence.

My questions are:

Does it matter if the inequality is not true over the interval or limits of integration $[a, \infty)$ or $x \ge a$ in order to apply the Comparison Theorem?

Were the solutions online correct?

答:
Considering the integrand, we have
$$\frac{x+1}{\sqrt{x^4 -x}}=\frac{2}{\sqrt{3} \sqrt{x-1}}-\frac{\sqrt{x-1}}{\sqrt{3}}+\frac{2
   (x-1)^{3/2}}{3 \sqrt{3}}+O\left((x-1)^{5/2}\right)$$ So, no problem around the lower bound.

On the other side
$$\frac{x+1}{\sqrt{x^4 -x}}=\frac{1}{x}+\frac{1}{x^2}+\frac{1}{2x^4}+O\left(\frac{1}{x^5}\right)$$ (in fact, in this last expansion, all coefficients are non-negative- so you can bound the integrand by wathever you want).

Asymptotically, we have
$$\int_1^n\frac{x+1}{\sqrt{x^4 -x}}=\log(n)+ C$$ where $C\sim 1.864$.

Title: Confusion Regarding Comparison Theorem


答:
For $x>1$,

$$\left\{\frac1x\right\}=\frac1x$$ and you are asking for $\left\{\frac1x\right\}$ as an expression involving $\{x\}$ and $x$ as a rational expression. I just wrote it above. Knowing $\{x\}$ is of no help.

For $0\le x<1$,

$$\{x\}=x$$ and you are in fact asking for $\left\{\frac1x\right\}$ as an expression involving $x$ alone as a rational expression. This is not possible due to the discontinuities in this function.

Title: {$\frac{1}{x}$} , $0<x\leq1$ in terms of {$x$}, $x\geq 1$, {$x$} is fractional part of $x$

Body:
$f(x)$ $=$ {$\frac{1}{x}$}, $0<x\leq1$ where {$x$} denotes the fractional part of $x$.

$g(x) =$ {$x$}, $x\geq 1$

I want an expression for $f(x)$ in terms of x and $g(x)$.

My try-

If $x\in \mathbb{Z} , x\geq 1$ , g(x)={x}=0

f(x)={$\frac{1}{x}$}=$\frac{1}{x}$
$$f(x)=\frac{1}{x}+g(x)$$
What happens when x is a fraction?
Jacobi theta function $\theta(x)= \sum_{n=-\infty}^\infty e^{-n^2\pi x}$ This function satisfies the functional equation $\theta(\frac{1}{x})=\sqrt{x}\theta(x)$. I want a similar functional equation of f(x) in terms of g(x)


问:
Compactness of Subset of $\ell^1$.
Let $X$ be a finite-dimensional normed space and let $\ell^1(X)$ denote the Banach space of all sequences in $X$ for which $\sum_{n=1}^{\infty} \|x_n\|_X<\infty$.  For $s\geq 0$, is the set
$
C = \left\{
(x_n)\in \ell^1(X):\, \sum_{n=1}^{\infty} n\|x_n\|\leq s
\right\}
$
compact?

What I've tried:

I tried using the Grothendieck compactness principle to show that $C$ is contained in the closed convex-hull of the sequence $(\frac{s}{n})$ but I'm not sure this is the case..

答:
I think it's important to understand the motivation behind why this result holds, so I'll be a little verbose with my answer, perhaps excessively so, but bear with me.

Since $\ell^1 (X)$ is a Banach space, by the Heine-Borel theorem we want to show $C$ is totally bounded and closed. Closed is obvious, since it is the pre-image under a continuous map of a closed set,
$$ C = \bigcap_{N \geq 1} C_N, \qquad C_N = \left\{ x \in \ell^1 : \sum_{n = 1}^N n |x_n| \leq s \right\}. $$
The idea behind showing $C$ is totally bounded: the factor of $n$ gives a quantitative decay condition on the tail of the $\ell^1$ norm of points in $C$, so we'd like to control the tail uniformly, and reduce to proving total boundedness on $X^M$ for some positive integer $M \in \mathbb N$, which is a nice finite dimensional Banach space.

To get a good intuition, think about the case $X = \mathbb R$. Compactness is easy to prove in $\mathbb R^M$ (closed and bounded), and $\mathbb R^M$ is embedded in $\ell^1 (\mathbb R)$ by just adding zeros or whatever. We are moving in the opposite direction, noting the tail contribution is small and projecting $C \subseteq \ell^1 (\mathbb R)$ down to $\mathbb R^M$. To put it as a moral, infinite dimensions bad (in particular, totally bounded $\neq$ bounded), but finite dimensions good (total bounded $=$ bounded)!

Controlling the tail uniformly: Let $\epsilon > 0$ and, without loss of generality, suppose $s = 1$. Then we can find $N_\epsilon \in \mathbb N$ such that $N_\epsilon \geq 2/\epsilon$. It follows that for any $x \in C$,
$$ \sum_{n > N_\epsilon} |x_n| \leq  \frac{1}{N_\epsilon}\sum_{n > N_\epsilon} n|x_n|  \leq \frac{1}{N_\epsilon} < \frac{\epsilon}{2}. $$
Reduce to finite dimensional case: We now project $C$ down to these first $N_\epsilon$ terms, namely
$$C^{N_\epsilon} = \{ (x_1, \dots, x_{N_\epsilon}) \in X^{N_\epsilon} : x \in C\}. $$
$C^{N_\epsilon} \subseteq X^{N_\epsilon}$ is bounded by definition of $C$ and, since $X^{N_\epsilon}$ is a finite dimensional vector space, therefore totally bounded, so there exist $k_\epsilon \in \mathbb N$ and $\widehat{x^{(1)}}, \dots, \widehat{x^{(k_\epsilon)}} \in X^{N_\epsilon}$ such that
$$ C^{N_\epsilon} \subseteq \bigcup_{k = 1}^{k_\epsilon} B_{\epsilon/2}  \left(\widehat{x^{(k)}}\right), \qquad B_{\epsilon/2}  \left(\widehat{x^{(k)}}\right) = \left\{ y \in X^{N_\epsilon} : |y - \widehat{x^{(k)}}| \leq \epsilon/2\right\} $$
Use the embedding of finite dimension into $\ell^1 (X)$ to conclude:
We extend each $\widehat{x^{(k)}}$ to an element $x^{(k)} = \{x_n^{(k)}\}_{n \geq 1} \in C$ be setting $x_n^{(k)} = \widehat{x_n^{(k)}}$ for $n \leq N_\epsilon$ and $x_n^{(k)} = 0$ for $n > N_\epsilon$. Let $x \in C$, then by total boundedness in $X^{N_\epsilon}$ we can find $1 \leq k \leq k_\epsilon$ such that
$$||x - x^{(k)}||_1 = \sum_{n = 1}^{N_\epsilon} |x_n - x_n^{(k)}| + \sum_{n > N_\epsilon} |x_n| < \epsilon.$$
We can conclude that $C$ is totally bounded. With appropriate modifications, you can extend this proof to $C$ defined similarly for $\ell^p (X)$ for any $1 \leq p < \infty$.

On the other hand, the closed ball $\{ x \in \ell^1(X) : ||x||_1 \leq 1 \}$ is not compact because we don't have a quantitative control over the tail, so you can always "move" mass to infinity, which is why total boundedness fails for balls in infinite dimensional Banach spaces. No matter how many balls you cover it with, these finitely many balls can't "capture" the mass at infinity.

Title: Compactness of Subset of $\ell^1$.


问:
Find an expression for $n$ in terms of $x$, $y$, and $z$.
At a certain rate of compound interest, $100$ will increase to $200$ in $x$ years, $200$ will increase to $300$ in $y$ years, and $300$ will increase to $1,500$ in $z$ years. If $600$ will increase to $1,000$ in $n$ year, find an expression for $n$ in terms of $x$, $y$, and $z$.

I know:

$600(1+i)^n = 1000$

I wrote:

$200 = 100(1+i)^x$

$300 = 100(1+i)^{x+y}$

$1500 = 100(1+i)^{x+y+z}$

Also, I know that :

$600=100(1+i)^{2x+y}$

Hence:

$1000=100((1+i)^{x+y+z}-(1+i)^{x+z}-(1+i)^x)$

$1000=100(1+i)^{2x+y}((1+i)^{-x+z}-(1+i)^{-x}-(1+i)^{-x-y})$

$1000=600((1+i)^{-x+z}-(1+i)^{-x}-(1+i)^{-x-y})$

$(1+i)^n=(1+i)^{-x+z}-(1+i)^{-x}-(1+i)^{-x-y}$

I need to get $n$ as a function of $x$,$y$ and $z$

I have a problem with my last line.
Does anyone knows a faster way to get the solution ?

答:
Let's make it clean and simple. Instead of $1+i$, I'll put $q$ as the rate. And I'll cancel out all the unnecessary terms. So we have
$$
2 = q^x \qquad \frac{3}{2}  = q^y \qquad 5 = q^z
$$
Now we want to make the ratio $\frac{10}{6}$ with these. We can 'massage' the expression a bit:
$$
\frac{10}{6}
= \frac{2 \cdot 5}{ \left( \frac{3}{2} \right)\cdot 2 \cdot 2}
= \frac{5}{ \left( \frac{3}{2} \right)\cdot 2} = 5 \cdot \left( \frac{3}{2} \right)^{-1} \cdot (2)^{-1}
$$
Now you can just plug in the known values and you're done.


问:
What exactly is this set?
The following set:

$\{X \in P(\mathbb{N}): |\complement{X}|< \infty)\}$

All subsets $X$ of $\mathbb{N}$ whose complement $\complement{}X$ is a finite set.

Is my interpretation in words correct?

Is it correct that the only infinite subset of $\mathbb{N}$ is $\mathbb{N}$ itself?

Therefore the set above includes all subsets of $\mathbb{N}$ except for the set whose complement is $\mathbb{N}$. Is $P(\mathbb{N})$ the universe and would the empty set then be the set whose complement is $\mathbb{N}$?

$\{X \in P(\mathbb{N}): |\complement{X}|< \infty)\} = \{X \in P(\mathbb{N}):X \neq \emptyset  \}$

答:
Is my interpretation in words correct?

Yes, the interpretation is correct.

Is it correct that the only infinite subset of $\mathbb{N}$ is $\mathbb{N}$ itself?

Most certainly not. For example, the set $\{2,3,4,5,\dots\}$ is an infinite subset of $\mathbb N$.

In fact, for a given subset $X\subseteq \mathbb N$, you have three options. You could have that:

$X$ is infinite, and $X^c$ is finite (for example, $X=\{2,3,4,\dots\}$ is infinite, and $X^c=\{1\}$ is finite) or

$X$ is finite, and $X^c$ is infinite (for example, $X=\{1\}$ is finite, and $X^c=\{2,3,4,\dots\}$ is finite) or

$X$ is infinite, and $X^c$ is infinite (for example, $X=\{2, 4, 6, 8, 10, \dots\}$ is infinite, and $X^c=\{1,3,5,7,\dots\}$ is infinite).

The only option that is impossible is to have $X$ finite and $X^c$ also finite (as this would mean that $\mathbb N=X\cup X^c$ is also finite).

Title: What exactly is this set?


问:
Zeckendorf with Negative Fibonacci Numbers
Zeckendorf : Every positive integer N can be expressed uniquely as a sum of distinct non-consecutive Fibonacci numbers

I was wondering if this theorem can be applied with the extended Fibonacci numbers, and especially I am looking for a way to find the Zeckendorf-like representation of $ N $, with Fibonacci numbers $ F_n $ having only negative indexes. (First for $ N \in \mathbb{N} $ then maybe $ N \in \mathbb{Z} $)

Mathworld states that this theorem only applies on positive numbers but do not says if Fibonacci numbers can be negative.
Online Zeckendorf Representation pages show positive indexes only and mostly use Binet's formula.

I've read that Binet's formula can be applied to generalized Fibonacci numbers (Proof), but it does not prove that Zeckendorf theorem can be used as well.

EDIT : I missed a paragraph in this Wikipedia page called Representation with Negafibonacci numbers that states this Zeckendorf-like representation exists and is unique. I probably can use NegaFibonacci coding method to find the indexes.

答:
This paper that proves the existence and shows a method

(we prove that) every integer can be represented uniquely as a sum of nonconsecutive Fibonacci numbers Fi where i < 0 and we specify an algorithm that leads to this representation


问:
summation of binomial series
Can someone pls provide the approach/solution to this problem
$$\sum_{r=1}^n \begin{pmatrix}n\\r\end{pmatrix}\sin(rx) \cos((n-r)x)$$
I have tried to simplify the sine and cosine terms and then arrived at this part but am unable to solve any further...
$$\sum_{r=1}^n \begin{pmatrix}n\\r\end{pmatrix}\frac {\sin(2rx-nx)}{2}+2^{n-1} \sin(nx)-\frac{\sin(nx)}{2}$$
Also it would be very helpful if you can advice on how to approach such questions.

答:
For answering the question first observe the fact that $\displaystyle \binom{n}{k} = \binom{n}{n-k}$

let $$\begin{align} S &= \sum_{r =\color{red}0}^{n}\binom{n}{r}\sin(rx)\cos((n-r)x)\\
S&=\sum_{r =\color{red}0}^{n}\binom{n}{n-r}\sin((n-r)x)\cos(rx) = \sum_{r=1}^{n}\binom{n}{n-r}\sin((n-r)x)\cos(rx) + \sin(0x)\cos(nx) \\
2S &= \sum_{r =\color{red}0}^{n}\binom{n}{r}\sin(rx + (n-r)x)\\
2S &= \sin(nx)\sum_{r = 0}^{n}\binom{n}{r}  \end{align}$$

Or $$\boxed{S = 2^{n-1}\sin(nx)} $$

Title: summation of binomial series


问:
Infinite subset of compact space has an accumulation point
Let $(K,\tau)$ be a compact space and $B$ be an infinite subset of $E$. Why can we conclude that $B$ has a $\tau$-accumulation point$^1$?

We may assume the contrary so that for every $x\in B$, there is a neighborhood $\Omega_x\in\tau$ of $x$ with $B\cap\Omega_x\setminus\{x\}\ne\emptyset$. If $F\subseteq B$ is finite, then $$\left|B\cap\bigcup_{x\in F}\Omega_x\right|\le|F|\in\mathbb N,\tag1$$ but since $B$ is infinite, this simplies $$B\not\subseteq\bigcup_{x\in F}\Omega_x\tag2.$$ In particular, $$K\not\subseteq\bigcup_{x\in F}\Omega_x\tag3.$$

However, I'm not able to conclude that this is a contradiction to the compactness of $(K,\tau)$. It is clearly a contradiction to the assertion that $(B,\left.\tau\right|_B)$, where $\left.\tau\right|_B:=\{\Omega\cap B:\Omega\in\tau\}$, is compact. But unless we assume that $B$ is $\tau$-closed, $(B,\left.\tau\right|_B)$ doesn't need to be compact.

What am I missing?

$^1$ $x\in E$ is called $\tau$-accumulation point of $B$ if for every $\tau$-neighborhood $N$ of $x$ it holds $B\cap N\setminus\{x\}\ne\emptyset$.

答:
What you’re missing is that an accumulation point of $B$ need not be in $B$. If $B$ has none, then each $x\in K$ has an open nbhd $U_x$ such that $U_x\cap B\subseteq\{x\}$. $\{U_x:x\in K\}$ is an open cover of $K$, so there is a finite $F\subseteq K$ such that $K=\bigcup_{x\in F}U_x$. Each $U_x$ contains at most one point of $B$, and $B\subseteq K$, so $B$ must be finite.

You could also work in $\operatorname{cl}B$, which is compact and must contain any accumulation points of $B$, but it’s simpler just to work in $K$.


问:
Solving inequality involving complex numbers
If z lies on the curve |z| = 1 such that $ a <|z + 1| + |1 + z^2 – z| <b $ then (a, b) can be

My approach

$Z=x+iy$

$X^2+y^2=1$

On arranging the given equation

a< $\sqrt{2+2x}+ \sqrt{2+x^2+2x^3-6x}$ <b

Further the only way I know is to differentiate to get min value which yields wrong answer . Any  better approach?

答: Write $z=e^{it},\,t\in\Bbb R$ and $c:=\cos(t/2)$. We wish to constrain$$|1+z|+\frac{|1+z^3|}{|1+z|}=2|c|+|4c^2-3|$$for $c\in[-1,\,1]$, or equivalently $2c+|4c^2-3|$ for $c\in[0,\,1]$. This function is $3+2c-4c^2$ (maximal at $c=1/4$ with value $13/4$) on $[0,\,\sqrt{3}/2]$, but $-3+2c+4c^2$ on $[\sqrt{3}/2,\,1]$. On $[0,\,\sqrt{3}/2]$, the minimum occurs at $c=\sqrt{3}/2$ with value $\sqrt{3}$; on $[\sqrt{3}/2,\,1]$, the maximum occurs at $c=1$ with value $3$. So $a$ can be any value less than $\sqrt{3}$, while $b$ can be any value greater than $13/4$.


问:
Doubt about conditions of divisibility
I was reading the solution of an exercise, which claims this:

$$4n^2 + 1 \equiv 0 \pmod{13} \iff n^2 \equiv 1 \pmod 5 \iff n \equiv
\pm 1 \pmod 5$$

and

$$4n^2 + 1 \equiv 0 \pmod{13} \iff n^2 \equiv 3 \pmod{13} \iff n
\equiv \pm 4 \pmod{13}.$$

Probably this is a stupid question, but i am new at this field so i am not sure how to do that, anyway, how do we go from the first equality to the second in both cases?

For example, maybe like this?:
$4n^2 + 1 = 13a \Longrightarrow n^2 = 5*(2a) + (1-3n^2+3a) ... $

But so we need to prove that $(1-3n^2+3a) = 1$?

答: Hint: $ 4n^2 \equiv -n^2 \pmod{5}$ and $ 12n^2 \equiv -n^2 \pmod{13}$

Title: Doubt about conditions of divisibility


问:
Not sure how to go about solving this integral
$\displaystyle \int \left( \frac{1}{x^2+3} \right)\; dx$

I've let $u=x^2+3$ but can't seem to get the right answer.

Really not sure what to do.

答:
Hint: Note that $$\frac{1}{x^{2}+3}=\frac{1}{3}\cdot \frac{1}{\left( \frac{x}{\sqrt{%
3}}\right) ^{2}+1}$$

and use the substitution $u=\frac{x}{\sqrt{3}}$.


问:
What is $-i$ exactly?
We all know that $i$ doesn't have any sign: it is neither positive nor negative. Then how can people use $-i$ for anything?

Also, we define $i$ a number such that $i^2 = -1$. But  it can also be seen that $(-i)^2 = -1$. Then in what way is $-i$ different from $i$? If it isn't, then why do we even have something like a  $-i$?

答:
I quickly scanned all the answers and didn't see anyone mention rotations, so I thought I would -- the simplest answer to your question is that $-i$ just represents a rotation of $\frac{\pi}{2}$ clockwise, whereas $i$ represents a rotation of $\frac{\pi}{2}$ counterclockwise. In other words, $i$ and $-i$ are inverses in the group $SO(2)\cong\left\{e^{i\theta}\,|\,\theta\in[0,2\pi)\right\}$ of rotations of the plane. From this perspective, $i$ and $-i$ are about as emphatically "different" as anyone could imagine.

EDIT: I thought I'd add that yet another way of saying what I've already said is that complex conjugation is just a way of swapping the orientation of the plane. (In particular, it is an orientation-reversing isometry.) The seemingly "sign-like" difference between $i$ and $-i$ is really just a difference in the choice of orientation.

Title: What is $-i$ exactly?


问:
Interchanging limits and little-o notation
If $f(n)=\displaystyle\sum_{l=1}^{\infty} o(1)$ converges for all finite $n,$ is it possible to say $\displaystyle\lim_{n\to\infty}f(n)=0$?

Thank you for your help.

答: If $a(n,\,l)=[n\ge l]/n$ in terms of Iverson brackets, then $f(n)=\sum_{l=1}^n1/n=1$.


问:
Proof that a collection of sets is a sigma-algebra
I need some help with this exercise:

Let $(X,\mathcal{M},\mu)$ a measure space, $Y$ a non-empty set and $\phi:X\rightarrow Y$ a function.

Define $\mathcal{N}=\{E\subseteq Y:\phi^{-1}(E)\in\mathcal{M}\}$ and prove that $\mathcal{N}$ is a $\sigma$-algebra.

Solution:

I have to prove the following:

$A\in\mathcal{N}\Rightarrow A^c\in\mathcal{N}$

$\{A_k\}_{k\geq 1}$ is a sequence of set of $\mathcal{N}$ then $\bigcup_{k=1}^{\infty}A_k\in\mathcal{N}$

For the first one I've done the following:

Let $A\in\mathcal{N}$ then for definition $\phi^{-1}(A)\in \mathcal{M}$ but since $M$ is a $\sigma$-algebra $\phi^{-1}(A)^c\in\mathcal{M}$. Now:
\begin{equation}
\phi^{-1}(A)=\{x\in X:\phi(x)\in A\}\Rightarrow\phi^{-1}(A)^c=\{x\in X:\phi(x)\notin A\}\Rightarrow\phi^{-1}(A)^c=\{x\in X:\phi(x)\in A^c\}=\phi^{-1}(A^c)
\end{equation}
and from the last we deduce that $A^c\in \mathcal{N}$.

For the second one I choose $\{A_k\}_{k\geq 1}$ a sequence of set in $\mathcal{N}$, then, for definition of $\mathcal{N}$ we have that $\phi^{-1}(A_k)\in\mathcal{M}$ $\forall k=1,...,n,...$ and that:

$\bigcup_{k=1}^{\infty}\phi^{-1}(A_k)\in\mathcal{M}$

Now:

\begin{equation}
\bigcup_{k=1}^{\infty}\phi^{-1}(A_k)=\bigcup_{k=1}^{\infty}\{x\in X:\phi(x)\in A_k\}=\{x\in X:\phi(x)\in\cup_{k=1}^{\infty}A_k\}=\phi^{-1}(\cup_{k=1}^{\infty}A_k)
\end{equation}

And the last one implies that $\bigcup_{k=1}^{\infty}A_k\in\mathcal{N}$

Is the proof correct?

答:
Yes the proof is fine. As noted in the comment, be careful and do not forget the empty set conditions, sometimes it seems trivial but it is a condition, and we love it nonetheless.

Note, just as a remark, how useful it is that the preimage commutes with the complement and the union operations (among others). As an example, this would not be the case for the “image” of a function, that does not commute with the complement operation.


问:
Show that $\mathbb F^{ \;m \times n}$ is isomorphic to $\mathbb F^{m n}$,where $\mathbb F$ is a field.
Show that $\mathbb F^{ \;m \times n}$ is isomorphic to $\mathbb F^{m n}$,where $\mathbb F$ is a field.

We can show that both have the same dimension:
$$\text{dim}(\mathbb F^{ \;m \times n})=mn=\text{dim}(\mathbb F^{m n})$$

On the other hand from this link we know there exists an isomorphism between the two vector spaces,hence the result.

I'm not sure if my argument is true.

答:
Your argument is correct.

If you denote $E_{ij}$ the matrix having all entries equal to zero except the one at $i$th-row and $j$th-column which is equal to one, then
$$\{E_{ij} \mid 1 \le i \le m, \, 1 \le j \le n\}$$ is a basis of $\mathbb F^{m \times n}$. Therefore $\dim \mathbb F^{m \times n} = mn.$


问:
show this $\lceil \frac{n}{1-a_{n}}\rceil =n+1$
let $a_{n}$ be squence such $a_{1}=2-\dfrac{\pi}{2}$, and
$$a_{n+1}=2a_{n}+\dfrac{a_{n}-1}{n},n\in N^{+}$$
show that
$$f(n)=\lceil \dfrac{n}{1-a_{n}}\rceil =n+1$$

I try:since
$$f(1)=\lceil \dfrac{1}{\dfrac{\pi}{2}-1}\rceil=2$$
and $$a_{2}=2a_{1}+a_{1}-1=3a_{1}-1=5-\dfrac{3}{2}\pi$$
so
$$f(2)=\lceil \dfrac{2}{\dfrac{3\pi}{2}-4}\rceil=3$$
and so on ,if we want prove

$$f(n)=\lceil \dfrac{n}{1-a_{n}}\rceil =n+1$$
or
$$n<\dfrac{n}{1-a_{n}}\le n+1$$
or
$$0<a_{n}\le\dfrac{1}{n+1}$$
if we use induction to prove it.then
$$a_{n+1}=(2+\dfrac{1}{n})a_{n}-\dfrac{1}{n}\le (2+\dfrac{1}{n})\cdot\dfrac{1}{n+1}-\dfrac{1}{n}=\dfrac{1}{n+1}$$
is wrong,becasue we want to prove $a_{n+1}\le\dfrac{1}{n+2}$

答:
This is only a partial solution, but it should be completed without too much extra effort (I currently have not sufficient time to work on it). As pointed out in a comment, it suffices to show $0<a_n\leq\frac1{n+1}$.

First, we let $a_1=t$ be an indeterminate, and write $a_n=x_nt+y_n$ as obtained by the recurrence. Then $$\begin{align}\begin{bmatrix}x_1\\y_1\end{bmatrix}&=\begin{bmatrix}1\\0\end{bmatrix},\\ \begin{bmatrix}x_{n+1}\\y_{n+1}\end{bmatrix}&=\frac{2n+1}n\begin{bmatrix}x_n\\y_n\end{bmatrix}+\begin{bmatrix}0\\-\frac1n\end{bmatrix}.\end{align}$$

Now easily, $$x_n=\prod_{k=1}^{n-1}\frac{2k-1}k=\frac{(2n-1)!}{2^{n-1}[(n-1)!]^2}.$$

Next, define $z_n=-\frac{x_n}{y_n}$. (Keep in mind that we want $x_n+ty_n\rightarrow0$, so $z_n$ should go to our desired initial value.) Substitute $y_n=-x_nz_n$ in the recurrence for $y_n$, and after some calculation, we get $$(z_{n+1}-z_n)x_{n+1}=\frac1n.$$

Hence, $\{z_n\}$ is an increasing sequence. Explicitly, $$z_n=\sum_{k=1}^{n-1}\frac1{kx_{k+1}}=\sum_{k=1}^{n-1}\frac{2^k(k!)^2}{k(2k+1)!}.$$

By comparison test ($\frac1{kx_{k+1}}<\frac1{k^2}$), $\lim_{n\rightarrow\infty}z_n$ exists. Write it as $a$: (We certainly want $a=2-\frac\pi2$.) $$a=\sum_{k=1}^\infty\frac{2^k(k!)^2}{k(2k+1)!}$$ Since the sequence is increasing, $a-z_n>0$ for all $n$.

Now take $a_1=a$ in the original sequence. Then $$a_n=ax_n+y_n=x_n(a-z_n)>0.$$

It remains to show two points: (1) $a=2-\frac\pi2$; (2) $x_n(a-z_n)\leq\frac1{n+1}$.


问:
What am I doing wrong in these quartic formula calculations?
I was a bit surprised that there is a general formula for the roots of a quartic equation, so I decided to test Wikipedia's version of it myself. To my surprise, I have arrived at a correct answer only once in about five attempts, using only integer coefficients of factorizable polynomials.

I am testing the formula for the equation $(x-1)(x-2)(x-3)(x-4)=0$, which should obviously return 1, 2, 3, and 4. However, my calculations tell a different story:

Expanding gives $x^4 - 10x^3 + 35x^2 - 50x + 24$. Therefore, $a = -10, b = 35, c = -50, d = 24$.

$u$ is therefore equal to $\frac{3(-10)^2 - 8(35)}{12} = \frac{300 - 280}{12} = \frac{20}{12} = \frac{5}{3}$.

$\Delta_0$ is $35^2 - 3(-10)(-50) + 12(24) = 1225 - 1500 + 288 = 13$.

$\Delta_1$ is $2(35^3) - 9(-10)(35)(-50) + 27(-10)^2(24) + 27(-50)^2 - 72(35)(24)$, which works out to 70.

The discriminant is $-\frac{\Delta_1^2 - 4\Delta_0^3}{27} = -\frac{-3888}{27} = 144$. Because it is positive, the equation must have either four real roots or four complex roots.

$Q$ is equal to $\sqrt[3]{\frac{70 + \sqrt{-3888}}{2}} = \sqrt[3]{\frac{70 + 36i\sqrt{3}}{2}} = \sqrt[3]{35 + 18i\sqrt{3}}$.

$v$ is $\frac{\left(\sqrt[3]{35 + 18i\sqrt{3}}\right)^2 + 13}{3\sqrt[3]{35 + 18i\sqrt{3}}}$. After rationalizing the denominator it becomes $\frac{1 + \sqrt[3]{35 - 18i\sqrt{3}}}{3}$.

$w$ might make things even more complicated, but mercifully its numerator (and therefore value) is zero.

Placing these values into one of the final expressions for the formula gives

$$\frac{1}{4}\left(10 + 2\sqrt{\frac{6 + \sqrt[3]{35 - 18i\sqrt{3}}}{3}} \pm 2\sqrt{\frac{9 - \sqrt[3]{35 - 18i\sqrt{3}}}{3}}\right),$$

a number which is neither rational nor even fully real, and the other two roots are the same.

I expected that the error would become self-evident as I typed this question (which took a while, as you can imagine) — but it has not. So what is wrong with these calculations?

答:
Everytime there is a complicated cube root or square root, there is a simplification that can occur.

For example, you can compute $(35+18i\sqrt 3)^{1/3}$ approximately in $\Bbb C$, or you can magically realize that $(-1-2i\sqrt 3)^3 = 35+18i\sqrt 3$ and so you can pick $Q = -1-2i\sqrt 3$ (don't try to find this cube root algebraically).

This should help you with the rest, and feel free to see what happens when you replace $Q$ with one of the other two cube roots.


问:
Showing that the sequence is bounded without finding its limit
I am given a sequence ${x_n}=1+\frac{1}{2}+...+\frac{1}{n}-\ln n$ and I have to show that it's bounded.

I tried doing this: $$1\leq 1+\frac{1}{2}+...+\frac{1}{n}\leq n$$
$$1-\ln n\leq 1+\frac{1}{2}+...+\frac{1}{n}-\ln n\leq n-\ln n$$
$$1-\ln n\leq{x_n}\leq n-\ln n$$
But I don't think I am allowed to do this since the boundaries depend on n.
Is there maybe another way to solve this? and could you please help me?
Thanks a lot!
I can't use integrals, derivatives or limits

答:
You have

$$x_{n+1} - x_n = \frac{1}{n+1} -\ln \left(1+ \frac{1}{n}\right)= -\frac{1}{2n^2} + O(\frac{1}{n^3})$$

As $\sum 1/n^2$ converges, you get the desired result.


问:
Domain of Double Integrals
I'm working on this double integral:

$ \int_\mathcal{D} \sqrt{x^4+1} dA \; \; \text{At the domaine       } \mathcal{D} = \{(x, y) \; |\; 0 \leq y \leq 8 \; \text{ and }\; \sqrt[3]{y} \leq x \leq 2 \} $

My questions, how to calculate the domain. As far as I can understand (please correct me if wrong!), the integral bounds for y is simply 0 to 8:

$ \int_0^8  $

But how about the domain for x? What confuses me, is that there is BOTH a y-variable and x-variable in

$ 3\sqrt(y) \leq x \leq 2 $

Hope you can help me!

答:
Taking a look at the graph of $x=y^{\frac{1}{3}}$ and the restrictions $0\le y\le 8,y^{\frac{1}{3}}\le x\le 2$, we can easily see that the integral can be written as
$$\int_0^2 \int_0^{x^3}\sqrt{x^4+1}dydx=\int_0^2\sqrt{x^4+1}x^3dx=\dfrac{(x^4+1)^{\frac{3}{2}}}{6}\bigg|_0^2=\dfrac{17\sqrt{17}-1}{6}.$$
Notice that $y\in [0,8]$, since the value of $f(x)=x^3$ has image in $[0,8]$ when $x\in [0,2]$.

Using this formulation of the integral it becomes easier to solve because we can make the substitusion $u=x^4+1$, whose derivative is exactly the term $x^3$ multiplied by a constant.

Title: Domain of Double Integrals


答:
You should perform this integral by switching the order of integration. The domain can also be described by $0 \leq x \leq 2$ and $0 \leq y \leq x^3$. Then,
$$I = \iint \sqrt{x^4+1 } \mbox{ dA} = \int_0 ^2 \int _0 ^{x^3}\sqrt{x^4+1 } \mbox{ dy dx}.$$
Evaluating the $y-$integral first gives
$$ I =  \int_0 ^2 x^3 \sqrt{x^4+1} \mbox{ dx} = \frac{1}{6}(x^4+1)^{3/2} |_0 ^2 = \frac{1}{6}(17\sqrt{17} - 1).$$

Title: Domain of Double Integrals

Body:
I'm working on this double integral:

$ \int_\mathcal{D} \sqrt{x^4+1} dA \; \; \text{At the domaine       } \mathcal{D} = \{(x, y) \; |\; 0 \leq y \leq 8 \; \text{ and }\; \sqrt[3]{y} \leq x \leq 2 \} $

My questions, how to calculate the domain. As far as I can understand (please correct me if wrong!), the integral bounds for y is simply 0 to 8:

$ \int_0^8  $

But how about the domain for x? What confuses me, is that there is BOTH a y-variable and x-variable in

$ 3\sqrt(y) \leq x \leq 2 $

Hope you can help me!


问:
How do I transform $\frac{1-\exp(-Vx^m)}{1-(1-x^m)^V}-1$ so I can evaluate numerically for $V$ and $m$ between $1$ and $30$, and $0\leq x\leq 1$?
I need to evaluate and plot the following expression:

$$
f(x,V,m)=\frac{1-\exp(-Vx^m)}{1-(1-x^m)^V}-1
$$

$V$ and $m$ are numbers from $1$ to $\sim30$ and $0\leq x\leq1$.

The issue is that for $x$ close to zero, the exponential, and the power term in the denominator take values very close to 1, making it difficult to evaluate $f$ numerically. I've tried MATLAB symbolic toolbox which, just to give one case, gives $f(10^{-3},1,6)=-1$. Mathematica also runs into similar problems. I know this to be incorrect because:

$$
\lim_{x\to 0} f(x,V,m)=0
$$

Is there a manipulation of the equation (the first term is the problem) that I am missing that would make it more amenable to evaluate numerically or simplify it? I want to know the behavior of the function for small values of $x$ but short of plotting I am struggling to conclude anything, because even though both numerator and denominator are monotonic, I cannot say the same about their ratio.

答:
Note that with $t=vx^m$, the expression is

$$\frac{1-e^{-t}}{1-\left(1-\dfrac tv\right)^v}-1=\frac{\left(1-\dfrac tv\right)^v-e^{-t}}{1-\left(1-\dfrac tv\right)^v}.$$

We know that $\left(1-\dfrac tv\right)^v$ is an approximation to $e^{-t}$, with the development

$$1-t+\left(1-\frac1v\right)\frac{t^2}2-\left(1-\frac3v+\frac2{v^2}\right)\frac{t^3}{3!}+\cdots$$

Your function is approximated by

$$\frac tv\,\frac{-\dfrac{1}2-\left(-3+\dfrac2{v}\right)\dfrac{t}{3!}+\cdots}{-1+\left(1-\dfrac1v\right)\dfrac{t}2-\left(1-\dfrac3v+\dfrac2{v^2}\right)\dfrac{t^2}{3!}+\cdots}$$


问:
Understanding Multiplication/Dot Product of Two Numbers/Two Vectors
This might be a silly question, but I just realized I don't think I understand what multiplication really is. For example I know $3\cdot4=12$ because that's what you learn in school. If we go with the idea of multiplication as repeated addition, $3\cdot4$ means add $3$ to $0$ (the zero elements) $4$ times. So here we have multiplication defined in terms of addition and the zero/identity element for natural numbers.

My confusion is around numbers that have units. Say we have oranges, how does one interpret $3$ oranges $\cdot 4$ oranges?

Does adding a unit (oranges) to the number make it a vector and therefore the multiplication operator has a different meaning because for two vectors it's dot products? If so, why do we interpret the multiplication of a basis vector with itself to be 1?

答:
I think learning a bit of Abstract Algebra (here's an introduction) will answer some of these questions. An operator (addition/multiplication/etc) and type of objects (integers/reals/vectors/etc) it acts on - come together. You need to define what a multiplication is for this type of objects. For instance:

Multiplication of 2 integers can be represented with repeated addition as you mentioned

(Scalar) Multiplication of a geometrical vector (an arrow) is stretching squeezing that vector. Imagine a physical stretching of an arrow without multiplication of its components. You just have an instruction to stretch 2x and that's what you do with a pen and paper.

You can come up with any mathematical operation which does craziest things. But most of such operations will be useless. E.g. they won't help finding unknown $x$ in the equation: $a*x=b$ where $*$ is your new mathematical operation, and $a, x, b$ are your new type of objects (I don't know.. let's say they are circles - you multiply circles). While existing, real maths creates these structures & operations in a way to be useful for some purpose.

Dot product is not a multiplication in the typical sense. E.g. you can't solve  $a\cdot x=b$. This operation has a completely different purpose - it's useful for different reasons. E.g. it can be used:

To define length of a vector: $|a|=\sqrt{a\cdot a}$

To find a unit vector (of length 1) that points to the same direction: $\frac{a}{|a|}$

To find out if vectors are perpendicular (when $a\cdot b=0$)

To find projections of one vector onto another ($b\cdot \frac{a}{|a|}$)

That's why Dot Products are useful - not because they can e.g. solve equations. Incidentally all this math can be generalized. Notice that e.g. functions like polynomials also have similar properties like geometrical vectors: you can multiply them by a scalar, you can add them. But then - what would be a length of such "vector"? What is a dot product? To answer these questions a list of requirements are introduced for such products taken from what we just saw like length of a unit vector $\frac{a}{|a|}$ is 1, and others. So if we could come up with a new operation that obeys these properties - we can use it the way we used dot products.

Such products are now called Inner Products and Dot Product is just one of the implementations. For functions for instance you could define a different implementation like $\int{f(x)f(x)dx}$ is going to be similar to $a \cdot a$. It's possible to define all sorts of implementations for Inner Product, but they still have to obey those roles. And now we can define what length is and it means to be perpendicular for functions and other entities that obey similar rules. And we can re-use all the maths from vectors with these entities - when used in such a way these entities and their respective operations are called Abstract Vectors.

Now units is about giving physical sense to numbers. Can you multiply oranges? Mathematically - the requirement is that you have right entities (e.g. integers) and appropriate operation (e.g. multiplication). But whether this has physical sense - that's not up to math to decide. If $orange^2$ (area of a field of oranges) makes sense - then go ahead and do the operation.

If you use oranges as one of the components of your vector e.g. then $|o|=\sqrt{o \cdot o}$ is still a length of a vector, so it gives you number of oranges in it (which you knew from the vector's component already).


问:
Show that the maximum of a set of convex functions is again convex
Let $f_1(x), f_2(x), \ldots, f_n(x)$ be a set of convex functions. We define $f(x)$ as

$$ f(x) = \underset{i}{\text{max}} \left\{ f_i(x) \right\}. $$

How do I show that $f(x)$ is also convex, and the domain of $f$ is intersection of the domains of $f_i$?

答:
Another way to do it: $y\geq\max f_i(x)$ iff $y\geq f_i(x)$ for each $i$;
so the epigraph of $\max f_i$ is the intersection of the epigraphs of $f_i$, so it is convex, as an intersection of convex is convex. This also shows that the family $f_i$ is arbitrary, ie a $\sup f_i$ is convex whenever the $f_i$s are.


问:
Which real functions have their higher derivatives tending pointwise to zero?
Let $\mathrm C^\infty\!(\Bbb R)$ be the space of infinitely differentiable functions $f:\Bbb R\rightarrow\Bbb R$, and define the subspace$$A:=\{f\in\mathrm C^\infty\!(\Bbb R):(\forall x\in \Bbb R)\lim_{n\rightarrow\infty} f^{(n)}(x)=0\},$$where $f^{(n)}$ is the $n$th derivative of $f\;(n=0,1,\dots).$ Clearly all polynomial functions are in $A$. Are any others?

Edit: Alfonso has answered this question well, but is there any characterization of $A$ in terms of familiar types of function?

Body:
$$f(x)=e^{x/2}$$
$$f^{(n)}(x)=2^{-n}e^{x/2}$$


问:
Derived tensor product , independence of resolution
Right after Lemma 20.26.13

We have the following paragraph of how to derive the tensor product.

It  claims that the end result is independent of choice of K-flat resolution
Suppose we take $L^\bullet \rightarrow F^\bullet$, then we have to show a quasi iso.  $$Tot(G^* \otimes K^*) \simeq Tot(G^* \otimes L^*)$$

This would follow if we can exhibit a map $K^* \rightarrow L^*$ (or vice versa) and apply 20.26.13, but this is unclear to me.

What am I missing?

Some explaination would be appreciated.

答:
Suppose $\alpha_K:K^*\to F^*$ and $\alpha_L:L^*\to F^*$ are two $K$-flat resolutions of $F^*$.

As one typically proves when constructing the derived category (see Prop. 10.4.1 of Weibel's Introduction to Homological Algebra, for example) there is a complex $G^*$ with quasi-isomorphisms $\beta_K:G^*\to K^*$ and $\beta_L:G^*\to L^*$ such that $\alpha_K\beta_K$ and $\alpha_L\beta_L$ are chain homotopic.

Let $\gamma:M^*\to G^*$ be a $K$-flat resolution of $G^*$. Then $\beta_K\gamma:M^*\to K^*$ and $\beta_L\gamma:M^*\to L^*$ are quasi-isomorphisms of $K$-flat complexes, so you can now apply Lemma 20.26.13 of the Stacks Project.


问:
Limit points of the set $\{\frac {\varphi(n) }n : n\in \mathbb{N}\}$
Let $\varphi(n)$ denotes the cardinality of the set $\{a | 1\le a\le n, (a,n)=1\}$ where $(a , n)$ denotes the gcd of $a$ and $n$. Which of the following is NOT true ?

$(a)$ There exist infinitely many $n$ such that $\varphi (n)\gt \varphi(n+1)$

$(b)$ There exist infinitely many $n$ such that $\varphi (n)\lt \varphi(n+1)$

$(c)$ There exist $N\in \mathbb{N}$ such that $N\gt 2$ and for all $n\gt N$,  $\varphi (N)\lt \varphi(n)$

$(d)$ The set $\{ \frac {\varphi (n)}{n} : n\in \mathbb{N} \} $ has finitely many limit points.

My Thinking:

$(a)$ is true  by selecting $n=p$  ( an odd prime greater than $3$)

Justification:

Since $p$ is odd, $p+1$ is even. Now $\varphi(p)=p-1$ and $\varphi(p+1)\le p+1-\frac{p+1}{2}=\frac{p+1}2$

So $\varphi(p+1)\le \frac{p+1}2 \lt p-1=\varphi(p)$

By the infinitudes of primes, the result follows.

$(b)$ Is true by taking $n=p-1$ where $p$ is an odd prime greater than $3$ and following the same line of reasoning.

$(c)$ This statement (if true ) means that the set $\{\varphi(n) : n\gt N\}$ is bounded below.

I think it is true

My argument :

Let $N=6$ . Then $\varphi(6)=2$

If $n\gt 6$ and is prime  , then clearly $\varphi(n)\gt 2$

Otherwise, if $n$ is composite, then let

$n=p_1^{r_1} p_2^{r_2}...p_k^{r_k}$ .

$\varphi(n)=p_1^{r_1-1}p_2^{r_2-1}...p_k^{r_k-1}(p_1-1)(p_2-1)..(p_k-1)$

If $\phi(n)=2$ , then the distinct primes can be just $2$ and $3$. So the above equality reduces to

$2=\phi(n)=2^{r_1}3^{r_2-1}$  when both the factors $2$ and $3$ are present .

This gives $r_1=r_2=1$ and so $n=6$

When only $3$ present ,  then we get $n=3$ and if the factor is only $2$ , then $n=4$.

So the result follows as $N=6$ works

$(d)$ This is NOT true.

Justification :- Let $\{p_i\}$ be the enumeration of primes.

Let an index $k$ be fixed and $i$ be the running index.

Then if the only prime factors of  $n_i$ are $p_k, p_i$ , then
$\frac{\varphi(n_i)}{n_i}=\big( 1-1/p_k\big)\big(1-1/p_i\big)$

So ,letting $i\to \infty$, we have

$\lim_{i\to \infty}\frac{\varphi(n_i)}{n_i}=1-1/{p_k}$

Since $p_k$ is arbitrary, there are infinitely many limit points (maybe uncountably many) limits points of the given set.

Please check if my work is correct or the answers can be shortened in any way.

Thanks for your time.

答:
(c) can be shortened by noting that if $n$ is $> 6$ we can find the following coprimes. If $n$ is even, $1$ $n-1$ are coprime with $n$. Also, one of $n/2 + 1$ and $n/2 + 2$ is coprime with $n$ (depending on whether $n/2$ is even), so $\phi(n) \geq 3$. If $n$ is odd, $1,2,n-1$ are coprime with $n$ and distinct so $\phi(n) \geq 3$.

(d) looks correct although the definition of $n_i$ is not very clear.


问:
Factorized Conditional Expectation w.r.t. a smaller domain
Let $X$ and $Y$ be real-valued random variables with joint density denoted by $f:=f^{(X,Y)}$ for sake of brevity.
Prove that
$$
E[X \mid \{Y=y\}] := \frac{1}{f^Y\!(y)} \int_\mathbb{R} x \cdot f(x,y) \, \lambda(dx)
$$
for all $y\in\mathbb{R}$ with $f^Y\!(y)>0$ is a factorized conditional $E[ X \mid Y = \cdot \,]$ with domain $\{f^Y > 0\}$.

I cannot understand the concept of Factorized Conditional Expectation. Here, the exercise asks to show that via the given formula a factorized conditional expectation is defined, not on the entire real numbers but on the smaller domain $\{f^Y>0\}$, where the formula actually makes sense.

How can I do that?

答:
This is indeed slightly weird.

Denote by $\mathcal{Y}$ the domain $\{f^Y > 0\}$. Clearly, $\mathcal{Y}$ is non-empty: otherwise, the probability density function $f^Y$ would not integrate to one. Fix $y_0 \in \mathcal{Y}$. Define a new random variable $\tilde{Y}$ according to
\begin{align*}
\tilde{Y}(\omega) = Y(\omega) \boldsymbol{1}_{\{Y(\omega) \in \mathcal{Y}\}} + y_0 \boldsymbol{1}_{\{Y(\omega) \notin \mathcal{Y}\}}.
\end{align*}
Then $\tilde{Y}$takes values in the smaller domain $\mathcal{Y}$. As far as I understand the question, it asks us to prove that the expression is a factorized conditional expectation of $X$ given $\tilde{Y}$. In other words, we have to show that $(Q_y)_{y\in\mathcal{Y}}$ given by
\begin{align*}
Q_y(E) = \frac{1}{f^Y\!(y)} \int_E f(x,y) \, \lambda(\mathrm{d}x)
\end{align*}
is a $(\mathcal{Y},\mathbb{B}(\mathcal{Y}))$-Markov kernel on $(\mathbb{R},\mathbb{B}(\mathbb{R}))$ which satisfies
$$
\mathbb{P}(X \in E, \tilde{Y} \in K) \overset{\star}{=} \int_K Q_y(E) \, \tilde{Y}(\mathbb{P})(\mathrm{d}y).
$$
for $E \in \mathbb{B}(\mathbb{R})$, $K \in \mathbb{B}(\mathcal{Y})$.

Checking that $(Q_y)$ is a Markov kernel is straightforward. The following argument verifies that $\star$ holds: Denote by $\tilde{f}$ the joint probability density function of $(X,\tilde{Y})$. By definition, $\tilde{f}^Y = f^Y$ and $f = \tilde{f}$ almost everywhere (on $\mathcal{Y}$ and $\mathcal{Y}\times\mathbb{R}$ w.r.t. the Lebesgue measure), thus
\begin{align*}
\int_K Q_y(E) \, \tilde{Y}(\mathbb{P})(\mathrm{d}y)
&=
\int_K \frac{\tilde{f}^Y\!(y)}{f^Y\!(y)} \int_E f(x,y) \, \lambda(\mathrm{d}x)\, \lambda(\mathrm{d}y) \\
&=
\int_{K \times E} \tilde{f}(x,y) \, \lambda(\mathrm{d}(x,y)) = \mathbb{P}(X \in E, \tilde{Y} \in K)
\end{align*}
as desired.


问:
Can the quotient set of a permutation equivalence relation be a vector space again?
Let $x,y\in\mathbb{R}^N$ and consider the equivalence relation $x\sim y\Leftrightarrow \exists \sigma\in\operatorname{Sym}(\mathbb{R}^N)\,\, x = \sigma(y)$ for some permutation $\sigma$. Let $A = \mathbb{R}^N/\sim$ be the quotient set. Is it possible to define an addition between elements of $A$ (+ inverse and scalar multiplication) such that $A$ becomes a vector space?

答:
Consider the vector $u\in\mathbb{R}^N$ with all entries equal to $1$. Then the map $a\mapsto [au]$ is an injection $\mathbb{R}\to\mathbb{R}^N$. By Cantor-Schröder-Bernstein, we conclude that $|\mathbb{R}^N|=|\mathbb{R}|$.

 This is because $|\mathbb{R}^N|=|\mathbb{R}|$ (assuming $N$ is a positive integer), so $|\mathbb{R}^N/{\sim}|\le|\mathbb{R}^N|=|\mathbb{R}|$

 Let $f\colon\mathbb{R}\to\mathbb{R}^N/{\sim}$ be a bijection. Use it to transfer the vector space structure.


问:
What is this so called "multiply-divide trick" used below for calculating the integral?
I am working through the exercises from this page. To solve exercise 3 they use a trick they call the "multiply-divide trick" to get $\int \frac{1}{\sin(2x)} \; dx = \int \frac{\sin(2x)}{1-\cos^2(2x)} \; dx$.

They have a short explanation of the "multiply divide trick":

Imagine that we have the expression $3x + 5$, but we would prefer to have the coefficent $7$ at $x$ and do not want other terms with $x$ there. This last requirement prevents us from adding and subtracting, but we can do this: $3x + 5 = 3⋅(x + 5/3) = (3/7)⋅(7x + 35/3)$.

However I don't get how this helps me with the integral above. So how do we get the integral equation above using the trick described here?

答:
Observe that your integral, after the "trick", is almost of the form

$$\int\frac{f'(x)}{1-f(x)^2}dx=\int\frac12\left(\frac{f'}{1-f}+\frac{f'}{1+f}\right)dx=\frac12\left(-\log|1-f|+\log|1+f|\right)+C=$$

$$=\log\sqrt{\frac{|1+f|}{|1-f|}} +C$$

and, thus, you can easily solve your integral


答:
This helps you because

$$\sin(2x) dx = -\frac{1}{2}d(\cos(2x))$$

Title: What is this so called "multiply-divide trick" used below for calculating the integral?

Body:
I am working through the exercises from this page. To solve exercise 3 they use a trick they call the "multiply-divide trick" to get $\int \frac{1}{\sin(2x)} \; dx = \int \frac{\sin(2x)}{1-\cos^2(2x)} \; dx$.

They have a short explanation of the "multiply divide trick":

Imagine that we have the expression $3x + 5$, but we would prefer to have the coefficent $7$ at $x$ and do not want other terms with $x$ there. This last requirement prevents us from adding and subtracting, but we can do this: $3x + 5 = 3⋅(x + 5/3) = (3/7)⋅(7x + 35/3)$.

However I don't get how this helps me with the integral above. So how do we get the integral equation above using the trick described here?


问:
How do you alternate positive and negative integers in set builder notation?
I want to rewrite this using set builder notation:

B = {−1, 1/2, −1/3, 1/4, −1/5, 1/6, · · · }

So far I can see its either alternating positive and negative integers to the power of -1. Or I could simply write it as 1/n where n is....

I can't work out how to express it.

Can anyone help?

would this work:

$n^{-1}$ if $n \in \mathbb{Z^+}$ and even

$n^{-1}$ if $n \in \mathbb{Z^-}$ and odd

答:
You have recognized that the $n$th term is $1/n$ when $n$ is an even positive integer and $1/n$ when $n$ is an odd negative integer.  However, we wish to index each term of the sequence with a positive integer.

Notice that
$$\frac{1}{-n} = -\frac{1}{n}$$
and that
$$(-1)^n = \begin{cases}
           1 & \text{if $n$ is even}\\
           -1 & \text{if $n$ is odd}
           \end{cases}
$$
Therefore, the $n$ term of the sequence is
$$\frac{(-1)^n}{n} = \begin{cases}\\
                     \dfrac{1}{n} & \text{if $n$ is even}\\[2 mm]
                     -\dfrac{1}{n} & \text{if $n$ is odd}
                     \end{cases}
$$
for each positive integer $n$.


问:
Number of non-congruent equilateral triangles which can be formed.
Two line segments $OA$ and $OB$ are inclined to each other at $120^\circ$ at point $O$. $
\quad P$ is a fixed point on the internal angle bisector of $\lt AOB$. $\quad OA = OB = 20cm, OP = 10cm$. Two points are chosen on the line segments $OA$ and $OB$ such that the two points and $P$ make an equilateral triangle. The two points maybe in the same line segment or in different line segments. If the number of such non - congruent equilateral triangles that can be made is $k$, then find the value of $[{200\over k}]$

My Work :-

According to me only $2$ non-congruent triangles can be formed;

$\ast$ One which would have vertices as $P, O$ and midpoint of $OA$

$\ast$ Second which would have vertices as $P$ and one vertex would be in $OA$ and another on $OB$

We also see another equilateral $\Delta$ which would have vertices as $P,O$ and midpoint of $OB$ however that triangle would be congruent to the first one; hence not considered.

Thus answer should be $100$.

答:
Let $C$ and $D$ be the midpoints of $AO$ and $OB$ respectively. Draw $PC$ and $PD$. Drop perpendiculars from point $P$ on $OC$ and $OB$ meeting them at point $E$ and $F$ respectively.

Notice that, if one vertex of any such equilateral triangle is between $C$ and $E$, then the other one(vertex $P$ is fixed ) has to be between $O$ and $F$. Every such triangle will have a congruent triangle which will have one of its vertices between $E$ and $O$ and the other one between $F$ and $D$. Also notice that there can't be two congruent triangles with both of them having one of their vertices between $C$ and $E$ and the other one between $O$ and $F$. Hence only those triangles are to be counted.
Choose a random point $Q$ on segment $CE$. Now choose a point $R$ on segment $OF$ such that $OR=10-OQ$. Draw $PQ$ and  $PR$.

Observe that, in $\triangle PCQ$ and $\triangle POR$,

$\left(i\right)$  $CQ=10-OQ=OR$

$\left(ii\right)$  $\angle PCQ=60^{\circ}=\angle POR$

$\left(iii\right)$  $PC=PO$ [ Since $\triangle PCO$ is equilateral ]

Hence, $\triangle PCQ\cong \triangle POR$ by $S-S-A$ criterion of congruence.

Thus, $\angle QPR=\angle QPO+\angle OPR=\angle QPO+\angle CPQ=\angle CPO=60^{\circ}$ and $PQ=PR$; Which indicates that $\triangle PQR$ is equilateral.

This is true for every point $Q$ on $CE$. Since there are an infinite number of such points on $CE$, $k\to\infty$.

Therefore, $\boxed {\lfloor\left(\frac{200}{k}\right)\rfloor \to 0}$


问:
Dimension of Bil(V)
Let $V$ be a vector space of finite dimension $n$, and let $\operatorname{Bil}(V)$ be the vector space of all bilinear forms on $V$.

In some notes by Keith Conrad, he says in an exercise that $\operatorname{Bil}(V)$ has dimension $n$ as well. I am confused by this, because it seems to me that the dimension ought to be $n^2$. Here is my reasoning: we have isomorphisms $\operatorname{Bil}(V)\cong (V\otimes V)^* \cong V \otimes V$ since $V$ is finite dimensional. The dimension of $V\otimes V$ is $n^2$. Where have I gone wrong here?

答: You have not gone wrong at all. More concretely, a bilinear form $B$ is associated with an $n\times n$ matrix [given a basis $\{v_i\}$, its $ij$-entry is, of course $B(v_i,v_j)$].</p>

Title: Dimension of Bil(V)


问:
Prove that $\triangle AMC$ is an isosceles right triangle

In $\triangle ABC$ construct squares $ABDE$ and $BCFG$. Let $M$ be the midpoint of $EF$ then, prove that $\triangle AMC$ is an isosceles right triangle.

I was able to prove that $\triangle AMC$ is an isosceles only and here is my solution:

The area of $\triangle BAF$ is equal to the area of $\triangle EBC$ so we get
\begin{equation}
AE×AF\sin \angle BAF=CF×CE\sin \angle ECB
\end{equation}
And using law of cosines on $EF$ we get
\begin{equation}
AE^2+AF^2=CF^2+CE^2
\end{equation}
By the length formula of median $AM$ and $CM$ we get that $AM=CM$. I know this solution is kinda rush and not give much detail but I'm not really good at Latex, please don't mind me. Please help prove that it is the right triangle. Thanks!

BTW, $N$ in the photo is the point where $AE$ meet $CF$

答:
We represent the points with complex numbers.

Let $ B$ be the origin, $ A = a, C = c$.

Show that

$E = (1+i) a $,

$F = (1-i) c$,

$ M = \frac{1}{2} \left[ (1+i) a + (1-i) c \right] $   .

$MA = - \frac{1}{2} \left[ (-1+i) a + (1-i) c \right]$.

$MC = - \frac{1}{2} \left[ (1+i) a + (-1-i) c \right] $.

$ MA = i MC$.

Thus, $MAC$ is an isosceles right angled triangle.


问:
Proving the Alternate Series Test
I am self-learning Real Analysis from Understanding Analysis by Stephen Abbott. I am finding it difficult to come up with a proof for the Alternating Series Test for an infinite series. I only have some initial ideas, so I'll put those down.

I'd like to ask if somebody could help, by providing any hints (please do not write the complete proof), on how to think about this.

Exercise 2.7.1

Proving the Alternating Series Test amounts to showing that the sequence of partial sums
\begin{align*}
 s_n = a_1 - a_2 + a_3 - \ldots \pm a_n
\end{align*}
converges. Different characterizations of completeness lead to different proofs.

(a) Produce the Alternating Series Test by showing that $(s_n)$ is a Cauchy sequence.

(b) Supply another proof for this result using the Nested Interval Property.

(c) Consider the sequences $(s_{2n})$ and $(s_{2n+1})$, and show how the Monotone Convergence Theorem leads to a third proof for the Alternating Series Test.

Proof.

(a)Let $(a_n)$ be a sequence satisfying

(i) $a_1 \ge a_2 \ge a_3 \ge \ldots \ge a_n \ge a_{n+1} \ge$

(ii) $(a_n) \to 0$

Our claim is that, the alternating series $\sum_{n=1}^{\infty}(-1)^n a_n$ converges.

Let $(s_n)$ be sequence of the partial sums, where
\begin{align*}
s_n = a_1 - a_2 + a_3 - a_4 + a_5 - a_6 + \ldots + (-1)^{n+1}a_n
\end{align*}

We are interested to show that $(s_n)$ is a Cauchy sequence. Let's see if can find a simple expression for $\lvert s_n - s_m \vert$. If $n > m$,

\begin{align*}
\lvert (s_n - s_m) \rvert &= \lvert (-1)^{m+2}a_{m+1} + (-1)^{m+3}a_{m+2} + (-1)^{n+1}a_{n} \rvert\\
&= \lvert {a_{m+1} - a_{m+2} + a_{m+3} - \ldots + (-1)^{n-m+1}a_n} \rvert
\end{align*}

A consequence of condition(i) is that $(a_n)$ is a decreasing sequence and has a lower bound zero, so $a_n \ge 0$ for all $n \in \mathbf{N}$. Since $a_{m+2} \ge 0$, $a_{m+4} \ge 0$, $a_{m+6} \ge 0, \ldots$ we can write,

\begin{align*}
\lvert (s_n - s_m) \rvert &= \lvert {a_{m+1} - a_{m+2} + a_{m+3} - \ldots + (-1)^{n-m+1}a_n} \rvert\\
&\le \lvert {a_{m+1} + a_{m+3} + a_{m+5} + \ldots + a_{n}} \rvert \\
&\le \vert {a_{m+1} + a_{m+1} + a_{m+1} + \ldots + a_{m+1}} \rvert
\end{align*}

But this gives me a variable number of terms (whereas I need to prove that the distance $\lvert s_n - s_m \rvert$ becomes smaller than fixed $\epsilon$ or $k\epsilon$. I am not quite sure, where to go from here. I know that the original sequence $(a_n)$ converges, so given any $\epsilon > 0$, $\exists N \in \mathbf{N}$, such that $\lvert a_n \rvert < \epsilon$ for all $n \ge N$.

(b) I am thinking of constructing a sequence of closed nested intervals $I_1 \supseteq I_2 \supseteq I_3 \supseteq \ldots \supseteq I_n \supseteq I_{n+1} \supseteq \ldots$, in such a way that, $I_1=\{s_1,s_2,s_3,\ldots\}$, $I_2 = I_1 - \{s_1\}$ and in general $I_{k+1} = I_k - \{s_k\}$, as I am interested in the tail of the sequence of partial sums. Then, I can apply the nested interval property to show that $\bigcup_{k\ge 1}I_k$ is not empty.

The one thing I don't know, is whether $(s_n)$ is bounded. If and only if this set is bounded, one may construct a closed interval. I think this is important for the construction.

(c) Again, I need some bounds for $(s_n)$, I guess.

答: Consider  $s_{2n} = a_1 - (a_2 - a_3) - \cdots -(a_{2n-2} - a_{2n-1} ) - a_{2n} \leq a_1 $ and it is easy to observe that $s_{2n}$ is increasing . So it is bounded and monotone hence converges and let the limit be $s$. Further we have $s_{2n+1} = s_{2n} + a_{2n+1}$. Since $a_n \to 0$ as $n \to \infty$ we have $\lim s_{2n+1} = \lim s_{2n} = s$.

Title: Proving the Alternate Series Test


问:
understand the relation between orientation , in circle test and determinant
i come from a software engineering background and I'm new to computational geometry ,
I found this picture online and can't manage to understand from where the logic comes from .
I know that determinants are used to know how a transform squishes space but i don't see the link here .

答:
$$\begin{vmatrix}
a_x - c_x & a_y -c_y\\
b_x - c_x & b_y -c_y
\end{vmatrix}$$ is equal to $\Vert ca \Vert \Vert cb \Vert \sin(ca, cb)$ where $\Vert \cdot \Vert$ stands for the norm. The sign of $\sin(ca, cb)$ depends on the fact that $c$ is on one side or the other of the line $ab$. This explains why the sign of the determinant enables to decide on the position of $c$ relatively to the line.

Similar considerations regarding the distance of $d$ vs. the radius of the circle defined by the three points $a,b,c$ enables to explain the second determinant. Have a look to this question.


问:
Origin of the name 'place' in algebraic number theory
I would like to explore the origin of the name  'place' in number theoretic sense.
A place of an algebraic number field $K$ is defined as an equivalence class of absolute
value functions on $K$. There are two types of places: the infinite and infinite
places.

It is well known that the 'finite' places of a number field are indeed in
one-to-one correspondence with the (maximal) prime ideals of $K$'s ring of integers $R$. See for example this discussion.
If we think about the variety/ scheme $V= \operatorname{Spec}(R)$ associated to $R$
then the finite places can be regarded as 'points' in Zariski topology
and this seems at least for me to provide a justification for the
geometrically afflicted notation 'place'.

First of all, is this exactly the original reason why (finite) places are called
'places'? i.e. that they are sitting somewhere in the associated topological space $V$.
The reason why I doubt and think that this is not the proper origin or the name 'place' is simply because the Grothendieck's fondation of algebraic geometry in language of schemes arises
later then the study of absolute values of algebraic number fields.

Does anybody know the true origin and motivation of the name 'place'?

In addition, there are also infinite places which cannot be associated with prime
ideals of ring of integers $R$. Is there also any reason why they carry
geometrically afflicted name 'places'?

答:
I checked first if the language of places appears in the Artin-Whaples paper in 1945 on axiomatizing number fields and function fields over finite fields in terms of a set of absolute values satisfying a product formula.  However, I did not find the technical term "place" there. It also is not in the Artin-Tate notes on class field theory (they refer to "primes" or "canonical absolute values") or Tate's thesis (he writes about "prime divisors"), which brings us to 1950.

In Edwin Weiss's Algebraic Number Theory, which was published in 1963 but was based on a course he gave at Harvard in 1958, he refers to an equivalence class of absolute values as a "prime divisor" and uses "place" to refer to the residue class field mapping $\mathcal O \to \mathcal O/\wp$ associated to a nonarchimedean prime divisor.

Bachman's 1964 book Introduction to $p$-adic Numbers and Valuation Theory also essentially uses "place" in the same sense as Weiss: the residue field mapping for an absolute value (same as mapping for any equivalence absolute value, since they all have the same valuation ring).

The terminology of places in its meaning today appears in Weil's Basic Number Theory, which was published in 1967 but was based on a course he gave at Princeton in 1961-1962. See Definition 3 in Chapter III. Weil defines a place as an equivalence class of completions rather than an equivalence class of absolute values, but that is morally the same idea, at least closer than the usage of Weiss and Bachman.

So if I had to bet on who finally gave the term "place" its standard meaning today in valuation theory, I'd go with Weil. But keep in mind that the term "place" was used earlier for related ideas in valuation theory.  The analogy with points on (compact) Riemann surfaces giving rise to the discrete valuations on their function fields (normalized to have value group $\mathbf Z$) was certainly in the back of the minds of people developing all this mathematics


问:
ODE and separations of variables: what about $0$?
Considering the following ODE:
$$ y'=f(x)g(y)$$ as long as $g(y) \neq 0$, I can divide by $g(y)$, and obtain
$$ \frac{y'}{g(y)}=f(x)$$
and by solving
$$ \int \frac{dy}{g(y)}= \int f(x)dx$$
I eventually find a solution $y$ on a domain $I$ (for the sake of clarity, let's put $I=(a,b)$) such that $g(y(I)) \neq 0$. But what if, for example, $y(a)=0$ is well defined, $y$ in $a$ is continuous and with a continuous derivative such that $y'(a)=0$, and $g(y(a))=0$? Can it even be possibile? In this case the method above is not working anymore in $a$, but since our original equation was $y'=f(x)g(y)$, would $y$ be a solution on $[a,b)$ and not only on $(a,b)$?

These may sound like trivial questions, but I'm currently studing ODE and their resolution methods and I'm having an hard time grasping all the concepts. Hopefully everything makes sense and it's written in proper English. Thanks everyone!

答:
With the method you describe, you can (in certain cases) find a solution $y(x)$ to your differential equation on an interval $I = (a,b)$ such that $g(y(x)) \neq 0$ for $x \in I$. After finding the solution, it may happen that there exists a larger interval $J \supset I$ such that we can extend $y$ to $J$, while $y$ still satisfies the differential equation $y'(x)=f(x)g(y(x))  $ (even though we may have $g(y(x))=0$ for points in $J$).

There is no problem with this: the requirement $g(y) \neq 0$ is neccesary to find a solution by seperation of variables, but there is no reason why a solution to the differential equation must always satisfy $g(y) \neq 0$.


问:
Two problems about 1-form
I am studying differential forms from Arnold's book Mathematics of Classical Mechanics and have two difficulty with two problems on page 175:

1.Show that every differential 1-form (not necessarily closed)on line is differential of some function.

2.Find differential 1-forms(not necessarily closed) on the circle and the plane which are not the differential of any function.

Can anyone help me?

答:
A differential $1$-form on the line is always closed. If $\alpha= f(x)dx$, it is the differential of $g(x)=\int_0^xf(t)dt$.

If $S^1$ is a circle, it is the quotient of $\mathbb{R}$ by the translation $t(x)=x+1$, consider the form $dx$ of $\mathbb{R}$ it is invariant by $t$ and induces a $1$-form on $S^1$ which is not exact, since there does not exist an element $x$ such that $\alpha_x=0$. If $\alpha=df$, $f$ has a maximum $x_0$ since $S^1$ is compact and $df_{x_0}=0$.

On $\mathbb{R}^2$, the form $xdy$ is not closed since its differential is $dx\wedge dy$.

Title: Two problems about 1-form


问:
$G$ is a group. $G$ contain element $a$, prove $a^n*a^{-n}=e$
I am a beginner for abstract algebra.
I have tried to solve this problem.Could anyone helps me to check
I have given a reason near each step.these are my steps</p>

答:
Hint:

Show that if two elements $\;a,b\;$ in any group $\;G\;$ commute (meaning that $\;ab=ba\;$) , then $\;(ab)^n=a^nb^n\;$ for all $\;n\in\Bbb Z\;$


问:
Intuition on the concept of bounding a sum
For example, in calculating the following limit

$$\lim_{n\to\infty}\sum_{k=1}^{3n}\frac{1}{\sqrt{n^2+k}}$$

I know (because my professor wrote) that one of the methods of solving this limit is via the sandwich rule since we know the lower and upper bound

$$\frac{3n}{\sqrt{n^2+3n}}<\sum_{k=1}^{3n}\frac{1}{\sqrt{n^2+k}}<\frac{3n}{\sqrt{n^2+1}}$$

But I don't understand why it's so, because I would have guessed it would be

$$\frac{1}{\sqrt{n^2+3n}}<\sum_{k=1}^{3n}\frac{1}{\sqrt{n^2+k}}<\frac{1}{\sqrt{n^2+1}}$$
instead

答:
Your guess is right in terms of bounding but you are forgetting about $\color{blue}{\text{summing}}$:
$$\color{blue}{\sum_{k=1}^{3n}}\frac{\color{red}1}{\sqrt{n^2+3n}}<\sum_{k=1}^{3n}\frac{1}{\sqrt{n^2+k}}<\color{blue}{\sum_{k=1}^{3n}}\frac{\color{red}1}{\sqrt{n^2+1}} \\
\frac{\color{blue}{3n}}{\sqrt{n^2+3n}}<\sum_{k=1}^{3n}\frac{1}{\sqrt{n^2+k}}<\frac{\color{blue}{3n}}{\sqrt{n^2+1}}$$


答:
A single term of the series obeys the inequality $$\frac{1}{\sqrt{n^2 + 3n}} < \frac{1}{\sqrt{n^2 + k}} \le \frac{1}{\sqrt{n^2 + 1}}$$ for any integer $k \in \{1, 2, \ldots, 3n\}$, but the sum does not.  Instead, you have to sum each of the expressions accordingly:
$$\sum_{k=1}^{3n} \frac{1}{\sqrt{n^2 + 3n}} < \sum_{k=1}^{3n} \frac{1}{\sqrt{n^2 + k}} \le \sum_{k=1}^{3n} \frac{1}{\sqrt{n^2 + 1}},$$ and because the leftmost and rightmost sums are independent of $k$, we get

$$\frac{3n}{\sqrt{n^2 + 3n}} < \sum_{k=1}^{3n} \frac{1}{\sqrt{n^2 + k}} \le \frac{3n}{\sqrt{n^2 + 1}}.$$

Title: Intuition on the concept of bounding a sum

Body:
For example, in calculating the following limit

$$\lim_{n\to\infty}\sum_{k=1}^{3n}\frac{1}{\sqrt{n^2+k}}$$

I know (because my professor wrote) that one of the methods of solving this limit is via the sandwich rule since we know the lower and upper bound

$$\frac{3n}{\sqrt{n^2+3n}}<\sum_{k=1}^{3n}\frac{1}{\sqrt{n^2+k}}<\frac{3n}{\sqrt{n^2+1}}$$

But I don't understand why it's so, because I would have guessed it would be

$$\frac{1}{\sqrt{n^2+3n}}<\sum_{k=1}^{3n}\frac{1}{\sqrt{n^2+k}}<\frac{1}{\sqrt{n^2+1}}$$
instead


问:
integration by substitution, using $\;t = \tan \left(\frac 12 x\right)$
$\displaystyle\int_0^\frac{\pi}{2}\frac{1}{2-\cos x} \, dx$ using the substitution $t=\tan\frac{1}{2}x$

$x=2\tan^{-1}t$

$\dfrac{dx}{dt}=\dfrac{2}{1+t^2}$

$dx=\dfrac{2}{1+t^2}\,dt$

$\displaystyle\int_0^1 \left(\frac{1}{2-\cos x}\right)\left(\frac{2}{1+t^2}\right)\,dt$

Is this the right idea? If so what do I do next?

$\displaystyle\int_0^1\left(\frac{1}{2-\frac{1-t^2}{1+t^2}}\right) \,\left(\frac{2}{1+t^2}\right)\, dt$

$\displaystyle\int_0^1\frac{2}{1+3t^2}\,dt$

$=2\left[\frac{\ln(1+3t^2)}{6t}\right]_0^1$

答:
If $\;t = \tan\left(\frac 12 x\right)$, i.e., $\,x = 2\tan^{-1}t,\,$ what should $\dfrac 1{2 - \cos x}$ then be?

We need to replace the function (integrand) of $x$ to one expressed as a function of $t$.

What are the new limits for $\,t\,$ if $\;t = \tan\left(\frac 12 x\right)$?

When $x = 0,\;$ $t = \tan\left(\frac 02\right) = 0$. Okay. But, when $x = \pi/2$, the upper limit of integration needs to be $t = \tan\left(\pi/4\right)$

See Weierstrass Substitution for why $\;\cos x = \dfrac{1-t^2}{1+t^2},\;$ and in general, for the logic of using "$t$-substitution": $t = \tan \frac x2$.

ADDED:

After substituting all of the above, we should have the integrand:

$$\frac{1}{2-\dfrac{1-t^{2}}{1+t^{2}}}\frac{2}{1+t^{2}}=\frac{2}{2(1+t^{2})-(1-t^{2})}=\frac{2}{1+3t^{2}}=\frac{2}{1+(\sqrt{3}t)^{2}}$$

So we have

$$\int_0^1 \frac{2}{1+(\sqrt{3}t)^{2}}\,dt = 2\int_0^1 \frac{1}{1+(\sqrt{3}t)^{2}}\,dt\tag{1}$$

Now, I'm afraid to say, the work isn't done yet. We cannot use $\ln|f(t)|$ where $f(t) =1 + \sqrt{3}t)^{2}$ because we do not have an integrand in the form of $\;\dfrac{f'(t)}{f(t)} \,dx$.

But we're all set up with $(1)$ to use the substitution $$u = \sqrt 3 t.\,\implies du = \sqrt 3 dt \implies dt = \dfrac{1}{\sqrt 3} du$$

Then we have an integrand of the form $$2 \int_0^{\sqrt 3} \dfrac{1}{\sqrt 3} \dfrac{1}{1 + u^2}\,du = \dfrac{2}{\sqrt 3}\int_0^{\sqrt 3}  \dfrac{1}{1 + u^2}\,du\tag{2}$$

Now, we recall that $$\int \dfrac {1}{1 + u^2} \,du = \tan^{-1}u + C\tag{$\star$}$$

Can you try and finish it from here? Apply $\star$ to the integral given by $(2)$

$(\star)$ See trigonometric substitution for integrals involving $a^2 + u^2$, where $a$ is a constant. Our integral is of the same form, with $a = 1$:

$$\int \frac{du}{a^2 + u^2} = \frac 1a\tan^{-1}\left(\frac ua\right)\,+ C$$


答:
Well, you need to consider the power laws in groups.

Let $G$ be a group and $a\in G$.

(1) Define $a^0 =e $ (unit element) and $a^{n+1} = a\cdot a^n$ for each integer $n\geq 0$. It is clear that $a^1=a$, since $a^1 = a\cdot a^0 = a\cdot e = a$.

(2) For each integer $n\geq 0$, define $a^{-n} = (a^{-1})^n$. This gives you the definition of negative powers.

(3) For all integers $m,n$, we obtain $a^m\cdot a^n = a^{m+n}$.

This gives you the solution of your problem by (3) and (1):

$a^n\cdot a^{-n} = a^{n+(-n)} = a^0=e.$

Commutativity is not necessary here.

Title: $G$ is a group. $G$ contain element $a$, prove $a^n*a^{-n}=e$

Body:
I am a beginner for abstract algebra.
I have tried to solve this problem.Could anyone helps me to check
I have given a reason near each step.these are my steps</p>


问:
How many of the committees have at least two members from the same party?

Q:

(a) A committee of 3 is to be chosen from a group of 9 politicians.
Suppose two of the group belong to Party I, three of the group belong
to Party II and four of the group belong to Party III. How many of the
committees have at least two members from the same party? Explain your
answer.

Attempt:

Firstly I did 9C3 to get 84. This is just to get things started. I then did 2C1 for party 1, 3C2 for party 2 and 4C3 for party 3 and my answers were 2,3,4. I feel like this is wrong though.

答:
The committee will have at least two members from the same party unless each member of the committee is drawn from a different party.  Therefore, we can find the answer by subtracting the number of committees in which each person is drawn from a different party from the total number of three-person committees which can be drawn from the nine politicians.

As you found, there are
$$\binom{9}{3}$$
ways to form a three-person committee drawn from the nine politicians.

The number of ways of obtaining a committee in which each person is drawn from a different party is
$$\binom{2}{1}\binom{3}{1}\binom{4}{1}$$
since one of the two members of Party I, one of the three members of Party II, and one of the four members of Party III must be selected.

Therefore, the number of three-person committees which have at least two members of the same party is
$$\binom{9}{3} - \binom{2}{1}\binom{3}{1}\binom{4}{1}$$


答:
A method that does not require a commutative property:

By definitions:

$a\cdot a^{-1}=e$

$a^{k+1}=(a^k)\cdot a$

$a^{-(k+1)}=(a^{-k})\cdot a^{-1}$ which also $=(a^{-1})\cdot a^{-k}$ by associativity

Now use associativity and mathematical induction to finish the proof.


答:
You can do it by induction. We have
$$ a^{n+1} \cdot (a^{-1})^{n+1} = a^n\cdot a \cdot a^{-1} \cdot (a^{-1})^n = a^n \cdot e \cdot (a^{-1})^n = a^n \cdot (a^{-1})^n$$
Therefore, if $a^n \cdot (a^{-1})^n = e$, then $ a^{n+1} \cdot (a^{-1})^{n+1} = e$. Since for $n=1$ we have $$a^1 \cdot a^{-1} = a \cdot a^{-1} = e$$ by induction it means that $a^n \cdot (a^{-1})^n = e$ for any $n\in\mathbb{N}$.


问:
Minimum value for A+B, given rounded percentages
The percentage of boys in a Math circle, rounded to an integer is equal to 51%. The percentage of girls in this Math circle, rounded to an integer is equal to 49%. What is the minimal possible number of participants in the circle?
(Source: "Formula of unity" contest, 2020 qualification rounds, R7, problem 2 - https://www.formulo.org/en/olymp/2020-math-en/)

What is a principled way to tackle such a problem?

Here was my sketch of solution:
It is easy to see 51 - 49 is a solution, although not minimal.

Assuming the girls percentage can be at least 48.501 and boys percentage 51.499, and that the difference between the two is at least 1 -- I could guess the solution is somewhere around 33 children in the class.
Then, 35 (17 + 18) is the smallest solution.

答:
Start with $ \frac{101}{200} < \frac{ B}{A } < \frac{103}{200}$.

This gives us $101 A < 200 B < 103 A $.

What's the smallest $A$ that this could happen?

Trial and error is a reasonable approach if you can't think of something creative.

As OP points out in the comments, if we wrote it as $ \frac{101}{200} < \frac{ B}{B+G } < \frac{103}{200}$, then we have

$101G < 99B$

$97B < 103 G \Rightarrow 97 (B-G ) < 6G$. Since $B-G \geq 1$, this gives us $ G > 16. $ We test $ G = 17$, and it works.


答: If $y$ is a zero of $g$, by Picard's method the solution is $y(x)=0$.

Title: ODE and separations of variables: what about $0$?

Body:
Considering the following ODE:
$$ y'=f(x)g(y)$$ as long as $g(y) \neq 0$, I can divide by $g(y)$, and obtain
$$ \frac{y'}{g(y)}=f(x)$$
and by solving
$$ \int \frac{dy}{g(y)}= \int f(x)dx$$
I eventually find a solution $y$ on a domain $I$ (for the sake of clarity, let's put $I=(a,b)$) such that $g(y(I)) \neq 0$. But what if, for example, $y(a)=0$ is well defined, $y$ in $a$ is continuous and with a continuous derivative such that $y'(a)=0$, and $g(y(a))=0$? Can it even be possibile? In this case the method above is not working anymore in $a$, but since our original equation was $y'=f(x)g(y)$, would $y$ be a solution on $[a,b)$ and not only on $(a,b)$?

These may sound like trivial questions, but I'm currently studing ODE and their resolution methods and I'm having an hard time grasping all the concepts. Hopefully everything makes sense and it's written in proper English. Thanks everyone!


问:
Is function has exactly one real root? If a function is injective, continuous and differentiable everywhere.
If a function is injective, continuous, and differentiable everywhere. Can we directly say the function has exactly one root? It seems function has exactly one root If a function is injective, continuous, and differentiable everywhere.
I think It's the intermediate value theorem.
But I think this injective graph hasn't a root?

when the graph goes minus infinity actually can we say the gradient is zero???

I haven't the tough idea If a function is injective, continuous, and differentiable everywhere, It has exactly real root.

Can you offer some assistance to clarify this mess.

Sorry for the poor English language manipulation

Thank you!

答: The function $f(x)=1+e^x$ is injective, continuous, and differentiable everywhere, yet it has no roots.


答: You're correct that these conditions don't guarantee a real root.  An easy example is $y=e^x$, which is injective, continuous, and differentiable everywhere, but has no real roots.  Of course, any injective function can only have at most $1$ real root.

Title: Is function has exactly one real root? If a function is injective, continuous and differentiable everywhere.

Body:
If a function is injective, continuous, and differentiable everywhere. Can we directly say the function has exactly one root? It seems function has exactly one root If a function is injective, continuous, and differentiable everywhere.
I think It's the intermediate value theorem.
But I think this injective graph hasn't a root?

when the graph goes minus infinity actually can we say the gradient is zero???

I haven't the tough idea If a function is injective, continuous, and differentiable everywhere, It has exactly real root.

Can you offer some assistance to clarify this mess.

Sorry for the poor English language manipulation

Thank you!


问:
Does Gauss-Bonnet theorem works with surfaces embedded in $\mathbb{R}^4$?
I was tryng to undertand if the Gauss-Bonnet theorem works for the real projective plane. I know that it does work for non orientable surfaces as long as they can be embedded in $\mathbb{R}^3$ (e.g. the Mobius strip) but the real projective plane is not.

So the real question is: does it work with surfaces embedded in $\mathbb{R}^4$?

By studying the problem I thing I have found a way to embed a sphere $S^2$ in $\mathbb{R}^4$ and twist it in such a way the total curvature become greater then $2\pi\chi$ but my example is quite involved and now I am not sure anymore if it is correct.

I have found very little on this on the internet.
Please help!

答:
The Gauss Bonnet Theorem works even for surfaces not embedded anywhere!

The theorem is about surfaces $S$ that are endowed with a Riemann metric $$ds^2=\sum_{i,k}g_{ik}(u_1,u_2)\, du_i\, du_k\ . $$ These surfaces can be closed or have boundary curves.

Many surfaces we meet are a priori embedded in some ${\mathbb R}^n$, $n\geq2$, and they obtain their $ds^2$ from this embedding (the $E$, $F$, $G$ in the "old" books). This is the case in your examples. Another example would be the hyperbolic plane, realized as half plane $H:=\bigl\{z=x+iy\in{\mathbb C}\bigm| y>0\bigr\}$. Here the $ds^2$ is given by
$$ds^2={1\over y^2}(dx^2+dy^2)\ .$$
The $dx^2+dy^2$ is just the euclidean $ds^2$ in the plane, but we have this additional scalar factor ${1\over y^2}$.

The Gauss Bonnet Theorem has various proofs. The geodesic curvature $\kappa_g$ can be measured "within" the surface, and no embedding or surface normal is needed. See, e.g., Wilhelm Klingenberg, A course in differential geometry, Springer Graduate Text (1978), pp. 138–141.


问:
Method to solve this indefinite integral $ \int \frac{x^{4}}{\sqrt{x^{2}+9}}dx $
I'm not able to solve this integral. Can anybody help me?
I tried doing it by parts noticing that
$$
\int \frac{x^{4}}{\sqrt{x^{2}+9}} \, dx \, = \int x^{3} \cdot \frac{x}{\sqrt{x^{2}+9}} .
$$
where we know that $$ D[\frac{x}{\sqrt{x^{2}+9}}] = \sqrt{x^{2}+9} $$

so that we can view the integral as

$$
\int \frac{x^{4}}{\sqrt{x^{2}+9}} \, dx \, = x^{3} \cdot \sqrt{x^{2}+9} - \int 3 \cdot x^{2} \cdot \sqrt{x^{2}+9} \, dx  .
$$
which led me in another integration by parts and in another one again, so i realized that maybe this wasn't the right way to do it

答:
Denominator term is to the power  -1.

Numerator multiplied by 1.

Use LIATE method to select first and 2nd terms for multiplication rule of Integration.

Use multiplication rule

Solve to get answer


答:
Since $k$ is in the denominator, the bigger $k$ the smaller the term of the sum. The biggest term of the sum is therefore $\frac 1 {\sqrt{n^2 + 1}}$. If you sum up this term $3n$ times, you get something that is definitely bigger than your original sum. But since this sum is independent of $k$, you can write it as a simple fraction with $3n$ in the numerator.

$$\sum_{k=1}^{3n}{\frac 1 {\sqrt{n^2 + k}}}<\sum_{k=1}^{3n}{\frac 1 {\sqrt{n^2 + 1}}}=\frac {3n} {\sqrt{n^2 + 1}}$$

Similarly, the smallest term of your sum will be $\frac 1 {\sqrt{n^2 + 3n}}$. Summing that up and simplifying you get:

$$\sum_{k=1}^{3n}{\frac 1 {\sqrt{n^2 + 3n}}}=\frac {3n} {\sqrt{n^2 + 3n}}<\sum_{k=1}^{3n}{\frac 1 {\sqrt{n^2 + k}}}$$

Hope this helps :)