问:
Prove that $p(n)$ is the number of ways of partition of $n$
For a positive integer $n$, define $p(n)$ to be the number of sets of positive integers $\{ x_1, x_2, \dots, x_k \}$ such that $x_1 > x_2 > \dots > x_k > 0$ and $n = x_1 + x_3 + x_5 + \dots$ (sum of all odd indices element of the set). For example, $p(6) = 11$ since we could have
$$ \{ 6 \}, \{ 6, 5 \}, \{ 6, 4 \}, \{ 6, 3 \}, \{ 6, 2 \}, \{ 6, 1 \}, \{ 5, 4, 1\}, \{ 5, 3, 1 \}, \{ 5, 2, 1 \}, \{ 4, 3, 2 \}, \{ 4, 3, 2, 1\} $$
Prove that $p(n)$ is the number of partition of integer $n$ for all $n \in \mathbb{N}$

This is from a selection test that already ended. I think it can be done by bijection (maybe generating function), but I have no idea how to continue.

答:
Suppose we are given an admissible set $\{x_1,\dots,x_k\}$. If $k$ is odd, add $0$ to the set so its cardinality is even. Then interpret this set as a strictly decreasing sequence $(a_1,b_1,a_2,b_2,\dots,a_{i+1},b_{i+1})$. There is a unique Young diagram associated to the sequence, constructible as follows:

Make an L of $a_{i+1}$ squares whose legs point down and right (like $\Gamma$), of which $b_{i+1}$ squares are right of the L's corner (white squares below).

For $j$ from $i$ to $1$, counting down:

Add $a_{j+1}+1$ compulsory squares to the top and left edges of the diagram (blue squares below). The diagram is now not in Young form because of a notch in the top-left, but this is only temporary.

Add an L of $a_j-(a_{j+1}+1)$ squares (white below): one square to fill the aforementioned notch, $b_j-(a_{j+1}+1)$ squares to the right of the newly added compulsory squares on top, the remaining squares below the newly added compulsory squares on the left. Young form is now restored.

In each loop iteration we add $a_j$ squares, so the total number of squares in the finished Young diagram is $\sum_{j=1}^{i+1}=n$. It is also easy to see that different sets for which the sum of odd-indexed terms is the same $n$ lead to distinct Young diagrams of $n$ cells. This establishes a bijection between admissible sets whose odd-indexed terms sum to $n$ and Young diagrams of $n$ cells, which themselves are in bijection with partitions of $n$. Thus $p(n)$ as defined in the question is identical to the unrestricted partition function (usually denoted $p(n)$ anyway).


问:
Calculating the number of 3*3 Matrices whose trace of $A^t$*$A$=6
Find the number of 3×3 matrices A such that the entries of A belong to the set Z of all integers, and such that the trace of $A^t$*$A$ is 6 . ($A^t$ denotes the transpose of the matrix A).

For this problem,I can't understand how to calculate the number of the 3*3 matrices satisfying the given condition. Any help would be appreciable.
A little wordy solution is appreciable.

Thank you.

答:
Let $A=\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$. Then by a matrix multiplication, one obtains that
$$\operatorname{tr}(A^t A)=\sum_{i,j} a_{ij}^2=6.$$
Hence all entries must be between $-2$ and $2$. First, consider the case where one of the entries has absolute value $2$. Then there must be $2$ other entries with absolute value $1$, and all other entries are zero. The number of such matrices is
$$\binom{9}{1}\cdot 2\cdot \binom{8}{2}\cdot 2^2,$$
where we have multiplied by $2^3$ to take into account that these entries may also be negative. Now consider the case where there does not exist an entry with absolute value $2$. Then there are $6$ entries with absolute value $1$ and $3$ are zero. The number of such matrices is
$$\binom{9}{6}\cdot 2^6.$$
The sum of these is the result
$$\binom{9}{1}\binom{8}{2}\cdot 2^3+\binom{9}{6}\cdot 2^6=7392.$$

ContentLicense: CC BY-SA 4.0


问:
Question about Bishop's "constructive analysis" for people familiar with bishop style constructive analysis only
It has been a while since i read Bishop's book "constructive analysis", recently I dug it out of my book shelve and started to read. I came around this observation on the top of page 85.

"A subset Y of $\mathbb{R}$ can be bounded as a
metric space but not bounded as a subset of $\mathbb{R}$."

I have tried to figure out how this can be and came up with nothing so far. Can anybody give me a hint?

PS: As far as I understand Bishop is referring to the metric induced by the absolute value.

PPS: The definition provided:
A metric space $(X, p)$ is called bounded if there exists a real number
$C > 0$, called a bound for $(X, p)$, such that $p(x, y) \leq C \forall x,y \in X$.
A subset $Y$ of a non void metric space $X$ is bounded if, for all
(equivalently, some) $x$ in $X$, the set $Y\cup\{x\}$ with the induced metric p* is a bounded metric space.

答:
Let $Y=[0,\infty[$ and define the distance $d(x,y) = |\tan^{-1}(x) - \tan^{-1}(y)|$ for any $x,y\in Y$. Then $d(x,y)\leq |\tan^{-1}(x)| + |\tan^{-1}(y)| \leq \frac{\pi}{2} + \frac{\pi}{2} = \pi$. So the metric space $Y$ is bounded.

Assume $Y$ is unbounded. Then there is a sequence $\{y_k\}_{k\in\mathbb{N}}$ such that $|y_k|\to\infty$. Assume the metric space $(Y,d)$ is bounded with $d(x,y) = |x-y|$ for any $x,y\in Y$ and with some center $a\in \mathbb{R}$. Take $d_k = d(a,y_k)$. We have $d_k \geq ||a|-|y_k|| \to \infty$. So $d_k\to\infty$.

Some definitions of bounded metric space don't take a center, that's fine too. Since $y_k\to \infty$, we can pick a subsequence $\{y_{k_j}\}_{j\in\mathbb{N}}$ such that $||y_{k_j}| - |y_{k_{j+1}}||\to\infty$ and redefine $d_j = d(y_{k_j},y_{k_{j + 1}})$.


问:
irreducibility of $x^n+y^n+z^n+w^n$ in $\Bbb C[x,y,z,w]$
I am trying to prove $x^n+y^n+z^n+w^n$ is irreducible in $\Bbb C[x,y,z,w]$. I am trying to apply the Esinsten's theorem, that some prime $p$ in $\Bbb C[y,z,w]$ divides $y^n+z^n+w^n$, but $p^2$ does not. I can find such prime $y-\frac{e^{i\pi}}{n} z$ in $\Bbb C[y,z]$, which divides $y^n+z^n$ since $(\frac{e^{i\pi}}{n}z)^n=-z^n $, but $(y-\frac{e^{i\pi}}{n} z)^2$ does not divides $y^n+z^n$(I cannot show this), therefore, Esinstein's theorem can be applyed to $x^n+y^n+z^n.$ But I cannot find such prime for $x^n+y^n+z^n+w^n$ to apply the Esinstein's theorem.

答: The element $w \in \mathbb C[x,y,z,w]$ is prime. It is well-known that if the image of $f = x^2 + y^2 + z^2 + w^2$ in $\mathbb C[x,y,z,w]/(w) = \mathbb C[x,y,z]$ is irreducible, then so is $f$. But you have already shown that $x^2+y^2+z^2$ is irreducible in $\mathbb C[x,y,z]$.


问:
Showing that a family is not normal
Suppose $f : \mathbb{D}\setminus \{0\} \rightarrow \mathbb{C} $ is analytic and $z=0$ is an essential singularity of $f$. Show that the family $\{f_n\} $ defined by
$$ f_n(z) = f \left( \frac{z}{2^n} \right), \quad z \in \mathbb{D}\setminus \{0\} $$
is not normal in $\mathbb{D}\setminus \{0\}$.

My attempt (inspired on this): Assume that $\{f_n\}$ is a normal family in $\mathbb{D}\setminus \{0\}$. So there exists from the sequence $(f_n)_n$ a convergent subsequence $(f_n)_{n_k}$ with limit $\hat{f}$. Because $f$ is analytic, it follows that $\hat{f}$ is analytic on $\mathbb{D}\setminus \{0\}$. We take an annulus $A$ of inner radius $1/3$ and outer radius $1/2$. Then we have that $\hat{f}(A)$ is bounded and because $f_{n_k} \rightarrow f$ as $k \rightarrow \infty$, we have that $f_{n_k} = f(\frac{A}{2^{n_k}})$ is bounded for $k$ large enough. By the Cauchy formula for Laurent series $$ |a_{-j}|  \leq r^j \max_{|z| = r}{|f|}.$$
We choose $r$ as the radius of the circle contained in $A/n_k$ and letting $k\rightarrow \infty$. We then have to show that $|a_{-j}|=0$ for every $j \leq -1$ such that we conclude that $f$ is an analytic function on $\mathbb{D}$ which gives us a contradiction.

But in the last part I got stuck. Because if $k \rightarrow \infty$ then $r \rightarrow 0$ and because $f$ has an essential singularity in $z=0$: $\max_{|z| = r}{|f|}  \rightarrow \infty $. So we have something of the form $0 \cdot \infty$ which is not defined.

Someone who can point out where I went wrong and help me out? Thanks!

答:
You have shown that $\max_{|z| = r_k}{|f|}$ is uniformly bounded for
$$
r_k = \frac{1}{2 \cdot 2^{n_k}}
$$
therefore choosing these radii in
$$
|a_{-j}|  \leq r^j \max_{|z| = r}{|f|}.
$$
implies that $a_{-j} = 0$ for $j \ge 1$.

Alternatively one can argue as follows: $\hat f$ is bounded on the circle $|z| = 1/2$ and $f_{n_k} \to \hat f$  uniformly on that circle. It follows that
$$
 |f_{n_k}(z)| \le M  \text{ for } |z| = \frac 12
$$
with some $M> 0$ and sufficiently large $k$. Translated back to $f$ this means that
$$
 |f(z)| \le M \text{ for } |z| = \frac{1}{2 \cdot 2^{n_k}} \, .
$$
So $|f|$ is bounded by $M$ on those concentric circles. Using the maximum modulus principle it follows that the same estimate holds in the annuli between those circles:
$$
 |f(z)| \le M \text{ for } \frac{1}{2 \cdot 2^{n_{k+1}}} \le |z| \le \frac{1}{2 \cdot 2^{n_k}} \, .
$$
and that implies that $f$ is bounded in a neighborhood of $z=0$.

Using Riemann's theorem it follows that $f$ has a removable singularity at $z=0$.

Title: Showing that a family is not normal


问:
Prove that $3^n>n^4$ if $n\geq8$
Proving that $3^n>n^4$ if $n\geq8$

I tried mathematical induction start from $n=8$ as the base case, but I'm stuck when I have to use the fact that the statement is true for $n=k$ to prove $n=k+1$.  Any ideas?

Thanks!

答:
HINT: Show that if $k\ge 8$, then

$$3\ge\left(\frac{k+1}k\right)^4=\left(1+\frac1k\right)^4\;.$$

When you go from $k$ to $k+1$, you’re multiplying $3^k$ by $3$, and you’re multiplying $k^4$ by $\left(\frac{k+1}k\right)^4$.

ContentLicense: CC BY-SA 3.0

Title: Prove that $3^n>n^4$ if $n\geq8$


问:
Function $f$ not in $L^1$ but $\hat{f} \in L^1$
Can you have a function $f \notin L^1$ but its Fourier transform $\hat{f} \in L^1$? Ive been playing around with examples and I cant find one, but I also cant prove one doesn't exist.

答:
More on the example of @md2pepe.
Line 18 on this page below from a textbook says that the Fourier transform of
$\frac{W}{\pi}\operatorname{sinc}(Wt)$ is $\operatorname{rect}(\frac{\omega}{2W})$

I just Googled "table of fourier transforms" and took the first image.

This table by itself does not show what definition of Fourier transform is used.  That is, "where is the $2\pi$?"  Possibilities are
$$
\widehat{f}(\omega) = \int_{-\infty}^\infty e^{-j\omega t}f(t)\;dt
\\
\widehat{f}(\omega) = \frac{1}{2\pi}\int_{-\infty}^\infty e^{-j\omega t}f(t)\;dt
\\
\widehat{f}(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-j\omega t}f(t)\;dt
\\
\widehat{f}(\omega) = \int_{-\infty}^\infty e^{-j 2\pi \omega t}f(t)\;dt
$$
and maybe more?


问:
Proof of Dedekind cuts.
This is my definition for Dedekind cuts:

A subset α of Q is said to be a cut if:

$α$ is not empty,$α\neq \mathbb{Q}$

If $p \in α,q\in\mathbb{Q}$,and $q<p$,then $q\inα$.

If $p\in α$,then $p<r$ for some  $r\inα$.

I've seen a specific example of Dedekind cut produced by $\sqrt{2}$:

$α={\{p∈Q:p<0\}}∪{\{p∈Q:p≥0 \:\text{ and  }\: p^2 <2\}}.$

To prove this subset is a cut, it needs to be shown it satisfies 1.2.3. For the proof of 3., the author constructed $r=\frac{2(p+1)}{p+2}$.

Now I'm working on a more general subset here: $a' = {\{x\mid x\geq 0 ,  x^2\leq p\}} \cup {\{x\mid x<0\}}$. $p$ is a positive integer but not a square of integer. I don't know how to prove 2. and 3. and construct such an $r$ here.

答:
I'll talk about Property 2 first, which boils down to simple casework. Suppose $t\in \alpha'$ and $s\in\mathbb{Q}$ satisfies $s<t$. If $s<0$, then $s\in\{x:x<0\}\subset \alpha'$. If, on the other hand, $0\leq s<t$, then properties of the ordered field $\mathbb{Q}$ imply $0\leq s^2<t^2$. Yet $t^2\leq p$, so $s\in\{x:x\geq 0, x^2\leq p\}\subset \alpha'$.

Now we'll prove Property 3. Let $t\in\alpha'$. If $t<0$, then $r=0$ works, so suppose $0\leq t$ and $t^2\leq p$. The square root of such a $p$ is irrational, so we actually have $t^2<p$. To find an $r$ that works we'll "rationally perturb" $t$; put $r=t+1/n$ where $n$ is some to-be-determined positive integer. This $r$ is rational and satisfies $t<r$. We would like to find conditions on $n$ which imply $r^2<p$. For this, note
$$r^2=\big(t+\frac{1}{n}\big)^2 = t^2+\frac{2t}{n}+\frac{1}{n^2}.$$
Supposing $r^2<p$ we may rearrange to find $2tn+1<n^2(p-t^2)$, or

$$1<n^2(p-t^2)-2tn = n[n(p-t^2)-2t].$$
Hence it is sufficient to choose $n$ so large that $n(p-t^2)-2t>1$, as this and the fact $n\geq 1$ ensure the above inequality is true. Choosing such an $n$ is possible by Archimedean property, so $r=t+1/n$ has $r^2<p$ and the proof is complete.

Title: Proof of Dedekind cuts.


问:
Is every integral finitely generated reduced $k$-algebra a domain?
I'm studying some classical algebraic geometry, and the book I'm using says that the contravariant functor from the category of algebraic affine sets ($k$ algebraically closed) to the category of finitely generated reduced $k$-algebraic that associates to $X \subseteq \mathbb{A}^n_k$ to the coordinate ring $k[X]$ is an equivalence of categories. Ok, no problem with this. But then, the book says that the restriction of this functor to irreducible algebraic sets gives an equivalence between the opposed category of irreducible algebraic sets and integral finitely generated reduced $k$-algebras. Since the coordinates ring of an irreducible algebraic set is always integral domain, this would imply that  integral finitely generated reduced $k$-algebras are integral domains. Is this the case? Here, I'm assuming that integral $k$-algebras are algebras such that the canonical morphism $k \to A$ is integral. How can this last equivalence of categories arise?

答:
It seems you confuse the notions "integral scheme" and "integral ring extension".

A scheme $(X,\mathcal{O}_X)$ is integral (this is Hartshorne's definition) iff for any open subset $U\subseteq X$ it follows the ring $\mathcal{O}_X(U)$ is an integral domain.

If $X:=\mathrm{Spec}(A)$ is an affine scheme it follows (this must be proved) $X$ is integral iff $A$ is an integral domain.

One implication is trivial. If $X:=\mathrm{Spec}(A)$ and $A$ is an integral domain you must prove that $\mathcal{O}_X(U)$ is an integral domain for any open subset $U \subseteq X$, and this follows from the fact that there is an inclusion $\mathcal{O}_X(U) \subseteq K(A)$ where $K(A)$ is the quotient field. Since $K(A)$ is a field, it follows any sub ring of $K(A)$ is an integral domain, in particular it follows $\mathcal{O}_X(U)$ is an integral domain.

Let $B:=k[x_1,...,x_n]$ be a polynomial ring in $n$ variables over a field $k$ and let $I \subseteq B$ be an ideal. Let $X:=\mathrm{Spec}(B/I)$. We may view $X$ as a closed sub-scheme of affine $n$-space $\mathbb{A}^n_k:=\mathrm{Spec}(B)$. It follows $\mathcal{O}_X(X)=A$ where $A:=B/I$. Noether normalization lemma says there is a finite set of elements $y_1,...,y_d \in A$ with $d=\dim(X)$ and the ring $k[y_1,...,y_d]\subseteq A$ generated by the elements $y_i$ is a non-trivial polynomial ring if $d \geq 1$. It follows the elements $y_i$ are transcendental over the field $k$ and do not satisfy a monic polynomial relation over $k$ in general, hence the ring extension $k \subseteq A$ is not an integral ring extension if $d \geq 1$. If $k \subseteq A$ is an integral ring extension it follows $\dim(X)=0$. It follows $A$ is an Artinian ring and there is a direct sum decomposition $A\cong A_1\oplus \cdots \oplus A_l$ with $A_i$ an Artinian local ring. Hence in this case $A$ is an integral domain iff $k \subseteq A$ is a finite field extension. $A$ is reduced iff $A$ is a finite direct product of finite field extensions of $k$.


问:
area of triangle
In $\triangle ABC$ points $D,E,F$ are on the sides $AB,BC,CA$, respectively, with $AD=DB$, $CE=3BE$ and $AF=2CF$. If the area of $\triangle ABC$ is $480 cm^2$, how do we find the area of $\triangle DEF$?</p>

答:
Hint: Calculate the area of $ADF, BDE, ECF$. Subtract their sum from $ABC$ to get $DEF$.

Further Hint: $\frac{\mbox{Area } ADF } { \mbox {Area } ABC } = \frac{ AD } { AB} \times \frac { AF} { AC} $.

Title: area of triangle


问:
Differntiability of a function $ F:\mathbb{R}^{n+n}\to\mathbb{R}^{n} $
This may sound very trivial but Im very confused.

Let $ f\left(\overrightarrow{x}\right):\mathbb{R}^{n}\to\mathbb{R}^{n} $ be a function such that   $ f\in C^{\left(1\right)}\left(\mathbb{R}^{n};\mathbb{R}^{n}\right) $ (differentiable and the partial derivatives are all continuous)

and define a function $ F\left(\overrightarrow{x},\overrightarrow{y}\right):\mathbb{R}^{n+n}\to\mathbb{R}^{n} $  by:

$ F\left(\overrightarrow{x},\overrightarrow{y}\right)=f\left(\overrightarrow{x}\right)-\overrightarrow{y} $

Why is it true that $ F\in C^{(1)}\left(\mathbb{R}^{n+n},\mathbb{R}^{n}\right) $ ?

my first intuition was to use the fact that the sum of two differntiable function is a differntiable function, but in this theorem the sum of the functions has to be a function from the mutual domain and here it is actually a "congrefate" of the domains so im it really confuses me.

Also I tried to think about it as composition of differntiable functions but also its not exactly a composition

If someone can help me understand how to think about it would be great, because the multivariable confuses me.

Thanks in advance

答: The product of differentiable functions is differentiable: consider $(x,y)\rightarrow (f(x),y)$ which is the product of the identity and $f$ composed with $(x,y)\rightarrow x-y$.


问:
How can I write these things formally?
Let's say we have $N$ computers of different kind given (every computer is different and only exists once). They are rankable, so you can order them from best to worst computer. There is only $1$ best computer.

How can this be written formally, as a set ?

I've thought about expressing these $N$ computers as an indexed set, so $\left\{1,2,..,N\right\}$

Then , we need our computer set: $$C=\left\{c_i \mid i \in \left\{1,..,N\right\}\right\}$$

But now I have problem writing the best computer, can it be done like this? : Let $c_i$ with $i=1$ be the best computer.

And also I'm having problem with my set above because I'm not sure if it's really rankable the way I have written it, and also I don't know how to write it such that each of its elements is distinct / different to all its other elements ? :/

答: I guess, you have a set $C$ of $|C|=N$ computers endowed with a total order $\le$.

Title: How can I write these things formally?


问:
Convergent Sequence + Limit is Compact using Sequential Compactness
Proposition: Let $(X,d)$ be a metric space and $\lbrace x_n \rbrace_{n=1}^\infty \subset X$ be a convergent sequence with $x_n \rightarrow x_0, n \rightarrow \infty$. Show that $K = \lbrace x_n \mid n \in \mathbb{N} \cup \lbrace 0 \rbrace \rbrace$ is a compact set.

Question: It is easy to see how to do this with open covers via the standard definition of compactness. What I am wondering is how one would prove this theorem using sequential compactness (since it is equivalent to regular compactness for metric spaces), if it is even possible to do so.

答:
I will rewrite my comment to egreg's solution as an answer to see if it raises some objections.

Suppose $(z_n)$ is a sequence in $K$. We only have two cases.

$(z_n)$ attains some value of $K$ infinitely many times, in which case it obviously has a convergent (constant) subsequence that converges in $K$, or

$(z_n)$ attains all values of $K$ a finite number of times. In this case $(z_n) \to x_0$.

To show 2., for $i\geq 1$, consider the finite set $$K_i = \{x_1,x_2,\dots,x_i\}.$$ Since none of the values of $K_i$ are reached infinitely many times by $(z_n)$, then there exists $N_i$ such that $z_n\not \in K_i$, for $n>N_i$.

Fix $\varepsilon > 0$. By convergence of $(x_n)$, there exists  $M_\varepsilon$ such that $d(x_n,x_0)< \varepsilon$, for all $n>M_\varepsilon$. Now, for $n>N_{M_\varepsilon}$ we have $d(z_n,x_0) < \varepsilon$, thus proving the convergence of $(z_n)$ to $x_0$. $\blacksquare$


问:
$\lim_{(x,y,z)\rightarrow(0,0,0)}((\frac{xyz}{x^2+y^2+z^2})^{(x+y)})$
I'm quite stumped. The answer to this question is it does not exist, but I've tried using $y=x$, $z=x$ and $y=x^2$ and $z=x^2$, or even $y=\sqrt{x}$ and $z=\sqrt{x}$, but they all give me the answer $1$. How can I find two diferent pathways to prove it does not exist?

答:
This may be wrong, but please comment kindly!

$$
\lim_{(x,y,z)\to(0,0,0)}\left(\frac{xyz}{x^2+y^2+z^2}\right)^{(x+y)} \tag{*}
$$
switch to spharical coordinates $x = r\sin\theta\cos\phi,\ y = r\sin\theta\sin\phi,\ z = r\cos\theta$
$$
\lim_{r\to0}\left(\frac{r^3\sin\theta\cos\phi\sin\theta\sin\phi\cos\theta}{r^2}\right)^{r\sin\theta(\cos\phi + \sin\phi)}  \\
= \lim_{r\to0}\left(\frac{r}{2}\sin^2\theta\cos\theta\sin2\phi\right)^{r\sin\theta(\cos\phi + \sin\phi)} \\
$$
take the logarithm of the expression (disregarding that the argument of the logarithm should be positive)
$$
r\sin\theta(\cos\phi + \sin\phi)\ln\left(\frac{r}{2}\sin^2\theta\cos\theta\sin2\phi\right)
$$
make the following abreviations
$$
A = \sin\theta(\cos\phi + \sin\phi),\qquad B = \frac{1}{2}\sin^2\theta\cos\theta\sin2\phi
$$
so we get
$$
\lim_{r\to0} Ar \ln(Br) = 0
$$
but!...this holds if (at least) $B \ne 0$, that is, for $(\theta,\phi) \notin \{0,\frac{\pi}{2},\pi\}\times\{0 ,\frac{\pi}{2},\pi,\frac{3\pi}{4}\}$. The last limit implies that the limit $(*)$ is $1$. What happens for $B = 0$ depends on what $0^0$ should mean.


答: The set of distinct computers could be denoted by $C \ =\ \{C_1,C_2,...,C_n\}$. The set of ranks has to have positive integers only. So instead of $R$, we could use positive integers $Z_{+}$. Since every computer is unique, it has a different rank than the other and let's suppose it is possible to rank them using some method. I am thinking of a one-to-one function (injection) from the set of computers to the set of positive integers. Then this system of computers and their ranking property could be written as $f:C\rightarrow Z_{+}$ where $f$ is a function to rank them and $C$ and $Z_{+}$ are the set of computers and positive integers respectively.

Title: How can I write these things formally?

Body:
Let's say we have $N$ computers of different kind given (every computer is different and only exists once). They are rankable, so you can order them from best to worst computer. There is only $1$ best computer.

How can this be written formally, as a set ?

I've thought about expressing these $N$ computers as an indexed set, so $\left\{1,2,..,N\right\}$

Then , we need our computer set: $$C=\left\{c_i \mid i \in \left\{1,..,N\right\}\right\}$$

But now I have problem writing the best computer, can it be done like this? : Let $c_i$ with $i=1$ be the best computer.

And also I'm having problem with my set above because I'm not sure if it's really rankable the way I have written it, and also I don't know how to write it such that each of its elements is distinct / different to all its other elements ? :/


问:
Finding integrating factor for non exact differential equation
I am trying to solve non exact differential equation but I am unable to find the integration factor.

I solve the following question
$$-3y\frac{dy}{dx}+2x=0 $$

and I get the integrationg factor $x^{-5/2}$ and I use the proper way to find the integrating factor, but in our text book the integrating facot for above differential equation is finded and that is $y/x^4$.

Now I am completely confused that why please help me out with this.

Body:
$$-3y\frac{dy}{dx}+2x=0$$
$$-\dfrac 32 2yy'+2x=0$$
Note that $$((y^2)'=2yy'$$
So that you have:
$$-\dfrac 32 (y^2)'+2x=0$$
Integrate:
$$-\dfrac 32 y^2+x^2=c$$


问:
multivariable limits with three variables
1)Find $\lim_{E\ni(x,y,z)\rightarrow(0,0,0)}x^{yz}$, where $E=\{(x,y,z):x,y,z>0\}$

2)Find $\lim_{(x,y,z)\rightarrow(0,1,0)}\dfrac{\sin(xyz)}{x^2+z^2}$

For(1), it seems to me that I should first try $\lim_{E\ni(x,y,z)\rightarrow(0,0,0)}yz\ln x$, but no matter what path I use, I'll always get the answer $0$. (The answer is that the limit does not exist, by the way.)

For (2), can I just simply ignore the $y\rightarrow1$? It seems rather....useless by itself. If I could change the limit to $\lim_{(x,z)\rightarrow(0,0)}\dfrac{\sin(xz)}{x^2+z^2}$, that would be way easier as I could just put $z=mx$ and say it does not exist. Is this rigorous?

Thanks for the help!

答:
1: Hint: $\lim_{x \to 0} x^x = 1$.

2: Yes. "$z=mx, y=1$" is a valid path passing through $(0,1,0)$.


答:
Hint:

Drop the perpendicular from $B$ to $AC$ and $D$ to $AC$, and use that to find the area of $\triangle DAF$.

Use same idea for others.

Title: area of triangle

Body: In $\triangle ABC$ points $D,E,F$ are on the sides $AB,BC,CA$, respectively, with $AD=DB$, $CE=3BE$ and $AF=2CF$. If the area of $\triangle ABC$ is $480 cm^2$, how do we find the area of $\triangle DEF$?</p>


问:
Local ring of a generic point on an integral scheme is a field

Let $X$ be an integral scheme and let $\eta \in X$ be its generic
point. Then the local ring $K(X) := \mathcal{O}_{X, \eta}$ is a field.
Moreover, if $U = \text{Spec} A$ is any open affine subset of $X$, then
$K(X) \cong \text{Frac} A$.

I would like to prove this proposition. I have already seen a proof that I understand, for example here. But I would like to see if my partial approach can somehow be extended to a full proof.

Since $X$ is integral, $A$ is an integral domain and $\text{Frac} A$ is well-defined and a field. So we only need to show the last part.

Let $U = \text{Spec} A$ be an affine open subset of $X$. Then since $\eta$ is the generic point, it is contained in all open subsets of $X$.

We have $A=\mathcal{O}_X(U)$ so $\text{Frac} A = \{ f/g \ | \ f, g \in \mathcal{O}_{X}(U), \ g\neq0 \}$. Define the map
\begin{align*}
   \phi: \text{Frac} A &\to K(X) = \mathcal{O}_{X, \eta}\\
   f/g & \mapsto (U \cap D(g), f/g).
  \end{align*}

Consider $(U \cap D(g), g) \in \mathcal{O}_{X, \eta}$. Provided $(U \cap D(g), g) \notin m_{\eta}$, then this element has an inverse $(U \cap D(g), 1/g) \in \mathcal{O}_{X, \eta}$ and we can multiply it with $(U \cap D(g), f) \in \mathcal{O}_{X, \eta}$. This is how our map $\phi$ is defined.

So we now show that $(U \cap D(g), g) \notin m_\eta$. How do we show that?

It is obvious that $\phi$ is a ring homomorphism.

Let $f/g \in \ker \phi$. Then $(U \cap D(g), f/g) =0 \in \mathcal{O}_{X, \eta}$ and so the product $(U \cap D(g), f)(U \cap D(g), 1/g) =0 \in \mathcal{O}_{X, \eta}$. Because $X$ is integral, $\mathcal{O}_{X, \eta}$ has no zero-divisors. Since $(U \cap D(g), 1/g)$ is a unit, it is not $0$, hence we must have $(U \cap D(g), f)=0$. So there is an open set $V \subset U \cap D(g)$ such that $f$ restricted to $V$ is $0$. So $f(\eta)=0$ and because $\eta$ is dense in $U$ and $\mathcal{O}_X(U)$ has no nilpotent elements we have $f=0$. So $\phi$ is injective.
How to prove surjectivity?

答:
If $(X,\mathcal{O}_X)$ is an integral scheme and $U \subseteq X$ is an open subscheme with $p\in U$ a point, it follows there is an isomorphism of local rings $\mathcal{O}_{X,p} \cong \mathcal{O}_{U,p}$. Here $\mathcal{O}_U$ is the restriction of the sheaf $\mathcal{O}_X$ to the open subscheme $U \subseteq X$.
Hence you may use the structure sheaf of any open subscheme $U$ containing $p$ to calculate the stalk at $p$. If $U=Spec(A)$ is an affine open subscheme containing $p$ it follows $\mathcal{O}_{X,p} \cong A_{\mathfrak{p}}$ where $\mathfrak{p}\subseteq A$ is the prime ideal in $A$ correspoding to the point $p$. Since $p$ is the generic point of $X$ it follows $\mathfrak{p}$ is the zero ideal in $A$. Hence there is an isomorphism $\mathcal{O}_{X,p} \cong A_{\mathfrak{p}} \cong K(A)$, where $K(A)$ is the quotient field of $A$ (since $X$ is integral it follows $A$ is an integral domain).

The map $\phi$ you write down above is the following "composed map":

Formula 1. $\phi: Frac(A) := A_{\mathfrak{p}} \cong \mathcal{O}_{U,p} := \lim_{\eta \in V}\mathcal{O}_U(V) \cong \lim_{\eta \in V}\mathcal{O}_X(V):=K(X)$

and all the maps in Formula 1 are canonical and isomorphisms. If you check the proof of this for all these maps you should get an answer to question 1 and 2  above.

Note: When I write $\mathcal{O}_{U,p}:=\lim_{p\in V}\mathcal{O}_U(V)$
using the sign $:=$ I means this is a definition. Hence here I define the local ring of the open scheme $U$ at the point $p$. The sign $\cong$ means "isomorphism".

The isomorphism $\mathcal{O}_{U,p} \cong A_{\mathfrak{p}}$ is in Hartshorne, Thm I.3.2. The definition $\mathcal{O}_{U,p} := \lim_{p\in V}\mathcal{O}_U(V)$ is explained in Chapter II.1 in Hartshorne. The isomorphism $\mathcal{O}_{X,p} \cong \mathcal{O}_{U,p}$ is an elementary result on limits, and you may find this in Atiyah-Macdonalds book on commutative algebra. To write this out in complete detail should be done on a blackboard in a classroom.


问:
Evalute this integral using Green's Thereom
Let C be the boundary of the half-annulus

$$1\leq(x^2+y^2)\leq4$$ where $$x\le0$$
  in the xy plane, traversed in the positive direction.

Evaluate  : $ \displaystyle \int_{c}(7\cosh^3(7x)-2y^3) dx - (3x^2+\sinh^4(5y)) dy$

I really have no idea where to start, could you please show the full working out if possible.

Note: We are working with parameterization, so I am looking for a method using that approach?

答:
Using Green's theorem you can evaluate the integral as
\begin{align*}
 & \; \; \; \;\int_{c}(7\cosh^3(7x)-2y^3) dx - (3x^2+\sinh^4(5y)) dy\\
 &= \iint_R \left( \frac{\partial }{\partial x } \left( - (3x^2+\sinh^4(5y))\right)   - \frac{\partial }{\partial y } \left( (7\cosh^3(7x)-2y^3)\right)\right) dx dy\\
 &= \iint_R (-6x + 6y^2) dx dy
\end{align*}

Here is a picture of region

\begin{align*}
\iint_R f(x,y) dx dy  &= \int_{-2}^{-1}\left( \int_0^{\sqrt{4-x^2}}f(x,y) dy + \int_{-\sqrt{4-x^2}}^0f(x,y) dy \right)+\\
 & \int_{-1}^{0}\left( \int_{\sqrt{1-x^2}}^{\sqrt{4-x^2}}f(x,y) dy + \int_{-\sqrt{4-x^2}}^{-\sqrt{1-x^2}}f(x,y) dy \right)
\end{align*}

I think you can evaluate those integrals.

Also you can parametrize your path as in given picture and evaluate the line integral along those paths.


问:
A question of Roots of unity

By considering the ninth roots of unity, show that: $\cos(\frac{2\pi}{9}) +
 \cos(\frac{4\pi}{9}) + \cos(\frac{6\pi}{9}) + \cos(\frac{8\pi}{9}) = \frac{-1}{2}$.

I know how to find the roots of unity, but I am unsure as to how I can use them in finding the sum of these $4$ roots.

答: Note that\begin{multline}\cos\left(\frac{2\pi}9\right)+\cos\left(\frac{4\pi}9\right)+\cos\left(\frac{6\pi}9\right)+\cos\left(\frac{8\pi}9\right)=\\=\frac12\left(e^{2\pi i/9}+e^{-2\pi i/9}+e^{4\pi i/9}+e^{-4\pi i/9}+e^{6\pi i/9}+e^{-6\pi i/9}+e^{8\pi i/9}+e^{-8\pi i/9}\right)\end{multline}But this is half the sum of all ninth roots of unity other than $1$. So, it's half the sum of the roots of$$x^8+x^7+x^6+x^5+x^4+x^3+x^2+x$$and that sum is $-1$.

Title: A question of Roots of unity


问:
According to Weierstrass theorem, a continuous function on a closed bounded interval is bounded.
What if the function is discontinuous at just one point? Is it bounded anyway?

Why is my example different from the one you proposed?

答:
If the function is discontinuous only at $c \in [a,b]$, and depending on the type of discontinuity, then you can say $f$ is bounded in $[a,c), (c, b]$ (but depending on the type of discontinuity!).

You're probably thinking on a removable discontinuity, but take $\tan x$ on $[0, \pi]$ (which is only discontinuous at $\frac{\pi}{2}$, but not bounded!)


问:
How to integrate these integrals from the free fall problem in classical mechanics?
How to integrate: $\int_{0}^{2}\dot x^2(t) dt$ and $\int_{0}^{2} 2\dot x(t)dt$?

I would be very grateful for any answers!

答:
I am guessing the dot means derivatve.

The second one is
$$
\int_0^2 2 \dot{x}(t)\;dt = 2x(t)\bigg\vert_0^2 = 2x(2)-2x(0)
$$
The reason is:
$$
\frac{d}{dt}\;x(t) = \dot{x}(t)
$$
so $x(t)$ is an anti-derivatvie of $\dot{x}(t)$.

For the first one here is no such simple evaluation.


问:
How do I represent the elements of $\mathbb{F}_9$ over $\mathbb{F}_3$
How do I represent the elements of $\mathbb{F}_9$ over $\mathbb{F}_3$?

The question hints to use the fact that $\mathbb{F}_9$ is the 8th cyclotomic field over $\mathbb{F}_3$, but I dont see how that helps. How does it help to know there are elements $a$ in $\mathbb{F}_9$ such that $a^8 = 1$ in $\mathbb{F}_3$?

答:
$\mathbb{F}_9$ is the splitting field of every quadratic polynomial irreducible over $\mathbb{F}_3$.

The simplest such polynomial is $x^2+1$.

Let $\alpha \in \mathbb{F}_9$ be a root of $x^2+1$. Then $\mathbb{F}_9=\mathbb{F}_3(\alpha)=\mathbb{F}_3[\alpha]=\{ a+b\alpha: a,b \in \mathbb{F}_3 \}$, where $\alpha^2=-1$.

Alternatively, $(\alpha+1)^4=-1$ and so $\mathbb{F}_9^* =  \{ (\alpha+1)^n : n=0,\dots,8 \}$


问:
How many ways are there to color a square prism with two colors?
How many ways are there to color a square prism with two colors (black and white) such that repetition is allowed and all faces can be the same color keeping in mind orientation?

I counted out the cases and got 18 but I wanted to know if there is an easier way to answer this.

Thanks!

答:
A square prism (not a cube) has $8$ rotational symmetries:

identity leaves $2^6$ colourings fixed

two $90^\circ$ rotations about the unique axis leave $2^3$ colourings fixed each

$180^\circ$ rotation about unique axis leaves $2^4$ colourings fixed

two $180^\circ$ rotations about axes perpendicular to the side faces leave $2^4$ colourings fixed each

two $180^\circ$ rotations about axes parallel to diagonals of the square top and bottom leave $2^3$ colourings fixed each

Thus, by Burnside's lemma, the number of colourings is
$$\frac{2^6+2×2^3+2^4+2×2^4+2×2^3}8=18$$


问:
USAJMO $2019$ problem $1$ Combinatorics

There are $a+b$ bowls arranged in a row, numbered $1$ through $a+b$, where $a$ and $b$ are given positive integers. Initially, each of the first $a$ bowls contains an apple, and each of the last $b$ bowls contains a pear.
A legal move consists of moving an apple from bowl $i$ to bowl $i+1$ and a pear from bowl $j$ to bowl $j-1$, provided that the difference $i-j$ is even. We permit multiple fruits in the same bowl at the same time. The goal is to end up with the first $b$ bowls each containing a pear and the last $a$ bowls each containing an apple. Show that this is possible if and only if the product $ab$ is even.

Now in this solution(3) how $A_1-A_2$ is invariant ?

答:
(Fill in the gaps as needed. If you're stuck, explain what you've tried.)

Hint: Show that for each legal move, an apple goes from an odd to even bowl iff a pear goes from an odd to even bowl.

 $i,j$ have the same parity.

  $i+1, j-1$ have the same parity.

Hence, $ A _ 1 - A_2$ is invariant.

 $A_1$ increases (resp decreases) by 1 iff $A_2$ increases (resp decreases) by 1.

FWIW Solution 2 and 3 are essentially the same. So if you understood solution 2, then you should understand solution 3.

Title: USAJMO $2019$ problem $1$ Combinatorics


答:
Denote
$$
\vec a = \vec{AB}, \qquad \vec b = \vec{AC}
$$
Then
$$
\vec c = -\vec a + \vec b
$$
and
$$
\vec{DE} = \frac12 \vec a + \frac14 (-\vec a + \vec b) = \frac14 \vec a + \frac14 \vec b
$$
$$
\vec{DF} = -\frac12 a + \frac23 \vec b
$$
Area of the $\triangle DEF$ can be calculated as
$$
\frac12 |\vec{DE} \times \vec{DF}| = \frac12 |(\frac14 \vec a + \frac14 \vec b) \times (-\frac12 a + \frac23 \vec b)| = \frac{7}{48} |\vec a \times \vec b|
$$
since $\vec a \times \vec a = \vec b \times \vec b = 0$ and $\vec a \times \vec b = \vec b \times \vec a$.

Area of $\triangle ABC$ is also
$$
\frac12 |\vec{a} \times \vec{b}| = 480
$$
implying
$$
|\vec{a} \times \vec{b}| = 960
$$
Area of $\triangle DEF$ is therefore
$$
\frac{7}{48} \cdot 960 = 140
$$

Title: area of triangle


问:
Calculus and the intermediate value theorem
Hi I am trying to understand and prove the fundamental theorem of calculus and I ran into some confusion understanding the intermediate value theorem . several sources online claim that if a function f(x) is continuous on [a,b] let s be a number such that f(a)<s<f(b) then there exists a number k in the open interval (a,b) such that f(k)=s my question is why do we only assume the open interval shouldn't it also include the closed interval [a,b] and also why does s have to be less than both f(a) and f(b)?</p>

答:
The assertion “there is some $k\in(a,b)$ such that $f(k)=s$” is stronger than the assertion “there is some $k\in[a,b]$ such that $f(k)=s$”. So, why would we state a weaker statement when we can as easiy prove a stronger one.

And, if you have in mind the statement “if $f(a)\leqslant s\leqslant f(b)$, then there is some $k\in[a,b]$ such that $f(k)=x$”, then that statement is trivial if $s=f(a)$ (just take $k=a$ then) or if $s=f(b)$ (just take $k=b$ then). So, the non-trivial part is when $f(a)<s<f(b)$.


问:
Find upper bound on number of grandparents
There is a group of $20$ children and a group of $n$ grandparents to these children. Each of these grandparents is either father's father or mother's father to at least one of these children. Now, following constraints are given:

Each pair of children has at least one grandparent in common who is also present in this group of grandparents.

Every grandparent has at least two grandchildren in this group of children.

Find upper bound on $n$, the number of grandparents present in the group such that above constraints can be satisfied.

答:
"is either father's father or mother's father to at least one of these children" -> this means we are talking only about grandfathers.

See this answer.

In short each child can have max 2 grandfathers and the upper bound for n is 3 in order for the first constraint to be satisfied.


问:
Laplace method for $\int_0^1 dx x^\gamma \frac{\partial}{\partial x} P\big(\frac{u}{x}\big) $ where $\gamma \gg 1$ and $P(\infty)\rightarrow 0$
I have a structure like
$$ F(u) = \int_0^1dx x^\gamma \frac{\partial}{\partial x}P\big(\frac{u}{x}\big) $$
where $\gamma\gg 1$ is a positive (noninteger) exponent and $P$ is an unknown function with $P(\pm \infty) = 0$. Since the integrand is dominated by the value $x=1$, I wonder if I can perform a Laplace-type asymptotic expansion of this integral despite not knowing anything about $P$.

Naively, I want to say something like $x^\gamma \approx 1$ so that I can trivially integrate for $F(u) = P(u) - P(\infty) = P(u)$, but I'm not sure this handwavy approach is correct and I'm unclear how to proceed. I am aware of some similar problems, but these do not involve unknown functions (like $\partial P(u/x)/\partial x$). I would be very happy to read any thoughts on if such a Laplace-like integral can be done with unspecified $P$.

答:
Though the major contribution to the integral comes from the region $x\simeq1$, this region becomes very narrow as $\gamma$ increases. The naive trick cannot apply.

To simplify the problem, an integration by part can be  performed first:
\begin{align}
 F(u) &= \int_0^1 x^\gamma \frac{\partial}{\partial x}P\left(\frac{u}{x}\right)\,dx\\
 &=P(u)-\gamma\int_0^1 x^{\gamma-1} P\left(\frac{u}{x}\right)\,dx
\end{align}
Now, by substituting $x=\exp(-t)$, the integral is transformed into a Laplace integral:
\begin{equation}
 F(u)=P(u)-\gamma\int_0^\infty e^{-\gamma t}P\left( ue^t \right)\,dt
\end{equation}
If, for $s\to 0^+$ the function $P(ue^t)$ can be expanded as
\begin{equation}
 P(ue^t)\sim\sum_{s=0}^{\infty}a_{s}t^{(s+\lambda-\mu)/\mu}
\end{equation}
where $\lambda$ and $\mu$ are positive constants, Watson lemma gives the asymptotic expansion
\begin{equation}
 \int_{0}^{\infty}e^{-\gamma t}q(t)\mathrm{d}t\sim\sum_{s=0}^{\infty}\Gamma\left(%
\frac{s+\lambda}{\mu}\right)\frac{a_{s}}{\gamma^{(s+\lambda)/\mu}}
\end{equation}
In particular, if $P(y)$ is analytical at $y=u$, then $\lambda=\mu=1$ and
\begin{equation}
 P(ue^{t})=P(u)+suP'(u)+\frac{s^2}{2}u\left( P'(u)+uP''(u) \right)+\cdots
\end{equation}
Thus
\begin{align}
 F(u)&=P(u)-\gamma\left( \frac{P(u)}{\gamma}+2\frac{uP'(u)}{\gamma^2} +3u\frac{ P'(u)+uP''(u)}{\gamma^2}+\cdots\right)\\
 &= -\frac{2uP'(u)}{\gamma}-3\frac{u\left( P'(u)+uP''(u) \right)}{\gamma^2}+\cdots
\end{align}


答:
For your first question, the statement could (not 'should') refer to the closed interval $[a,b]$ but when $s=f(a)$ or $s=f(b)$ we can obviously take $k=a$ or $k=b$  and this part does not require the continuity assumption at all. Thus, the "interesting" part of the theorem, which does require the continuity assumption, is about an $s$ which is not equal to $f(a)$ or $f(b)$, and consequently, we don't need the endpoints $a,b$ anymore.

For your second question, note that "$s$ has to be less than both $f(a)$ and $f(b)$" is a wrong formulation. What we require of $s$ is that it be an "intermediate value". Think about an intermediate value as some number "intermediate" between $f(a)$ and $f(b)$. In other words, if $f(a)<f(b)$ then for $s$ to be between them means $f(a)<s<f(b)$, and if $f(b)>f(a)$ then for $s$ to be between them means $f(b)<s<f(a)$. Note that if $f(a)=f(b)$ the theorem says nothing - you have to assume that $f(a)\neq f(b)$ in order to get proper intermediate values.

Title: Calculus and the intermediate value theorem

Body: Hi I am trying to understand and prove the fundamental theorem of calculus and I ran into some confusion understanding the intermediate value theorem . several sources online claim that if a function f(x) is continuous on [a,b] let s be a number such that f(a)<s<f(b) then there exists a number k in the open interval (a,b) such that f(k)=s my question is why do we only assume the open interval shouldn't it also include the closed interval [a,b] and also why does s have to be less than both f(a) and f(b)?</p>


问:
Equality in non commutative field
Let $K$ be a non commutative field and $a,b \in K$ such that $ab \neq ba$.
Show that :
$$ a=\big[b-(a-1)^{-1}b(a-1)\big]\big[a^{-1}ba-(a-1)^{-1}b(a-1)\big]^{-1}. $$

I tried to move the second term of the right hand side to the left and develop but it seem to go nowhere.

答:
Multiply on both sides on the right by $[a^{-1}ba-(a-1)^{-1}b(a-1)]$:

$a[a^{-1}ba-(a-1)^{-1}b(a-1)] = [b-(a-1)^{-1}b(a-1)]$

Distribute the $a$:

$aa^{-1}ba-a(a-1)^{-1}b(a-1) = ba-a(a-1)^{-1}b(a-1) = b-(a-1)^{-1}b(a-1)$

where $aa^{-1}ba=eba=ba$ where $e$ is the identity of multiplication in the field

Group together on both sides of the '$=$' that have common left or right factors:

$ba-b=a(a-1)^{-1}b(a-1)-(a-1)^{-1}b(a-1)$

Factor out the $b$ on the left and $(a-1)^{-1}b(a-1)$ on the right:

$b(a-1)= (a-1)[(a-1)^{-1}b(a-1)]$

then $b(a-1)=b(a-1)$ because $(a-1)(a-1)^{-1}=e$, the multiplicative identity.

Title: Equality in non commutative field


答:
Like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$,

The roots of $$\dfrac{z^9-1}{z-1}=0$$ are $e^{2\pi ir/9}; r=1,2,3,4,5,6,7,8$

Dividing both sides $z^4,$

$$z^4+\dfrac1{z^4}+z^3+\dfrac1{z^3}+z^2+\dfrac1{z^2}+z+\dfrac1z+1=0$$

$z+\dfrac1z=2\cos\dfrac{2\pi r}9=2y$(say)

Use
$\left(z+\dfrac1z\right)^2=z^2+\dfrac1{z^2}+2$

$\left(z+\dfrac1z\right)^3=z^3+\dfrac1{z^3}+3\left(z+\dfrac1z\right)$

$z^4+\dfrac1{z^4}=\left(z^2+\dfrac1{z^2}\right)^2-2=\left(\left(z+\dfrac1z\right)^2-2\right)^2-2$

to form a bi-quadratic equation in $y$

Now use Vieta's formula

See also: factor $z^7-1$ into linear and quadratic factors and prove that $ \cos(\pi/7) \cdot\cos(2\pi/7) \cdot\cos(3\pi/7)=1/8$

Title: A question of Roots of unity

Body:
By considering the ninth roots of unity, show that: $\cos(\frac{2\pi}{9}) +
 \cos(\frac{4\pi}{9}) + \cos(\frac{6\pi}{9}) + \cos(\frac{8\pi}{9}) = \frac{-1}{2}$.

I know how to find the roots of unity, but I am unsure as to how I can use them in finding the sum of these $4$ roots.


问:
Complex number polar forms
$(1-i\sqrt{3})^{50}$ in the form x + iy

I have tried this out but seem to be missing something

答: Hint: First show $(1-i \sqrt(3))^6=64$ and then use $50=6 \cdot 8+2.$

Title: Complex number polar forms


问:
To find a vector perpendicular to a vector with variable values
Let A vector = iA cos$\theta$ j + A sin$\theta$ be any vector. Another vector B which is normal to A is

How I solved it:

I thought that there can be two possibilities in which A vector is perpendicular to two other vecotrs in which I choose either of them to be our B vector.if we make the figure look like image 1.

OD AND OE are iAcos $\theta$ and iBsin$\theta$.
Then I thought for others and made the angles equal according to image 2.

For that answer should be Bcos $\theta$(- i ) and B sin $\theta$ j

Image 2

Image 1

My answer did not come right.Pls help where am I wrong and how should I approach it.

Please let me know where you find it understand my answer or the question.

Thank you

答:
The given vector $\, \vec{OA}$ is $(A \cos \theta, A \sin \theta) \,$.

Unit vector in the direction of $\vec{A}$ is $ \,\hat{a} = (\cos \theta, \sin \theta)$.

Now unit vector orthogonal to $\vec{a}$ will be $(\cos (90^0 + \theta), \sin (90^0 + \theta)) = (-\sin \theta, \cos \theta)$. Please see the diagram. In your image $1$, please note that $\angle GOI = (90^0 - \theta) \, $ and not $\theta$.

So we have a unit vector $(-\sin \theta, \cos \theta) \,$ in the orthogonal direction of $\vec{OA}$. We will also have another orthogonal vector in the direction of $\vec{OC}$ and that is just opposite direction of $\vec{OB}$ and so it is $(\sin \theta, -\cos \theta)$.

Now if you want a vector of magnitude of $B$, just multiply the unit vector by the magnitude. So orthogonal vector $\vec{OB} = (-B \sin \theta, B \cos \theta)$.

As we know the dot product of two orthogonal vectors is zero, the easiest way to find orthogonal vector to a given vector in $2D$ space is to find the unit vector and then interchange values of $i$ and $j$ position with one of them being negative sign.

So if you have a unit vector $(x,y)$, the orthogonal vectors will be $(y,-x)$ and $(-y, x)$. You can then assign the required magnitude.


问:
Canonical log structure defined by a normal crossing divisor
Given a locally noetherian scheme $X$ and $D \subset X$  a normal crossing divisor.
 Let $j: U=X-D \hookrightarrow X$ the open complement immersion. Define the log structure $(M_X,\alpha_X)$ on $X$ as the direct image by $j$ of the trivial log structure on $X$.
 Can someone explain to me why the morphism $\mathbb{N}_U^r \to M_U$ defined by $e_i \mapsto t_i$ where $t_i$ are part of a regular system of parameters of $\mathcal{O}_{X,x}$ and $e_i$ are basis of $\mathbb{N}^r$ is chart of $X$ on $U$; and especially the fact that at a geometric point $x$ of $U$ every element $a$ of $M_x$ can be uniquely written as $a=u \prod_{i \in I} t^{n_i}_i$ with $n_i \in \mathbb{N}$, $u \in \mathcal{O}_x^*$, and $I$ is the set of $I$ such that $(t_i)_x \in \mathfrak{m}_x$?

Also, it results that if $D$ is normal crossing divisor sum of regular divisors $D_i$ ($1 \leq i \leq m$), then the characteristic sheaves are given by:
 $$ C_X= \oplus_{1 \leq i \leq m} \mathbb{N}_{D_i}, C_X^{gp} = \oplus_{1 \leq i \leq m} \mathbb{Z}_{D_i}.$$

All this is in example 3.4 of the manuscript "Géométrie logarithmique" of L. Illusie and A. Ogus.

Could you explain this to me?

Thank you

答:
Here is the basic idea modulo some details. I'm sorry, but somebody else will have to fill in the missing parts about characteristic sheaves.

If we put the trivial log structure on $U,$ then the direct image log structure has sections which are (locally) regular functions on $X$ whose restriction to $U$ is invertible. This follows by definition of $j_*^{log}(\mathcal O_U^\times)=\mathcal M_X$ as the fibre product of the morphisms $\mathcal O_X\to j_*\mathcal O_U$ and $j_*\mathcal M_U\to j_*\mathcal O_U,$ where $\mathcal M_U=\mathcal O_U^\times.$

Now if $D$ is a (reduced) normal crossings divisor, then by definition there exists locally in a neighbourhood of $x$ a (regular) system of parameters $\{t_i\}_{i\in J}$ such that $D=V(\prod_{i\in I} t_i)$ where $I\subseteq J.$ Thus, the local sections of $\mathcal M_X$ are those regular functions $f=ug,$ where $u$ is invertible and $g=\prod_{i\in I}t_i^{n_i}$ vanishes exactly along $D.$

So, for example, if $X=\mathbb A^2_K$ and $D=V(xy),$ then $U=D(xy)=\operatorname{Spec}(k[x,y]_{xy})$ is the plane minus the two axes, and $\mathcal M_U$ is the sheaf associated to $(k[x,y]_{xy})^\times.$ To get $\mathcal M_X$ we consider the (associated sheaf to the) fibre product over $k[x,y]\to k[x,y]_{xy}$ and $(k[x,y]_{xy})^\times\to k[x,y]_{xy},$ which has for example global sections $(a,b)\in k[x,y]\times(k[x,y]_{xy})^\times$ such that $a$ is invertible as an element of $k[x,y]_{xy}$ (equal to $b$), that is, there exists some $c\in k[x,y]$ which we can assume is not divisible by $xy$ such that $ac=(xy)^m$ for some nonnegative power $m$. Then, in say the local ring at any point $x\in D,$ we have $a=u(xy)^m$ with $u=c^{-1}$ a unit, since $c\notin\mathfrak m_x.$

In this case, we see that locally the map $\mathbb N_X\to \mathcal M_X,m\to (xy)^m$ essentially determines $\mathcal M_X,$ except that we are missing the invertible $u.$ This is because $\mathbb N_X$ is only a pre-log structure, i.e., the associated log structure $\mathbb N_X^a$ adds the units of the structure sheaf back in for a complete description.


问:
Mulitplicativity of $\Vert \cdot\Vert_{L^\infty}$
Let $(X,\mathscr{A},\mu)$ be some measure space and let $L^\infty(X)$ denote $L^\infty(X,\mathscr{A},\mu)$.
Furthermore let $f\in L^\infty(X)\cap L^1(X,\mathscr{A},\mu)$ i.e. $\int_X \vert f\vert \ d \mu <\infty$ and $\operatorname{ess sup}_X \vert f \vert <\infty$. For some $k\in \mathbb{R}^+$  is the following true:

$\Vert f^k\Vert_{L^\infty(X)} \le \Vert f\Vert_{L^\infty(X)}^k$? Where $\Vert g\Vert_{L^\infty(X)}:= \operatorname{ess sup}_X \vert g\vert$ for any $g\in L^\infty(X)$.

答: Yes.  We may assume $\|f\|_{\infty}=1$, and since $|x|^k\leq 1$  $\forall x\in [0,1]$ and every $k>0$, we get $\|f^k\|_{\infty}\leq 1$. You don't need to assume $f\in L_1$


问:
Find $ \int_0^\infty \frac{x}{x^3 +1} dx$
Find $ \int_0^\infty \frac{x}{x^3 +1} dx$

What cantor should I choose?

答: Contour integration isn't the best approach. It's better to take $x=\tan^{2/3}t$, so the integral is$$\int_0^{\pi/2}\frac23\tan^{1/3}tdt=\frac{\pi}{3}\csc\frac{2\pi}{3}=\frac{2\pi}{3\sqrt{3}}.$$


问:
Prove that $3^n>n^4$ if $n\geq8$
Proving that $3^n>n^4$ if $n\geq8$

I tried mathematical induction start from $n=8$ as the base case, but I'm stuck when I have to use the fact that the statement is true for $n=k$ to prove $n=k+1$.  Any ideas?

Thanks!

答: First show the base case. Then assume $3^k>k^4$. Hence, $3^{k+1}=3*3^k>3k^4>(k+1)^4$</p>

Title: Prove that $3^n>n^4$ if $n\geq8$


问:
Is the ring of the polynomials on a finite algebraic set always a product of local rings?
A proposition from Fulton's book 'Algebraic Curves':

Let $I$ be an ideal in $k[X_1, ..., X_n]$ ($k$ algebraically closed), and
suppose $V(I) = \{ P_1, \ldots, P_N \}$ is finite. Let
$\mathcal{O}_i = \mathcal{O}_{P_i}(\mathbb{A}^n)$. Then there is a natural
isomorphism of $k[X_1, \ldots, X_n] \mathbin{/} I$ with
$\prod_{i = 1}^{N} \mathcal{O}_i \mathbin{/} I\mathcal{O}_i$.

But if $k$ is not algebraically closed, is it still true that, if $V(I)$ is
finite, $k[X_1, \ldots, X_n] \mathbin{/} I$ is a direct product of local
rings? And if so, then some factors are most likely given by localizations at
the points of $V(I)$, whereas other factors are not; how to describe these
latter factors?

The reason for this question is the observation that field extensions of $k$
are fields and thus local rings. For instance,
$\mathbb{Q}[X] \mathbin{/} (X^2 + 1)$ is a field, but $V(I)$ is empty and
$\mathbb{Q}$ is not algebraically closed. Hence, by the Chinese remainder
theorem, it seems to me that, for any $f$ in $k[X]$, $k[X] \mathbin{/} (f)$ is
the direct product of local rings anyway. Is it at least that correct?

答:
Let $B:=k[x_1,..,x_n]$ with $k$ any field and let $I \subseteq B$ be any ideal.
Let $X:=Spec(A)$ with $A:=B/I$. By the Noether normalization lemma it follows there is a finite set of elements $y_1,..,y_d \in A$ with $d=dim(X)$ and the property that the sub-ring $k[y_1,..,y_d] \subseteq A$ generated by the elements $y_i$ is a polynomial ring. The ring extension $k[y_1,..,y_d] \subseteq A$ is an integral extension of rings. If $d=0$ it follows from the same lemma the ring extension $k \subseteq A$ is integral and in particular it follows $A$ is an artinian ring. In particular it follows $A\cong R_1 \prod \cdots \prod R_l$ is a direct product of artin local rings $R_i$ with $\mathfrak{p_i} \subseteq R_i$ the maximal ideal (this is Atiyah-Macdonald's book Thm 8.7). Hence $Spec(A) \cong \prod Spec(R_i)$ is a disjoint union of the schemes $U_i:=Spec(R_i)$. We may write $X:=\{ \mathfrak{p_1}, \ldots , \mathfrak{p_l}\}$ as a disjoint union of $l$ distinct points, with $\mathfrak{p_i} \subseteq  R_i$ the unique maximal ideal in the local ring $R_i$. Since $U_i \subseteq X$ is an open subscheme it follows there are isomorphisms $\mathcal{O}_{X,\mathfrak{p_i}} \cong \mathcal{O}_{U_i,\mathfrak{p_i} } \cong (R_i)_{\mathfrak{p_i}} \cong R_i$ since $R_i$ is a local ring and every element in $R_i - \mathfrak{p_i}$ is a unit. Here $\mathcal{O}_{U_i}$ is the restriction of $\mathcal{O}_X$ to the open set $U_i$. Hence for any base field $k$ it follows from Atiyah-Macdonalds (AM) book "Commutative Algebra" that your ring $A:=B/I$ decompose as a finite direct sum of Artinian local rings $R_i$. At the level of sheaves you get the following formulas:

$\mathcal{O}_X \cong  \mathcal{O}_{U_1} \oplus \cdots \oplus \mathcal{O}_{U_l}$

and

$  \mathcal{O}_{X,\mathfrak{p_1}} \oplus \cdots \oplus \mathcal{O}_{X,\mathfrak{p_l}} \cong  \mathcal{O}_{U_1,\mathfrak{p_1}} \oplus \cdots \oplus \mathcal{O}_{U_l,\mathfrak{p_l}} \cong \mathcal{O}_{U_1}(U_1) \oplus \cdots \oplus \mathcal{O}_{U_l}(U_l) .$

These formulas answers the following questions:

Question 1: "But if k is not algebraically closed, is it still true that, if V(I) is finite, k[X1,…,Xn]/I is a direct product of local rings?"

Answer 1: Yes, this is true by the above argument.

Question 2: "And if so, then some factors are most likely given by localizations at the points of V(I), whereas other factors are not; how to describe these latter factors?"

Answer 2: All factors are given as localizations at points by the above argument.

Consider your example with $A:=\mathbb{Q}[x,y]/(x^2+y^2-3)$ and let $C:=Spec(A)$. By Zorn's Lemma (if you accept this) there is a maximal ideal $\mathfrak{m} \subseteq A$, and the residue field $\kappa(\mathfrak{m})$ is a finite extension of $\mathbb{Q}$. Hence $C$ - which as a set is the set of prime ideals in $A$ - is non-empty. If $k$ is the algebraic closure of $\mathbb{Q}$ and you pass to the fiber product $C\otimes k$ you get a curve. By Hartshorne Prop I.1.13 it follows $dim(C\otimes k)=1$ since it is defined by a hypersurface in $k[x,y]$.

Let $k$ be any field and let $A:=k[x]$ be the polynomial ring in the variable $x$. Let $f(x) \in A$ be any polynomial. We may write $f(x)=\prod_{i=1}^l p_i(x)^{d_i}$ where $p_i(x)$ is an irreducible polynomial and  $d_i \geq 1$ is an integer for every $i$. Here $p_i(x)$ and $p_j(x)$ are distinct for $i\neq j$. Let $B_i:=k[x]/(p_i(x)^{d_i})$ and let $B:=A/(f(x))$.
There is a canonical isomorphism

$\phi: B \rightarrow B_1\oplus \cdots \oplus  B_l$

defined by

$\phi(x + (f(x))) :=(x+(p_1(x)^{d_1}),..,x+(p_l(x)^{d_l}))$.

This is Prop 1.10 in the AM-book. The ideals $(p_i(x)^{d_i})$ and $(p_j(x)^{d_j})$ are coprime for $i\neq j$. Since the ideals $(p_i(x))$ are maximal, it follows the rings $B_i$ are Artinian local rings, hence the decomposition given by $\phi$ decompose the Artinian ring $B$ into a direct sum of Artinian local rings $B_i$. Hence in this case Prop 1.10 in AM implies Thm 8.7 in the AM book (the "stucture theorem on Artin rings").

Question 3: "And is there an higher-level explanation why it (apparently, not quite sure) works in one variable, that is, k[X]/f is a direct product of local rings for any f∈k[X], even if k is not algebraically closed, and even if V(f) is empty?"

Here you must understand the fact that any polynomial $f(x) \in k[x]$ factors as a product of powers of irreducible polynomials, and that an irreducible polynomial defines a maximal ideal in $k[x]$. Moreover if $p(x)$ is an irreducible polynomial, it follows the ring $k[x]/((p(x)^d)$ for $d\geq 1$ is an Artinian ring. Then you must read AM, Chapter1 where it is proved that the map $\phi$ is an isomorphism of rings.


问:
Can matrices do cross product with vectors?
this is a practical problem raised from my chemistry research, apologize if somewhere I used incorrect words.

I'm dealing with the derivatives of a vector with respect to another vector. After some searching, I found the rule for differentiating a column vector with respect to a row vector is (all these vectors are 3 dimensional in Cartesian):

$$
\mathbf{e}=\begin{bmatrix}e_1\\e_2\\e_3\end{bmatrix}
$$
$$
\mathbf{P}^T=\begin{bmatrix}P_1 & P_2 & P_3 \end{bmatrix}
$$
$$
\frac{\partial \mathbf{e}}{\partial \mathbf{P}^T}=
\begin{bmatrix}
\frac{\partial e_1}{\partial P_1} & \frac{\partial e_1}{\partial P_2} & \frac{\partial e_1}{\partial P_3} \\
\frac{\partial e_2}{\partial P_1} & \frac{\partial e_2}{\partial P_2} & \frac{\partial e_2}{\partial P_3} \\
\frac{\partial e_3}{\partial P_1} & \frac{\partial e_3}{\partial P_2} & \frac{\partial e_3}{\partial P_3} \\
\end{bmatrix}
$$

Here my $\mathbf{e}$ is determined from a cross product, that is, $\mathbf{e}=\mathbf{n}\times\mathbf{m}$. So I tried to do this:
$$
\frac{\partial \mathbf{n}\times \mathbf{m}}{\partial \mathbf{P}^T}
$$

From wiki, it says the product rule can be applied, so I arrived here:
$$
\frac{\partial \mathbf{n}\times \mathbf{m}}{\partial \mathbf{P}^T}=\frac{\partial \mathbf{n}}{\partial \mathbf{P}^T}\times\mathbf{m}+\mathbf{n}\times\frac{\partial \mathbf{m}}{\partial \mathbf{P}^T}
$$
Now I'm confused because $\frac{\partial \mathbf{n}}{\partial \mathbf{P}^T}$ and $\frac{\partial \mathbf{m}}{\partial \mathbf{P}^T}$ are matrices, and they do cross product with vectors.

I didn't found anywhere saying the cross product of matrix, so I'm concerned perhaps there is something wrong in the derivation.

Hope you can give some help. Thanks in advance!

答:
$\def\e{\varepsilon}\def\p#1#2{\frac{\partial #1}{\partial #2}}$
The Levi-Civita symbol $(\e_{ijk})$ can be used to write the cross product of two vectors as
$$\eqalign{
a\times b &= -a\cdot\e\cdot b \\
 &= +b\cdot\e\cdot a \;\;=\; -b\times a
}$$
Replacing either vector with a matrix effectively defines the cross product for matrices, e.g.
$$\eqalign{
A\times b &= -A\cdot\e\cdot b \\
a\times B &= -a\cdot\e\cdot B \\
A\times B &= -A\cdot\e\cdot B \\
}$$
Having defined the matrix cross product, the naive product rule
$$\eqalign{
\p{(a\times b)}{p}
\;\ne\; a\times\left(\p{b}{p}\right) \;+\; \left(\p{a}{p}\right)\times b \\
}$$
is still incorrect $\,\big({\rm NB}:$ It is correct when $p$ is a scalar instead of a vector$\big).$

However, the following rule is compatible with the matrix cross product
$$\eqalign{
\p{(a\times b)}{p}
\;=\; a\times\left(\p{b}{p}\right) \;-\; b\times\left(\p{a}{p}\right) \\\\
}$$
Another approach is to define a skew matrix associated with each vector
$$\eqalign{
{\cal A} &\doteq {\rm Skew}(a) \;=\; (a\times I) \;=\; -a\cdot\e \\
{\cal B} &\doteq {\rm Skew}(b) \;=\; (b\times I) \;=\; -b\cdot\e \\
}$$
Then the cross product can be replace by a normal matrix-vector product
$$\eqalign{
a\times b \;=\; {\cal A}b \;=\; -{\cal B}a \\
}$$
and the gradient can use standard matrix-matrix products
$$\eqalign{
\p{(a\times b)}{p}
  &=  {\cal A}\left(\p{b}{p}\right) \;-\; {\cal B}\left(\p{a}{p}\right)\\
}$$
$$\eqalign{
}$$


问:
Description of curve arising from rotation of a cube
Take a unit cube, and place it so that one of its body diagonals lies along the $z$-axis.  For symmetry, assume that the vertices of the cube are at $(0,0,\pm\sqrt{3}/2)$.  Then rotate it about the z-axis, forming a solid of rotation.  The resulting solid is two cones joined by a "waist".

This same solid could be formed by rotating the following piecewise-defined curve in the $xz$-plane about the $z$-axis:

A straight line from $(0,0,\sqrt{3}/2)$ to $A$, defined as $(\sqrt{2/3}, 0, \sqrt{3}/2 - 1/\sqrt{3})$.

A curve from $A$ to $B$, defined as defined as $(\sqrt{2/3}, 0, -\sqrt{3}/2 + 1/\sqrt{3})$.

A straight line from $B$ to $(0, 0, -\sqrt{3}/2)$.

Question:  What is the equation of the curve between $A$ and $B$ so defined?

This was a question that occurred to me recently, but I don't even have a good idea how to begin solving it.  I have a vague sense that this curve can be thought of as some kind of 3D envelope of the cubes as they rotate, but since a cube is not a smooth surface, I'm not sure how to apply standard techniques.

答:
That curve is an arc of hyperbola $2x^2-4z^2=1$, the section of a hyperboloid with equation $2x^2+2y^2-4z^2=1$.

That surface is generated by the rotation of an edge of the cube, connecting a vertex in the positive $z$ half-space (for instance $P=(1/\sqrt6,1/\sqrt2,\sqrt3/6)$) with a neighbouring vertex in the negative $z$ half-space (for instance $Q=(-1/\sqrt6,1/\sqrt2,-\sqrt3/6)$). A generic point on that edge can be written as
$$
M=Pt+Q(1-t)=\left({1\over\sqrt6}(1-2t),{1\over\sqrt2},{\sqrt3\over6}(2t-1)\right)
$$
and the square of its distance from $z$-axis is
$$
x_M^2+y_M^2={1\over6}(1-2t)^2+{1\over2}=2z_M^2+{1\over2}.
$$
Point $M$ and all its rotated images lie then on the hyperboloid given above.


问:
Laminar family (of pairwise distinct non-empty sets) has at most $2n-1$
Show that a laminar family (of
   pairwise distinct non-empty sets) has at most $2n-1$ members, where $n$ is number of elements of a family of subsets.

I tried to get only at most $n-1$ members, but we are asked to prove it is at most $2n-1$ members. Any help is appreciated. Source of the question:https://books.google.hu/books?id=CR_oBwAAQBAJ&pg=PA543&lpg=PA543&dq=laminar+family+of+at+most+2n-1+members&source=bl&ots=IV0IYEiNF7&sig=ACfU3U1Iaj_uouxermw4336gxqj4q_yb_w&hl=en&sa=X&ved=2ahUKEwjRivC32ajmAhXil4sKHRScDnsQ6AEwAXoECAoQAQ#v=onepage&q=laminar%20family%20of%20at%20most%202n-1%20members&f=false

答:
I know this question is old, but maybe someone stumbles across it.

We can generate the maximal laminar family with $2n-1$ sets by the following procedure:

Start with a set including all points.

Remove one point from this set and create a new set containing only this point.

The new set with the points removed also are a set for your family.

repeat from 2

For me it's easiest to visualize this with a matrix where each column corresponds to a set and each row to a point. If the entry is a $1$, it means the point is in this set, 0 otherwise.

Here is an example with 4 Points and 7 sets.
$$
\mathcal{L} = \begin{pmatrix}
 1 & 1 & 1 & 1 & 0 & 0 & 0 \\
 1 & 1 & 1 & 0 & 0 & 0 & 1 \\
 1 & 1 & 0 & 0 & 0 & 1 & 0 \\
 1 & 0 & 0 & 0 & 1 & 0 & 0
\end{pmatrix}
$$

As you can see, we always end up with this pyramid-form ($n\times n$) and an identity matrix, missing the last column ($n \times n-1$). Therefore we have a laminar family $\mathcal{L}$ with $|\mathcal{L}|=2n -1$.


问:
An "exchanging $\sup$ and $\limsup$" question
Let $a_{n,k}\in \mathbb{R}$ with:

$0\leq  a_{n,k}\leq M$

$\displaystyle a_{n,k} \rightarrow_k a_n$, uniformly with $n$

Prove: $$\displaystyle \sup_n\lim_ka_{n,k}\geq\limsup_k\sup_na_{n,k}$$

Does anybody know of a result that would help me out?

If I knew $\displaystyle a_{n,k} \rightarrow_n a_k$ I could conclude the existence of the double limits, and their equality, however here I cannot suppose this.

答:
Take $\epsilon \gt 0$. By hypothesis on the uniform convergence of $\{a_{n,k}\}$ it exists $K \in \mathbb N$ such that for $k \ge K$ and $n \in \mathbb N$

$$ a_n - \epsilon \lt a_{n,k} \lt a_n + \epsilon.$$

As $0\leq  a_{n,k}\leq M$ for all $n,k \in \mathbb N$ $\sup_n a_n$ exists and for $k \ge K$ we have

$$\sup_n a_{n,k} \le \sup_n a_n + \epsilon$$ which means that

$$\sup_n a_{n,k} \le \sup_n \lim_k a_{n,k} + \epsilon$$ as $a_n$ is defined by $a_n = \lim_k a_{n,k}$. As the sequence in $k$ $\{\sup_n a_{n,k}\}$ is less than the real number $\sup_n \lim_k a_{n,k} + \epsilon$ for $k \ge K$ we obtain

$$\limsup_k\sup_n a_{n,k} \le \sup_n \lim_k a_{n,k} + \epsilon.$$ As this is true for any $\epsilon \gt 0$ we finally get the desired result

$$\limsup_k\sup_n a_{n,k} \le \sup_n \lim_k a_{n,k}.$$


问:
Sum of random number of Bernoulli
Let $T$ be a non-negative, integer valued random variable, and $X_i$ independent Bernoulli s.t.
$P(X_i=0)=p$. Of $T$, I know only that it has mean $\mu$ with $1 \lt \mu \lt \infty$. How do I find the distribution of the random sum $\sum_{n=1}^T X_n$? I know it will be a binomial sum, but how explicit can I be when I don't know the distribution of $T$?

答:
Set

$$Y=\sum_{i=1}^T X_i$$

The conditional density $Y|T=t$ is a binomial $Bin(t;q)$ where $q=1-p$.

Thus the pmf of Y (marginal is the following)

$$P(Y=y)=\Sigma_t p_T(t)p_{Y|T}(y|t)$$

If one wanted $E(Y)$ he can use the conditional expectation properties

$$E(Y)=E[E(Y|T)]=E[T(1-p)]=(1-p)E(T)=\mu(1-p)$$

Title: Sum of random number of Bernoulli


答:
Hint:

Look at $(n+1)^4=n^4+4n^3+6n^2+4n+1$.

Notice that $4n^3 +6n^2\leq (n/2)n^3+n^3\leq n^4/2+n^4/2=n^4$

and $4n+1\leq n^4$.

Thus you get $(n+1)^4\leq n^4+n^4+n^4$.

Now what? (You should be able to take it from here.)

Title: Prove that $3^n>n^4$ if $n\geq8$

Body:
Proving that $3^n>n^4$ if $n\geq8$

I tried mathematical induction start from $n=8$ as the base case, but I'm stuck when I have to use the fact that the statement is true for $n=k$ to prove $n=k+1$.  Any ideas?

Thanks!


问:
Some doubts on a proof in the representation theory of finite groups
I'm learning repensentation theory by the book A Journey Through Representation Theory written by Caroline Gruson and Vera Serganova. I do not quite understand the proof of following lemma:

Lemma 6.4 (a) If k is algebraically closed, then every non-zero invariant bilinear form on an irreducible representation $\rho$ is either symmetric or skew-symmetric.

(b) Define $m_\rho=\frac{\sum\chi_\rho(g^2)}{|G|}$ (where g runs over
all elements of G). Then $m_\rho=1,0,-1$.

(c) If $m_\rho=0$, then $\rho$ does not admit an invariant form. If $m_\rho=1$ (resp. $m_\rho=-1$), then $\rho$ admits a symmetric (resp. skew-symmetric) invariant form.

Proof of (b) and (c): Recall that $\rho\otimes\rho=\rho_{alt}\oplus\rho_{sym}$. We obatin $(\chi_{sym},\chi_{triv})=\frac{1}{|G|}\sum\frac{\chi_{\rho}(g)^2+\chi_{\rho}(g^2)}{2}$, $(\chi_{alt},\chi_{triv})=\frac{1}{|G|}\sum\frac{\chi_{\rho}(g)^2-\chi_{\rho}(g^2)}{2}$. Note that $\frac{1}{|G|}\sum\chi_{\rho}(g)^2=(\chi_{\rho},\chi_{\rho^*})$. Therefore $(\chi_{sym},\chi_{triv})=\frac{(\chi_{\rho},\chi_{\rho^*})+m_{\rho}}{2}$, $(\chi_{alt},\chi_{triv})=\frac{(\chi_{\rho},\chi_{\rho^*})-m_{\rho}}{2}$. We have the following trichotomy:

$\rho$ does not have an invariant form iff $\rho$ is not isomorphic to $\rho^*$. In this case $(\chi_{\rho},\chi_{\rho^*})=0$ and
$(\chi_{sym},\chi_{triv})=(\chi_{alt},\chi_{triv})=0$. Therefore
$m_{\rho}=0$.

$\rho$ has a symmetric invariant form iff $(\chi_{\rho},\chi_{\rho^*})=1$ and $(\chi_{sym},\chi_{triv})=1$. This
implies $m_{\rho}=1$.

$\rho$ has a skew-symmetric invariant form iff $(\chi_{\rho},\chi_{\rho^*})=1$ and $(\chi_{alt},\chi_{triv})=1$. This
implies $m_{\rho}=-1$.

Here $\chi_{sym}$ and $\chi_{alt}$ denotes the characters of the symmetric square and alternating square of the representation $\rho$, respectively. A bilinear form $B$ on $V$ is invariant if $B(\rho_gv,\rho_gw)=B(v,w)$ for any $v,w\in V$, $g\in G$.

I don't undertstand the three claims the author made after saying "We have the following trichotomy:...". I only know the fact that an irrecducible representation admits a non-zero invariant form iff $\rho$ is isomorphic to $\rho^*$. But I can't find how the existence of a symmetric or a skew-symmetric invariant form is related to $(\chi_{sym},\chi_{triv})$ or $(\chi_{alt},\chi_{triv})$, repectively. Can someone give me some hints? Thanks in advance.

答:
By exercise 6.1(3) in the book you mentioned, it says that if $B$ is an invariant form then $B^+$ and $B^-$ are invariant (Here $B^+$ and $B^-$  denote the symmetric and skew-symmetric  forms s.t $B^\pm (v,w) = \frac{1}{2}(B(v,w)\pm B(w,v))$).

It is knows (Lemma 4.6 in your book) that $(\chi_\sigma,\chi_{triv})=\dim V^G$ for $\sigma:G\rightarrow \text{GL}(V)$ and $V^G=\{v\in V|\sigma_g(v)=v, g\in G\}$. Thus $(\chi_{sym},\chi_{triv})=\dim S^2V^{\text{sym}}$, where $S^2V=\{u\otimes v|u\otimes v=v\otimes u\}$.

If, for instance, $\rho$ does not admit a symmetric invariant form, then $(\chi_{sym},\chi_{triv})=\dim S^2V^{\text{sym}}=0$. Otherwise, by the irreducibility of $\rho$, $(\chi_{sym},\chi_{triv})=1$.

Same goes for $(\chi_{alt},\chi_{triv})$.


问:
How to solve this bi-quadratic example?
$$5\left(\frac{x+1}{x}\right)^2 - 16\left( \frac{x-1}{x}\right) - 52 = 0$$

I learned how to use a swap for instance I'd say that

$$\frac{x+1}{x}=t$$ and then swap throughout the equation, but I'm stuck here, I can't get rid of x completely.

Someone edited my question, I don't know how to fix it, I had x + 1/x not (x+1)/x

Title: How to solve this bi-quadratic example?

Body: You can write $(x + \frac{1}{x})^2$ as $(x - \frac{1}{x})^2 + 4$.


问:
There are three shooters A, B, C who are shooting a target D at the same time.What's the probability the target doesn't get hit?
There are three shooters $A$, $B$, $C$ who are shooting a target $D$ at the same time. Consider the following three events:

A)$E_a$=Shooter $A$ hits the target

B) $E_в$=Shooter $B$ hits the target

C) $E_c$=Shooter $C$ hits the target

Assume $P(E_a) =0.9$, $P(E_в) =0.8$, $P(E_c)=0.7$ and these three events are mutually independent.

What is the probability that target D was hit by some players?

答:
The probability that none of the shooters hit the target is

$$(1-0.9)(1-0.8)(1-0.7)=0.006$$

Thus the requested probability is its complement: $0.994$


问:
Limit $\lim_{x\rightarrow0}\frac{1}{x}\int_{0}^{\sin x}\sin \frac{1}{t} \cos t^{2}\mathrm{d}t$
Determine whether or not the limit below exists.
$$\lim_{x\rightarrow0}\frac{1}{x}\int_{0}^{\sin x}\sin \frac{1}{t} \cos t^{2}\mathrm{d}t$$
I tried to use the Mean value theorem integrals to prove the limit exists, but it does not exist for $\lim\limits_{x\rightarrow0}\sin \frac{1}{x}$. So I guessed the limit does not exist and used the Cauchy principle to prove it, but I failed.

Any idea will be helpful.

答:
Well, here is a slightly different approach.

Note that $(\sin x) /x\to 1$ as $x\to 0$ and hence (add missing steps here if you need) the desired limit equals the limit of $$\frac {1}{x}\int_{0}^{x}\sin(1/t)\cos t^2\,dt$$ and we can write the above expression as $$\frac{1}{x}\int_{0}^{x}\sin(1/t)(\cos t^2-1)\,dt+\frac{1}{x}\int_0^x\sin(1/t)\,dt$$ The integrand in first integral tends to $0$ as $t\to 0$. Therefore by the fundamental theorem of calculus (part 1) the first term above tends to $0$.

The desired limit thus equals $$\lim_{x\to 0}\frac {1}{x}\int_0^x\sin(1/t)\,dt$$ This is a famous question on this website and the desired limit is $0$. You can prove this by using the auxiliary function $$g(x) =x^2\cos(1/x),x\neq 0,g(0)=0$$ We have $$g'(x) =2x\cos(1/x)+\sin(1/x),x\neq 0,g'(0)=0$$ By fundamental theorem of calculus (part 2) we have $$g(x) =g(x) - g(0)=\int_0^x g'(t) \, dt$$ which means that $$\frac {1}{x}\int_{0}^{x}\sin(1/t)\,dt=\frac{g(x)}{x}-2\cdot\frac{1}{x}\int_0^x t\cos(1/t)\,dt$$ Now the first term on right tends to $g'(0)=0$ and second term on right also tends to $0$ (via fundamental theorem of calculus (part 1) as the integrand tends to $0$ as $t\to 0$).

You should also observe that the above derivation works in case $\cos t^2$ is replaced by any Riemann integrable function $\phi(t)$ which tends to $1$ as $t\to 0$.

For those wondering about the version of FTC used in above I state them explicitly.

FTC Part 1: Let $f:[a, b] \to\mathbb {R} $ be a function which is Riemann integrable on $[a, b]$. Then the function $F:[a, b] \to\mathbb {R} $ defined by $$F(x) =\int_a^x f(t) \, dt$$ is continuous on $[a, b] $. Further if $c\in[a, b] $ is a point such that right hand limit of $f$ at $c$ is $L$ then the right hand derivative of $F$ at $c$ exists and equals $L$ (the statement holds if right is replaced by left).

FTC Part 2: Let $f:[a, b] \to\mathbb {R} $ be Riemann integrable on $[a, b] $ and let $F:[a, b] \to \mathbb {R} $ be continuous on $[a, b] $ and differentiable on $(a, b) $ with $F'(x) =f(x)\, \forall x\in(a, b) $ then $$\int_a^b f(x) \, dx=F(b) - F(a) $$


问:
Getting wrong answer in a rate of change problem

In the figure, a metal ball is placed at the bottom of a spherical container with water. The radii of the container and the metal ball are $8$ cm and $4$ cm respectively. Let $h$ cm be the depth of water.

(a) It is given that the volume of a portion of a sphere cut off by a horizontal plane is $\pi \left(rH^2 - \frac{H^3} 3 \right) \ \text{cm}^3$, where $r$ cm is the radius of the sphere and $H$ cm is the height of the portion. Show that the volume of water in the container is

(i) $\frac{\pi}{3}(-h^3 + 24h^2 - 256) \ \text{cm}^3$ for $8 \leq h \leq 16$,

(ii) $4\pi h^2 \ \text{cm}^3$ for $0 \leq h \lt 8$

(b) When the container is fully filled up with water, its bottom cracks suddenly, and the water leaks at a rate of $\pi \ \text{cm}^3/s$. Find the rate of decrease of the depth of water after the container has been leaking for 6 minutes. (Give your answer correct to 3 significant figures.)

I have done (a) part but got the wrong answer in (b).   Here is how I do it:

Let V denote the volume of the sphere, $v(h)$ denotes the volume of water and $h$ denotes the depth of water,

After $6$ minutes, the volume of water is
$$\frac{4}{3} \pi 8^3 - 360 \pi = \frac{968 \pi}{3} \lt \frac{1}{2}V$$
Then find $h_0$ which is the final depth of water,
\begin{align}
4 \pi h_0^2 &= \frac{968 \pi}{3} \\
h_0 &= \sqrt{\frac{242}{3}}
\end{align}
Then,
\begin{align}
\frac{dv}{dt} &= \frac{dv}{dh} \frac{dh}{dt} \\
-\pi &= 8 \pi h_0 \frac{dh}{dt} \\
\frac{dh}{dt} = -0.139
\end{align}

But my answer was wrong, the suggested one was $-0.0162 \ \text{cm/s}$, where did I go wrong? Thanks in advance.

答:
After 6 minutes, the volume of water is actually:

$$\frac{4}{3} \pi (8)^3 - 6(60) \pi - \frac{4}{3}\pi(4)^3= \frac{712 \pi}{3} \lt \frac{1}{2}V$$

because you have forgotten to subtract the volume of the steel ball.

Then repeating your steps (which are correct), $h_0 = \frac{\sqrt{534}}{3}$, and so $\frac{dh}{dt} = -\frac{1}{8h_0} = -0.0162 \ \text{cm}/\text{s}$.


问:
Are the sets $\mathbb X=\{0,1,4,15,56,...,x_h,...\} $ and $\mathbb Y=\{0,2,12,70,408,...,y_i,...\}$ (excepting the elements $x_0=y_0=0$) disjunct?
In trying to answer another question I came to the problem as written in the title:

Q1: Are the sets $\mathbb X=\{0,1,4,15,56,...,x_h,...\} $ and $\mathbb Y=\{0,2,12,70,408,...,y_i,...\}$ (excepting the elements $x_0=y_0=0$) disjunct?

and how to approach a proof.

I have two ways to describe the sets as sequences with their elements depending on their indexes:
 $$x_0=0, x_1=1, x_{h+1}=4x_h-1x_{h-1}\\
 y_0=0, y_1=2, y_{i+1}=6y_i-1y_{i-1}\\ \tag 1$$
and using $p=2+\sqrt3$ and $q=3+\sqrt8$
 $$ x_h = f(h)= \sinh (h \cdot \ln p)/\sqrt 3 \\
 y_i =g(i) = \sinh (i \cdot \ln q)/\sqrt 8  \tag 2$$
(The latter is a compacted version of a Binet-like formula (as it is known for instance for the Fibonacci numbers))

What I've tried was ("ansatz 1") to look at the composition of $i=g°^{-1}(f(h))$ and see, whether I can prove, that $i$ is never integer when $h$ is integer, but looking at heuristics and working with the results a bit I do not see any further forcing and/or reliable way to proceed.

Next ("ansatz 2") I've looked at the prime-factorizations of $f(h)$ and $g(i)$ in terms of $h$ resp. $i$. This gives at least insight to some basic facts, for instance that $h$ must be even, then divisible by $3$ and from this $i$ must be divisible by $3$ and so on, and at least leading to a proof for small $h$ and $i$. But this is so far only reasoning for finitely many classes of cases.

Q2: Can my ansatz 1 or ansatz 2 be improved? Or not?

  Q3: Is there any differnt route to approach this?

答:
A simple observation seems to contain the potential of a more immediate proof than that provided in the answer of @RenéGy.

The set $X$ is the set of solutions of the Pell-equation (taken from the referred question):
$$ 3x^2+1 = a^2 \tag {1.1}$$
This gives the set of solutions, depending of index $h \in \{0,1,2,3,..\}$
$$ x_h = \{0,1,4,15,...\}_h \\
   a_h = \{1,2,7,26,...\}_h\\ \tag {1.2}
$$
both with the same recursion $x_{h+1}=4x_h-x_{h-1}$ resp. $a_{h+1}=4a_h-a_{h-1}$

The set $Y$ is the set of solutions of the Pell-equation:
$$ 2y^2+1 = b^2 \tag {2.1}$$
This gives the set of solutions, depending of index $i \in \{0,1,2,3,..\}$
$$ y_i = \{0,2,12,70,...\}_i \\
   b_i = \{1,3,17,99,...\}_i\\\tag {2.2}
$$
both with the same recursion $y_{i+1}=6y_i-y_{i-1}$ resp. $b_{i+1}=6b_i-b_{i-1}$

If we assume some element $x_h$ exists as well as element $y_i$ then we could equate
$$ 3x_h^2 + 1=  a_h^2 \qquad \text{  and at the same time  } \qquad  2x_h^2 + 1= b_i^2 \\
\implies x_h^2 = a_h^2-b_i^2 \tag 3$$
This is in the form of a pythagorean triple, and solutions for that must have the form
(rewritten from the formula in wikipedia)
$$ x_h=(m^2-n^2) \qquad b_i=2mn \qquad a_h = (m^2+n^2) \qquad \text{ with } m \gt n \gt 0 \tag 4
$$
This formula shows, that the parity of $x_h$ and $a_h$ must be of the same type since $m^2-n^2$ and $m^2+n^2$ have the same parity.

Indication of nonexistence: From the table in $(1.2)$ we see, that $x_h$ and $a_h$ for $h \gt 0$ have always opposite parity, contradicting $(4)$. Thus no solution $x_h = y_i$ for $h,i \gt 0$ should exist.

So if I didn't mess something we need only the proof of the opposite parity of $x_h$ and $a_h$ and this seems really simple because of the definitions by the recursion-formula.


答: You want to show $3^n>n^4$. This i.e. to showing $e^{n\ln3}>e^{4\ln n}$. This means you want to show  $n\ln 3>4\ln n$. It suffices to show $\frac{n}{\ln n }>\frac{4}{\ln 3}$. Since $\frac{8}{\ln 8}>\frac{4}{\ln 3}$ and since $f(x)=\frac{x}{\ln x}$ has a positive first derivative for $x\geq 8$, the result follows. </p>

Title: Prove that $3^n>n^4$ if $n\geq8$


问:
Module of differentials of a finitely generated $R$-algebra
I'm reading Commutative Algebra with a View Toward Algebraic Geometry written by Eisenbud. In chapter 16(Module of Differentials) there is a proposition:

If $S$ is a finitely generated $R$-algebra, say $S=R[x_1,...,x_r]/I$,
and if $I=(f_1,...,f_s)$, then $S
 \otimes_R\Omega_{R[x_1,...,x_r]/R}=\oplus_iSdx_i$.

My attempt: Let $T=R[x_1,...,x_r]$, we have the following map with $I$ as its kernel: $T\to S\to 0$. $\{d(x_i+I)\}_{I\in\{1,...,r\}}$ generate $\Omega_{S/R}$ therefore, $\Omega_{S/R}=\oplus_iSd(x_i+I)$. Using the conormal sequence, there is a surjective morphism $D\pi$ from $S \otimes_R\Omega_{R[x_1,...,x_r]/R}$ to $\Omega_{S/R}$. $D\pi$ is not necessarily an isomorphism...

答:
If $A:=R[x_1,..,x_r]$ is a polynomial ring over $R$ in $r$ variables it follows there is an isomorphism of $A$-modules $\Omega_{A/R} \cong \oplus_i Adx_i$. The sum is a direct sum, hence $\Omega_{A/R}$ is a free $A$-module of rank $r$ on the elements $dx_i$. It follows $S\otimes_R \Omega_{A/R} \cong \oplus_i S\otimes_R Adx_i$ and $S\otimes_R A \neq S$ are non-isomorphic rings, hence it seems your formula is incorrect. If the elements $f_1,..,f_l$ generate your ideal $I$, and you want a formula for the $S$-module $\Omega_{S/R}$ the following holds:

Formula 1. $\Omega_{S/R} \cong \oplus_i Sdx_i/C,$

where the module $C$ is the left $S$-module generated by the differentials $df_i$ for all $i=1,..,l$. You may find details on this construction in a commutative algebra book where they discuss the module of derivations and differentials. There are "fundamental exact sequences" involving the module of differentials, and these sequences can be used to prove such results.
Mastumuras book "Commutative Ring Theory" studies derivations and differentials and gives explicit proofs of properties of this construction. Formula 1 is proved.


问:
$\sum_{k=1}^{n} \arcsin(\sin(k))= a_n+b_n \pi$, for $a_n,b_n\in\mathbb{Z}$; what can be said about the sequence $(a_n,b_n )$?
We have $\sum_{k=1}^{n} \arcsin(\sin(k))= a_n+b_n \pi$ for some integers $a_n,b_n$. I have several questions about the behavior of these numbers:

For which $n$ does $b_n=0$ and $a_n\ne 0$, i.e. the sum is a non-zero integer?

For which $n$ does $(a_n,b_n)=(0,0)$, i.e. the sum vanishes?

Aside from these subsequences, based on numerical evidence I conjecture that $\lim_{n\to\infty} a_n/b_n=-\pi$, but I don't know how to prove it.

Recall the convergents of $\pi$ are $3, 22/7, 333/106, 355/113, 103993/33102$, etc. I computed $(a_n,b_n)$ for $1\le n\le 120000$ and here's what I found:

As expected, I saw 'increased aberrations' for $n$ near the numerator of the numerators of the convergents. For example, $b_n=0$ and $a_n\ne 0$ when $$n=\{1,8,16,52,60,96,104,140,148,184,192,228,236,272,280,316,324,360,103632,103668,103676,103712,103720,103756,103764,103800,103808,103844,103852,103888,103896,103932,103940,103976,103984\}$$ and $(a_n,b_n)=(0,0)$ when $$n=\{24,44,68,88,112,132,156,176,200,220,244,264,288,308,332,352,103640,103660,103684,103704,10372
   8,103748,103772,103792,103816,103836,103860,103880,103904,103924,103948,103968,103992\}.$$ In particular, when $b_n=0$ and $a_n \ne0$, aside from $n=1$ I found that $a_n=2$.

Aside from these values, the ratio is quite close to $\pi$. We have $a_{120000}/b_{120000} =-\frac{248162}{78993}$, which differs from $-\pi$ by about $0.0000231474$. Here is a plot:

Any information or insight would be much appreciated. I've heard of several continued fractions for $\pi$ and was wondering if perhaps this has to do with that.

Update: I checked oeis.org for both the numerator and denominator and found nothing. Since $\arcsin(\sin(k))=x_k+y_k \pi$, $x_k,y_k\in\mathbb{Z}$, is essentially the signed distance between $k$ and its nearest multiple of $\pi$, it is is clear that $x_k/y_k\approx -\pi$. Then perhaps using the CLT or another probabilistic argument, one could examine the convergence of the random variable $a_n/b_n$.

答:
This may help:
$$\arcsin(\sin(x)) = |((x-\pi/2) \bmod (2\pi))-\pi|-\pi/2$$
or equivalently
$$\arcsin(\sin(x)) = \left|x-3\pi/2 -2\pi\left\lfloor\frac{x-\pi/2}{2\pi}\right\rfloor\right|-\pi/2$$

If we write $\arcsin(\sin(k)) = x_k + y_k \pi$ with $x_k, y_k \in \mathbb Z$ we have $$x_k =\operatorname{sgn}(\cos(k))\,k$$ and $$y_k=-\operatorname{sgn}(\cos(k))\left(\frac{3}{2}+2\left\lfloor\frac{k-\pi/2}{2\pi}\right\rfloor\right)-\frac{1}{2}$$


Title: How to solve this bi-quadratic example?

Body:
$$5\left(\frac{x+1}{x}\right)^2 - 16\left( \frac{x-1}{x}\right) - 52 = 0$$

I learned how to use a swap for instance I'd say that

$$\frac{x+1}{x}=t$$ and then swap throughout the equation, but I'm stuck here, I can't get rid of x completely.

Someone edited my question, I don't know how to fix it, I had x + 1/x not (x+1)/x

Body:
$$5(\frac{x+1}{x})^2 - 16( \frac{x-1}{x}) - 52 = 0$$

$$5(1+\frac{1}{x})^2 - 16( 1-\frac{1}{x}) - 52 = 0$$

$$5+\frac{10}{x}+\frac{5}{x^2} - 16+\frac{16}{x} - 52 = 0$$

$$\frac{5}{x^2}+\frac{26}{x} - 63 = 0$$

$$(\frac{5}{x}-9)(\frac{1}{x}+7)=0$$

OR

$$5t^2+26t-63=0$$

where $t=\frac{1}{x}$


问:
Prove with induction that $b_{n+1} \geq ({3 \over 2})^{n-1}$
I need to prove with induction that $b_{n+1} \geq ({3 \over 2})^{n-1} $ $∀ n ∈ \mathbb N$

Sequence is given as $b_{1}=b_{2}=1$ and for every $n \gt 2$ $,b_{n}=b_{n-1} + b_{n-2}$ so basically fibonacci sequence

I tried to prove it with strong mathematical induction

Base:$\quad n=1: 1\geq ({3 \over 2})^{1-1}$ applies

$\quad\quad\quad  n=2: 2\geq ({3 \over 2})^{2-1}$ applies

Step: Let $b_{k+1} \geq ({3 \over 2})^{k-1} $ be correct for every k ∈ $ {1,...,m}$

We have to show that $b_{m+2} \geq ({3 \over 2})^{k} $ is also correct.

So we get that $b_{m-1} + b_{m} \geq ({3 \over 2})^{m} $

Solving this gives me $({3 \over 2})^{m} \geq 0$

So for every natural number this inequality holds

Is this correct?

Any help is appreciated!

答:
Your induction step seems questionable. You did not use the induction hypothesis (or at least it's not clear where you used it). For example, the induction step using the induction hypothesis explicitly my look like this (just a possibility):

By the induction hypothesis, we have $b_m\ge (\frac32)^{m-2}, b_{m-1}\ge(\frac32)^{m-3}$, so $b_{m+1}=b_{m-1}+b_m\ge (\frac32)^{m-3}+(\frac32)^{m-2}=(\frac32)^{m-3}(\frac32+1)=(\frac32)^{m-3}(\frac52)>(\frac32)^{m-3}(\frac32)^2=(\frac32)^{m-1}$, since $\frac52 > \frac94=(\frac32)^2$.


问:
Can an expected value (mean) be higher than the values used to create it?
I have this distribution, where '$x_i$' is shorthand for $X=i$, where $i=0,1,2,3,4$ and where $X$ is a random variable with possible values $0,1,2,3,4$

$$
P(x_0) = 0.2 \\
P(x_1) = 0.25 \\
P(x_2) = 0.3 \\
P(x_3) = 0.15 \\
P(x_4) = 0.1
$$

Using the Expected Value formula:

$$
\mu = (0)0.2 + (1)0.25 + (2)0.3 + (3)0.15 + (4)0.1 \\
\mu = 1.7
$$

How does this make sense? How can the expected value be LARGER than any probability in my distribution? Am I using the wrong formula?

答: Expected value $E=\sum_{j=1}^k p_j x_j$ is the inner-product of two vectors: a vector of values $x=(x_1,\ldots,x_k)$ and a vector of probabilities. $(p_1,\ldots,p_k)$. You can make $E$ as small or as large as you want by fixing $p$ and playing with $x$, or vice-versa. However, by the Cauchy-Schwarz inequality for vectors in $\mathbb R^k$, we have the bound $|E| \le \|x\|_\infty\|p\|_1 = \|x\|_\infty := \max_j |x_j|$.


Title: How to solve this bi-quadratic example?

Body:
Let $\frac 1 x=a$

we get:

$5a^2+26a-63=0$

$a=\frac 95$, so $x=\frac 59$

$a=-9$, $x=-\frac 1 9$


问:
What is actually the dot product of function and its total derivative?
If $f$ is a differentiable function from ℝ to the unit sphere $S^{n-1} \subset ℝ^n$, how we define the dot product of $f$ and $Df$?

答:
We have a function $t\mapsto f(t)\in{\mathbb R}^n$ satisfying $$\langle f(t),f(t)\rangle=|f(t)|^2\equiv1\ .$$ Since
$${d\over dt}\langle f(t),f(t)\rangle=\langle f'(t),f(t)\rangle+\langle f(t),f'(t)\rangle=2\langle f(t),f'(t)\rangle$$
this implies
$$\langle f'(t),f(t)\rangle\equiv0\ .$$


问:
According to Weierstrass theorem, a continuous function on a closed bounded interval is bounded.
What if the function is discontinuous at just one point? Is it bounded anyway?

Why is my example different from the one you proposed?

答:
It is not necessarily bounded anymore. For instance, take the function
$$f:[-1,1]\to\mathbb R,~f(x)=\begin{cases}0&x\leq0\\\frac{1}{x}&x>0\end{cases}$$
It is defined on a closed, bounded interval (this is important: $\frac1x$ or $\tan$ are not counterexamples, since they are continuous, they are just not defined everywhere), has one discontinuity, and it is not bounded.


答:
Your base step is problematic.  Please pay attention to the indices.

Base Step:

When $n=0$, we have
$$b_{n+1} = b_1 = 1 \ge \left(\frac32\right)^{n-1} = \left(\frac32\right)^{-1} = \frac23.$$

When $n=1$, we have
$$b_{n+1} = b_2 = 1 \ge \left(\frac32\right)^{n-1} = \left(\frac32\right)^{0} = 1.$$

Induction Step:

When $n \ge 2$, we have
\begin{align*}
b_{n+1}
&= b_{n-1} + b_n \\
&\ge \left(\frac32\right)^{n-3} + \left(\frac32\right)^{n-2} \\
&= \left(\frac32\right)^{n-3} \left(1 + \frac32\right) \\
&= \left(\frac32\right)^{n-3} \frac{10}{4} \\
&\ge \left(\frac32\right)^{n-3} \frac94 \\
&= \left(\frac32\right)^{n-1}.
\end{align*}

This completes the proof.

Title: Prove with induction that $b_{n+1} \geq ({3 \over 2})^{n-1}$

Body:
I need to prove with induction that $b_{n+1} \geq ({3 \over 2})^{n-1} $ $∀ n ∈ \mathbb N$

Sequence is given as $b_{1}=b_{2}=1$ and for every $n \gt 2$ $,b_{n}=b_{n-1} + b_{n-2}$ so basically fibonacci sequence

I tried to prove it with strong mathematical induction

Base:$\quad n=1: 1\geq ({3 \over 2})^{1-1}$ applies

$\quad\quad\quad  n=2: 2\geq ({3 \over 2})^{2-1}$ applies

Step: Let $b_{k+1} \geq ({3 \over 2})^{k-1} $ be correct for every k ∈ $ {1,...,m}$

We have to show that $b_{m+2} \geq ({3 \over 2})^{k} $ is also correct.

So we get that $b_{m-1} + b_{m} \geq ({3 \over 2})^{m} $

Solving this gives me $({3 \over 2})^{m} \geq 0$

So for every natural number this inequality holds

Is this correct?

Any help is appreciated!


问:
On the definition of adjoint maps on inner product spaces

Definition: Let $A:V\to V$ be linear, where $V$ is an inner product space over $\mathbb{F}$. Then $A^*:V\to V$ is defined by the equation: $$\langle y,A^*x\rangle = \langle Ay,x\rangle \text{ for all }x,y\in V$$
Here $A^*$ is called the adjoint operator with respect to $A\in\mathcal{L}(V)$.

So, it is clear that $A^*$ is defined on inner product spaces only. However, we see that the conjugate transpose of a matrix is defined regardless of whether $V$ is an inner product space or not (right?), i.e. if $A\in M_n(\mathbb{F})$, we define $A^* = (\bar A)^T$.

I suspect a possible correlation between the conjugate transpose defined for matrices, and the adjoint operator defined on inner product spaces, because of results such as:

Let $\mathcal{B}$ be an orthonormal ordered basis of $V$, then $A:V\to V$ is self adjoint (i.e. $A^* = A$) iff $[A]^*_{\mathcal{B}} = [A]_{\mathcal{B}}$ (matrices of linear maps w.r.t. basis $\mathcal{B}$).

However, it seems weird that $A^*$ (adjoint) in the context of linear maps is defined only on inner product spaces, while the conjugate transpose of a matrix $A\in M_n(\mathbb{F})$ is defined more generally. Am I missing something here? I was hoping for a stronger connection between the two definitions.

Thanks!

P.S.

As mentioned in the comments, it is required to assume $\mathbb{F} = \mathbb{C}$ for $\bar A$ to make sense.

答:
The point is that $\Bbb F^n$ carries a natural inner product, when $\Bbb F=\Bbb R$ or $\Bbb C$, namely the ordinary one:
$$\langle x,y\rangle:=\sum_k\overline{x_k}y_k\,.$$
And one can prove that $A^*={\bar A}^T$ using this inner product, when identifying a matrix $A$ with the linear map $x\mapsto A\cdot x$.

 Let $e_1,\dots,e_n$ be the standard basis, then $\langle e_i,Ae_j\rangle=A_{i,j}$ is the $i,j$th matrix entry, so $$(A^*)_{i,j}\ =\ \langle e_i,A^*e_j\rangle\ =\ \overline{\langle A^*e_j,e_i\rangle}\ =\ \overline{\langle e_j,Ae_i\rangle}\ =\ \overline{A_{j,i}}\,.$$

Note that over $\Bbb R$, conjugation is the identity, we can omit it, so in that case the adjoint of $A$ is simply its transpose $A^T$.


问:
Equivalence between $\| \cdot \|_1$ and $\|x\|=\max_t |x(t)|$
Let $C[0,1]$ be a Banach space with norm $\| \cdot \|_1$, and when $\|x_n-x\|_1\to 0$, for any $t\in[0,1]$, $|x_n(t)-x(t)|\to 0$ holds. Prove that, $\| \cdot \|_1$ is equivalent to the norm $\|x\|=\max_t |x(t)|$.

I don't have any ideas about the problem. Can anyone help?

答:
The identity map
$$
  \text{id}: (C[0, 1], \|\cdot\|_1) \to  (C[0, 1], \|\cdot\|_\infty )
  $$
is continuous by the Closed Graph Theorem.  Therefore it is a homeomorphism by the Open Mapping Theorem, so the norms
are equivalent.


问:
Infinite sums and intersections of subsets of $\mathbb{R}^2$
In this exercise I am asked to find $\bigcup_a \bigcap_b X_{a,b}$ (as well as other variations of sums and intersections of $X_{a,b}$) where $X_{a,b}=\{(x,y)\in \mathbb{R}^2 \vert y \leq ax(x-b)\} a,b \in \mathbb{R}, a,b >0 $. I drew the graph of $y =ax(x-b)$ but it's really hard for me to visualize what $\bigcup_a \bigcap_b X_{a,b}$ is going to look like by sheer looking at the graph. I also plugged the formula into desmos to see how it behaves when $a$ and $b$ are changing, but this didn't help either as the results I got were different from the answers provided by the teacher. Is there any systematic approach to a problem like that? Working on exercises of this type becomes dreadful without any good method as the sets get more complicated - what about this monstrosity for instance:
$X_{a,b}=\{(x,y)\in \mathbb{R}^2 \vert ax^2 < y \leq \sqrt[3]{ab^2} a,b \in \mathbb{R}^2 a,b>0\}$?!

答:
I would first try to determine what $S_a = \bigcap_b X_{a,b}$ looks like for a fixed $a$.

Then you can consider how $a$ affects its shape, and try to determine what $\bigcup_a S_a$.

Is that helpful?


问:
The sum of in-degrees $\sum_{v\in V} id(v)$ and out-degrees $\sum_{v\in V} od(v)$ are always equal?
I'm working on Problem #24 of section 11.3 of Ralph P. Grimaldi's Discrete and Combinatorial Mathematics, an Applied Introduction, fifth edition.

Question:

Let $G=(V,E)$ be a directed graph, where $|V|=n$ and $|E|=e$. What are the values for $\sum_{v\in V} id(v)$ and $\sum_{v\in V} od(v)$?

$id(v)$ and $od(v)$ are the in-degrees and out-degrees.

In and out degrees are mentioned in passing at the end of section 11.3, so one is left on their own to answer this question.

I tried to deduce the facts necessary about $\sum_{v\in V} id(v)$ and $\sum_{v\in V} od(v)$ and I'de like to know if my logic is correct:

Given a finite number of vertices $n$ for every 'directed' edge you add you are adding $1$ to $\sum_{v\in V} id(v)$ and $\sum_{v\in V} od(v)$ respectively, and they must always be equal?

If so, then $\sum_{v\in V} id(v)=\sum_{v\in V} od(v)=e?$

答: To make your idea rigorous, you need to use induction. Let $Q(k)$ be the statement that every directed graph with exactly $k$ edges satisfies the desired claim. Then use induction on $Q$. When doing so, be careful that you do not do anything like "If $Q(k)$ is true, then let $G$ be any directed graph with $k$ edges and add an edge..."! Read this explanation of induction to make sure you do induction correctly.


答:
This can't be true. Consider the identity map
$$ id \colon (X,|| \cdot||_1) \to (X,||\cdot ||_\infty) .$$
Evidently, $id$ has a closed graph but it is not continuous (take $u_n(t)=t^n$, then $||u_n||_1=\frac{1}{n+1} \to 0 $ but $||u_n||_\infty=1$). As a result, $(X,|| \cdot||_1) $ is not a Banach space.

Title: Equivalence between $\| \cdot \|_1$ and $\|x\|=\max_t |x(t)|$

Body:
Let $C[0,1]$ be a Banach space with norm $\| \cdot \|_1$, and when $\|x_n-x\|_1\to 0$, for any $t\in[0,1]$, $|x_n(t)-x(t)|\to 0$ holds. Prove that, $\| \cdot \|_1$ is equivalent to the norm $\|x\|=\max_t |x(t)|$.

I don't have any ideas about the problem. Can anyone help?


问:
a question which seem related to derivation of argument princple
This question is from ponnusamy and silvermann Complex variables ( Subsection : Argument Principle)and I am unable to solve it.

Question: Let f(z) be analytic inside and on a simple closed contour except for a finite number of poles inside C. Denote the zeroes by $z_1, z_2,...,z_k$ ( none of which lies on C)and the poles by $w_1, w_2,...,w_m$. If g(z) is analytic inside and on C, prove that

$\frac {1}{2\pi i} \int_C g(z) \frac{f'(z) }{f(z)} dz = \sum_{j=1}^n g(z_j) -\sum_{j=1}^m g(w_j)$,

Attempt : I assumed $f(z)= \frac{(z-z_1)^{k_1} ... (z-z_n)^{k_n} F(z)}{(z-w_1)^{l_1} ...(z-w_m)^{l_m}}$, where F(Z) is analytic in and on C.
and then I Took Logarithm and then 1st order derivative both sides and then multiplying by g(z) and then using cauchy's formula I got LHS equal to $1/ 2\pi i \int_{C} g(z) \frac{ f'(z)}{f(z)} dz$ and RHS equal to and RHS equal to $\sum_{i=1}^n k_i g(z_i) -\sum_{j=1}^{m} l_j g(z_j)$ so, I am getting an extra $k_i $ and $l_j$ in each sum.

For more clarity I am adding 2nd last step which is $1/2\pi i \int_{C} \frac{g(z) f'(z)} {f(z)}dz =$1/2\pi i$[\sum_{i=1}^n k_i \int_{C} \frac{g(z)}{z-z_i} dz -\sum_{j=1}^m l_j \frac{g(z}{z-w_j}dz+ \int_{C} \frac{F'(z)}{F(z)} g(z)dz]$.

Can you please tell what mistake I made in 2nd last step so that I got an extra factor of $k_i$ and $l_j$ in each of th e terms?

Thanks!

答:
There is no mistake. If $z_j$ and $w_j$ are the distinct zeros and poles of $f$ with respective multiplicities $k_j$ and $l_j$ then
$$
\frac {1}{2\pi i} \int_C g(z) \frac{f'(z) }{f(z)} dz = \sum_j k_j g(z_j) -\sum_j  l_j g(w_j) \, .
$$

Another convention is to list multiple zeros/poles multiple times, according to their multiplicity, and then
$$
\frac {1}{2\pi i} \int_C g(z) \frac{f'(z) }{f(z)} dz = \sum_j g(z_j) -\sum_j  g(w_j)
$$
because for a $k_j$-fold zero at $z_j$ the term $g(z_j)$ occurs $k_j$ times in the sum (and similarly for the poles).