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H: How to obtain the Klein bottle as a product of manifolds?
I know the Klein bottle $K$ is a fiber bundle over $S^1$, but my question is: is it possible to find a manifold $M$ such that $K = S^1 \times M$ without the need to take an equivalence relation afterwards?
My thoughts are, maybe something like taking $M$ as the image of $S^1$ under $z \to z^2$ or something in that direction. Do you think I will be able to get somewhere? Thanks in advance.
AI: The Klein bottle cannot be expressed as a product of one dimensional manifolds. There are a few ways of showing this. One useful starting point is the result that there are only two connected one dimensional manifolds (without boundary, up to homeomorphism), $S^1$ and $\mathbb{R}$. Products of non-connected manifolds are non-connected, so if the Klein bottle is a product of two one dimensional manifolds, they must each be either $S^1$ or $\mathbb{R}$. However, none of the products, $S^1\times S^1$, $\mathbb{R}^2$ or $S^1\times\mathbb{R}$ are homoemorphic to the Klein bottle (e.g. because all three are orientable, while the Klein bottle is not).
In general, being a nontrivial product of lower dimensional manifolds is a very strict condition, and excludes a lot of interesting cases, such as the Klein bottle. It's for precisely this reason that constructions like fiber bundles exist, allowing for more general decompositions of manifolds into simpler pieces. |
H: The function $f(x)=|x|^p,$ $x\in \mathbb{R}^{n}$ is strictly convex for $p>1$?
Let $p>1$. In the paper [1] below, it says that The function $f(x)=|x|^p,$ $x\in \mathbb{R}^{n}$ is strictly convex.
I would like to prove that. By definition, We need to show that the Hessian matrix
$H=\left(\frac{\partial^2 f}{\partial x_{i} \partial x_{j}}\right)$ is positive definite (i.e, $x^{T}Hx>0$ for every $x\in \mathbb{R}^{n}\setminus\{0\}.$) But this looks really tedious. So I am trying to verify the alternative equivalent definition:
$$|(1-t)w_{1}+tw_{2}|^{p}<(1-t)|w_{1}|^p+t|w_{2}|^{p}$$ for all $t\in]0,1[$ and $w_{1}\neq w_{2}$. Not much of trying because I don't know where to begin.
[1] P. Lindqvist, On the equation $div(|\nabla u|^{p-2}\nabla u)+\lambda |u|^{p-2}u=0$, Proc. Amer. Math. Soc. 109, (1990), 157 - 164.
AI: The real variable function $\phi(t)=|t|^p$ is strictly convex in $\mathbb{R}$ and strictly increasing in $[0,\infty)$. The norm function $\mathbf{x}\mapsto \|\mathbf{x}\|_r$, $\|\mathbf{x}\|_r=\Big(\sum^n_{j=1}|x_j|^r\Big)^{1/r}$ is convex
$\|(1-\alpha)\mathbf{a}+\alpha\mathbf{b}\|_r\leq (1-\alpha)\|\mathbf{a}\|_r + \alpha\|\mathbf{b}\|_r$. Applying $\phi$ leads to
$$
\|(1-\alpha)\mathbf{a}+\alpha\mathbf{b}\|^p_r\leq \Big((1-\alpha)\|\mathbf{a}\|_r + \alpha\|\mathbf{b}\|_r\Big)^p\leq (1-\alpha)\|\mathbf{a}\|^p_r+\alpha\|\mathbf{b}\|^p_r$$
with equality only if $\|\mathbf{a}\|_r=\|\mathbf{b}\|_r$. |
H: Correct my intuition: every Galois group is $S_n$, and other obviously incorrect statements
(I hope that this question is acceptable and within the rules of math.stackexchange. If not, mods should edit at will and let me know if this question must be broken into several different questions. I ask these all together at once because they seem crucially tied together insofar as answers would correct my misunderstandings.)
I am currently studying Galois Theory, but I am unable to get a handle on the subject. My intuition leads to me conclusions which are obviously incorrect, so I will ask a brief series of questions which I think will help me correct my course.
Let $Q$ be the rationals. Let $f$ be an irreducible polynomial (hence separable) over $Q$ with degree $n$. Let $F$ be the splitting field of $f$. So $F/Q$ is Galois. Let $G$ be the Galois group of $F/Q$. Let $a_1,\dots ,a_n$ be the distinct roots of $f$. My understanding is that $F=Q(a_1,\dots ,a_n)$.
Question 1: Since every automorphism of $F/Q$ permutes $a_1,\dots ,a_n$, it is clear that $G$ can be embedded into $S_n$ as a subgroup. Why is it not the case that $G$ is automatically all of $S_n$? Surely any permutation of the roots gives an automorphism of $F$ preserving $Q$? If not, what might be an instructive minimal example?
Question 2: Conceptually speaking, what exactly prevents certain permutations from being acceptable automorphisms of $F/Q$?
Question 3: Alternatively, if $f$ is not irreducible, then $F$ is the splitting field of some polynomial which is not irreducible. In this case, I believe that roots from different irreducible components cannot jump to roots of other irreducible components. Why is this the case?
Question 4: Again we assume that $f$ is irreducible. What must be unique about the situation in order for $G$ to really be all of $S_n$?
Question 5: Now set $n=4$. I know that $A_{4}$ is the only subgroup of $S_4$ with order $12$. Suppose that $G=S_4$. Suppose $K=Q(a_1)$. Why is it not the case that $F/K$ has order $12$ and hence has Galois group $A_4$? It seems that the Galois group of $F/K$ could include all permutations of $a_1,\dots ,a_4$ that map $a_1$ to itself.
Question 6: Suppose we are in the case of Question 5. Why is it the case that the Galois group of $F/K$ has a transposition?
AI: Question 1: Consider $f(x) = x^4 + 1$, which is irreducible over $\mathbb{Q}$. Then we may write all the roots of $f(x)$ as $\zeta_8^i$ where $i = 1, 3, 5, 7$. In particular $[F : \mathbb{Q}] = [\mathbb{Q}(\zeta_8) : \mathbb{Q}] = 4$. Thus the galois group cannot be all of $S_4$ - if you write out the automorphisms you will see that saying where $\zeta_8$ goes is the same as saying where all the $\zeta_8^i$ go.
Question 2: This was answered above, sometimes there is dependencies between the roots, so because each element of the galois group is a field automorphism, some will be disallowed.
Question 3: If $F$ is the splitting field of $f(x)$ and $f(x) = g(x)h(x)$ then if $\sigma \in Gal(F / \mathbb{Q})$ we have $\sigma(g(x)) = g(\sigma(x))$. In particular if $\alpha$ is a root of $g(x)$ then so is $\sigma(\alpha)$.
Question 4: This is a somewhat difficult question to answer. There are many criteria for when a subgroup of $S_n$ might be the symmetric group (e.g., if it is a transitive subgroup containing an $(n-1)$-cycle and a transposition). In general if I were trying to answer this in some situation I would do some algebraic number theory and look mod $p$.
Question 5: The degree $[\mathbb{Q}(a_i): \mathbb{Q}] = 4$ ($a_i$ a root of an irreducible polynomial of degree $4$, in particular $[F : \mathbb{Q}(a_i)] = 24/4 = 6$ by the tower law. You are correct, the elements are the permutations that fix $a_i$, but that is just $S_3$!.
Question 6: The element fixing $a_1$ and $a_2$ but permuting $a_3$ and $a_4$ is in $Gal( F / \mathbb{Q}(a_1))$ and is clearly a transposition. |
H: What should I do if I can't see through abstractions in complex shapes
I am recently solving Problems & Solutions in Euclidean Geometry. I have learnt a lot of facts regarding to parallelogram, triangles, areas etc etc and solve quite a lot of questions. For the most part of questions, I can see through it either immediately or drawing all given information in detail. At any rate, the theorems can help me solve the problems.
But recently, I have come across some questions that after applying all theorems at first glance, I go nowhere. I have been sitting there for 5 hours per question. And right after I give up and look for the answer, all of my prior observations based on what I learnt from the theorems aren't even brought up in the solution!
For example,
The question asked me to prove $\Delta$ABC = $\Delta$ACE + $\frac{1}{2}\Delta$ABC
I didn't realize $\Delta$EAB$\cong \Delta$CAD when I was given $\Delta$ACE and $\Delta$ABD are equilateral. Then I wasted like an hour by struggling whether $\Delta$ACE $\cong\Delta$ACF. Turns out it simply isn't.
Another example, when only given G and H the mid-point and asked to prove Area $\Delta$EHG = $\frac{ABCD}{4}$. I can only observe 4ABHG = ABCD and achieve nothing for 3 hours again. Then when I read the solution, I understand it but I just can't come up with the theorems I learnt. It manipulated $\Delta$ABE, $\Delta$BGE and $\Delta$AHE, which are things that I have never thought of.
So my question is, how to train myself seeing through questions with several layer of abstractions, which obviously cannot be inferred by simply knowing the facts? To some extent, it seem beyond using the fact I have learnt in this book.
I have seen answer referring to more practices. But the problem more practices mean more frustration. I just can't see what I can't see.
AI: My answer would be still "more practice". With more examples previously analyzed, your intuition is likely to improve, so you know what makes sense to look for. But don't burn out on problems where you've spent too much time already. After a reasonable attempt, find a solution from some source, analyze it, and go on to the next problem.
It's worth noting that complicated plane geometry problems relate very little to most parts of advanced math, so if you have goals beyond plane geometry, don't worry too much about it. |
H: Prove $T \mathrm{ker}(A T) = \mathrm{ker}(A)$.
Let $T \in \mathbb{R}^{n\times n}$ (not necessarily invertible), $A \in \mathbb{R}^{m\times n}$. Prove the following equation
$$T \mathrm{ker}(A T) = \mathrm{ker}(A),$$
where $\mathrm{ker}(\cdot)$ gives the null space.
Here is my proof:
Let $\mathrm{ker}(A T) = \{x\colon ATx = 0\}$. Then,
$$T \mathrm{ker}(A T) = \{y = T x\colon x \in \mathrm{ker}(A T)\} = \{y = T x\colon ATx = 0\} = \{y\colon Ay = 0\} = \mathrm{ker}(A).$$
Am I right? Thanks!
Conclusion: the proof holds when $T$ is non-singular; for general $T$, we have $T \mathrm{ker}(A T) \subseteq \mathrm{ker}(A)$; see the accepted answer provided by Azif00.
AI: In order to ensure that $$\{Tx : x \in \Bbb R^n \textrm{ and } A(Tx)=0\} = \{y \in \Bbb R^n : Ay = 0\},$$ $T$ needs to be invertible. What happens if $T = 0$ and $A$ is not invertible? |
H: Wronskian of two linearly independent differential functions. Show $c$ in [a,b] such that $g(c) = 0$ exists
Let $f,g: [a,b] → R$, two differential functions and suppose $f(a) = f(b) = 0$.
If $W(f,g): [a,b] → R$ and $W(f,g)(x) = f(x)g'(x) - g(x)f'(x)$ doesn't equal 0 for all $x$ in $[a,b]$, show that a $c$ in $[a,b]$ must exist such that $g(c) = 0$.
I thought about using the Rolle Theorem to say that there exists a $k$ in [a,b] such that $f'(k) = 0$ and then use the Intermediate Value Theorem to show that g(a) and g(b) are the opposite signs therefore a c such that g(c)=0 must exist, but I get lost at that last part.
Can anyone help me? Thank you!
AI: If possible, take $g\neq 0 $ for all $x\in [a,b] $
Then we can easily define the differentiable function $\frac{f}{g} $ on $[a,b]$.
Clearly, $(\frac{f}{g})(a) = 0 = (\frac{f}{g})(b) $.
Then by Rolle's theorem, there exist at least one $c\in (a,b) $ such that $(\frac{f}{g})' (c) = \frac{gf'-fg'}{g^2}=\frac{-W(f,g)}{g^2} = 0 \implies W(f,g)=0 $ , and this is a contradiction .
So, there exist $c\in (a,b) $ such that $g(c)=0$
Edit: Here, we shouldn't use the closed interval $[a,b]$, for this statement, $\text{there exist $c\in (a,b) $ such that g(c)=0 } $.
Since at $a,b$, if either of any $g(a)=0$ or $g(b)=0$ true,
Then , $W(f,g)$ becomes $0$, which makes the fact " $f,g$ linearly independent " wrong.
So, we use $(a,b)$ instead of $[a,b]$, for the conclusion of the proof. |
H: Different approaches in evaluating the limit $\frac{(x^3+y^3)}{(x^2-y^2)}$ when $(x,y)\to(0,0)$.
Note that this question has been previously asked here. I understood the solutions available there but I have two different approaches to this problem, I'm not sure whether they are correct.
I need to know whether both of these solutions are correct and complete. If not, why are they incorrect?
Approach 1
Take path 1 as $y=3x$, hence limit goes to $0$.
Take path 2 as $y=(-x^3+x^2-y^2)^{1/3}$, hence limit goes to $1$.
Therefore, the limit does not exist.
Is the second path a valid path cause $y$ is not necessarily $0$ when $x=0$?
Approach 2
Take $x=r\cos\theta$ and $y=r\sin\theta$.
We have $r\frac{\cos^3\theta+\sin^3\theta}{\cos^2\theta-\sin^2\theta}$.
Take path $r = \cos^2\theta-\sin^2\theta$ hence limit goes to $\cos^3\theta+\sin^3\theta$ which is different for every $\theta$ and hence limit cannot exist.
Is this choice of $r$ allowed?
AI: The first one is valid because when $x\to 0$ also $y \to 0$ (see the plot).
The second one is valid also because $r = \cos^2\theta-\sin^2\theta \to 0$ when $\theta \to \frac \pi 4+k\frac \pi 2$ with the limitation that $\cos^2\theta>\sin^2\theta$ (see the plot).
To avoid these kind of check we can use simpler parametric path as for example $x=t$ and $y=t-t^2$ to obtain
$$\frac{x^3+y^3}{x^2-y^2}=\frac{2t^3}{2t^3-t^4}\to 2$$ |
H: Show that there exists a subsequence $\{E_{n_k}\}$ of $\{E_n\}$ such that $m(\cap_{k=1}^\infty E_{n_k})>\epsilon$ under these conditions....
Question: Let $\{E_n\}$ be a sequence of nonempty Lebesgue measurable subsets of $[0,1]$ such that $\lim_{n\rightarrow\infty}m(E_n)=1$. Show that for each $0<\epsilon<1$ there exists a subsequence $\{E_{n_k}\}$ of $\{E_n\}$ such that $m(\cap_{k=1}^\infty E_{n_k})>\epsilon$.
My Thoughts: I am a bit stumped on this one. I am sure there is a technical way of doing it, but in my head, and this could be completely wrong, was thinking about taking each $E_n$ and, say, cut it in half and take the half such that the distance between any point on the interval of the cut and $0.5$ is smallest. If the half cut overlaps $0.5$, then choose that one. If the middle of the subset is exactly on $0.5$, then make the cut, and shift the subset to the left (or right) the length of $\frac{\epsilon}{2}$. Then, we get a bunch of subsequences that all lay "on top" of $0.5$, so they don't have an empty intersection, and that would solve our problem (I think). But, I am a bit worried about the "for each $0<\epsilon<1$, because, say, if $\epsilon=0.8$, then my method wouldn't necessarily work, but really only works for a "small enough" $\epsilon$. Maybe a more technical approach would be best...
Any thoughts, suggestions, etc. are greatly appreciated! Thank you.
AI: Note that $m(E_n^{c}) \to 0$. Choose an increasing sequence of integers $n_k$ such that $m(E_{n_k}^{c}) <\frac {1-\epsilon} {2^{k}}$. Then $m(\cup_k E_{n_k}^{c}) \leq \sum_k m(E_{n_k}^{c}) <1-\epsilon$ and $m(\cap_k E_{n_k})=1-m(\cup_k E_{n_k}^{c}) >\epsilon$ |
H: Does Riemann integrability implies integral mean value theorem?
We know that if $f$ is continuous on [a,b] and $f:[a,b] \to \mathbb{R}$, then there exists $c \in [a,b]$ with $f(c)(a-b) = \int_a^bf(x)dx$
If we change ''f is continuous on [a,b]'' to ''f is Riemann integrable'', does the mean value theorem for integral still holds? If not, Can you give me a counter-example?
I know that the first Mean-Value Theorem for Riemann-Stieltjes does not requires continuity, but that is still different from this statement.
reference:http://mathonline.wikidot.com/the-first-mean-value-theorem-for-riemann-stieltjes-integrals
AI: What about $f(x) = x$ for $x \in [0,1]\setminus \{1/2\}$ and $f(1/2) = 1$? |
H: Changing limit and derivative operator
I was trying to solve the following problem:
Let $f:\mathbb R\rightarrow \mathbb R$ be a differentiable function such that
$\lim_{x \to \infty} f(x)=1$ and
$\lim_{x \to \infty}f'(x)=a$.
Then find the value of $a$.
My brother who is not a pure math student suggested that take derivative on first equation you will get $a=0$.
I am not sure whether we can do that or not. Can we do that? I don't think it's the correct way of solving this.
AI: You can't take the derivative in a limit. But the way a continuously differentiable map can have a limit while the derivative doesn't. Example: $f(x) = 1+ \frac{\sin x^2}{x}$ (defined for $x>0$).
However, you can solve the problem in following way.
Suppose that $a >0$, then for $x$ large enough, say $x\ge M >0$ you have $f^\prime(x) \ge a/2$. And applying Mean Value Theorem, $f(x) \ge a/2(x-M) + f(M)$ for $x \ge M$. Which implies that $ \lim\limits_{x \to \infty} = \infty$. In contradiction with $\lim\limits_{x \to \infty} f(x)=1$ .
You can proceed in a similar way for $a<0$.
Hence the only option is $a=0$. |
H: Possibly a variation of the increment theorem for functions of multiple variables
Suppose that $F:\mathbb R^n\to\mathbb R$ is $C^\infty$. I'd like to prove that $\forall a=(a_1,\ldots,a_n)\in\mathbb R^n$, there exist $C^\infty$ functions $G_i,i=1\ldots,n,$ such that $\forall x=(x_1,\ldots,x_n)\in\mathbb R^n$, we have
$$F(x)=F(a)+\sum_{i=1}^n(x_i-a_i)G_i(x).$$
In fact, we have
$$G_i(a)=\frac{\partial F}{\partial x_i}(a)$$
for all $i$. This theorem is quoted in the appendix to my physics textbook, but I'm not sure about its legitimacy. According to my experience in the calculus course, the change in the value of $F$ from $a$ to $x$ should take the form
$$\Delta F=\sum_{i=1}^n[(x_i-a_i)\frac{\partial F}{\partial x_i}(a)+\epsilon_i(x_i-a_i)],$$
where each $\epsilon_i$ goes to zero as all $x_i-a_i$ tend to zero. Are these two statements consistent with each other? Thank you.
AI: Yes, the theorem is true as stated. It's all a matter of how much regularity you want to impose on your functions. Here's the general theorem:
Let $n,k\geq 1$ be integers (with $k=\infty$ also allowed), and let $F:\Bbb{R}^n \to \Bbb{R}$ be a $C^k$ function. Then, for each $a\in \Bbb{R}^n$, there are functions $G_1,\dots, G_n:\Bbb{R}^n\to \Bbb{R}$ of class $C^{k-1}$ such that for every $x\in \Bbb{R}^n$,
\begin{align}
F(x) -F(a) &= \sum_{i=1}^n (x_i-a_i)\cdot G_i(x),
\end{align}
and such that $G_i(a) = \dfrac{\partial F}{\partial x_i}(a)$.
The proof is actually pretty simple: fix a point $a \in \Bbb{R}^n$, and given any $x\in \Bbb{R}^n$, we consider the parametrized line $\gamma(t) = a + t(x-a)$, $t\in \Bbb{R}$ Clearly, $\gamma$ is $C^{\infty}$; now notice that
\begin{align}
F(x)-F(a) &= F(\gamma(1)) - F(\gamma(0)) \\
&= \int_0^1 (F\circ \gamma)'(t)\, dt \tag{by FTC} \\
&= \int_0^1 \sum_{i=1}^n (x_i-a_i) \dfrac{\partial F}{\partial x_i}(\gamma(t))\, dt \tag{chain rule} \\
&= \sum_{i=1}^n (x_i-a_i) \cdot \underbrace{\int_0^1 \dfrac{\partial F}{\partial x_i}(a + t(x-a))\, dt}_{G_i(x)}
\end{align}
Now, all you have to do is prove that $F$ being $C^k$ implies each $G_i$ is $C^{k-1}$; this is only slightly technical. Finally, it is obvious that $G_i(a) = \dfrac{\partial F}{\partial x_i}(a)$.
Note that the second equation you wrote down is nothing more than the definition of differentiability at a single point $a$. We're not assuming anything more than differentiability at a single point. In the theorem above we're making the much stronger assumption that of $F$ being $C^k$ on the whole of $\Bbb{R}^n$, which is why you can show the existence of functions $G_i$ with such and such property. |
H: the slower grape crusher
Two Grape Crushers take 4 days to crush certain amount of grapes.If one of them crushed half the grapes and the other crushed other half , then they complete the job in 9 days. How many days will it take slower crusher to do the job alone?
my solution :
the time taken when they both do 1/2 the work individually is 9 days given which means the slower guy took 9 days because consider A took 1 day he sits and waits for B to complete which he does on the 9th day right? so B will take double the time to complete double the work so 18 days is the answer according to me. But it doesnt make use of the first part of the Q. hence i am doubtful.
what do you guys think ? what is the correct method to solve this?
AI: The jobs are done parallelly but not serially, one guy need not wait for the other to finish.
Wlog take the total weight of grapes to be 2 kg, their crushing speeds v1 and v2.
$$\dfrac{1}{v1}+\dfrac{1}{v2}=9$$
$$\dfrac{2}{v1+v2}= 4 $$
has solutions for speeds
$$\dfrac{1}{6},\dfrac{1}{3}; $$
For the slower crusher take the reciprocal, so the slow guy takes 6 days,
and for the faster one take reciprocal, so faster guy takes 3 days. |
H: counting words with a condition
Suppose we are given a sequence $x_1x_2x_3x_4x_5x_6$ where the $x_i$ are digits 0 to 9, and we want to know how many of them do we have that satisfy $x_1<x_2<x_3<x_4<x_5<x_6$?
$discussion:$
Notice that $x_1$ can only be a number betwwen $0$ to $4$ so if $x_1=0$, then we reduce our problem to count number of strings $x_2<x_3<x_4<x_5<x_6$ where $x_i$ are digits $\geq 1$. And here notice that $x_2$ must be between $1$ and $5$. So, If $x_2 = 1$ now we have another subproblem... in this counting those that satisfy $x_3<x_4<x_5<x_6$ and now $x_3$ must be between $2$ and $6$ and so if you let $x_3=2$ then $x_4$ can be betwwen $3$ and $7$ and we see that each time we have $4$ choices for the $x_i$
So, we see that there $4 \times 4 \times ... \times 4 = \boxed{4^6}$ such sequences.
Now, my question, in general, if we have sequence $x_1x_2...x_n$ there are ${\bf no}$ sequences that satisfy $x_1<x_2 < ... <x_n$ but if $n=9$ say, then we only have $1^9$ choices. if $n=8$, then we have $2^8$ choices. If $n=7$, we have $3^7$ choices. if $n=6$, we have $4^6$ choices and so on...
Is this a correct generalization?
AI: To build a sequence of length $n$ in ascending order, you need to choose $n$ different digits. Once the digits are chosen, you have only one order in which you can place them in the sequence. Hence, the number of sequences that satisfy the condition $x_1<\ldots<x_n$ is ${10 \choose n}$.
So, for $n=10$ you have only one option: 0123456789. For $n=9$ you have $10$ options: just choose which digit not to include. For $n=6$ you have ${10 \choose 6}=210$ options and not $4^6=4096$. Looking at it as many sub-problems is very problematic, as the number of choices in each problem strongly depends on previous choices. |
H: Is there a finite group which has two isomorphic maximal subgroups such that no automorphism can map one to the other?
Let $G$ be a finite group. Suppose $H$ and $K$ are two isomorphic maximal subgroups of $G$, then can we claim that there must be an automorphism $\alpha\in {\rm Aut}(G)$ such that $\alpha(H)=K$?
If $H$ and $K$ don’t have to be maximal subgroups, then I’ve already found a counterexample of order $8$. In $G:=C_2\times C_4$, there are three subgroups isomorphic to $C_2$ and only one of them is characteristic in $G$. Therefore no automorphism of $G$ can map the characteristic one to the others.
I failed to find a counterexample in the case where $H$ and $K$ are maximal subgroups of $G$. Could you give me some ideas? Any help is appreciated. Thank you!
AI: There are many such examples, I'm afraid. One source of examples is the maximal subgroups of classical groups. The smallest example I could find were two classes of $\mathrm{PSL}_2(7)$ in $\mathrm{PSp}_6(q)$, for $q\equiv \pm 7\bmod 16$, $q\neq 7$.
More generally, almost every irreducible character of a simple group $G$ yields a maximal embedding of $G$ into some classical group. (This isn't quite right, but it's good enough for these purposes.) The multiplicities of irreducible character degrees of finite groups tend to infinity as the group order tends to infinity, so there are unboundedly many such classes in general. |
H: Interesting question regarding evolution of algorithm under certain conditions
On a random string of six digits containing numbers of the set $S=\{1,2,3,4,5,6\}$, repeat the following operation:
If $k$ is the first number of the string, then reverse the order of first $k$ numbers of the string.
For example: $342561\rightarrow 243561\rightarrow 423561\rightarrow 532461\rightarrow 642351\rightarrow 153426$
Prove that any such string would terminate with $1$ at the first position.
Now on checking a few examples, I have made a few observations:
$1.$ The last digit of the string is an increasing monovariant.
$2.$ The first digit of string is always a different number unless the last digit changes. This means that until the last digit of the string change, no number that appeared in the first position in the string will ever repeat.
$3.$ When the last digit reaches its maximum value (which is $6$ here), the observation $2$ works the same but for th second last digit now instead of the last one.
Now the first observation can be explained by the fact that the last digit will only change when we would have $6$ in the first position and no number in the string is greater than $6$.
The third observation can be explained by the fact that when the last digit reaches the value of $6$, this value would never change as all the numbers are less than $6$. Thus, the observation $2$ would now work with the second last digit instead of the last digit since it would be the same as applying the algorithm to a string of length $5$ instead of $6$.
Now if I can prove observation $2$, then I can say that since no number occuring in the first position repeats, we would have either $6$ or $1$ in the first position sometime in atmost $6$ steps. If we obtain $6$ before $1$ in the first position, then $6$ would move to the last position. Similarily after $6$ have moved to the last position, we would either have $1$ or $5$ in the first position after atmost $5$ steps and so on.
This indicates that after some finite steps, the sequence is bound to have $1$ at the first position.
Now I need help to prove observation $2$. Please help.
THANKS
AI: The idea is as follows:
6 can be the first digit in the sequence at most once. After that, it's permanently stuck as the last digit of the sequence.
5 can be the first digit in the sequence at most twice. After 5 is the first digit, it'll be the 5th digit and can only move downwards if 6 becomes the 1st digit. But that can only happen once.
In general, $6 - n$ can be the first digit at most $2^n$ times. For after each time $6-n$ is the first digit, it only be dislodged from its position by a higher number; this dislodging can occur at most $\sum\limits_{i = 0}^{n - 1} 2^i = 2^n - 1$ times. You can use well-founded induction to make this argument rigorous.
Thus, the process will eventually terminate with 1. |
H: Bernoulli's Inequality for $-1 \leq x\leq 0$
My original goal was to prove that
$$\lim_{x\to 0}\frac{e^x-1}{x}=1$$
using the squeeze theorem as we haven't seen differentiability yet and thus I cannot use arguments such as Taylor series nor Bernoulli's theorem, nor can I use induction. For that I wanted to find a lower and upper bound for $e^x$ in order to apply the squeeze theorem.
For the upper bound I used the fact that $x^n\leq x^2$ for $-1\leq x\leq 1$ and $n\geq 2$ thus one has that
\begin{align*}e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n&=\lim_{n\to\infty}\sum_{k=0}^n{n\choose k}\frac{x^k}{n^k}\\
&=\lim_{n\to\infty}1+x+\sum_{k=2}^n{n\choose k}\frac{x^k}{n^k}\\
&\leq \lim_{n\to\infty}1+x+\sum_{k=2}^n{n\choose k}\frac{x^2}{n^k}\\
&= \lim_{n\to\infty}1+x+\left(\sum_{k=2}^n{n\choose k}\frac{1}{n^k}\right)\cdot x^2\\
&= \lim_{n\to\infty}1+x+\left(\left(1+\frac{1}{n}\right)^n-2\right)\cdot x^2\\
&= 1+x+\left(e-2\right)\cdot x^2
\end{align*}
I could now potentially bound $x^n\geq -x^2$ in the same interval and obtain the bound
\begin{align*}e^x\geq 1+x-\left(e-2\right)\cdot x^2
\end{align*}
but I am not happy with it as I know that Bernoulli's inequality is stronger and gives
\begin{align*}e^x\geq 1+x.
\end{align*}
For $x\in (0,1)$ it's rather trivial to prove as
$$e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=\lim_{n\to\infty}1+x+\underbrace{{n\choose 2}\frac{x^2}{n^2}+\cdots+\frac{x^n}{n^n}}_{\geq 0}\geq 1+x$$
but for $x\in(-1,0)$ the same argument does not apply straightforwardly due to the changing signs. So I modified it as follows:
For $-1\leq x\leq 0$ one has that $x^3\leq x^n$ ($x^3$ is in particular negative)
\begin{align*}
e^x&=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=\lim_{n\to\infty}1+x+{n\choose 2}\frac{x^2}{n^2}+\sum_{k=3}^n{n\choose k}\frac{x^k}{n^k}\\
& \geq \lim_{n\to\infty}1+x+\frac{(n-1)x^2}{2n}+\sum_{k=3}^n{n\choose k}\frac{x^3}{n^k}\\
& = \lim_{n\to\infty}1+x+\frac{(n-1)x^2}{2n}+\left((1+\frac{1}{n})^n-\frac{n-1}{2n}-2\right)x^3\\
& \geq \lim_{n\to\infty}1+x+\frac{(n-1)x^2}{2n}+\left(e-\frac{n-1}{2n}-2\right)x^3\\
&= \lim_{n\to\infty}1+x+x^2\left(\frac{n-1}{2n}+\left(e-\frac{n-1}{2n}-2\right)x\right)
\end{align*}
Now we not that the cubic function $x^2\left(\frac{n-1}{2n}+\left(e-\frac{n-1}{2n}-2\right)x\right)$ has a double zero at $x=0$ and the remaining zero is at
$$x=-\frac{\frac{n-1}{2n}}{e-\frac{n-1}{2n}-2}\overset{n\to \infty}{\longrightarrow} -\frac{1/2}{e-1/2-2}\cong -2.29$$
thus for $n$ sufficiently large the last zero is to the left of $-1$ and hence the function $x^2\left(\frac{n-1}{2n}+\left(e-\frac{n-1}{2n}-2\right)x\right)$ is positive on $(-1,0)$ thus
\begin{align*}
e^x& \geq \lim_{n\to\infty}1+x+\underbrace{x^2\left(\frac{n-1}{2n}+\left(e-\frac{n-1}{2n}-2\right)x\right)}_{\geq 0,\quad x\in(-1,0)}\\
&\geq 1+x
\end{align*}
Since I wrote up this proof I ask: could please give it a look and tell me if there are any mistakes or if there is a shorter solution which I overlooked?
Many thanks in advance!
AI: As regards your original goal, there is a shorter way. We have that
$$e^x-1-x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n-1-x=\lim_{n\to\infty}\sum_{k=2}^n{n\choose k}\frac{x^k}{n^k}.$$
Hence, for $x\in [-1,1]$, $$|e^x-1-x|=\left|\lim_{n\to\infty}\sum_{k=2}^n{n\choose k}\frac{x^k}{n^k}\right|\leq x^2\lim_{n\to\infty}\sum_{k=2}^n{n\choose k}\frac{|x|^{k-2}}{n^k}\\\leq x^2\lim_{n\to\infty}\sum_{k=2}^n{n\choose k}\frac{1}{n^k}
\leq x^2\sum_{k=2}^{\infty}\frac{1}{k!}\leq e x^2$$
and the given limit follows as $x\to 0$ by the squeeze theorem.
Along the same argument we show that for $x\in [-1,1]$,
$$\left|e^x-\sum_{k=0}^{n}\frac{x^{k}}{k!}\right|<e|x|^{n+1}$$
which implies that $e^x=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}$.
Bernoulli inequality for $-1<x<0$:
$$e^x-1-x=\sum_{k=2}^{\infty}\frac{x^{k}}{k!}=\sum_{k=1}^{\infty}\frac{x^{2k}}{(2k)!}\underbrace{\left(1+\frac{x}{2k+1}\right)}_{\geq 0}\geq 0.$$ |
H: How is $\mathbb{R}^n$ a compact subset of $\mathcal{H}(\mathbb{R}^n)$ (the set of all compact subsets of $\mathbb{R}^n$)?
I'm reading Essential Real Analysis by Michael Field. There's a definition of a metric space $(\mathcal{H}(\mathbb{R}^n),h)$, in which the set $\mathcal{H}(\mathbb{R}^n)$ is the set of all non-empty compact subsets of $\mathbb{R}^n$. Field writes on page 330:
"... In particular, we can regard $\mathbb{R}^n$ as a subset of $\mathcal{H}(\mathbb{R}^n)$ by the map $(x_1,...,x_n)\rightarrow\{(x_1,...,x_n)\}$".
I have difficulties in understanding this. Does he mean that $\mathbb{R}^n$ can be divided into closed and bounded (thus compact) sets and $\mathbb{R}^n\subset\mathcal{H}(\mathbb{R}^n)$ because those compact sets are all in $\mathcal{H}(\mathbb{R}^n)$? But then what is the role of the map he gives?
(h is the Hausdorff distance. It doesn't seem to have anything to do with the claim, tough.)
AI: Points are compact. This makes allows us to consider every subset of $\Bbb R^n$ a subset of $\mathcal H(\Bbb R^n)$ by identifying the point $(x_1,\ldots,x_n)\in\Bbb R^n$ with the point $\{(x_1,\ldots,x_n)\}\in\mathcal H(\Bbb R^n)$.
Note that this map $\Bbb R^n\to \mathcal H(\Bbb R^n)$ is not only injective, but is in fact an isometry! So we obtain $\Bbb R^n$ in a very nice manner as a subspace of $\mathcal H(\Bbb R^n)$. |
H: Proof for an adapted version of Gauss' Lemma
I am self learning abstract algebra. Today I watched a youtube video explaining the proof for an adapted version of Gauss' Lemma
In his proof, he claims that for any polynomial $f = gh \in \mathbb{Z}[x]$ and for any prime $p$, where $g,h \in \mathbb{Z}[x].$ If $p$ divides all the coefficients of $f$, then $p$ must divide all the coefficients of $g$ or $h$ .
His argument is simply as following :
After taking mod $p$, $f=gh$ will turn into $0=g'h'$(mod$p$). Then $p$ must divide all the coefficients of $g$ or $h$.
I am very doubtful about his argument. This works well if $f$ is a integer, but seem not to work if $f$ is a polynomial. Since both $g$ and $h$ are in $\mathbb{Z}[x]$, we can expand $g$ and $h$as $\sum_{i=0}^{n}a_{i}x^i$ and $\sum_{j=0}^{m}b_{j}x^j$ respectively. Then $f = \sum_{i=0}^n \sum_{j=0}^m a_ib_jx^{i+j}$.
Then we can see If $p$ divides all the coefficients of $f$, then $p$ must divides $\sum_{i+j=q}a_ib_j$ for any $q\in[0,m+n]\cap\mathbb{Z}$
But this will not give us the desired result, which states $p$ has to divide either {${a_i}$}$_{i=0}^n$ or {${b_j}$}$_{j=0}^m$
My question is how could one derive his claim or is his claim correct at all?
AI: If you accept that $(\Bbb Z/p\Bbb Z)[X]$ is an integral domain, then $f'g'=0$
in $(\Bbb Z/p\Bbb Z)[X]$ implies either $f'=0$ in $(\Bbb Z/p\Bbb Z)[X]$
or $g'=0$ in $(\Bbb Z/p\Bbb Z)[X]$. (Here, $f'$ and $g'$ denote the images
of $f$ and $g$ in $(\Bbb Z/p\Bbb Z)[X]$.)
For a more naive approach, if $f$ and $g$ are both non-zero modulo $p$,
each has a "first" coefficient that is nonzero modulo p. So
$p\nmid a_r$ and $p\mid a_i$ for $i<r$, and
$p\nmid b_s$ and $p\mid b_j$ for $j<s$. Then the coefficient
of $x^{r+s}$ in $fg$ equals $a_rb_s$ plus some multiples of $p$
and so is not a multiple of $p$. |
H: A doubt about $\int_{0}^{1} f(x)~ \left(\int_0^x |f(t)| dt \right)~ dx=7$, if $\int_{0}^{1}f(x) dx=2, \int_{0}^{1} |f(x)| dx=4$
A question gives $f(x)$ as continuous and $f'(x)>0$ for all real values
of $x$ such that $\int_{0}^{1} f(x) dx=2, \int_{0}^{1} |f(x)| dx=4.$
So it is good to conclude that $f(x)=0$ will have exactly one real root in $(0,1)$, let us call it $x=a$.
The question then asks one to show that $$\int_{0}^{1} f(x)~ \left( \int_0^x |f(t)| dt \right)~ dx=7.$$
My doubt/question is: Why the said integral is constant independent of the value of $a$ (that is unknown)? Please show it with necessary steps in any case.
AI: Although the integrand $(x,t)\mapsto f(x)\lvert f(t)\rvert$ depends on $a$, it will disappear when you do the integral on the triangle and leave you with dependence only via $\int_0^1 f$ and $\int_0^1\lvert f\rvert$, both of which you know (although they do depend on $a$ morally). To spell it out completely:
Let $F(x):=\int_0^x f$. Then
$$
\int_0^x \lvert f(t)\rvert\,\mathrm{d}t =
\begin{cases}
-F(x) & x\in[0,a]\\
F(x)-2F(a) & x\in[a,1]
\end{cases}
$$
and we have $F'=f$ by FTC. Imposing our given conditions $F(1)=\int_0^1 f=2$ and $4=\int_0^1\lvert f\rvert=F(1)-2F(a)$ gives $F(a)=-1$. So
\begin{align*}
&\int_0^1 f(x)\int_0^x \lvert f(t)\rvert\,\mathrm{d}t\,\mathrm{d}x\\
&=\int_0^a f(x)\int_0^x \lvert f(t)\rvert\,\mathrm{d}t\,\mathrm{d}x
+\int_a^1 f(x)\int_0^x \lvert f(t)\rvert\,\mathrm{d}t\,\mathrm{d}x\\
&=\int_0^a -f(x)F(x)\,\mathrm{d}x
+\int_a^1 f(x)\Big(F(x)-2F(a)\Big)\,\mathrm{d}x\\
&=\int_0^a -f(x)F(x)\,\mathrm{d}x
+\int_a^1 f(x)F(x)\,\mathrm{d}x
-2F(a)\int_a^1 f(x)\,\mathrm{d}x\\
&=-\frac12F(a)^2
+\frac12\Big[F(1)^2-F(a)^2\Big]
-2F(a)\Big[F(1)-F(a)\Big]\\
&=7.
\end{align*} |
H: Equivalent definition of Cohen-Macaulay Ring
We suppose all rings are commutative and unital. The most general defition for Cohen-Macaulayness goes as follows: A Noetherian local ring $R$ is $\textit{Cohen-Macaulay}$ if its depth is equal to its Krull dimenion. More generally a ring is called Cohen–Macaulay if it is Noetherian and all of its localizations at prime ideals are Cohen–Macaulay.
In Richard Kane's 'Reflection Groups and Invariant Theory', the definition of Cohen-Macaulay is given as: Let $A$ be an algebra over $k$, the ring structure on $A$ is Cohen-Macaulay if there exists a polynomial subalgebra $k[a_1,a_2,\ldots,a_n]$ such that $A$ is free and finite over $k[a_1,a_2,\ldots,a_n]$, where $a_1,a_2,\ldots,a_n \in A$, in algebraic terms we can choose elements $b_1,\ldots,b_m \in A$ such that $$A = \bigoplus_{i=1}^mk[a_1,a_2,\ldots,a_n]b_i ,$$
then let $G$ be a finite group and $V$ be a linear representation of $G$, then the algebra of invariants $S(V^*)^G$ is Cohen-Macaulay, this is the Hochster-Roberts theorem. https://en.wikipedia.org/wiki/Hochster%E2%80%93Roberts_theorem
Can someone enlighten me as how these two definitions are equivalent?
AI: The answer is given in the paper by Hochster and Roberts, https://core.ac.uk/download/pdf/82276959.pdf, page 14. And one must assume that the algebra $A$ is graded. |
H: Product of sines when the angles form a sequence
I am wondering how to find the following value of $x$
$$x=\prod_{n=1}^{9}\sin\left(\frac{n\pi}{10}\right)$$
I notice that it has something to do with the de moivre's theorem as the angles are root angles of $1^\frac{1}{10}$
To my surprise, the value of the above product seems to be a rational number.
The solution is given as
$$x=\frac{10}{2^9}$$
I attempted to solve this product in different ways but none seems to give out such an elegant solution. Thank you for any help!
AI: $x = (\prod\limits_{n = 1}^4 \sin(\frac{n \pi}{10}))^2$
$\prod\limits_{n = 1}^4 \sin(\frac{n \pi}{10}) = \prod\limits_{n = 1}^2 \sin(\frac{n \pi}{10}) \prod\limits_{n = 1}^2 \cos(\frac{n \pi}{10}) = \prod\limits_{n = 1}^2 \sin(\frac{n \pi}{10}) \cos(\frac{n \pi}{10}) = \prod\limits_{n = 1}^2 \frac{\sin(\frac{n \pi}{5})}{2} = \frac{1}{4} \sin(\frac{\pi}{5}) \sin(\frac{2 \pi}{5})$
$x = \frac{1}{16} \sin^2(\pi/5) \sin^2(2\pi/5) = \frac{1}{64} (1 - \cos(2\pi / 5))(1 - \cos(4\pi / 5))$
Now consider the fact that $\sum\limits_{j = 0}^4 \cos(2j \pi / 5) = 0$. In particular, we have $2 \cos(2 \pi / 5) + 2 \cos(4 \pi / 5) = -1$. Then, letting $w = \cos(2\pi / 5)$, we have $2 w + 4 w^2 - 2 = -1$. Then $2w + 4w^2 = 1$.
We have $x = \frac{1}{64} (1 - w)(2 - 2w^2)$. Now we have $4(1 - w)(2 - 2 w^2) = 2(1 - w)(4 - 4w^2) = 2(1 - w)(3 + 2w) = 2 (3 - 2w^2 - w) = 6 - 4w^2 - 2w = 6 - 1 = 5$. Then $x = \frac{1}{64} \frac{5}{4} = \frac{5}{2^8}$. |
H: Product decomposition of a nonnegative matrix
Suppose $A\neq 0$ is a nonnegative matrix which can be decomposed as $BC$ with $B\neq 0$ nonnegative and $C$ orthogonal. Then, is $C$ also nonnegative?
I think yes. The case when $B$ is invertible is trivial. But, if $B$ is not invertible, it does not seem obvious. Any hints? Thanks beforehand.
AI: No. For example, if $A = B = 0$, then $C$ can be anything. |
H: Minimum spanning tree formulation
I'm writing up a report for a solution of an energy grid problem for school, and after reading the CS-book multiple times (and a bit of googling), I can't seem to find the mathematical definition of the MST-problem. Is this correct?
Problem: Given an undirected weighted graph $G(V,E, w)$, $\text{vertex set } V$, $\text{edge set } E$, $w$ an edge weight function $w: E \rightarrow \mathbb{R_{+}}$, produce a subgraph $H(V^{*},E^{*})$, without cycles, such that the cost function $C: E \rightarrow \mathbb{R_{+}}$, $C(E)=\sum_{e \in E}{f(e)}$ is minimal (i.e. the smallest possible scalar value among all constructible spanning trees).
AI: Your definition is almost correct, just that the graph should be spanning, that is, $V=V^*$ and that $H$ should be connected, trivially. |
H: Show that the determinant of a matrix is nonzero
Suppose $u,v,w \in \mathbb{Q}$ with $u,v,w \neq 0$. Show that the
determinant of the following matrix is nonzero.
$$M = \begin{bmatrix} u & 2w & 2v \\ v & u & 2w \\ w & v & u \end{bmatrix}$$
Hint: Argue by contradiction, reduce to the case when $u,v,w$ are integers and use some number theory over $\mathbb{Z}$.
I know that the determinant is given by
$$ \det M = u^{3} + 2v^{3} + 4w^{3} - 6 \,u\,v\,w $$
It is unclear to me how to utilize the hint. How to convert the problem over $\mathbb{Z}$?
AI: As $u,v,w$ are rationals, you can just multiply every entry with a sufficiently large integer so that every entry becomes an integer. This will not change the fact whether the determinant is zero or not. Therefore, w.l.o.g. $u,v,w$ can be assumed to be integers. Similarly, we can assume that $u,v,w$ are not all even (otherwise just divide all entries by $2$ and repeat). Now observe that if $\det M = 0$, then $u$ must be even. But that means that $u^3+4w^3-6uvw = -2v^3$ is divisible by $4$. Hence $v$ is even, too. Thus $u^3+2v^3-6uvw = -4w^3$ is divisible by $8$, and so $w$ is even as well. Contradiction. |
H: Proof that an equation is irrational
I hope you're keeping safe and well. I stumbled across this problem and wondered whether you could help.
Show that $\left(a+\sqrt b\right)\left(a-\sqrt b\right)^3$ is irrational if $a$ and $b$ are NOT square numbers.
Thank you so much for your help in advance.
Pac-Man
AI: Let's develop the expression:
$\left(a+\sqrt b\right)\left(a-\sqrt b\right)^3=\left(a+\sqrt b\right)\left(a-\sqrt b\right)·\left(a-\sqrt b\right)^2=(a^2-b)(a^2-2a\sqrt b+b)$
We can see that the left term is rational if (but not only if) $a$ and $b$ are rational. On the other hand, the right term is rational if (again, not only if) "$-2a\sqrt b$" is rational. That holds if b is a square. As you are supposing that neither $a$ nor $b$ are squares, this term is irrational, what implies that all the expression is irrational. HOWEVER, that only holds if we suppose that both $a$ and $b$ are rational numbers.
Hope it was useful :) |
H: Extracting the diagonal terms of a square matrix.
For a given square matrix $A\in\mathbb{R}^{m\times m}$ does there exist a matrix $B\in\mathbb{R}^{m\times m}$ such that for the product $C:=AB$ we have $C_{ii}=A_{ii},$ $1\leq i\leq m$, and $C_{ij}=0$ if $i\neq j$ ?
I think the question can be also restated in the following way:
for a given square matrix $A\in\mathbb{R}^{m\times m}$ does there exists a vector $V\in\mathbb{R}^{m\times 1}$ such that for the product $W:=AV$ we have $W_i=A_{ii},$ $1\leq i\leq m?$
AI: For $m > 1$, there does not necessarily exist such a matrix. For example, take $A$ to be the matrix whose entries are all $1$. Such a matrix $B$ would satisfy $AB = I$, which means that $B$ would be the inverse of $A$. However, the matrix $A$ is not invertible (and in fact has rank $1$).
We also see that your second condition is not equivalent. In particular, if we take $V = (1,0,\dots,0)^T$, then we see that we can positively answer the second question for this $A$, but not the first.
It is true, however, that a matrix that fulfills the first requirement automatically fulfills the second, which is to say that the second condition is weaker.
Note that for any invertible matrix $A$, such a $B$ can necessarily be found. |
H: An invertible linear map over C has a square root (Linear Algebra Done Right 8.33)
$T \in \mathcal{L}(V)$ is a complex finite-dim linear operator. 8.33 proves that if $T$ is invertible, it must have a square root.
It shows that $T |_{G(\lambda_i, T)} = \lambda_i (I + \frac{N_i}{\lambda_i})$ has a square root, where $N_i \in \mathcal{L}(G(\lambda_i, T))$ is a nilpotent, $G(\lambda_i, T)$ is generalized eigenvectors space and $\{ \lambda_i \}$ is a set of distinct eigenvalues of $T$.
Finally, the proof finishes by saying that the operator $Rv = R_1 u_1 + ... + R_m u_m$, where $R_i = \sqrt{T |_{G(\lambda_i, T)}} = \sqrt{\lambda_i (I + \frac{N_i}{\lambda_i})}$, is indeed the square root of $T$ and $u_i \in G(\lambda_i, T)$.
I'm having trouble verifying this.
By applying $R$ expansion to the right side I get
$R_1 (R_1 u_1 + ... R_m u_m) + ... + R_m(R_1 u_1 + ... + R_m u_m) = R_1^2u_1 + ... + R_m^2u_m + R_1(R_2u_2 + ... + R_m u_m) + ... + R_m(R_1 u_1 + ... + R_{m-1}
u_{m-1})$
$R^2_i u_i = \lambda_i (I + \frac{N_i}{\lambda_i}) u_i = T|_{G(\lambda_i, T)} u_i$, is what I need, but the rest of the terms are excessive.
So I need to show that $R_1(R_2u_2 + ... + R_m u_m) + ... + R_m(R_1 u_1 + ... + R_{m-1} u_{m-1}) = 0$.
Q1) $R_i$ is defined on the elements of $G(\lambda_i, T)$, but in the equation above, $R_1$, for instance, needs to map an element of $G(\lambda_2, T) + ... + G(\lambda_m, T)$. It's probably going to map to $0$, but I'm confused how a linear map restricted to $G(\lambda_i, T)$ is to deal with elements outside this set.
Q2) If $R_i u_j \neq 0$ for $i\neq j$, then what is it equal to?
Edit: I realized I applied $R$ incorrectly. Applying it correctly makes the verification simple.
AI: Are you familiar with the Jordan normalform of a matrix? This states that any complex (square) matrix $T$ can be written as $ T = S \Lambda S^{-1} $, where $\Lambda$ is block diagonal and $S$ consists of the (generalized) eigenvectors of $T$. Hence changing the basis to $S$, the matrix $T$ will be block diagonal. This means that the vector space $V$ can be written as $V = \oplus_{i=1}^m V_i $, where $V_i$'s are invariant under the action of $T$. Thus for any $u\in V$
$$ Tu = T_1 u_1 + T_2 u_2 + \ldots + T_m u_m $$
holds,
where $T_i T_j = T_jT_i = 0$ for $i\neq j$. This also implies that, due to the definition of the $R_j$s, $R_i R_j = R_j R_i = 0$ for $i\neq j$. |
H: Upper bound for the sinc function
Show that there exists a constant $0<c<1$ such that
$$
\frac{\sin x}{x} < c,\quad\textrm{for all }x\ge1.
$$
--Context--
In Probability Theory Lévy's Theorem is crucial to uncover probability measures from certain functions which are obtained as limits of characteristic functions (Fourier transforms). The hard load of the work comes from Prokhorov's Theorem plus one key inequality involving the sinc function. In the 1-dimensional case it is enough to observe that ${\rm sinc}(x) \le \min(1,1/|x|)$. However, in the multivariate case a more accurate upper bound is needed, namely ${\rm sinc}(x) < c$ for some positive constant $c<1$.
It was in the middle of trying to adapt a proof of single-dimensional Lévy's Theorem to the multidiminsional case that I came across the need for this "detail". Since this function seemed related to the Laplace and Fourier transforms and I had found it in other contexts I decided to google it so to learn a little bit more. Here was where I learned how relevant this function is in e.g., Signal Processing (Kalman filters?). So I thought than instead of trying to proof the inequality myself it would be good to have it in SE so that others could start a "relationship" with this interesting function and its many properties. Obviously, I was plain wrong and I regret being so naïve.
AI: $\frac {\sin x} x \to \sin 1 <1$ as $x \to 1$. Hence there exists $r>0$ such that $1 \leq x \leq 1+r$ implies $\frac {\sin x} x <a$ where $a=\frac {1+\sin 1} 2$. Also $\frac {\sin x} x \leq \frac 1 {1+r}$ for $x \geq 1=r$. Can you finish? |
H: Topology question about a special subset in $\mathbb R^2$
Problem Statement:
Let $X = (\bigcup \limits_{n \in \mathbb N} \{\frac{1}{n}\} \times [0,1] ) \cup \{(0,0),(0,1)\}$ have a subspace topology as a subspace of $\mathbb R^2$. For any separation $U$ and $V$ of $X$, if $(0, 0) \in U$, then $(0, 1) \in U$ as well.
My attempt:
By the result from Munkres, if $U$ and $V$ are a separation of $X$ and $Y$ is a connected subspace of $X$, then $Y$ is completely contained in either $U$ or $V$. Hence, to show that $(0, 0) \in U$ would imply $(0, 1) \in U$, it suffices to show that there exists some connected subspace of $X$ that contains both $(0, 0)$ and $(0, 1)$.
From here, I am having trouble finding some connected subspace of $X$ that contains both points.
AI: Your solution will not work. Any neighbourhood $W$ of $X$ which contains both $(0,1)$ and $(0,0)$ can be disconnected either by the open sets $\{(x,y)|y>\frac12\}$ and $\{(x,y)|y<\frac12\}$ or if there exists $n\in\mathbb{N}$ with $(\frac1n,\frac12)\in W$, by the open sets $\{(x,y)|x<\frac1n\}$ and $\{(x,y)|x>\frac1{n+1}\}$.
Instead you can use the following argument:
If $U$ and $V$ separate $X$, then each vertical line $L_n=\{\frac1n\}\times [0,1]$ is connected, so by the result you mentioned completely contained in $U$ or $V$.
Any neighbourhood $U$ of $(0,0)$ will intersect all vertical lines $L_n$ for all $n>m_1$ for some $m_1$.
Similarly, any neighbourhood $V$ of $(0,1)$ will intersect all vertical lines $L_n$ for all $n>m_m$ for some $m_2$.
Thus if $U,V$ separate $X$, they will each contain all $L_n$ for $n>\max(m_1,m_2)$, yielding the desired contradiction. |
H: Question on poset of positive semi-definite matrices
Let $\Omega$ be a subset of the partially ordered set (poset) of $n\times n$ positive semi-definite matrices. I know that $\inf \Omega\in \bar{\Omega}$, where $\bar{\Omega}$ denotes the closure of $\Omega.$
If $\inf \Omega=X$, can I say that
\begin{array}{ll} \text{inf} & \operatorname{trace}(Y'\Omega Y)=\operatorname{trace}(Y'XY)?\\ \Omega\end{array}
I think trace is monotone increasing on the cone of positive definite matrices, it should hold?
AI: I believe that you are trying to ask the following:
If there exists an $X \in \bar \Omega$ such that $X \preceq Z$ (Loewner order) for all $Z \in \Omega$, then does it necessarily hold that
$$
\inf_{Z \in \Omega} \operatorname{trace}(Y'ZY) = \operatorname{trace}(Y'XY)?
$$
The answer to this is yes. In particular, if such an $X$ exists, then for all $Z \in \Omega$, we have
$$
\operatorname{trace}(Y'ZY) =
\operatorname{trace}(Y'(Z - X)Y) + \operatorname{trace}(Y'XY) \geq
\operatorname{trace}(Y'XY).
$$ |
H: Exterior power on short exact sequence of modules with free middle term
Let $(R,\mathfrak m,k)$ be a Noetherian local ring. For a finitely generated $R$-module $M$, let $\wedge^j(M)$ denote its $j$-th exterior power. Recall that $\wedge^j(R^{\oplus j})\cong R,\forall j\ge 1$.
Now suppose we have an exact sequence of finitely generated $R$-modules
$0\to M \xrightarrow{f} R^{\oplus n}\xrightarrow{g} N\to 0 $ . So we have an induced map $\wedge^n(f): \wedge^n(M)\to \wedge^n(R^{\oplus n})\cong R $ . Let $a\in \operatorname{Im}(\wedge^n(f))\subseteq R$ . Then how to prove that $aN=0$ ?
(If needed , I'm willing to assume that $f(M)\subseteq \mathfrak m R^{\oplus n}$ . )
My thoughts: Since $g$ is surjective, we have $\wedge^n(g)$ is surjective. Since $g\circ f=0$, we also have by functoriality that $\wedge^n(g)\circ\wedge^n(f)=0$ . We also have an exact sequence
$R^{\oplus n}\otimes M\cong \wedge^{n-1}(R^{\oplus n})\otimes \ker g \to \wedge^n(R^n)\xrightarrow{\wedge^n(g)} \wedge^n(N)\to 0$ .
Apart from this, I can't think of anything else. Please help.
AI: We can think of $M$ as a submodule of $R^n$. Then the image of $\bigwedge^n M$
in $R$ is the ideal generated by all $\det A$ where $A$ runs
through the $n$ by $n$ matrices whose columns are in $M$. So the result boils down
to the assertion that $(\det A) R^n\subseteq M$ whenever $A$ is such a matrix.
But for $u\in R^n$, $Au\in M$, and for $u\in R^n$
$$(\det A)v=A(\text{adj}\,A)v=Au\in M$$
where $u=(\text{adj}\,A)v$.
This argument works over all commutative rings. |
H: What do you call an L1 regularization term involving a matrix vector product
Often times I have difficulties finding certain things when I do not use the correct terminology. Or there are many different terms for the same thing.
I would like to know what you would call the following optimization problem, so I have an idea what to look for:
$
\begin{align}
\min\limits_{x} ||Ax-b||^2_2 + \lambda \mathcal{R}(x)
\end{align}
$
where the regularizer is of the form
$
\mathcal{R}(x) = ||Dx||_1
$
What if $D$ is just a positive diagonal matrix? Would it be called differently?
Thank you very much for your answers.
AI: If by positive you mean strictly positive, so that all the diagonal elements are $>0$, then $D$ is invertible and your problem is equivalent to
$$\min_{y}\|By-b\|^2_2+\lambda\|y\|_1\,,$$
where $B = AD^{-1}$ and you're using the standard $L^1$-regularisation.
If you're interested in the $x^\star$ achieving the minimum in your original problem, it is simply given by
$$x^\star = D^{-1}y^\star\,,$$
where $y^\star$ achieves the minimum in the standard-regularised problem.
EDIT
Note that $D$ does not even need to be diagonal or positive defined, just invertible, in order to go back to the original $L^1$-regularisation. |
H: Why $\omega<2^{\omega}$ but $\omega^{\omega}\sim\omega$?
I understand the Cantor's proof for why $S<2^S$, but also we know that ordinals $\omega\sim\omega^2\sim\omega^3...$. This approaches to $\omega^{\omega}$, what should be at least not less than $2^{\omega}$. What do I miss with this logic?
AI: You are confusing two different notions of exponentiation: the cardinal exponentiation and the ordinal exponentiation.
$2^\omega$ can mean either
The cardinality of the power set of $\omega$ (or the set of functions $\omega\to 2$), or
The ordinal number corresponding to the order type of finite binary sequences, ordered lexicographically.
The latter is a countably infinite ordinal (which is, in fact, equal to $\omega$, so it also happens to be cardinal). The former is an uncountable cardinal, the continuum.
Likewise, $\omega^\omega$ can mean one of the two:
The cardinality of the set of functions $\omega\to \omega$,
The ordinal number corresponding to the order type of finite sequences with values in $\omega$.
Here, the former is a cardinal number (in fact, equal to the continuum), while the latter is a countable ordinal (which is strictly between $\omega$ and the continuum).
Especially when there is risk of confusion, it is good practice to use $\aleph_0$ when talking about $\omega$ as a cardinal (in particular, when you mean to use cardinal exponentiation), $\omega$ when you mean to talk about it as an ordinal, and $\mathbf N$ when you mean to talk about it as a set of integers. Then it would be pretty clear that $\aleph_0=\omega=2^\omega<\omega^\omega<2^{\aleph_0}=\aleph_0^{\aleph_0}$ and $\omega\sim 2^\omega\sim \omega^\omega\not\sim 2^{\aleph_0}\sim \aleph_0^{\aleph_0}$. |
H: Sign of the line integral ($\int_{\vert z\vert=1} {1 \over z^2} \tan({\pi \over z}) dz$)
Find the value of the $$\int_{\vert z\vert=1} {1 \over z^2} \tan\left({\pi \over z}\right) dz$$
When we substitute $\omega = {1 \over z}$, then $d\omega = - {1 \over z^2}dz$, hence $\int_{\vert z\vert=1} {1 \over z^2} \tan({\pi \over z}) dz = -\int_{\vert \omega \vert=1} $$\tan({\omega \pi}) d\omega = -4i$. (By Either residue thm on $\pm {1 \over 2}$ or argument principle).
But the problem is the answer was $4i$
Why does the sign have to be positive?
AI: $\int_{\vert z\vert=1} {1 \over z^2} tan({\pi \over z}) dz = \int_{\vert \omega \vert=1} $$tan({\omega \pi}) d\omega $. Use the definition of contour integral to justify the formula I have written. |
H: Is $\mathbb Q \times \{0\}$ an integral domain?
Is $\mathbb Q \times \{0\}$ an integral domain?
I understand that $\mathbb Q \times \{0\}$ is a commutative ring with unity.
But there was no clear proof that it has no zero divisor.
How do I prove $\mathbb Q \times \{0\}$ is a zero divisor, if it's true that $\mathbb Q \times \{0\}$ is an integral domain?
AI: If $(a,0)\ne (0,0)$ and $(b,0)\cdot (a,0)=(ba,0)=(0,0)$, then $ba=0$ and $a\ne 0$, and therefore $b=0$. |
H: Indefinite integral of $\frac{\sec^2x}{(\sec x+\tan x)^\frac{9}{2}}$
$$\frac{\sec^2x}{(\sec x+\tan x)^\frac{9}{2}}$$
My approach:
Since it is easy to evaluate $\int{\sec^2x}$ , integration by parts seems like a viable option.
Let $$I_n=\int{\frac{\sec^2x}{(\sec x+\tan x)^\frac{9}{2}}}$$
$$I_n=\frac{\tan x}{(\sec x+\tan x)^\frac{9}{2}} + \frac{9}{2}\int{\frac{\sec x \tan x}{(\sec x+\tan x)^\frac{9}{2}}dx}$$
Evaluating the new integral again using by parts yields
$$\frac{\sec x}{(\sec x+\tan x)^\frac{9}{2}}+\frac{9}{2}\int{\frac{\sec^2x}{(\sec x+\tan x)^\frac{9}{2}}\,dx}$$
$$=\frac{\sec x}{(\sec x+\tan x)^\frac{9}{2}} + \frac{9}{2} I_n$$
Plugging it back, we obtain
$$I_n=\frac{\tan x}{(\sec x+\tan x)^\frac{9}{2}} + \frac{9}{2}\frac{\sec x}{(\sec x+\tan x)^\frac{9}{2}} + \frac{81}{4}I_n $$
$$\frac{-77}{4}I_n=\frac{\tan x}{(\sec x+\tan x)^\frac{9}{2}} + \frac{9}{2}\frac{\sec x}{(\sec x+\tan x)^\frac{9}{2}}$$
This obviously doesn't match with bprp's answer. Help!
Edit:
How do I convert my answer to the answer obtained by him, if mine is correct
AI: $-77I_n=4 \times \frac{\tan x}{(\sec x+\tan x)^\frac{9}{2}} + 18 \times \frac{\sec x}{(\sec x+\tan x)^\frac{9}{2}}$
$-77I_n= 11 \times \frac{\tan x}{(\sec x+\tan x)^\frac{9}{2}} - 7 \times \frac{\tan x}{(\sec x+\tan x)^\frac{9}{2}} + 11 \times \frac{\sec x}{(\sec x+\tan x)^\frac{9}{2}} + 7 \times \frac{\sec x}{(\sec x+\tan x)^\frac{9}{2}}$
$-77I_n= 11 \times \frac{\sec x+\tan x}{(\sec x+\tan x)^\frac{9}{2}} + 7 \times \frac{\sec x - \tan x}{(\sec x+\tan x)^\frac{9}{2}}$
First part is straightforward. The second part can be simplified multiplying both numerator and denominator by $(\sec x+\tan x)$. We know, $(\sec x+\tan x) (\sec x-\tan x) = 1$.
You can take it from here. |
H: Evaluate $\lim_{h\to 0}\frac{1}{h^2}\begin{vmatrix}\tan x&\tan(x+h)&\tan(x+2h)\\\tan(x+2h)&\tan x&\tan(x+h)\\\tan(x+h)&\tan(x+2h)&\tan x\end{vmatrix}$
Evaluate
$$
\lim_{h\to 0}\frac{\Delta}{h^2}=\lim_{h\to 0}\frac{1}{h^2}\begin{vmatrix}
\tan x&\tan(x+h)&\tan(x+2h)\\
\tan(x+2h)&\tan x&\tan(x+h)\\
\tan(x+h)&\tan(x+2h)&\tan x
\end{vmatrix}
$$
Attempt
$$
\lim_{h\to 0}\frac{\Delta}{h^2}=\begin{vmatrix}
\lim_{h\to 0}\tan x&\lim_{h\to 0}\dfrac{\tan(x+h)-\tan x}{h}&\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan(x+h)}{h}\\
\lim_{h\to 0}\tan(x+2h)&\lim_{h\to 0}\dfrac{\tan x-\tan(x+2h)}{h}&\lim_{h\to 0}\dfrac{\tan(x+h)-\tan(x+2h)}{h}\\
\lim_{h\to 0}\tan(x+h)&\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan(x+h)}{h}&\lim_{h\to 0}\dfrac{\tan x-\tan(x+2h)}{h}
\end{vmatrix}\\
=\begin{vmatrix}
\lim_{h\to 0}\tan x&\lim_{h\to 0}\dfrac{\tan(x+h)-\tan x}{h}&\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan(x+h)}{h}\\
\lim_{h\to 0}\tan(x+2h)&-2.\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan x}{h}&-1.\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan(x+h)}{h}\\
\lim_{h\to 0}\tan(x+h)&\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan(x+h)}{h}&-2.\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan x}{2h}
\end{vmatrix}\\
$$
$$
\lim_{h\to 0}\dfrac{\tan(x+h)-\tan x}{h}=\frac{d}{dx}\tan x=\sec^2x\\
\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan x}{2h}=\frac{d}{dx}\tan x=\sec^2x\\
\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan(x+h)}{h}=\frac{d}{dx}\tan(x+h)=\sec^2(x+h)
$$
But my reference gives the solution $9\tan x.\sec^4x$, I think by taking $\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan(x+h)}{h}=\sec^2x$. Will that make a difference ?
It might be silly but could anyone clarify this confusion in my attempt ?
AI: It is easy to see $\lim_{h\to 0}\frac1h(\tan(x+2h)-\tan(x+h))$ is the derivative of $\tan$ at $x$ if you already know Taylor expansion, but if you probably haven't encounter them yet, here is a way without explicit expansion:
Note that
\begin{align*}
\frac{\tan(x+2h)-\tan(x+h)}{h}&=
\frac{\tan(x+2h)-\tan x}{h}-\frac{\tan(x+h)-\tan x}{h}\\
&=2\frac{\tan(x+2h)-\tan x}{2h}-\frac{\tan(x+h)-\tan x}{h}.
\end{align*}
So, taking limit $h\to 0$, we have
$$
\lim_{h\to0}\frac{\tan(x+2h)-\tan(x+h)}{h}=2\frac{\mathrm{d}}{\mathrm{d}x}\tan x-\frac{\mathrm{d}}{\mathrm{d}x}\tan x=\frac{\mathrm{d}}{\mathrm{d}x}\tan x.
$$ |
H: 2-stable subsets of groups
In this post, multiplication of subsets of a group is defined by
$$ST= \{st| s\in S, t \in T\}$$
A subset of a group is called $n$-stable if there exists a natural number $n$ s.t. $$S = S\cdot \overbrace{S\cdot S\cdot S...}^{n \text{ times}}$$
The minimal $n$ for which it’s true determines it’s name.
Does there exist a group who’s subset is 2-stable? i.e, $$S=SSS\neq SS$$
Of course, any singleton of a order 2 element suffices, but I’m looking for subsets of size 2 or more.
If we take a pair of elements $S=\{a,b\}$, $SS=\{a^2,ab,ba,b^2\}$...
Any help?
AI: Without loss of generality, we may assume that $G=\langle S\rangle$. I claim that $S$ is $2$-stable if and only if there is an index-$2$ subgroup $H$ such that $S=G\setminus H$.
Suppose that $H$ has index $2$, and let $S=G\setminus H$. Certainly $SS=H$, and $SSS=HS=S$, and so the result holds in this case. Thus we need to establish the converse. Let $H$ denote the set of all even-length products of elements of $S$. This is a subgroup, since products of even-length products are even-length. We must prove that $H\cap S=\emptyset$, and then that $G\setminus H=S$. Notice that all odd-length products of elements of $S$ are elements of $S$.
Notice that this means that all even-length products appear in $SS$. If $t$ has even-length, then $t=s's$ for $s\in S$ and $s'$ of odd length, hence in $S$. Thus $G=S\cup SS$. But $SS$ is a subgroup, and therefore has at most half of the elements of $G$ in it. But it also has $Ss$ for a fixed $s\in S$ in it, and these must all be distinct. Thus $|S|=|SS|$ and $H$ has index $2$. |
H: Uniqueness of the derivative of a differentiable function on a non-open set
If $E_i$ is a $\mathbb R$-Banach space and $\Omega_1\subseteq E_1$, then $f:\Omega_1\to E_2$ is called $C^1$-differentiable at $x_1\in\Omega_1$ if $$\left.f\right|_{O_1\:\cap\:\Omega_1}=\left.\tilde f\right|_{O_1\:\cap\:\Omega_1}$$ for some $\tilde f\in C^1(O_1,E_2)$ for some $E_1$-open neighborhood $O_1$ of $x_1$.
Is $${\rm D}f(x_1):={\rm D}\tilde f(x_1)\tag1$$ well-defined, i.e. independent of the choice of $\tilde f$?
I would say, it obviously should be, but the discussion below this question raised some doubts.
Maybe I'm missing subtlety but if $\hat O_1$ and $\hat f$ are other choices for $O_1$ and $\tilde f$, then there is a $\varepsilon>0$ such that the $E_1$-open ball around $x$ with radius $\varepsilon$ is contained in $O_1\cap\hat O_1$ and it should clearly hold \begin{equation}\begin{split}&\left\|\left({\rm D}\tilde f(x_1)-{\rm D}\hat f(x_1)\right)h_1\right\|_{E_2}\\&=\left\|\frac{\hat f(x_1+h_1)-\hat f(x_1)-{\rm D}\hat f(x_1)h_1}{\left\|h_1\right\|_{E_1}}-\frac{\tilde f(x_1+h_1)-\tilde f(x_1)-{\rm D}\tilde f(x_1)h_1}{\left\|h_1\right\|_{E_1}}\right\|_{E_2}\left\|h_1\right\|_{E_1}\end{split}\end{equation} for all $h_1\in E_1\setminus\{0\}$ with $\left\|h\right\|_{E_1}<\varepsilon$, which tends to $0$ as $h\to0$. So, it should hold $${\rm D}\tilde f(x_1)={\rm D}\hat f(x_1)\tag2.$$ What am I missing?
AI: Consider $f\colon\mathbb{R}\times\{0\}\rightarrow\mathbb{R}^2,\,(x,0)\mapsto(x,0)$. Consider $\operatorname{id}\colon\mathbb{R}^2\rightarrow\mathbb{R}^2$ and $\pi\colon\mathbb{R}^2\rightarrow\mathbb{R}^2,\,(x,y)\mapsto(x,0)$. Both of these are differentiable (smooth even) functions and clearly $\operatorname{id}\vert_{\mathbb{R}\times\{0\}}=f=\pi\vert_{\mathbb{R}\times\{0\}}$. However, $D\operatorname{id}(x)=\operatorname{id}\neq\pi=D\pi(x)$ for any $x\in\mathbb{R}^2$. |
H: Polynomial olympiad problem
Let $p(x)$ be a monic polynomial of degree four with distinct integer roots $a, b, c$ and $d$. If $p(r)=4$ for some integer $r$, prove that $r=\frac{1}{4}(a+b+c+d)$
My only idea was to let $p(x)=(x-a)(x-b)(x-c)(x-d)$, so that:
$4=(r-a)(r-b)(r-c)(r-d)$. But the casework here, looking for $4$ factors of $4$, seems too tedious
AI: As you said, we have $4=(r-a)(r-b)(r-c)(r-d)$. Therefore, as $a,b,c,d$ are integers. the absolute values of any two of the factors of $4$ above should be $2$ and two of them should be $1$. Now, the numbers which have the same absolute values must have different sign, as otherwise, the numbers would become equal. Thus, let us assume, without loss of generality $r-a=2, r-b=-2,r-c=1,r-d=-1$, which gives us $a+b+c+d=4r$, from which the desired conclusion follows. |
H: To prove that an operation is well-defined in modular arithmetic
I started to study the relation of congruence modulo n and a big important question came to me. In the book Poofs and Fundamentals, by Ethan D. Bloch, we have the definition:
Definition: Let $n \in \mathbb{N}$. Define operations $+$ and $\cdot$ on $\mathbb{Z}_{n}$ by letting $[a] + [b] = [a + b]$ and $[a] \cdot [b] = [ab]$ for all $[a], [b] \in \mathbb{Z}_{n}$.
Next, Bloch consider the following problem: Let $n \in \mathbb{N}$, and let $[a], [b], [c], [d] \in \mathbb{Z}_{n}$. Suppose that $[a] = [c]$ and $[b] = [d]$. Do $[a + b] = [c + d]$ and $[ab] = [cd]$ necessarily hold?
Bloch also states that if this doesn’t hold, then both operations are not well-defined. Reading this made me think of the following questions:
Why proving that if $[a] = [c]$ and $[b] = [d]$ then $[a+b] = [c+d]$ shows that $+$ is well-defined in $\mathbb{Z}_{n}$?
If I show that $\mathbb{Z}_{n}$ is closed under $+$, am I automatically showing that $+$ is well-defined in $\mathbb{Z}_{n}$? (If yes, what’s the relation between these two?)
Thank you so much for your attention!
AI: Question 1. We say that a definition is well-defined when, even though there is an apparent ambiguity in the definition, in fact there is not.
In your case, the definition $[a]+[b]=[a+b]$ is ambiguous because equivalence classes of different elements can coincide. For example, in $\mathbb{Z}_5$, $[3]=[8]$. So it is not immediately clear that if you take different representatives for $a$ or $b$ will give the same result for $[a+b]$. That is, one needs to prove that if $[a_1]=[a_2]$ and $[b_1]=[b_2]$ then $[a_1+b_1]=[a_2+b_2]$.
For example, suppose we define for $\mathbb{Z}_5$, $[a]^{[b]}:=[a^b]$. It looks fine but it is not really because $[3]=[8]$, $[2]=[7]$, but $[3^2]=[4]\ne[8^7]=[2]$, so the mapping is not well-defined.
Question 2. An operation is a well-defined function mapping $X^2\to X$. Showing closure and well-defined are not the same. An operation that is not well-defined is not normally called an operation, so in this strict sense, it does not make sense for an operation to be closed but not well-defined.
But even if one is generous with what makes an operation, showing closure does not automatically imply it is well-defined. For example, one can argue that the example above $[a]^{[b]}:=[a^b]$ is 'closed' in the sense that it gives an equivalence class as output, but it is still not well-defined. |
H: interior and closure in metric spaces
let say we have $(\ell^{1}(\Bbb{N}),d_{1})$ as a metric space with $d_{1}((x_{n})_{n},(y_{n})_{n})=\sum_{n=0}^{\infty}|x_{n}-y_{n}|$. If $$D=\left\{x \in \ell^{1}(\Bbb{N}) \,\,\Big|\, \sum_{n=1}^\infty n|x_{n}|<\infty \right\}$$
I'm looking for the interior of $D$ and the closure of $D$.
I thought that the interior of $D$ was empty.
This is my attempt to prove it:
Let's say that the interior isn't empty
take an random $x$ element of $l^{1}(N)$ that belongs to the interior of $D$. Then there is a $\delta >0$ so that $B(x,\delta)$ belongs to $D$.
Now take $y=\left(\frac{x_n}n\right)_n$ then $y$ isn't an element of $D$ but $d_{1}(x,y)<\delta$.
So the interior must be empty.
Now i'm not sure if my proof is correct and i find this kind of excercises really difficult. I hope someone can help me to explain this to me and help me to do it right.
For the exterior i thougt that it was just D but i can't even start to prove it.
AI: You are correct that $\operatorname{Int} D = \emptyset$ but for your $y$ I can't see why would $d_1(x,y) < \delta$.
The idea is similar, though. Assume that $B(x,\delta) \subseteq D$. Notice that $D$ is a vector subspace of $\ell^1$, meaning it is closed under addition and scalar multiplication. Pick some $y \in \ell^1 \setminus D$ such as $y = \left(\frac1{n^2}\right)_n$ and notice that
$$d_1\left(\frac{y}{\|y\|_1}\frac{\delta}2 - x,x\right) = \left\|\frac{y}{\|y\|_1}\frac{\delta}2\right\| = \frac\delta2 < \delta$$
so $z := \frac{y}{\|y\|_1}\frac{\delta}2 - x$ is contained in $B(x,\delta) \subseteq D$. Now we have
$$y = \frac{2\|y\|_1}{\delta}(x+z) \in D$$
because $D$ is a subspace. But this is a contradiction since $y \notin D$. Therefore interior of $D$ is empty.
Regarding the closure, notice that $D$ contains the set $c_{00}$ of all finitely supported sequences in $\ell^1$ and recall that $c_{00}$ is dense in $\ell^1$. Threfore
$$\overline{D} \supseteq \overline{c_{00}} = \ell^1$$
so the closure is $\overline{D} = \ell^1$. |
H: What does dimension in integer programing problem mean
In the problem P1 below described in a research paper I am reading, the authors say that the problem P1 below is a three dimensional integer programing programing problem. Can I ask what does 3-dimension means here? does it mean that it has three min/max functions or three or the fact that it depends on three variables $a\in A, n\in N,$ and $, m \in M$.
Explanation of Problem P1
AI: It is a 3D problem because you have three variables here, $n, m, k$. |
H: convex sets in convex optimization
How to prove that the following set is not convex?
$$M = \left\{ \mathbb{R}^{3}: x_{1}x_{2}x_{3}\le 1,x_{1}+x_{3}\ge 2,x_{1} \ge 0 \right\}$$
Thanks for any help.
I tried to write it down as intersection of two sets $\{x_1x_2x_3 \le 1\}$
and $\{x_1+x_3 \ge 2,x_1 \ge 0\}$. The second set is convex (it is clear) and the first set is not convex. I have used theorem about level sets, the function $x_1x_2x_3$ is not convex so the level set is not convex. This is problem, it does not mean that the intersection is not convex
AI: Guide:
It's important to analyze where is the "weakness" of a problem. The last two constraints are linear. If all the constraints are convex, then the intersection must be convex. The first constraint is the weakness, attack it.
Set $x_3=1$, now visualize the projected set. Prove that the projected set is not convex by constructing a pair of points in the set where its midpoint is not in the set.
It might help if you to visualize the set $xy \le 1$. Give it a try.
Please try the following question first:
Prove that $M_2 = \{ \mathbb{R}^2: x_1 x_2 \le 1, x_1 +1 \ge 2, x_1 \ge 0\}$ is not convex. |
H: Groups of order $252 = 4 \cdot 7 \cdot 9$ are solvable
The goal is to prove that any group of order $252 = 36 \cdot 7$ is solvable, and because I managed to confuse myself, I'm asking here.
Let $G$ be a group of order $252$. By Sylow's Theorems, the number of $7$-Sylow subgroups of $G$ is either $1$ or $36$. If it is $1$, we are done, because the quotient then has order $36$, and groups of order $7$ and $36$ are solvable.
Hence we are left with the much more interesting case in which the number of $7$-Sylow subgroups is $36$. One proof to show solvability is the following:
By the orbit-stabilizer theorem (since $G$ acts transitively on the set of its $7$-Sylow subgroups), the normalizer $N_G(P)$ of a $7$-Sylow $P$ of $G$ has order $7$, hence
$$N_G(P) = Z_G(P) = P,$$
where $Z_G(P)$ is the centralizer of $P$. By Burnside's Transfer Theorem, we obtain that $G$ contains a normal subgroup $N$ of order $36$. Since $|G/N| = 7$, we are done.
Questions to the second case (number of $7$-Sylows is $36$):
I checked with GAP and saw that there is no group of order $252$, whose $7$-Sylow is not normal. Is there an easy way to see this without invoking a computer algebra system?
Can one prove in a more elementary way that there is a normal subgroup of order $36$? Indeed, there are exactly $36 \cdot 6$ elements of order $7$, thus there are $36$ elements, whose order is coprime to $7$. How does one see that these $36$ elements form a subgroup? If we could see that in an elementary way, there is of course a unique subgroup of order $36$, hence a normal one, and there is no need to invoke Burnside's Transfer Theorem.
AI: I suppose the more elementary way is to look at the Sylow $3$s next.
We know $n_3=1,4,7,28$. If $n_3=1$ we are done, and $n_3=4$ we have a homomorphism $G\to S_4$ with nontrivial kernel. So $n_3=7$ or $n_3=28$. But since we only have $36$ elements left, there must be two Sylow 3s, say $H_1,H_2$ that intersect nontrivially.
So $P=H_1\cap H_2$ has order $3$, whose centralizer (since $H_i$ are abelian) $C_GP$ contains at least the set $H_1H_2$ of $27$ elements. Therefore $\lvert C_GP\rvert$ has to be a factor of $252$ that is at least $27$ and divisible by $9,$ so must be $36$ (the other choices, $63$, would be an index-$4$ subgroup so again we have a nontrivial homomorphism to $S_4$, or $126$ which is index 2 hence normal). So $C_GP$ is every element with order prime to $7$.
But that is enough for contradiction. $C_GP$ contains all Sylow $3$s since we basically used up those elements, but the group generated by all Sylow $3$s is normal in $G$. |
H: If $A^2=\mathbb{I} (2\times 2$ identity) then $\mathbb{I} + A$ is invertible only if $A=\mathbb{I}$
I want to show that if $A^2=\mathbb{I}$ ($2\times2$ identity) then $\mathbb{I} + A$ is invertible only if $A=\mathbb{I}$.
I know that $A^2=\mathbb{I}$ means that $A=A^{-1}$
I've started by letting $A=\begin{bmatrix} a&b\\c&d \end{bmatrix}$ and then $A+1=\begin{bmatrix} a+1&b\\c&d+1 \end{bmatrix}$ so it's invertible only when $(a+1)(d+1)-bc\neq 0$ and then I was going to see when this would be the case based on what we know about $A$ but this seems over complicated and I feel like there's a better way that I'm missing.
AI: Hint: $$A(I+A)=I+A\,\,\,\,\,\,$$ |
H: If $|f'(c)|
We have a derivative function $f$ with for every $c$ element of $\mathbb{R}: |f'(c)|<M$. I tried to prove that
prove that $\displaystyle \left|\int_{0}^{1}f(x)\mathrm{d}x-\frac{1}n \sum_{k=0}^{n-1}f\left(\frac{x}n\right)\right|\leq\frac{M}{n}$.
I really have no idea how to start. I'm trying to use integral , derivative, sums ... so my paper is full of definitions but i can't use one. Can someone give me a hint how to start with this question so that i can move on.
I'm really sorry i can't give a proper prove that i have already found but i'm stuck at the beginnin already.
AI: Your hypothesis on $f'$ implies that $f$ is Lipschitz
$$|f(x)-f(y)|< M|x-y|\,.$$
You can rewrite the integral splitting the domain in $n$ subinterval of length $1/n$, i.e.
$$\int_0^1 f(x)dx = \sum_{k=0}^{n-1}\int_{\frac{k}{n}}^{\frac{k+1}{n}}f(x)dx\,.$$
So you get
$$\left|\int_{0}^{1}f(x)dx-1/n \sum_{k=0}^{n-1}f(k/n)\right|\leq \sum_{k=0}^{n-1}\left|\int_{\frac{k}{n}}^{\frac{k+1}{n}}f(x)dx-\frac{1}{n}f(k/n)\right|\leq\sum_{k=0}^{n-1}\int_{\frac{k}{n}}^{\frac{k+1}{n}}\left|f(x)-f(k/n)\right|dx\,.$$
Using the Lipschitz property you have
$$\left|f(x)-f(k/n)\right|< M|x-k/n|$$
and since $x\in[k/n,(k+1)/n]$ it follows that
$$\left|f(x)-f(k/n)\right|< M/n\,.$$
So you conclude
$$\left|\int_{0}^{1}f(x)dx-1/n \sum_{k=0}^{n-1}f(k/n)\right|< \sum_{k=0}^{n-1}\int_{\frac{k}{n}}^{\frac{k+1}{n}}\frac{M}{n}dx=\sum_{k=0}^{n-1}\frac{M}{n^2}=\frac{M}{n}\,.$$ |
H: Topological groups vs regular groups
I know group theory and I'm familiar with the concept and definition of Group.
Today I was reading an article about topology and discoverer the concept of "topological group". I read the definition and immediately the following question came to my mind:
What is the purpose of topological groups? What can we do with topological groups that we can't do with regular groups? What are topological groups used for? What can topological groups do that groups without a topology can't?
I barely know any topology and perhaps that's why I can't see any utility of adding a topology to a group.
AI: Well... you clearly know some topological groups: Take the fields $\mathbb{Q},$ $\mathbb{R}$ and $\mathbb{C}$ for instance, or the multiplicative group $(0,\infty)$. Now ask yourself what sort of homomorphisms you know involving these groups. The first to come to mind are probably linear maps, the exponential function or the logarithm. Now, these all have the property of being more than continuous - they're even smooth.
Do discontinuous homomorphisms exist? Well, yeah, just check https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation#Existence_of_nonlinear_solutions_over_the_real_numbers. However, you've never heard of those, because it's not clear that they're interesting. It's something of a pathology of the real numbers that its endomorphisms aren't all $\mathbb{R}$-linear (although, they are all $\mathbb{Q}$-linear).
Why are they not interesting? Well, the same reason why you probably cannot name many maps $\mathbb{Z}\to \mathbb{Z}$ that aren't either group homomorphisms or have some combinatorial interpretation. Generally, we care about the mathematical structure of our objects. Thus, when looking at maps of groups, we are mostly interested in the maps that can see the group structure - these are the homomorphisms.
Similarly, if you are looking at a topological space, the most natural maps to care about are those which respect the topological structure - i.e., the continuous maps. So I would say that the point is more so that a bunch of groups that we naturally study also happen to carry natural topologies - see the above number fields and all manners of matrix groups. Since they are both groups and topological spaces in a natural fashion, it makes sense to study those two properties at the same time.
Now, of course, there is strength to be had in topology. For instance, you get the intermediate value theorem for the real numbers and more generally, that continuous homomorphisms have to respect connectivity properties. The regularity of continuous maps is very rich, I just don't think that that in and of itself is the reason why topological groups are interesting. They are just interesting because lots of interesting groups happen to be topological. |
H: prove $\left(3, 1+\sqrt{-5}\right)$ is prime ideal of $\mathbb{Z}\left[\sqrt{-5}\right]$
How to prove that $(3, 1+\sqrt{-5})$ is prime ideal of $\mathbb{Z}[\sqrt{-5}]$?
attempt 1: use definition
Consider $a, b, c, d, k_1, k_2 \in \mathbb{Z}$ s.t. $$ac-5bd=3k_1+k_2,\, \, ad+bc=k_2.$$ To prove $\exists j_1, j_2 \in \mathbb{Z}$ s.t. $3j_1+(1+\sqrt{-5})j_2=a+b\sqrt{-5}$ or $=c+d\sqrt{-5}$.
This is a bad way.
attempt 2:
To prove $\dfrac{\mathbb{Z}\left[\sqrt{-5}\right]}{\left(3, 1+\sqrt{-5}\right)}$ is integral domain. I know how to work with quotient of polynomial ring but not how to work with quotient of $\mathbb{Z}\left[\sqrt{-5}\right]$.
attempt 3:
$$\mathbb{Z}\left[\sqrt{-5}\right]\cong \mathbb{Z}/\left(x^2+5\right)$$
When we have $\mathbb{Z}/\left(x^2+5\right)$, converting into $\mathbb{Z}\left[\sqrt{-5}\right]$ simplifies the problem. May be the other way round is useless.
Please give a hint. Please do not give solution. Thanks!
AI: $$
\frac{\mathbb{Z}\left[\sqrt{-5}\right]}{\left(3, 1+\sqrt{-5}\right)}
\cong \frac{\mathbb{Z}[x]}{\left(3,1+x,x^2+5\right)}
\cong \frac{\mathbb{Z}_3[x]}{\left(1+x,x^2-1\right)}
\cong \cdots
$$ |
H: Maps Preserving Arc Connectedness?
We know that continuous functions do not preserve arc-connectedness (for an example, see this question I asked previously). So, the natural question that comes next is - which maps preserve arc-connectedness?
That is, if $X$ is arc connected, and $f:X\to Y$, then what are the weakest properties that $f$ should have so that $Y$ is arc connected. (Obviously, homeomorphisms are enough. I'm asking if there are weaker conditions that are enough.)
AI: Every path-connected Hausdorff space is arc-connected so if you assume $X$ is arc-connected and $Y$ is Hausdorff, then $f(X)$ will be arc-connected.
The real line with "two origins" is path-connected and $T_1$ but is not arc-connected and you can construct it as the quotient of a path-connected metric space and the quotient map will even be a perfect map. This should convince you that you won't be able to do a whole lot better than this. |
H: Find the number of other values of n for which $S_{n}$ = r is
Consider an arithmetic progression whose first term is $4$ and
the common difference is $-0.1.$ Let $S_{n}$ stand for the sum of the first
n terms. Suppose r is a number such that $S_{n}$ = r for some n. Then
the number of other values of n for which $S_{n}$ = r is
I am getting $S_{n}$ as $$\frac{n}{2}(8.1-0.1*n)$$
Next how to process further?
AI: Let's rewrite your expression as $$S_n=\frac{81n-n^2}{20}$$ Now if the sum of the first $m$ terms equals the sum of the first $n$ terms, we have $$\frac{81n-n^2}{20}=\frac{81m-m^2}{20}$$
Can you finish it now?
EDIT
We have $n+m=81$ or $m=81-n$. But $m$ must be a positive integer for this to make sense. So there is $1$ solution if $1\leq n\leq80$ and no solution if $n\geq81$.
P.S.
To be absolutely complete, I should note that the case $m=n$ is not possible. |
H: Validation of my proof for proving that the Sorgenfrey Line does not satisfies the second axiom of countability
In an exercice I am asked to prove the following:
Prove that the Sorgenfrey Line does not satisfies the second axiom of contability.
This is my second proof for this exercise because the first one was wrong. Because I'm studying topology by my self and the book has no solutions I'm share my proof hopping that someone could correct me if I made some mistake or if the proof is not valid.
My proof:
Let $\cal B$ be a basis for the Sorgenfrey Line.
Then we have that $\forall U \in \tau, \forall x \in U, \exists B \in \mathcal B: x \in B \subseteq U$.
So let $U_a = [a,2)$, with $a \in [0,1]$, then $U \in \tau$
So $\exists B_a \in \mathcal B: a \in B_a \subseteq U$.
Now let's define the set $S = \{B_a:a \in [0,1]\}$, Then we can define a bijective function $f: [0,1] \to S$ with $f(x) = B_x$, so we have that $S \sim [0,1]$, which means that $S$ is uncountable.
We know that if a set $A$ is countable then every subset $B$ of $A$ is also countable. We know that $S \subset \mathcal B$, and $S$ is uncountable, which means that $\mathcal B$ cannot be countable, thus proving that does not exists a countable basis for the Sorgenfrey Line.
Is this proof correct?
AI: You have to actually show $f: [0,1]\to \mathcal{B}, a \to B_a$ is an injection (the set $S$ is unnecessary ): suppose $a \neq a'$. Then $a < a'$ WLOG. But then $a \in f(a)$ and $a \notin f(a')$ shows that $a \in f(a)\setminus f(a')$ so that $f(a) \neq f(a')$ (as sets).
And the existence of an injection from $[0,1]$ into $\mathcal{B}$ is a direct proof that $|\mathcal{B}| \ge |[0,1]|$ and so all bases of the Sorgenfrey line have size at least continuum.
BTW, minor point, we don't need to use $[0,1]$ and sets $[a,2)$ but $\Bbb R$ and taking $B_x$ with $x \in B_x \subseteq [x,x+1)$ is simpler to me and more "uniform". The same arguments apply to this choice as well. |
H: If $\left(\sqrt2\right)^x + \left(\sqrt3\right)^x = \left(\sqrt{13}\right)^{\frac{x}{2}}$ then the number of values of $x$ is?
If $\left(\sqrt2\right)^x + \left(\sqrt3\right)^x = \left(\sqrt{13}\right)^{\frac{x}{2}}$ then the number of values of $x$ is?
It is an exponent topic question tried squaring method but couldn't get to the right answer. The answer is $1$.
AI: We have $$(4/13)^{x/4}+(9/13)^{x/14}=1$$
Now clearly $x/4=1$ is a solution
As the left hand side is monotonic decreasing, there can be no more solution
This approach can be applied for
$$(\cos^2t)^n+(\sin^2t)^n=1$$ |
H: Notation for a k-partite graph
I am reading Graph Theory by Bondy and Murty. I always see the notation for a complete $k$-partite graph, but what is the notation for a $k$-partite graph? For example, my partite sets are of order $2$, $3$, and $4$, and it is not a complete $k$-partite graph. How can I write this graph?
AI: I think you can just say "$3$-partite graph" or "tripartite graph", omitting the word "complete". "Bipartite graph" is certainly standard.
I assume that when you say
my partite sets are of order $2$, $3$, and $4$
you mean a graph with $9$ vertices, $2$ red, $3$ blue and $4$ yellow, all of whose edges connect vertices of different colors. |
H: How do you find the Taylor series of an indefinite integral?
I am given the following problem :-
I know that I have to first find the Taylor series of the polynomial and then integrate each term, however I am having trouble finding the Taylor series before integrating because the derivative of $\frac{(e^t-e)}{(t-1)}\to \frac{e^t(t-1)-1(e^t-e)}{(t-1)^2}$ at $t=1$ does not exist $\frac{0}{0}$. How do I go about finding the Taylor series of $\frac{(e^t-e)}{(t-1)}?$
Thanks for any and all help!
AI: Hint:
Set $t=1+u\enspace(u\to 0)\,$ first. The integrand becomes
$$\frac{\mathrm e^t-\mathrm e}{t-1}=\mathrm e\,\frac{\mathrm e^u-1}u.$$
Can you take it from here? |
H: How to compute this integral over a small ball?
$$\int_{{\sqrt{x^2+y^2}}<1}x^2dxdy$$
I only know that $$z=(x, y)^T$$ is a small, directional vector.
AI: Not sure why $z$ would help here, but one way would be to do this directly, i.e.
$$
\int_{x=-1}^{x=1} \int_{y = -\sqrt{1-x^2}}^{y = \sqrt{1-x^2}} x^2 dy dx
= \int_{x=-1}^{x=1}
\left[ \int_{y = -\sqrt{1-x^2}}^{y = \sqrt{1-x^2}} dy \right]
x^2 dx
= 2 \int_{-1}^1 x^2 \sqrt{1-x^2} dx
$$
It may be easier to switch to polar
$$
\int_0^1 \int_0^{2\pi} \left(r^2 \cos^2 \theta\right) rdr d\theta
= \left(\int_0^1 r^3 dr\right) \left( \int_0^{2\pi} \cos^2 \theta d\theta \right)
$$
Can you finish? |
H: If $x^2+y^2+xy=1$ then find minimum of $x^3y+xy^3+4$
If $x,y \in \mathbb{R}$ and $x^2+y^2+xy=1$ then find the minimum value of $x^3y+xy^3+4$
My Attempt:
$x^3y+xy^3+4$
$\Rightarrow xy(x^2+y^2)+4$
$\Rightarrow xy(1-xy)+4$ (from first equation)
$\Rightarrow xy-(xy)^2+4 =f(x)$
For minimum value, $\frac{df(x)}{dx}=0$.
$\Rightarrow \frac{df(x)}{dx}=(y-2xy²) + \frac{dy}{dx}(x-2yx²)=0$
How should I proceed from here?
AI: By AM-GM, $x^2+y^2=|x|^2+|y|^2\ge2|xy|$ so $r^2=x^2+y^2$ has extrema $\tfrac23,\,2$ respectively achieved by$$2xy=x^2+y^2\implies1=(1+1/2)(x^2+y^2)$$and$$-2xy=x^2+y^2\implies1=(1-1/2)(x^2+y^2).$$Note the minimum of $r^2(1-r^2)$ on $[\tfrac23,\,2]$ occurs at $r^2=2$. In particular, $x=-y=1$ minimizes $r^2(1-r^2)+4$ as $2\times-1+4=2$. |
H: Counting the number of integers with given restrictions
Question: Consider the numbers $1$ through $99,999$ in their ordinary decimal representations. How many contain exactly one of each of the digits $2, 3, 4, 5$?
Answer: $720$.
Attempt at deriving the answer:
We have two cases: four digit numbers and five digit numbers.
Five digit numbers:
Let $x \in \{2, 3, 4, 5\}$. If the first position in a five digit number is not $x$, then there are $5$ possibilities for this position as there are four values for $x$ and $0$ is inadmissable. The rest of the four positions will have the various permutations of four values of $x$. There are $5 \times 4!$ such numbers. If the non-$x$ value is in the second position, then there are $4$ ways to choose an $x$-value for the first position, $6$ integers for the second position and $3!$ permutations for the rest of the positions. There are $4\times 6 \times 3!$ such numbers. If the non-$x$ value is in the third position, then there are $\binom 42$ ways to choose two $x$-values, $2!$ ways to permute them and $6$ integers for the third position meaning there are $6 \times 2 \times 6$ such numbers. When the non-$x$ value is in the fourth position, there are $4 \times 3! \times 6$ such numbers. Finally, if non-$x$ is in the fifth position, there are $4! \times 6$ such numbers.
Four digit numbers:
We just need to permute the number $2345$. There are $4!$ such permutations.
Thus the number of numbers with the given restrictions is $5\times 4! + 4\times 3!\times 6 + 2\times 6 \times 6 + 4 \times 3! \times 6 + 6 \times 4! + 4! = 648$.
What did I forget to take into account? Thanks.
AI: To answer the question you asked: In the case where the non-$x$ value is in the third position, you missed permuting the fourth and fifth digits of the number, so that term should be $6\cdot 2\cdot 6\cdot 2$ (rather than $6\cdot 2\cdot 6$). |
H: Cannot solve linear system of equations
It would be nice if somebody could find my mistake for the following linear system of equations:
$$
\left\{\begin{array}{rcrcrcr}
-2x & - & 4y & - & z & = & -21
\\
-3x & + & y & + & 2z & = & -14
\\
x & - & 2y & - & z & = & a
\end{array}\right.
$$
It is solvable for all $a \in \Bbb R$.
I know that $y= 3- a$ , however when I try to solve this system I get a different result.
The first thing I did was to subtract the first equation from the third so that I can eliminate $z$. The third equation now would be $$3x+2y=a+21.$$
After that I wanted to eradicate the $z$ in the second equation so I multiplied the first equation by two and added it to the second equation. My second equation now is: $$-7x-7y=-56.$$
I then went on to eliminate the $y$ in the third equation so I multiplied the second equation by 3 and the third equation by 7 and added the second equation to the third. The third equation would now be: $$-7y=7a-33$$ which is equivalent to $$y=-a+\frac{33}{7}$$ which is unequal to the actual solution. Where is my mistake?
AI: Adding 3 times the second and 7 times the third equation produces $-7y = 7a - 168 + 147 = 7a - 21$. |
H: Probability of bit strings
Suppose you pick a bit string of length $10$. Find the probability that the bit string has exactly two $1$'s, given that the string begins with a $1$.
Can someone please explain to me how to do it?
AI: Since you know that the first bit is $1$, you just want to find the probability that exactly one of the remaining $9$ digits is $1$. The total number of possible $9$-bit sequences is $2^9$. Of these only $9$ have exactly a single $1$, one for each possible position of the $1$. So the probability is $\frac{9}{2^9}$. |
H: How to efficiently sample edges from a graph in relation to its spanning tree
Consider a connected, unweighted, undirected graph $G$. Let $m$ be the number of edges and $n$ be the number of nodes.
Now consider the following random process. First sample a uniformly random spanning tree of $G$ and then pick an edge from this spanning tree uniformly at random. Our process returns the edge.
If I want to sample many edges from $G$ from the probability distribution implied by this process, is there a more efficient (in terms of computational complexity) method than sampling a new random spanning tree each time?
AI: While the other answer is correct, it requires the computation of $|E| + 1$ many determinants. There is a faster route when $|E|$ is large. The first thing to note is Kirchoff's theorem which states that if $T$ is a uniform spanning tree then
$$P(e \in T) = \mathscr{R}(e_- \leftrightarrow e_+)$$
where $e = \{e_-, e_+\}$ and $\mathscr{R}(a \leftrightarrow b)$ is the effective resistance between $a$ and $b$ when each edge is given resistance $1$. This implies that the probability an edge is sampled in your process is $$\mathscr{R}(e_- \leftrightarrow e_+)/(|V| - 1).$$
Thus we only need to compute the effective resistance.
If we let $L$ denote the graph Laplacian and $L^+$ to be its Moore-Penrose pseudoinverse, then
$$\mathscr{R}(a \leftrightarrow b) = (L^+)_{aa} + (L^+)_{bb} - 2 (L^+)_{ab}. $$
(See this master's thesis for some nice discussion and references.)
Thus, the only computational overhead for computing the marginals is computing a single psuedoinverse. Depending on how large $|E|$ is, this may be faster than computing $|E|$ many determinants.
EDIT: some discussion on complexity
The Pseudoinverse of an $n \times n$ matrix can be done in $O(n^3)$ time. So computing $L^+$ takes $O(|V|^3)$ time. We have to compute this for $|E|$ many edges, so the above computes all marginals in $O(|E| |V|^3)$ time. Conversely, a determinant can be done in, say, $O(n^{2.3})$ time. So the other answer has complexity $O(|E|^2 |V|^{2.3}).$ Since $G$ is connected, $|E| \geq |V|-1$ and so this algorithm is always faster (asymptotically, at least). |
H: Mixed partials on the diagonal
Let $I\subseteq\mathbb{R}$ be an open interval and $\varphi\in C^2(I^2;\mathbb{R})$ be twice continuously-differentiable on $I^2:=I\times I$.
We call $\varphi$ separable if it can be written $\varphi\equiv\varphi(x,y) = \alpha(x) + \beta(y)$ for some $\alpha,\beta: I\rightarrow\mathbb{R}$, and we call $\varphi$ non-separable if for each open $\mathcal{O}\subseteq I^2$ the restriction $\varphi_{\mathcal{O}}:=\left.\varphi\right|_{\mathcal{O}}$ is not separable.
Remark: For any $U\subseteq I^2$ open and connected, the function $\varphi_U$ is separable iff $\left.(\partial_x\partial_y\varphi)\right|_U = 0$; consequently $\varphi$ is non-separable if and only if the set $\{z\in I^2\,|\,\partial_x\partial_y\varphi(z)\neq 0\}$ is dense in $I^2$.
Question: Is it true that: If $\varphi$ is non-separable, then $\{x\in I \,|\,\partial_x\partial_y\varphi(x,x)\neq 0\}$ is dense in $I$?
AI: No!
Example: Take $I=(0,1)$, and
$$
\varphi(x,y)=
\begin{cases}
0 & x=y\\
\exp(-1/(x-y)^2) & x<y\\
\exp(-2/(x-y)^2) & x>y.
\end{cases}
$$
By construction, $\varphi$ is smooth with derivatives of all order vanishing on the diagonal. But $\varphi$ isn't separable. Indeed, the mixed partial does not vanish anywhere else. |
H: Improper integral of periodic function
I was given the following question:
Find all continuous periodic functions $f(x)$ for which the integral $ \int^\infty_0f(x)dx $ converges.
Now, I have a feeling this is only the function $ f(x) = 0$ , yet I have a problem formally proving it.
Any help will be appreciated!
AI: Hint: Suppose $f$ is periodic with period $P$. Find an equation relating $\int_0^{t+nP} f(x)\; dx$ to $\int_0^t f(x)\; dx$ and $\int_0^P f(x)\; dx$. What does that tell you $\int_0^t f(x)\; dx$ must be? |
H: Can I use induction with increments higher than 1?
Say I want to prove:
$a$ is odd $<=>$ $a^2$ is odd
However, instead of proving this in both directions, I want to show that this statement is true for all odd numbers.
Can I use induction with increments of 2 to get only the odd numbers? Does this require additional proof?
Is this proof correct?
Base: for $k=1$, $1$ is uneven and $1^2=1$ is odd.
Step: Assume the claim is true for some odd $k$, prove for $k+2$
Assumption: $k$ and $k^2$ are odd.
Proof:
1. $(k+2)^2 = k^2 + 4k + 4 = k^2 + 4(k+4)$
Since $k^2$ is odd, and $4(k+4)$ is even, then (odd+ even) = odd.
Therefore $(k+2)^2$ is odd.
2. $k+2$ = odd + even = odd
Therefore, $k+2$ is odd.
We have shown that $k+2$ is odd and $(k+2)^2$ is odd, completing the proof for all odd numbers.
AI: The axiom of induction doesn't let you do that directly, no.
But you can do the following: you make the claim (call it $P(k)$) that for every nonnegative integer $k$, $(2k+1)^2$ is odd.
$P(0)$ says that $1$ is odd, which is true.
If you suppose $P(k)$ is true, then you can show $P(k+1)$ is true using exactly your argument.
And then induction tells you that $P(k)$ is true for all integers $k \ge 0$.
====
You can also prove, using the axiom of induction, a little theorem, namely that...
If $H$ is a set of odd numbers containing $1$, and with the property that whenever $n \in H$, you also have $n+2 \in H$, then $H$ contains all positive odd numbers.
You can then apply this little theorem in doing the proof the way you wrote it in the first place.
In practice, any working mathematician would have no hesitation about writing your kind of proof, because the "little theorem" would be pretty obvious. But as you're learning, the obligations are stronger than when you've gained some expertise. :( |
H: An Application Kolmogorov's Three Series Theorem
I want to prove the following question, which is found in this practice exam:
My attempt so far is as follows - I just can't show that the $\sum E(Y_i)$ converges.
AI: Write $$0 = E X_i = E[ X_i 1_{|X_i| \leq 1}] + E[ X_i 1_{|X_i| > 1}] = E[Y_i] + E[ X_i 1_{|X_i| > 1}]. $$
Thus $$|E[Y_i]| \leq E[|X_i| 1_{|X_i| > 1}] \leq E[\psi(X_i)].$$ |
H: Determine $\sqrt{1+50\cdot51\cdot52\cdot53}$ without a calculator?
I've tried a lot of things but failed to do it, I've calculated the result inside the square root which is $7027801$ using substitution and factoring but $\sqrt{7027801}$ isn't possible to simplify.
AI: I used the following remarkable identity: $$(a+b)^2 = a^2 + 2ab+b^2.$$
So $1+50\cdot 51 \cdot 52 \cdot 53=1 + 50\cdot53 \cdot (50+1) \cdot (53-1) = 1+ (50\cdot 53)^2 + 2 \cdot 50\cdot53 = (1+50\cdot53)^2$.
And hence the answer is $50\cdot 53 +1 = 2651$. |
H: toom-cook algorithm matrix G
For this toom-cook algorithm at https://arxiv.org/pdf/1803.10986v1.pdf#page=6 , how do I get the value 4/2 in the matrix G ?
AI: We have
$$G = \begin{bmatrix} x_0^0 N^0 & x_0^1 N^0 & x_0^2 N^0 \\ x_1^0 N^1 & x_1^1 N^1 & x_1^2 N^1 \\ x_2^0 N^2 & x_2^1 N^2 & x_2^2 N^2 \\ x_3^0 N^3 & x_3^1 N^3 & x_3^2 N^3 \end{bmatrix} = \begin{bmatrix} (1)( -\dfrac{1}{6}) & (1)(-\dfrac{1}{6}) & (1)(-\dfrac{1}{6})\\ (1)(\dfrac{1}{2}) & (2)(\dfrac{1}{2}))& (4)(\dfrac{1}{2})) \\ (1)(-\dfrac{1}{2}) & (3)(-\dfrac{1}{2}) & (9)(-\dfrac{1}{2}) \\ (1)(\dfrac{1}{6}) & (4)(\dfrac{1}{6}) & (16)(\dfrac{1}{6}) \end{bmatrix}$$ |
H: How did the variance get calculated?
The Elm Tree golf course in Cortland, NY is a par 70 layout with 3 par
fives, 5 par threes, and 10 par fours. Find the mean and variance of par on this
course.
Mean was calculated as follows: Mean = 70/18 = 3.8888
Variance was found to be: second moment = (75 + 160 + 45)/18 = 280/18 = 15.555
variance = (5040 − 4900)/324 = 140/324 = 0.432
I am not sure what happened from the second moment part up to the calculation of variance. I never calculated variance like this. Can someone elaborate how second moment was found and how it led to the calculation of variance?
AI: The second moment is $$\mathbb E(x^2)=\frac1{n}\sum_{i=1}^m n_i\cdot x_i^2$$
with $m=3, n=18, n_1=3, n_2=5, n_3=10, x_1=5, x_2=3, x_3=4$.
And we have the relation $\sum\limits_{i=1}^3 n_i=n$
And then the variance is $Var(x)=\mathbb E(x^2)-[\mathbb E(x)]^2$ |
H: Banach space of continuous and discontinuous functions on R
The set $C(\mathbb{R})$ of bounded continuous functions on $\mathbb{R}$ is a Banach space when equipped with the sup norm. In my understanding, it just follows from the fact that a Cauchy sequence of continuous functions converges uniformly to a continuous function.
How about the set, which I will denote $C_{\rm{d}}(\mathbb{R})$, of bounded functions, which are continuous except at $x=0$, where a jump discontinuity is allowed, i.e. $\displaystyle{\lim_{x\rightarrow 0^\pm}f(x)=f^\pm}$ both exist. Is that a Banach space under the sup norm?
It seems like yes, since I can apply the Cauchy argument to both intervals $(-\infty,0]$ and $[0,\infty)$ and conclude that a Cauchy sequence on $C_{\rm{d}}(\mathbb{R})$ will uniformly limit to a function, which is continuous on both intervals individually.
Am I right or completely wrong? If I am right, is there a name for such a space?
AI: You're correct. Try showing that $C_d(\Bbb R)$ is isomorphic(as a normed linear space) to $C((-\infty,0]) \oplus C([0,\infty))$. Since the spaces $C((-\infty,0])$ and $C([0,\infty))$ are individually Banach spaces, so is their direct sum. As far as I know, there is no special name for this space. |
H: Is $(\mathbb{Q},+,\cdot)$ a divisible semifield?
I know that $(\mathbb{Q},+,\cdot)$ is a semifield. But I would like to know that whether it is divisible with respect to the usual addition and multiplication. Please any idea or help? Thanks.
AI: The additive group $(\mathbb{Q},+)$ is divisible: $q \in \mathbb Q, n \in \mathbb N^* \implies q=ny$ for $y=q/n \in \mathbb Q$.
The multiplicative group $(\mathbb{Q}^*,\cdot)$ is not divisible: $2 \in \mathbb Q$, but there is no $y \in \mathbb Q$ such that $y^2=2$. |
H: Matrix multiplied by its pseudo-inverse doesn't give the identity matrix. Why?
Using Matlab, I randomly generate matrix $A \in \Bbb C^{2 \times 1}$ and compute its pseudo-inverse $A^{+}$. I notice that $AA^{+} \neq I$, and yet $\mbox{Tr}(AA^{+}) = 1$.
For other sizes it seems like the trace is equal to the smaller dimension of $A$. I couldn't find this property explained. Could anyone help me understand these two facts?
AI: Suppose you generate $A\in\mathbb{C}^{m\times n}$ randomly, as you specified. Then $\text{rank}(A)=\min\{m,n\}$ with high probability (depending on the probability distribution you used, of course). Now, if $A$ has rank $m$, then $AA^*\in\mathbb{C}^{m\times m}$ has rank $m$ and is invertible. Therefore, the Moore-Penrose pseudoinverse $A^\dagger = A^*(AA^*)^{-1}$ can be used to actually "obtain" $I$ through right-inversion, i.e., $AA^\dagger=I_m$. Similarly, if $\text{rank}(A)=n$, then $A^*A\in\mathbb{C}^{n\times n}$ is invertible, and $A^\dagger = (A^*A)^{-1}A^*$ is a left inverse of $A$; $A^\dagger A = I_n$.
Note that if $\text{rank}(A)=n<m$, then for all $B\in\mathbb{C}^{n\times m}$ it holds that
$$\text{rank}(AB)\le \min\{\text{rank}(A),\text{rank}(B)\}=\min\{n,\text{rank}(B)\}\le n < m$$
and therefore the matrix $AB\in\mathbb{C}^{m\times m}$ cannot possibly equal the identity (which has rank $m$). A similar remark holds in the case that $\text{rank}(A)=m<n$, namely, that there is no left inverse.
Therefore, for your example of $A\in\mathbb{C}^{2\times 1}$, you cannot expect there to be a right inverse of $A$, since $AB$ will be of rank-1 for all $B\in\mathbb{C}^{1\times 2}$. However, you should be able to find a left inverse (a vector that has inner product equal to one with $A$).
As per your question regarding trace: are you using MATLAB's pinv function? If so, it will compute whichever Moore-Penrose inverse "makes sense". In other words, for your matrix $A\in\mathbb{C}^{2\times 1}$, the MATLAB function will compute a left pseudo-inverse $A^\dagger\in\mathbb{C}^{1\times 2}$. In this case, $A^\dagger A$ should equal to $I_1=1$. Furthermore, by the cyclic property of the trace, using the left inverse on the "wrong side" (the right), you will still get a trace of one, since $\text{tr}(AA^\dagger)=\text{tr}(A^\dagger A) = \text{tr}(1)=1$.
More generally, suppose $A\in\mathbb{C}^{m\times n}$ has rank $n<m$ (and therefore is left-invertible). Then the left Moore-Penrose pseudoinverse $A^\dagger=(A^*A)^{-1}A^*$ when used on the right will give
$$\text{tr}(AA^\dagger) = \text{tr}(A^\dagger A) = \text{tr}((A^*A)^{-1}A^*A)=\text{tr}(I_n) = n$$
as per your observation. |
H: Inverse of cumulative distribution function
Let $F(x)$ is the cumulative distribution function and $P(x)$ is the (given) probability distribution function and $X$ is a random variable.
Can anybody please intuitively explain,
Why can the inverse of the CDF give us the random variable $X$?
Why can't we find the random variable $X$ from the PDF?
I wonder what $F$ and $F^{-1}$ get and return?
Thanks.
AI: What do you mean 'find the rv X from the PDF'? If you know pdf, you 'know' rv (whatever you mean by that). The main property of CDF is that it is a non-decreasing function. Specifically, for continuous rvs, you can find $x_0$, s.t.:
$$
p=P(X \leq x_0)=F_{X}(x_0) \Leftrightarrow F^{-1}_{X}(p) = x_0
$$
also referred to as a percentile. |
H: Unclear limit in showing that $\ell^2$ spaces are complete
I'm working my way through the book Introduction to Hilbert Spaces with Applications and trying to follow the early example (1.4.6) showing that $\ell ^2$ spaces are complete (Cauchy sequences $(a_n) \in \ell ^2$ have their converge in $\ell ^2$). At this point only basic results of normed vector spaces are shown, neither the Dominated nor Monotone convergence theorems are presented.
They start by declaring the Cauchy sequence $(a_n) \in \ell ^2$ with elements $a_n = (\alpha_{n1}, \alpha_{n2}, ...)$. By the definition of Cauchy sequences and the $\ell ^2$ norm they get that there exists $N$ such that
\begin{equation}
\sum_{k=1}^{\infty}|\alpha_{mk} - \alpha_{nk}|^2 < \epsilon^2 \tag{1}
\end{equation}
for $m, n > N$. They conclude that this means that for each $k$, $|\alpha_{mk} - \alpha_{nk}| < \epsilon$ so $(\alpha_{nk})$ is a Cauchy sequence in $\mathbb{C}$ and therefore converges to a limit $\alpha_k \in \mathbb{C}$. Now to the part that is unclear to me. They state that from (1), by letting $m \rightarrow \infty$, we get
\begin{equation}
\sum_{k=1}^{\infty}|\alpha_{k} - \alpha_{nk}|^2 \le \epsilon^2 \tag{2}
\end{equation}
for $n > N$, without mentioning how. It seems like this limit is supposed to be obvious, but the only methods I can figure out or find rely on being able to state that ($\stackrel{!}{=}$)
\begin{equation}
\lim_{m \rightarrow \infty} \sum_{k=1}^{\infty}|\alpha_{k} - \alpha_{nk}|^2 = \lim_{m \rightarrow \infty} \lim_{K \rightarrow \infty} \sum_{k=1}^{K}|\alpha_{k} - \alpha_{nk}|^2 \stackrel{!}{=} \lim_{K \rightarrow \infty} \sum_{k=1}^{K}\lim_{m \rightarrow \infty}|\alpha_{mk} - \alpha_{nk}|^2 = \sum_{k=1}^{\infty}|\alpha_{k} - \alpha_{nk}|^2
\end{equation}
using later results like the aforementioned convergence theorems. Is there some better method I'm completely missing here?
AI: Do it in a few steps. Take any constant $n>N$ and $K\in\mathbb{N}$. For all $m>N$ we have:
$$
\sum_{k=1}^K |\alpha_{mk}-\alpha_{nk}|^2\leq\epsilon^2
$$
Now, this is a finite sum so there is no problem to use arithmetic of limits. By taking $m\to\infty$ we get the inequality $\sum_{k=1}^K |\alpha_k-\alpha_{nk}|^2\leq\epsilon^2$. And since this is true for all $K\in\mathbb{N}$ (since $K$ was an arbitrary natural number), by taking $K\to\infty$ we get $\sum_{k=1}^\infty |\alpha_k-\alpha_{nk}|^2\leq\epsilon^2$.
By the way, it is important to write non strict inequalities. After all, strict inequalities are not always preserved when taking a limit. |
H: Prove that for all $x, y>0$, $\ln \left(\frac{x}{y}\right) \geqslant 4 \frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}$
Prove that for all $x, y>0$,
$\ln \left(\frac{x}{y}\right) \geqslant 4 \frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}$
Is there any role for
mean value theorem in the proof?,
Can we use the fact
$\ln (x)<\sqrt{x},~\forall x>0$
AI: Use the substitution as mentioned in the comments. Let $u=\sqrt{x/y}$.
Then we have to prove that $\ln (u)-2*\frac{u-1}{u+1} \geq 0$.
Let $p(u)=\ln (u)-2*\frac{u-1}{u+1}$
Then $$p(u+\delta)-p(u)=\ln \left(\frac{u+\delta}{u}\right)+\frac{2\delta}{(u+\delta+1)(u+1)} \geq 0$$. Hence $p(u)$ is an increasing function and hence greater than $0$. Please note that the condition is $u \geq 1$ or $x \geq y$. |
H: How do we further simplify this expression involving a complex number?
I will state the problem here:
Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Compute
$\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3}.$
W is obviously a complex number. I've tried arranging and rearranging the terms in a way to further simplify the expression, but I'm having trouble getting anywhere with it. Anyone have ideas on how to proceed and solve the problem?
AI: The first and last terms are equal, as can be seen by multiplying the latter's numerator and denominator by $\omega^2$. A similar treatment, with a factor of $\omega$, shows the middle terms are equal too. So we want to double$$\frac{\omega}{1+\omega^2}+\frac{\omega^3}{1+\omega}=\frac{\omega+\omega^2+\omega^3+\omega^5}{(1+\omega)(1+\omega^2)}=1,$$giving $2$ as the final answer. We actually don't need to exclude $\omega=1$ for this to work. |
H: How to account for increasing probability in odds of winning a competition when each round there is a random single decrease in the non-winner pool?
My Example:
Players 1 through n are playing a game and in each round there is one winner and after the winner is determined, one of the non-winners gets randomly taken out of the total players pool (i.e. round 1 = n total players, round 2 = n-1, round 3 = n-2, etc).
I am trying to show that if we look at a large sample of these games, a strong player (potentially the winner) should typically have a higher amount of competition round wins at the end of each game. However, I do not understand how to go about adjusting for the increased probability that a player might win each round when a non-winner is eliminated [i.e. round 1 = $\frac{1}{n}$ % chance of winning that round, round 2 = $\frac{1}{n-1}$, round 3 = $\frac{1}{n-2}$, etc]. This is assuming of course that each player has an equal chance of winning each round, which is obviously what i'm trying to get around.
Does anyone know how I might account for this increased probability and possibly how to also take into consideration as the game goes on that a stronger player who has won in the first round (or other future rounds) is given a higher probability than the rest to win the next rounds?
Basically, how do I calculate the strength of a player and add it into a predictive model on how they will do based on past games with other players data.
Thank you for any help in advance.
AI: There are multiple ways to solve this, but you can give each player the following score :
$\mathbf{S}(x) = \underset{\{i\ :\ \text{x won round i}\}}{\sum \frac{1}{p_i}}=\underset{\{i\ :\ \text{x won round i}\}}{\sum (n- i)}$
With $p_i = \frac{1}{n -i}$ being the probability of winning the round $i$. So the earlier rounds count for more than the later ones.
If you want the effect of the round won to be smaller, you can take
$\mathbf{S}(x) = \underset{\{i\ :\ \text{x won round i}\}}{\sum 1 + \frac{1}{k}\frac{1}{p_i}}= \text{number of wins} + \underset{\{i\ :\ \text{x won round i}\}}{\sum \frac{1}{k}(n- i)}$
So you are still counting all the wins of the player, and then adding extra points for winning in the earlier rounds. |
H: If a.i=4, then ,what is the value of (axj).(2j-3k) , where a is a vector
This is a question I saw in a question paper of a competitive exam but I was unable to solve it. Can anyone please assist me with any sort of hint to solve this problem and any type of explanation if needed in the hint. Any help will be highly appreciated.
AI: $a\cdot i=4$ means that $a=4i+sj+tk$ where $s$ and $t$ are real, and so
$a\times j=4k-ti$. Now take the dot product with $2j-3k$; I think that pesky $t$
will obligingly vanish. |
H: Column-sum and row-sum for fat matrices
For an $m \times n$ fat matrix ($m<n$)
$$A = \begin{pmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n}\\ a_{21} & a_{22} & \ddots & \cdots & a_{2n} \\ a_{31} & \cdots & \ddots& \ddots & \vdots \\ \vdots & \cdots & \cdots & \ddots & \vdots \\ a_{m1} & \cdots & \cdots & \cdots & a_{mn} \end{pmatrix}$$
with $a_{ij} \geq 0$ and at least one positive entry in every column, why is the smallest column sum always smaller than the largest row sum?
AI: The $n$ column sums add up to $\displaystyle\sum_{i = 1}^{m}\sum_{j = 1}^{n}a_{ij}$, so the smallest column sum is at most the average column sum, which is $\dfrac{1}{n}\displaystyle\sum_{i = 1}^{m}\sum_{j = 1}^{n}a_{ij}$. The $m$ row sums add up to $\displaystyle\sum_{i = 1}^{m}\sum_{j = 1}^{n}a_{ij}$, so the largest column sum is at least the average row sum, which is $\dfrac{1}{m}\displaystyle\sum_{i = 1}^{m}\sum_{j = 1}^{n}a_{ij}$.
Hence, the smallest column sum is less than or equal to $\dfrac{1}{n}\displaystyle\sum_{i = 1}^{m}\sum_{j = 1}^{n}a_{ij}$, which is less than $\dfrac{1}{m}\displaystyle\sum_{i = 1}^{m}\sum_{j = 1}^{n}a_{ij}$, which is less than or equal to the largest row sum. |
H: probability question : what are the chances that at least 1 out of 150 passengers are infected
There are $10{,}000$ corona patients in a city of $7{,}500{,}000$ people.
$150$ people fly on a plane, what are the chances that at least $1$ person is infected?
I solved it using the below approach
select 1 infected patient / select 150 out of total population.
$$\dfrac{{10000 \choose 1}}{{7500000 \choose 150}}$$ which is approximately equals 0,
My friend suggested a different solution and came to a value of 19.3% probability
His approach:
pct of patient infected is $10000/7500000 = 0.0013$
and pct of 1 person infected = 1 - no person infected
= $1-( (1 - 0.0013)^{150} )=0.19$
What is the correct answer? and why is any of the approach incorrect?
AI: Neither.
The probability of at least one person is infected is one minus the probability that all are healthy, which is
$$1-\tfrac{{7500000-10000 \choose 150}}{{7500000 \choose 150}}=0.18$$
Your friend assumes people are sick independently. This would have been true if the wording was "each person is sick with probability 0.0013" or something similar. As you phrased it, there are exactly 10,000 people who are sick in the city. Since the sample (150) is small relative to the population, there is no much difference.
Your approach is wrong as you only consider 1 sick person and not at least one. Moreover, you don't account for the number of options to choose the other 149 passengers out of the healthy population.
Finally, please use MathJax to typset your question. |
H: $(0)$ is the only minimal prime of $k[x,y]$
We define a prime ideal to be minimal if it is minimal with respect to inclusion. Given this, why $(0)$ is the only minimal prime of $k[x,y]$?, where $k$ is any field.
In particular, is there an easy way to see this? If I pick a nonzero prime ideal of $k[x,y]$, how do I know no other prime ideal is contained in it?
AI: If I pick a nonzero prime ideal of k[x,y], how do I know no other prime ideal is contained in it?
I think you have this backwards. A minimal prime is a prime $\mathfrak{p}$ such that, for all primes $\mathfrak{q}$, if $\mathfrak{q} \subseteq \mathfrak{p}$, then $\mathfrak{p} = \mathfrak{q}$. Thus, a nonzero prime ideal of $k[x,y]$ which contains no other prime ideal inside it would be minimal, and you need to show that there are no nonzero minimal primes. In other words, you need to show that if $\mathfrak{p}$ is a nonzero prime ideal of $k[x,y]$, then $\mathfrak{p}$ contains a prime ideal (besides itself) – but this is easy because $(0)$ is a prime ideal of $k[x,y]$!
In any case, $(0)$ is contained in every ideal of any ring (in other words, it is the minimal ideal of any ring), so if $(0)$ is a prime ideal of any ring, it must be the unique minimal prime of that ring
Edit: Just to solidify this idea, make sure the following makes sense to you. For any commutative ring $R$, the following are equivalent:
$R$ is a domain
$(0)$ is a prime ideal of $R$
$(0)$ is a minimal prime ideal of $R$
$(0)$ is the unique minimal prime ideal of $R$ |
H: How to find $a$, $b$, $c$ such that $P(x)=ax^3+bx^2+cx$ and $P\left(x\right)-P\left(x-1\right)=x^2$
I'm trying to find $a$, $b$ and $c$ such that $P(x)=ax^3+bx^2+cx$ and $P\left(x\right)-P\left(x-1\right)=x^2$.
After expanding the binomial in $P(x-1)$, I end up getting
$3ax^2-3ax+2bx+a-b=x^2$. What next? Using $3a = 1$ doesn't work.
AI: Sure it does work. As you let $P(x)=ax^3+bx^2+cx$, we have $P(x)-P(x-1)=3ax^2-(3a-2b)x+a-b+c$
Now just compare the coefficients. That is, $3a=1$, $3a-2b=0$, $a-b+c=0$ which after solving gives $a=\frac13 , b=\frac12 , c=\frac16$. |
H: If $X$ is homeomorphic to subset of set $Y$ and $Y$ is homeomorphic to subset of $X$ then are $X$ and $Y$ homeomorphic?
Let $(X, \tau_1)$ and $(Y, \tau_2)$ be topological spaces. Is it true that if $X$ is homeomoprhic to a subset of $Y$ and $Y$ is homeomorphic to a subset of $X$ then $X$ and $Y$ are homeomorphic spaces?
I'm trying to find a counter-example cause I don't believe that the statement is correct, however I'm struggling since I don't know that many homeomorphisms. An idea I had is to prove that a subset of $\mathbb{R}$ is homeomorphic to a subset of $\mathbb{R^2}$ and the other way around but $\mathbb{R}$ and $\mathbb{R^2}$ are not homeomorphic.
AI: $\;[0.2,\,0.7]\subseteq (0,1)=:X\;$ is homemomorphic to $\;Y:=[0,1]\;$ , and $\;(0.2,\,0.7)\subseteq Y\;$ is homeomorphic to $\;X\;$ , but ... |
H: Proving a open and bounded subset of $\mathbb{R}$ is a union of disjoint open intervals
Below is the proof I have written up. I would love to get some feedback on it as I have not been able to readily spot any logical holes, but that might be because I'm missing something. What's more, I would love to know if there is any way I can achieve more brevity in my argument!
Statement: An open and bounded subset of $\mathbb{R}$ is a union of disjoint open intervals.
Proof:
Let $S$ denote an open and bounded subset of $\mathbb{R}$. The empty set is both bounded and open, yet can impossibly be written as a union of disjoint open intervals, since each such is non-empty. So the statement is not true for $\varnothing$.
So suppose then $S \neq \varnothing$. Our $S$ is bounded and therefore bounded above and below. This furnishes $\sup S$ and $\inf S$. Neither of these are contained in $S$, since otherwise greater/lower elements than these would lie in $S$ and thus contradict their property of being extremities of $S$.
Take now some $x \in S$. Since $S$ is open, there is some positive real number $r(x)$ such that the open interval $\big(x - r(x), x + r(x)\big) \subset S$. In particular both halves of this open interval lie in $S$, which we state as $\big( x - r(x), x] \subset S$ and $[x, x + r(x)\big) \subset S$.
We have found an open interval containing $x$ and would now like to investigate how big we can make it before it leaves $S$.
Let's begin looking for its left extremity. The set $L_x = \{ r \in \mathbb{R} : (r, x] \subset S \}$ is not empty, owing to $\big( x - r(x), x] \subset S$. And if $r$ is a real number smaller than $\inf S$, then $(r, x] \not \subset S$. Therefore each element in $L_x$ is bounded below by $\inf S$. This furnishes $\inf L_x$.
The real number $\inf L_x$ has two properties:
$\inf L_x \in L_x$
If not, then $( \inf L_x, x] \not \subset S$. So there is a $t \in (\inf L_x, x]$ such that $ t \not\in S$. But since $\inf L_x < t$, there is some $\alpha \in L_x$ such that $ \alpha < t$. All together: $\inf L_x < \alpha < t \leq x$, so for that reason $t \in (\alpha, x] \subset S$. But this contradicts $t \not \in S$.
$\inf L_x \not \in S$
If not, then there is some real $\epsilon > 0$ such that $(\inf L_x - \epsilon, \inf L_x] \subset S$ (since $S$ is open). By property 1, $[\inf L_x, x] \subset S$ and therefore $[\inf L_x - \frac {\epsilon}2, x] \subset S$. But then $\inf L_x - \frac {\epsilon}2 \in L_x$, which contradicts the greatest lower bound property of $\inf L_x$.
So we have found the left extremity and conclude $(\inf L_x, x] \subset S$.
For the right extremity, the argument is analogous. We get $[x, \sup R_x) \subset S$, where $R_x = \{ r \in \mathbb{R} : [x, r) \subset S \}$.
Thus the largest interval containing $x$ is $I_x = (\inf L_x, \sup R_x) \subset S$.
We now know how to create a maximally large interval around any point in $S$. What remains to be checked is if these intervals are disjoint. If $I_x$ contains all elements of $S$, there is nothing left to prove. So suppose there is some other $y \in S$ such that $ y\not \in I_x$.
Then, without loss of generality, say $y > \sup R_x$. Then $L_y = \{r \in \mathbb{R} : (r, y] \subset S\}$ is bounded below by $\sup R_x$. From this it follows $\inf L_y \geq \sup R_x$. Thus the interval $I_y = (\inf L_y, \sup L_y)$ has no elements from $I_x$.
We can continue generating disjoint intervals in this fashion until we have captured all elements of $S$. Thus $S$ is a union of disjoint open intervals. $\blacksquare$
If it is logically correct: Is there anything I could improve, or perhaps shorten?
Thanks!
AI: Technically $\varnothing$ is an open interval, e.g., $(0,0)$, though some definitions of open interval exclude this case. At the end of your argument you can’t properly speak of continuing to generate these intervals unless you describe some recursive procedure for doing so. Fortunately, it’s not necessary to do so: your $x$ was an arbitrary element of $S$, so you have in fact constructed $I_x$ for each $x\in S$, and you’ve shown that these intervals are pairwise disjoint. Thus, your final paragraph is superfluous: you were done at the end of the penultimate paragraph.
Apart from those two reservations, the proof is fine.
There is an alternative approach that is a little shorter. Define an equivalence relation $\sim$ on $S$ by setting $x\sim y$ iff $x\le y$ and $[x,y]\subseteq S$, or $y\le x$ and $[y,x]\subseteq S$, and for each $x\in S$ let $I_x$ be the $\sim$-equivalence class of $x$. The family $\{I_x:x\in S\}$ is automatically pairwise disjoint, so we need only show that each $I_x$ is an open interval. It’s clear from the definition of $\sim$ that $I_x$ is order-convex, i.e., that if $a,c\in I_x$ and $a\le b\le c$, then $b\in I_x$, and the boundedness of $S$ ensures that it is not a ray or the whole line, so $I_x$ must be an interval. Your argument that $\inf L_x\notin S$ is easily adapted to show that the interval $I_x$ does not contain its endpoints and is therefore open. |
H: How to correctly compate $f(n)$ and $g(n)$ when working through $O(n)$ notation?
Going through theory, missing the idea, need a bit of help. So, the initial state is:
$$f(n) = O(g(n))$$
Assume that $f$ and $g$ are both nondecreasing and always bigger than 1. And, from my understanding, $f$ must be less or equal to $g$, because it must represent the maximum complexity. But, if we add non-equal elements like:
$$f(n)*log_2(f(n)^c) = O(g(n)*log_2(f(n)))$$
it becomes a bit messy for me. Assume that $c$ is some positive constant.
As I see, if, for example, previously $f(n) = n$ and $g(n) = n$, by adding $c$ power (even inside the $log$), it could make $f(n)$ bigger.
But, because $O(n)$ focuses on the highest order terms, I need to exclude $log_2$ from it anyway, so the whole statement stays equal. Is it equal and do I understand the logic correctly?
AI: First things first: writing $f(n) = O(g(n))$ is a shortcut for $f\in O(g)$, i.e. $O(g)$ is a set of functions. But that is a little pedantry we will not bother with.
Now, the above does not imply that $f\leq g$! It means that there exists $N\in\mathbb{N}$ and some $C > 0$ such that
$$
f(n) \leq C\cdot g(n)
$$
for all $n \geq N$. That means that for $n$ large, $f$ is at most a constant bigger than $g$.
Next, you need to use the properties of the logarithm: for any $x, c > 0$, we have
$$
\log(x^c) = c\cdot\log x.
$$
This is true for all logarithms. Now, assume that $f(n) = O(g(n))$, i.e. we have the above inequality for $n$ large enough. Then
$$
f(n)\cdot \log(f(n)^c) = c\cdot f(n)\cdot \log(f(n)) \leq c\cdot C g(n)\cdot \log(f(n))
$$
for $n$ large enough. Taking $c\cdot C > 0$ as the a new constant $C'$, we get that $f(n) = O(g(n)\cdot \log(f(n)))$.
Also, you need to be very careful: a logarithmic term is asymptotically smaller than a linear term, i.e.
$$
\log(n) = O(n),
$$
BUT if you multiply with a logarithmic term, you cannot simply forget it, i.e.
$$
n + \log n = O(n)\qquad\text{ but }\qquad n\cdot \log n = O(n\cdot \log n) \neq O(n).
$$
You would, at most, have $n\cdot \log n = O(n^2)$.
Perhaps, it could be interesting for you to take the above definition and to prove basic properties of the Landau notation, e.g.
$$
f\in O(g)\text{ and } f'\in O(g')\qquad\text{ then }\qquad ff'\in O(gg') \subseteq O(\max\{g,g'\}^2).
$$ |
H: Prove/Disprove an inner product on a complex linear space restricted to its real structure is also an inner product
Let $V$ be an $n$-dimensional linear space and $(\cdot, \cdot)$ be an inner product on it. Define the conjugation map $\sigma: V \to V$ such that for any $\alpha, \beta \in V$ and $\lambda \in \mathbb{C}$, $\sigma(\alpha + \beta) = \sigma(\alpha) + \sigma(\beta)$, $\sigma(\lambda\alpha) =
\bar{\lambda}\sigma(\alpha)$, $\sigma^2(\alpha) = \alpha$. The space
\begin{align*}
R_\sigma(V) = \{\alpha \in V: \sigma(\alpha) = \alpha\}
\end{align*}
is known as the real structure of $V$. In the link, it has been shown $R_\sigma(V)$ is an $n$-dimensional real linear space. I felt the inner product $(\cdot, \cdot)$, which is originally defined on $V$, when restricted to $R_\sigma(V)$,
is also an inner product.
While the positiveness and bilinearity are easy to prove, it seems difficult to show the symmetry. In particular, how to show $(\alpha, \beta)$ is a real number when $\alpha, \beta \in
R_\sigma(V)$?
AI: This is not true in general. For instance, let $V=\mathbb{C}^2$. The vectors $v=(1,0)$ and $w=(i,1)$ are a basis for $V$, and there is a unique inner product $(\cdot,\cdot)$ on $V$ for which they are orthonormal. Now let $\sigma:V\to V$ be given by $\sigma(a,b)=(\overline{a},\overline{b})$. This is a conjugation, and $R_\sigma(V)=\mathbb{R}^2$. However, the inner product is not real-valued when restricted to $\mathbb{R}^2$: for instance, $v=(1,0)$ and $w-iv=(0,1)$ are both in $\mathbb{R}^2$, but $(v,w-iv)=i$ since $v$ and $w$ are orthonormal. |
H: Negative exponential of an exponential random variable is a uniform random variable?
I know that the negative log of a uniform random variable is an exponential random variable. I am trying to prove the reverse, but I seem to have arrived at something that doesn't make sense.
Define $Y \sim \text{Exponential}(\lambda)$, and $X = -\exp(Y)$.
\begin{align}
P(X \leq x) = P(-\exp(Y) \leq x) = P(Y \geq -\ln(x)) \\
= \int_{-\ln(x)}^{{\infty}} \lambda \exp (-\lambda y) dy
= -\exp(- \lambda y) \big|_{-\ln(x)}^{\infty} \\
= \exp(\lambda \ln(x)) \\
= x^{\lambda}
\end{align}
If you take the derivative of this to get the pdf of $x$, you see that it is a function of $x$, which isn't constant, hence $X$ isn't a uniform random variable. So I am confused, is the negative exponential of a exponential random variable not a uniform random variable, or did I make a mistake in my derivation?
AI: A fix in red: what does have a $U(0,\,1)$ distribution is $\color{blue}{1-}\exp(-\color{red}{\lambda}Y)$, where the blue part is unnecessary but makes the transformation order-preserving. In fact, it's $Y$'s CDF, say $F$. This is no coincidence:$$P(F(Y)\le f)=P(Y\le F^{-1}(f))=F(F^{-1}(f))=f\implies F(Y)\sim U(0,\,1).$$ |
H: Logarithm over complex numbers
The logarithm function is not certainly defined for every $\text{Re}(z)\leq0$, but the question is where is it defined?
I also know $\displaystyle \int_{C} \frac{1}{z} dz\neq0$ where $C$ is the unit circle defined by $ \gamma(t)=e^{it} $ for $0\leq t\leq 2\pi$, which implies that $\frac{1}{z}$ has no antiderivative.
Is this true because any set $U\supset C$ contains $z\in \mathbb{C}:Re(z)\leq0$ ?
If $U$ does not contain $z\in \mathbb{C}:Re(z)\leq0$ , is $\log(z)$ "well behaved" in $U$?
AI: One thing you should do is to remove from your mind the concept of THE logarithm function, and replace it with A logarithm function. Once you've done that, then you can formulate a sensible question:
On what (open) subsets $U \subset \mathbb C - \{0\}$ does there exist a logarithm function $l : U \to \mathbb C$?
And you can even say precisely what you mean by this, namely that $l$ is holomorphic on $U$ and the composition $z \mapsto \exp(l(z))$, when restricted to $U$, is the identify function on $U$.
To start, here's some examples: for each angle $\theta \in \mathbb{R}$ a logarithm is defined on the domain $U_\theta$ obtained by removing the ray of angle $\theta$ from $\mathbb C$, which is given by the formula
$$U_\theta = \{r e^{2 \pi i \phi} \mid r > 0, \,\, \theta - 2\pi < \phi < \theta\}
$$
For more examples, every simply connected open subset $U \subset \mathbb C - \{0\}$ has a logarithm.
And here is a reasonably simple general answer to this question: a logarithm is defined on $U$ if and only if for every smooth Jordan curve $C$ in $U$, the curve $C$ does not wind around the origin, equivalently $\int_C \frac{dz}{z}=0$. |
H: The Nash equilibrium, an existence proof
I do not follow here in the -4th line
that the equality is achieved only if
$$p_i(s,\alpha)\leq 0$$
for every $s$.
AI: If $p_i(s,\alpha)\leq 0$ for every $s$, then $\alpha'_{i} = \alpha_{i}$, thus we get equality (proving "if" but not "only if").
Suppose there exists an $\hat{s}$ where $p_i(\hat{s},\alpha)> 0$, then for any $s$ where $p_i(s,\alpha)\leq 0$ and $\alpha_{i}^{(s)} > 0$ we have $\alpha_{i}^{\prime(s)} < \alpha_{i}^{(s)}$, which implies:
$$\sum_{s:p_{i}(s, \alpha)\leq0} \alpha_{i}^{\prime(s)} < \sum_{s:p_{i}(s, \alpha)\leq0} \alpha_{i}^{(s)}$$
(Note, there must exist such an $s$ because $\alpha_{i}$ cannot result in strictly lower payoff than all pure strategies in its support.)
$$\Rightarrow \sum_{s:p_{i}(s, \alpha)>0} \alpha_{i}^{\prime(s)} > \sum_{s:p_{i}(s, \alpha)>0} \alpha_{i}^{(s)}$$
The last implication follows because $\sum\limits_{s}\alpha_{i}^{\prime(s)}=\sum\limits_{s}\alpha_{i}^{(s)}=1$.
Hence, we can say equality is achieved if and only if $p_i(s,\alpha)\leq 0$ for every $s$. |
H: Calculate $\int_{0}^{1} \sin(x^2)$ with an error $\le 10^{-3}$
Calculate $\int_{0}^{1} \sin(x^2)$ with an error $\le 10^{-3}$
Let $f(x)= \sin(x^2) $ continuous in [0,1] so by the MVT for integrals we know $\int_{0}^{1} \sin(x^2) = \sin(c^2) \; \text{for} \; c \in [0,1]$. I don't really know if this is of any help. Another thing that I know is that $\int_{0}^{1} \sin(x^2) \le \int_{0}^{1} x^2 \; \text{for} \; x \in [0,1]$. Any hints on how to resolve this ? Thanks in advance.
AI: Use the Maclaurin series for $\;\sin x^2\;$, which converges absolutely at every point and thus you can integrate/differentiate it termwise:
$$\sin x^2=\sum_{n=0}^\infty\frac{(-1)^n (x^2)^{2n+1}}{(2n+1)!}=\sum_{n=0}^\infty\frac{(-1)^n x^{4n+2}}{(2n+1)!}\implies$$
$$\int_0^1\left(\sum_{n=0}^\infty\frac{(-1)^n x^{4n+2}}{(2n+1)!}\right)dx=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}\int_0^1 x^{4n+2}dx=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}\cdot\frac1{4n+3}$$
The above is a Leibniz series and we can estimate its value in a reasonably easy way:Leibniz theorem telss us that if $\;S\;$ is the series sum and $\;S_n\;$ the $\;n\,$- th term of its partial sums sequence, and $\;a_n\;$ is its general term sequence, then
$$|S-S_n|<|a_{n+1}|=\frac1{(2n+3)!(4n+7)}\stackrel{\text{we want!}}<\frac1{1000}$$
Well, you can now even try a very few times and get the correct value of $\;n\;$ |
H: Determining group structure after computing homology: $\langle b,c \mid 2(b+c)=0, b+c=c+b \rangle.$
I am trying to determine what group this is a presentation of:
$$\langle b,c \mid 2(b+c)=0, b+c=c+b \rangle.$$
I am pretty sure it is $\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$, but am stuck on how to show it.
AI: Put multiplicatively, your presentation is
$$\langle b,c\mid (bc)^2, bc=cb\rangle.$$
By Tietze transformations, introduce $a=bc$, so that $c=b^{-1}a$, giving the presentation
$$\langle a,b\mid a^2, a=(b^{-1}a)b\rangle,$$
which simplifies to
$$\langle a,b\mid a^2, ba=ab\rangle;$$
and this presentation yields, multiplicatively, the group $\Bbb Z_2\times \Bbb Z$. This is isomorphic to $\Bbb Z\oplus \Bbb Z/2\Bbb Z$. |
H: Finding the function that satisfies to condition that the length of the curve is the same as the volume of rotation around the x-axis
I want to find the function that satisfies the following DE:
$$\pi y(x)^2=\sqrt{1+(y'(x))^2}$$
This comes from the fact that the left-hand side gives the volume of radiation around the x-axis and the right-hand side gives the length of the curve. I also know that the initial condition is given by $y(a)=b$ where $a$ and $b$ are real values.
The questions I have are:
What is the function $y(x)$?
In what range of $x$ is the function $y(x)$ real-valued?
AI: “The” function?
Take any function $f_1$ whose domain is an interval $[a,b],$ subject only to the restrictions that $(f_1(x))^2$ has a finite integral on the domain of $f_1$ and the length of the curve $y=f_1(x)$ is finite.
Now for any positive real number $t,$ define $f_t(x)=tf(x/t)$ on the interval $[ta,tb].$
That is, $f_t$ is $f_1$ “scaled up” in both the $x$ and $y$ dimensions.
For any choice of $f_t,$ the curve is $t$ times as long as the curve for $f_1$ but the volume of the solid of rotation is $t^3$ times as large. So if $f_1$ has a volume smaller than the curve's length, increase $t$ until the length and volume match. If the volume is larger than the length, decrease $t.$
You can make many very different functions that satisfy your requirements in this way.
A simple example is the function $y(x)=1/\sqrt\pi$ on any finite interval. |
H: Equivalence of two 1-form defining the same hyperplane field on a manifold.
I have tried my best to explain my question.
Suppose M is n dimensional smooth manifold and $\xi$ is a smooth hyperplane field define on it. In addition, M has two 1-form define on it, $\alpha_1$ and $\alpha_2$ such that kernel($\alpha_1$)=kernel($\alpha_2$)= $\xi$.
Since $\xi$ can also be thought as smooth subbundle of the smooth tangent bundle, we can (pointwise) factor the tangent bundle $TM = \xi \oplus \xi^{\bot}$. Therefore, these 1-form has action (effective) on the 1-dimensional bundle $\xi^{\bot}$. At a point $p \in M$ one can associate a real values depending on $p$ such that ${\alpha_1}_p = \lambda(p) {\alpha_2}_p$ and since p is arbitrary choosen. We define a non-zero real valued function $\lambda: M \to \mathbb{R}-\{0\}$.
Now I dont know, how to prove that this function $\lambda$ is smooth.
AI: You can find a section $X$ of $\xi^{\perp}$ on an open subset which contains $U$. The function $f_i(x)=\alpha_i(X(x))$ is differentiable and not zero ${{f_1}\over{f_2}}$ is differentiable. |
H: Convergence Range of $\sum\limits_{n=1}^{\infty} \frac{\sin(2n-1)x}{(2n-1)^2}$
$$\sum\limits_{n=1}^{\infty} \frac{\sin(2n-1)x}{(2n-1)^2}$$
My Attempt: I realize that $-1 \leq \sin(2n-1) \leq 1$. If I take the absolute value I can create the inequality:
$$\sum\limits_{n = 1}^{\infty} | \frac{\sin(2n-1)x}{(2n-1)^2}| \leq (1)\sum\limits_{n=1}^{\infty} \frac{x}{(2n-1)^2}$$
I know $(1)$ will converge regardless of what $x$ is which makes the original series converge as well. Hence, the range of convergence would be $-\infty <x <\infty$ Is this correct, and if not could someone provide a better justification?
AI: I support George Coote's comment: it should be $\sin[(2n-1)x]$ rather than $[\sin(2n-1)]x$. However, the crucial estimation works regardless of this issue. In fact
$$\sum_{n\ge1}\left|\frac{\sin[(2n-1)x]}{(2n-1)^2}\right|=\sum_{n\ge1}\frac{|\sin[(2n-1)x]|}{(2n-1)^2}\leqslant\sum_{n\ge1}\frac1{(2n-1)^2}$$
Thus, the series converges absolutely independent of the actual value of $x$ and therefore the radius of convergence (assuming $x$ is real) is in fact the entire real line. |
H: Comparison of sequence of functions and function on $\mathbb{R}^2$
Let $\left(f_n\right)_{n\in\mathbb{N}}$ be a sequence of functions, where $f_n:\mathbb{R}\to\mathbb{R}$ and be $g:\mathbb{R}^2 \to \mathbb{R}$ some function. I know that if I plug in a sequence $\left(x_n\right)_{n\in\mathbb{N}}$ into $g$ I can construct a sequence of functions, $\left(g_n\right)_{n\in\mathbb{N}}$, defined by $g_n:=g(x_n,y)$.
I am wondering if a sequence of functions, $\left(f_n\right)_{n\in\mathbb{N}}$, can be interpreted as a function of two variables?
In this case $F:\mathbb{N}\times \mathbb{R} \to \mathbb{R}$, where $F(n,x)=f_n(x)$. I am a little bit unsure because sometimes we have sequences of functions where $n$ doesn't appear in the expression of a member $f_n(x)$ but it changes the domain of $f_n$, e.g. $$f_n(x) = \begin{cases} x &, x \leqslant \frac{1}{n}\\ 2 - x &, \frac{1}{n} < x \leqslant \frac{2}{n}\\ 0 &, x > \frac{2}{n} .\end{cases}$$
AI: In your example, n does not change the domain of the function.
If the domain of $f_n$ is $\mathbb{R}$ $\forall n$ then the domain of F is $\mathbb{N} \times \mathbb{R}$ and F can be interpreted as you require. If the domain of $f_0$ differs from that of $f_1$, for example, then F is only defined for values of x that are present in both domains. |
H: Heine Borel Theorem statement (a)
I have been following Prof Winston Ou's course on analysis on Youtube.
In the lecture on Heine Borel theorem, he mentioned that a set $E$ in $\mathbb R$ is closed and bounded implies that $E$ is a k-cell (hence $E$ is compact).
I don't understand how he came to this conclusion. For instance, I imagine $E$ could be a discrete set (and still being closed and bounded), but it will not be a k-cell. I would be very grateful if anyone could enlighten me.
AI: He means $D$ will be contained in a $k$-cell, and as $k$-cells are compact and $E$ is still a closed subset of it, it will also be compact.
The boundedness in Euclidean space/metric forces the set inside a product of compact intervals.. |
H: Integral to Gamma function
For this function i need to convert it to either a Gamma or a Beta function
$$\int_0^1 \frac{3}{(1-x^3)^\frac{1}{3}} \,dx$$
I know I need to make the substitution $x^3=z$ but I am unsure where to go from here
AI: For simplicity, define
$$\mathcal I := \int_0^1 \frac{3}{(1-x^3)^\frac{1}{3}} \,dx$$
Make the $u$-substitution $u=x^3$ as you prescribed. Then:
$x = u^{1/3}$
$dx = \frac 1 3 u^{-2/3} du$
Insert both of these definitions into the integral and pull out constants via linearity. (They happen to cancel.) We find that
$$\mathcal I = \int_0^1 u^{-2/3} (1-u)^{-1/3}du$$
Recall that the definition of the beta function is given by
$$\text B(x,y) := \int_0^1 t^{x-1}(1-t)^{y-1}dt$$
$\mathcal I$ now matches this same form for appropriate $x,y$:
$-2/3 = x-1 \implies x = 1/3$
$-1/3 = y-1 \implies y = 2/3$
Thus,
$$\mathcal I = \text B \left( \frac 1 3, \frac 2 3 \right) \approx 3.6276$$
WolframAlpha numerically confirms this answer: their answer for the integral and for the beta function are at the links. If you want to convert to a gamma function version, note the identity
$$\text B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$
In this case, then (since $\Gamma(1/3 + 2/3) = \Gamma(1) = 1$),
$$\mathcal I = \Gamma \left( \frac 1 3 \right) \Gamma \left( \frac 2 3 \right)$$
We can even simplify this result further by use of Euler's reflection formula and get a better closed form. The reflection formula is given by
$$\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}$$
Take $z = 1/3$ and how it might be utilized for $\mathcal I$ is clear. Then
$$\mathcal I = \frac{\pi}{\sin(\pi/3)} = \frac{\pi}{\sqrt{3} / 2} = \frac{2 \pi}{\sqrt 3}$$
(Credits to user Ty who pointed out the simplification owing to Euler's reflection formula in the comments!) |
H: Importance of Tannery's theorem
Tannery's theorem:
Let $S_n=\sum_{k=0}^\infty a_k(n)$ and $\lim_{n\to\infty}a_k(n)=b_k$. If $|a_k(n)|\le M_k$ and $\sum_{k=0}^\infty M_k\lt\infty$, then $\lim_{n\to\infty}S_n=\sum_{k=0}^\infty b_k$.
This can be used to prove that
$$\lim_{n\to\infty}\sum_{k=0}^n \frac{x^k}{k!}\prod_{m=1}^{k-1}\left(1-\frac{m}{n}\right) =\sum_{k=0}^\infty \frac{x^k}{k!}.$$
But I've never really understood why is the theorem "needed", as $\prod_{m=1}^{k-1}\left(1-\frac{m}{n}\right)$ tends to $1$ with increasing $n$, so the result should be obvious.
What could go wrong here? Could someone provide a counterexample to this reasoning?
AI: This is just a special case of a combination of (1) the Weierstrass M-test for uniform convergence and (2) the result that the limit can be exchanged with the sum when convergence is uniform.
(1) Suppose we have a sequence of functions $x \mapsto a_k(x)$ where $x \in D \subset \mathbb{R}$ and $|a_k(x)| \leqslant M_k$ for all $k$ and for all $x \in D$. If $\sum_{k=0}^\infty M_k < \infty$, then the series $\sum a_k(x)$ is unformly convergent for $x \in D$.
(2) If $x_0 \in D$, then under the conditions of (1) where $\sum a_k(x)$ is uniformly convergent, we have $\lim_{x \to x_0} \sum_{k=0}^\infty a_k(x) = \sum_{k=0}^\infty \lim_{x \to x_0} a_k(x)$. This holds when $x_0 = +\infty$ as well.
Tannery's theorem can be seen to be a special case of these results by taking $D = \mathbb{N}$ and noting that
$$S_n = \sum_{k=0}^na_k(n) = \sum_{k=0}^\infty a_k(n) \mathbf{1}_{\{k \leqslant n\}},$$
since $|a_k(n)| \leqslant M_k$ implies that $ |a_k(n) \mathbf{1}_{\{k \leqslant n\}}| \leqslant M_k$.
What can go wrong is that without sufficient conditions as in uniform / monotone / dominated convergence it is not always possible to exchange the limit and the sum. |
H: Exponentials of the roots of polynomials $P(x)$ are always the roots of $P^*(x)?$
For any polynomial $P(x)$ with rational coefficients and no constant terms, are the exponentials of the roots of $P(x)=c_1x^n+c_2x^{n-1}+\cdot\cdot\cdot~+ c_kx$ always equal to the roots of $P^*(x)=e^{c_1\log^n(x)}+e^{c_2\log^{n-1}(x)}+ \cdot\cdot\cdot~+c_kx?$
There's probably an obvious way to see this but I haven't been able to find a proof. I've tried calculating many examples and so far I haven't found any counter-examples. For example I calculated that $p(x)=x^2-x^3$ has roots @ $x=0,1$ and $p^*(x)=e^{\log^2(x)}-e^{\log^3(x)}$ has roots @ $x=1,e.$
AI: When $P(x)=x$, we get $P^{*}(x)=x$. Note that $0$ is a root of $P$ but $e^{0}=1$ is not a root of $P^{*}$.
Another example: $P(x)=x+x^{2}$. the roots are $0$ and $-1$. But $e^{-1}$ is not a root of $P^{*}$ in this case.
In your example the calculation of $P^{*}$ is wrong. The minus sign goes to the exponent. |
H: Elementary combinatorics: how many lunch salads?
Here is the problem: a restaurant offers salads with the following options: choose five ingredients from a list of eight, plus two dressings from a list of four, but do not choose radishes (one of the ingredients) with peanut butter dressing (you can have radishes without peanut butter dressing or peanut butter dressing without radishes, but not both). How many salads are possible?
I get (eight choose five) times (four choose two) minus (seven choose five) times (three choose two) or
$\frac{8!}{5!3!} \frac{4!}{2!2!} - \frac{7}{5!2!} \frac{3!}{2!1!}=273$
The text, Gerstien's Introduction to Mathematical Structures and Proofs, gives 231. Which is correct and why?
AI: You should be subtracting the cases where you choose radishes and peanut butter dressing. That leaves you only four vegetables to choose out of seven and one dressing to choose out of three, so it should be
$$\frac{8!}{5!3!} \frac{4!}{2!2!} - \frac{7}{4!3!} \frac{3!}{1!2!}=231$$ |
H: Bounding sum by (improper) integral
I am trying to verify the following inequality that I came across while reviewing some analysis exercises online:
$$
\sum_{n=1}^{k} \left(1-\frac{n}{k}\right)n^{-1/7}\leq \int_{0}^{k}\left(1-\frac{x}{k}\right)x^{-1/7}\,dx, \hspace{3mm} k>1
$$
$\textbf{My question:}$ Why does the above inequality hold? Isn't the integral on the right an improper integral?
My idea was to justify the inequality by replacing $0$ with a small positive number and then using concavity of the function $f(x)=(1-x/k)x^{-1/7}$ but since the function is concave up I am having a hard time justifying it.
AI: Setting the scene: define $f_k(x) = (1-\frac xk)x^{-1/7}$, so that the inequality in question is $\sum_{n=1}^k f_k(n) \le \int_0^k f_k(x)\,dx$. This inequality is of course equivalent (for $k\ge1$) to
$$
f_k(1) + \sum_{n=2}^k f_k(n) \le \int_0^1 f_k(x)\,dx + \int_1^k f_k(x)\,dx.
$$
Since the OP indicated that they're comfortable with the inequality $\sum_{n=2}^k f_k(n) \le \int_1^k f_k(x)\,dx$ (since $f_k$ is decreasing), let's focus on the inequality
$$
f_k(1) \le \int_0^1 f_k(x)\,dx
$$
that contains the improper integral.
The main point: I claim that the fact that the integral is improper is really a red herring. Indeed, this last inequality is equivalent to
$$
f_k(1) - f_k(1) \le \int_0^1 f_k(x)\,dx - f_k(1),
$$
or simply
$$
0 \le \int_0^1 \big( f_k(x) - f_k(1) \big) \,dx.
$$
And, improper or not, this inequality is obvious because the integrand is nonnegative (again since $f_k$ is decreasing). |
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