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H: Partial fractions for geometric probability-generating function wrong
Let $X\sim \text{Geo}(1/4), Y\sim \text{Geo}(1/2)$ be given. First I have to compute $\mathbb{E}[z^{X+Y}]$:
$$\mathbb{E}[z^{X+Y}]=\mathbb{E}[z^{X}]\cdot\mathbb{E}[z^{Y}]=\frac{\frac{1}{4}z}{1-\left(1-\frac{1}{4}\right)z}\cdot\frac{\frac{1}{2}z}{1-\left(1-\frac{1}{2}\right)z}=\frac{z^2}{8\left(1-\frac{1}{2}z\right)\left(1-\frac{3}{4}z\right)}.$$
The next task is to find $\alpha,\beta,\gamma\in\mathbb{R}$ such that
$$\mathbb{E}[z^{X+Y}]=\alpha+\frac{\beta}{1-\frac{1}{2}z}+\frac{\gamma}{1-\frac{3}{4}z}.$$
Here is my procedure (which seems to be wrong...):
$$\frac{z^2}{8\left(1-\frac{1}{2}z\right)\left(1-\frac{3}{4}z\right)}=1+\frac{10z-8}{8\left(1-\frac{1}{2}z\right)\left(1-\frac{3}{4}z\right)}$$
$$\frac{10z-8}{8\left(1-\frac{1}{2}z\right)\left(1-\frac{3}{4}z\right)}=2\cdot\frac{5z-4}{(z-2)(3z-4)}=\frac{A}{z-2}+\frac{B}{3z-4}$$
$$\frac{A(3z-4)+B(z-2)}{(z-2)(3z-4)}=\frac{(3A+B)z+(-4A-2B)}{(z-2)(3z-4)}$$
$$\left.\begin{cases}
3A+B&=5 \\ -4A-2B&=-4
\end{cases}\right\}\implies A=3,\;B=-4\implies 2A=6,\;2B=-8$$
$$\rightsquigarrow \mathbb{E}[z^{X+Y}]=1+\frac{6}{z-2}-\frac{8}{3z-4}=1-\frac{3}{1-\frac{1}{2}z}+\frac{2}{1-\frac{3}{4}z}\implies\alpha=1,\;\beta=-3,\;\gamma=2$$
However it seems that the values for $\alpha,\beta,\gamma$ should each be just a third of my result! Can anyone explain me, what I did wrong?
AI: Due to anons hint (thanks!) I found the mistake.
$$\frac{z^2}{8\left(1-\frac{1}{2}z\right)\left(1-\frac{3}{4}z\right)}=1+\frac{\frac{10}{3}z-\frac{8}{3}}{8\left(1-\frac{1}{2}z\right)\left(1-\frac{3}{4}z\right)}$$
This leads to $\alpha=1/3,\;\beta=-1,\;\gamma=2/3$ which is correct. |
H: Area of a revolution of $x=\frac{1}{3}\left(y^2+2\right)^\frac{3}{2}$
I think my biggest problem here is I can not find a good way to find the square root in this problem
$$x=\frac{1}{3}\left(y^2+2\right)^\frac{3}{2} \ \ \ \ 1 \le x \le 2$$
$$\int_1^2 2 \pi \cdot {\frac{1}{3}\left(y^2+2\right)^\frac{3}{2}} \sqrt{1 + (1/3 (y^2 + 2)^{\frac{3}{2}})^2}$$
$$\int_1^2 2 \pi \cdot \frac{1}{3}\left(y^2+2\right)^\frac{3}{2} \sqrt{1 + y^3 + 2y}$$
Here is where I am stuck, I have tried u-substitution and trig and I can not make this work. I have two pages of notes but I expect that they would be useless to type up.
AI: My suggestion is: take it easy and take it step by step. Don't throw everything in one expression, because it is easy to make a slip:
$$x=\frac{1}{3}\left(y^2+2\right)^\frac{3}{2}$$
$$x'=\frac{1}{2}\left(y^2+2\right)^\frac{1}{2}\cdot\underbrace{ 2y}_{\text{chain rule}}=y\left(y^2+2\right)^\frac{1}{2}$$
$$x'^2=y^2\left(y^2+2\right)$$
$$1+x'^2=1+y^2\left(y^2+2\right)=y^4+2y^2+1=\left(y^2+1\right)^2$$
Now calculating the differential it is useful to take it a step too far to simplify calculations in the end:
$$ds=2\pi x\sqrt{1+x'^2}=2\pi y\left(y^2+1\right)dy=\pi\left(y^2+1\right)\cdot 2ydy=\pi\left(y^2+1\right)d\left(y^2\right)=\pi\left(y^2+1\right)d\left(y^2+1\right)$$
Where I am "pushing expressions under $d$": $f'(x)dx=d\left(f(x)+C\right)$
$$s=\pi\int_1^2\left(y^2+1\right)d\left(y^2+1\right)=\left.\frac{\pi(y^2+1)^2}{2}\right|_1^2=12\pi$$ |
H: Double integrals: finding a volume of a solid
I am trying to determine the bounds given a solid bounded by the plane $y+z=4$ and the cylinder $y=x^2$, and $xz$ and $yz$ planes in the first octant. I am not sure if I am on the right track, but to find my bounds for $y$, I set $z$ equal to zero and calculated $y=4$. This gives a region enclosed by a parabola with the upper y-bound at 4. So my outer integral (assuming type I region), I believe, is $x^2 \leq y \leq 4$. And since there is a vertical asymptote at $x=1$, I am assuming that my inner integral is bounded by $0 \leq x \leq 1$.
I know that I am going to need to integrate xz and yz as the functions, but I am not sure how to express these in terms of $z$, which is what I assume I am supposed to do since all problems I have encountered up until this point that are in 3D have done this. However, all of those problems didn't have z as a coefficient of either the x or y terms (e.g. they were in a form like $x + y + z$). Am I on the right track? And if so, how should I approach this?
AI: Your first paragraph is almost spot on. I cannot make sense of your second paragraph. The mistake in the first paragraph is that you are making up the limit for $x$. The intersection between $y=x^2$ and $y=4$ (the intersection of the plane $y+z=4$ and the $x$-$y$ plane).
Since we have to stay in the first octant, the solid is bounded below by the $x$-$y$ axis.
The volume of your solid is
$$
\int_0^2\int_{x^2}^4\,(4-y)\,dy\,dx.
$$
Properly speaking, your integrand is $(4-y)-0$, as you integrate the difference between the "ceiling" and the "floor". |
H: How can I alternately solve or otherwise optimize the solution to finding paths up a ladder 1 or 2 steps at a time?
I got thrown off by an interview question recently,
What structure do you use to find all the ways up a ladder with N
steps in 1 or 2 steps moves?
It occured to me today that a graph expansion could be used but it's not optimal because when N = 30, it evalates almost 700,000 nodes on a single round of expansion.
Here's the output of solving for N = 5.
2 2 1
2 1 2
1 2 2
2 1 1 1
1 2 1 1
1 1 2 1
1 1 1 2
1 1 1 1 1
My instinct is that maybe this problem can be solved with regular methods from Discrete Math that I'm just not remembering.
Is there a way to get the set of sequences containing (1, 2) where the sum of the elements is N?
If anyone is interested here's the code of the program I wrote.
AI: If $N=0$ there is one way: $\langle\rangle$. (No steps at all.) If $N=1$, there is one way, $\langle 1\rangle$. For $N>1$, take each way up a ladder of $N-2$ steps and append a $2$, and each way up a ladder of $N-1$ steps and append a $1$. Thus, for $N=2$ you get $\langle 2\rangle$ and $\langle 1,1\rangle$. The first few stages, with semicolons separating the paths derived from two steps back from those derived from one step back:
$$\begin{align*}
&N=0: \langle\rangle\\
&N=1: \langle 1\rangle\\
&N=2: \langle 2\rangle;\langle 1,1\rangle\\
&N=3: \langle 1,2\rangle;\langle 2,1\rangle,\langle 1,1,1\rangle\\
&N=4: \langle 2,2\rangle,\langle 1,1,2\rangle;\langle 1,2,1\rangle,\langle 2,1,1\rangle,\langle 1,1,1,1\rangle
\end{align*}$$
You never need to have more than two consecutive levels in storage at once, and it’s a straightforward generation with no trial and error. By the way, the number of paths for a given $N$ is $F_{N+1}$, the $(N+1)$-st Fibonacci number, |
H: If $X$ is a connected subset of a connected space $M$ then the complement of a component of $M \setminus X$ is connected
I have an exercise found on a list but I didn't know how to proceed. Please, any tips?
Let $X$ be a connected subset of a connected metric space $M$. Show that for each connected component $C$ of $M\setminus X$ that $M\setminus C$ is connected.
AI: Here is the theorem found on Kuratowski's book. Thanks for the reference, it is a very excellent book.
The Theorem II.4 cited above: |
H: A simple Integral Question
What are the steps to calculate the value of $c$ in the following integral equation?
$$
\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty}c.e^{-(x_1+2x_2+3x_3)}\,dx_1 \, dx_2
\, dx_3 = 1
$$
AI: Since $e^{-(x_1+2x_2+3x_3)}=e^{-x_1}e^{-2x_2}e^{-3x_3}$, we have
$$
\int_0^\infty\int_0^\infty\int_0^\infty e^{-x_1+2x_2+3x_3}=\left(\int_0^\infty e^{-x_1}\right)\left(\int_0^\infty e^{-2x_2}\right)\left(\int_0^\infty e^{-3x_3}\right)=1\,\frac12\,\frac13=\frac16.
$$
So $c=6$. |
H: Closure of the interior of another closure
Let $X$ be a topological space and let $A \subset X$.
Is it true that $\overline{\rm{Int}(\overline{A})}=\overline {A}$?
This question arose when I try to show$\overline{X-\overline{\rm{Int}(\overline{A})}}=\overline{X-\overline{A}}$
AI: The statement is false in general. Take $A$ to be a closed, nowhere dense set; then $\operatorname{cl}A=A$, but $\operatorname{cl}\operatorname{int}\operatorname{cl}A=\operatorname{cl}\operatorname{int}A=\operatorname{cl}\varnothing=\varnothing.$ In the space $\Bbb R^n$, for instance, any closed, discrete set provides a counterexample, as does any Cantor set.
Added: To show that
$$\overline{X-\overline{\rm{Int}(\overline{A})}}=\overline{X-\overline{A}}\;,$$
note first that you already know that
$$\overline{X-\overline{\rm{Int}(\overline{A})}}\supseteq\overline{X-\overline{A}}\;.$$
Suppose that $x\notin\operatorname{cl}\left(X\setminus\operatorname{cl}A\right)$, and let $V$ be an open nbhd of $x$ disjoint from $X\setminus\operatorname{cl}A$; $V\subseteq\operatorname{cl}A$, so $V\subseteq\operatorname{int}\operatorname{cl}A\subseteq\operatorname{cl}\operatorname{int}\operatorname{cl}A$, and therefore $V\cap (X\setminus\operatorname{cl}\operatorname{int}\operatorname{cl}A)=\varnothing$, i.e., $x\notin\operatorname{cl}(X\setminus\operatorname{cl}\operatorname{int}\operatorname{cl}A)$. It follows that $\operatorname{cl}(X\setminus\operatorname{cl}\operatorname{int}\operatorname{cl}A)\subseteq\operatorname{cl}\left(X\setminus\operatorname{cl}A\right)$, and we’re done. |
H: Algebraic proof of a binomial sum identity.
I came across this identity when working with energy partitions of Einstein solids. I have a combinatorial proof, but I'm wondering if there exists an algebraic proof.
$$\sum_{q=0}^N\binom{m + q - 1}{q}\binom{n + N - q - 1}{N - q} = \binom{m + n + N - 1}{N}$$
I've tried induction, but Pascal's Identity cannot simultaneously reduce the top and bottom argument for an inductive proof.
For those interested, a combinatorial proof of the identity can be given as follows: Consider the ways of distributing $N$ quanta of energy to a system of $n + m$ oscillators (where each oscillator can have any number of quanta). This is equivalent to the question of asking how many ways of putting $N$ objects into $n + m$ boxes. From the traditional stars and bars method, the total is given by
$$\binom{m + n + N - 1}{N}$$
which is the right-hand side. Alternatively, consider partitioning the units of energy between the first $m$ and the last $n$ oscillators. Give $q$ units of energy to the first $m$ oscillators. Then there remains $N - q$ units of energy for the latter $n$. Together, the number of states for this particular partition is
$$\binom{m + q - 1}{q}\binom{n + N - q - 1}{N - q}$$
Summing over all partitions gives the left-hand side.
Thanks for your time.
AI: Let $$f_n(z)=\sum_{q=0}^\infty \binom {n+q-1}q z^q$$
Claim: This is $(1-z)^{-n}$.
Then $f_n(z)f_m(z) = f_{n+m}(z)$. But the left hand side of your formula is the coefficient of $z^N$ in $f_n(z)f_m(z)$, and the right hand side is the coefficient of $f_{n+m}(z)$.
The proof of the claim is the generalized binomial expansion, and uses the fact that $$\binom {-n} q = (-1)^q \binom {n+q-1}{q}$$
Alternatively, you can rewrite your statement as:
$$\sum_{q=0}^N \binom {-m} q \binom {-n}{N-q} = \binom {-(m+n)}q$$ |
H: Square three digit numbers, the efficient way
I would like to square a three digit number in my head.
Now I know that the formula is
$$ ( X + r ) ( X - r ) + r^2 = X^2 - rX + rX - r^2 + r^2 = X^2
$$
Where $\,r\,$ is a number such that $\,X + r\,$ is divisible by $10$ and/or $100$
Now the problem is that, I would like to first figure out whether finding the $r$ such that the $3$ digit number is divisible by $100$ or $10$ would be more efficient.
If I choose $r < 10$ then the $\,( X + r ) ( X - r )\,$ becomes quite unpleasant. If I choose $r < 100$ then the $r$ itself will have to be broken into pieces such that
$\begin{array}\\
( X + r ) ( X - r ) + [ ( r + j ) ( r - j ) + j^2 ]
&= ( X + r ) ( X - r ) + ( r^2 - jr + jr - j^2 + j^2 )\\
&=( X + r ) ( X - r ) + r^2\\
&= X^2 - rX + rX - r^2 + r^2\\
& = X^2\\
\end{array}
$
This kind of yields recursive solution where any number squared works. The job is much more easier, but the person has to memorize more numbers while continuing to do more math.
So which method would you recomend? Or if you would have some other method I would love to hear it.
If you choose recursive method how do you memorize numbers in an efficient way without having numbers collide in your brain?
Thanks to anyone for their response, and sorry if tags are off, I am unsure which tag fits my question.
AI: Many mental calculations are easier the more facts you memorize. Knowing the squares up to $31^2=961$ can make it easier, reducing the need to go all the way to single digits. Also the fact that $(10n+5)^2=100n(n+1)+25$, for example $65^2=(6\cdot7)25=4225$. If you break it all the way to individual digits, I find it easier to keep track of my place if I start with the most significant digits and add as I go, so $359^2=300^2+2\cdot300\cdot50+\ldots =90000+30000+\ldots$
$=120000+ 50^2+\ldots = 122500+2\cdot300\cdot9+\ldots=127900+2\cdot50\cdot9+9^2$ where for many purposes you can quit early when you have the precision needed. |
H: Extreme boundary of a compact, convex, metrizable set is $G_\delta$
Let $X$ be a topological vector space (no assumptions about local convexity are made in the question, though I am worried they might be required). Suppose $K\subset X$ is a compact, convex, metrizable subset of $X$, and denote by $\partial_e K$ the extreme boundary of $K$. We wish to show that $\partial_e K$ is $G_\delta$ in $K$.
It seems that this would boil down to writing a correct, alternate definition of being an extreme point as an intersection of bunch of open or $G_\delta$ sets, and using separability (of $K$ or of $\mathbb{R}$) to reduce that intersection to a countable one. I had two possible approaches:
(1): For each rational $\lambda$ with $0<\lambda<1$, let $$U_\lambda = \{x\in K : \forall y,z\in K (x=\lambda y + (1-\lambda z) \implies y=z)\}.$$ I believe $\partial_e K = \bigcap_\lambda U_\lambda$, and would like to claim that each $U_\lambda$ is open.
(2): Let $D$ be a countable dense set in $K$ (which is separable). For $y\neq z$ in $D$, let $$G_{y,z}=\{x\in K:x \text{ is not in the interior of the line segment between $y$ and $z$}\}.$$ It is easy to see that each $G_{y,z}$ is $G_\delta$, and $\partial_e K\subset \bigcap_{y\neq z \in D} G_{y,z}$. There is an obvious problem if $D$ is dense in the interior of $K$ while containing no boundary points of $K$, but we could throw in a countable dense subset of the boundary of $K$ as well. Under the assumption of local convexity, we need only show that every boundary point of $K$ which is in $\bigcap_{y\neq z \in D} G_{y,z}$ is in $\partial_e K$, which seems geometrically clear to me, but I do not see the details.
Any help?
AI: Let $K$ be compact, convex and metrizable in the topological vector space $X$ (local convexity isn't required for the argument). Fix a metric $d$ on $K$ compatible with the topology. Let $$F_n = \left\{x \in K\,:\, \text{there are } y,z \in K\text{ such that }x = \frac{1}{2}(y+z)\text{ and }d(y,z) \geq \frac{1}{n}\right\}.$$
Then $F_n$ is closed and a point is non-extremal if and only if it is in the $F_\sigma$-set $F = \bigcup_n F_n$. Thus the set of extremal points $\partial_e K = K \smallsetminus F$ is a $G_\delta$.
Note: I posted the above as a comment in this related MO thread where it is also mentioned that this may fail if $K$ is compact convex but not metrizable (even if $X$ is assumed to be locally convex): the set $\partial_e K$ need not even be a Borel set in $K$ by an example described by Bishop–de Leeuw in section VII of The representations of linear functionals by measures on sets of extreme points. Annales de l'institut Fourier, 9 (1959), p. 305–331.
Added:
Some more details:
The set $C_n = \{(y,z) \in K \times K\,:\,d(y,z) \geq 1/n\}$ is closed since it is the preimage of $[1/n,\infty)$ under the continuous function $d\colon K \times K \to [0,\infty)$. Therefore $C_n$ is compact. The set $F_n$ is the image of the compact set $C_n$ under the continuous function $(y,z) \mapsto \frac{1}{2}(y+z)$, hence $F_n$ is compact and thus closed.
We have $\partial_e K = K \smallsetminus F$ where $F = \bigcup_n F_n$ as above.
If $x$ is not an extremal point, then $x \in F_n$ for some $n$:
By definition $x = (1-\lambda) y + \lambda z$ for some $0 \lt \lambda \lt 1$ and $y \neq x \neq z$. If $\lambda = \frac{1}{2}$ then $x \in F_n$ where $n$ is so large that $\frac{1}{n} \leq d(y,z)$. If $\lambda \neq \frac{1}{2}$ we may (possibly after switching $y$ and $z$) assume that $0 \lt \lambda \lt \frac{1}{2}$ and write $x = \frac{1}{2}(y + z_\lambda)$ where $z_\lambda = (1-2\lambda)y + 2\lambda z \in K$. Since $y \neq z$ we have $y \neq z_\lambda$ and thus $d(y,z_\lambda) \geq \frac{1}{n}$ for large enough $n$, so $x \in F_n$.
Conversely, if $x \in F_n$ for some $n$ then $x$ is clearly not extremal.
Later:
Let me add some remarks on your approaches:
The problem with idea (1) is that $U_{\lambda}$ is not open. In fact, I showed in point 2. above that $U_{1/2} = \partial_e K$ and a small modification of that argument gives that $U_{\lambda} = \partial_e K$ for all $\lambda \in (0,1)$, so you're right that $\partial_e K = \bigcap_{\lambda} U_{\lambda}$, but proving that $U_{\lambda}$ is a $G_\delta$ is the same as the original problem, so that idea won't lead anywhere.
The second idea looks much better, however I doubt that exploiting separability of $K$ only is enough (that is: I doubt that the set of extremal points of a compact convex separable but non-metrizable set is a $G_\delta$, but I haven't checked this thoroughly). I think the argument I gave is one way to get around the difficulties.
Another point I'd like to add is that the distinction of boundary points and interior points you seem to be making does not work for infinite-dimensional compact convex sets. In fact, the Hilbert cube $C$ is homogeneous in the sense that its homeomorphism group acts transitively (see here for a good write-up of the non-trivial proof), so no point is distinguished by topological properties alone. This property is generic in the sense that every compact convex metrizable set in a locally convex vector space is either contained in a finite-dimensional subspace or homeomorphic to $C$ by a theorem of Klee. See my answer here for more on this.
Finally, we can't hope to do much better than $G_\delta$. In three dimensions take the double cone $K$ obtained by taking the convex hull of a circle $C$ of radius $1$ around $(1,0,0)$ in the $(x,y)$-plane and the two points $p_\pm=(0,0,\pm1)$ on the $z$-axis. Then $\partial_e K = \{p_\pm\} \cup C\smallsetminus \{(0,0,0)\}$ is not closed. [In two dimensions the non-extremal points are open in the boundary, hence the set of extremal points of a compact convex set is closed, the one-dimensional case is trivial.] |
H: If $p:E\to B$ is a covering space and $p^{-1}(x)$ is finite for all $x \in B$, show that $E$ is compact and Hausdorff iff $B$ is compact and Hausdorff
I can show that if $E$ is compact and Hausdorff $B$ has the same properties, also I can show that if $B$ is compact and Hausdorff $E$ is Hausdorff, but I have troubles trying to prove that $E$ is also compact. Any suggestions would be appreciated.
I would like to know if there is a short way or at least a simple way to show that if E is Hausdorff so is B, I can prove it but I have to make a lot of observations and I get a really really long demostration.
This is an exercise in Hatcher (Algebraic Topology) Section 1.3, exercise 3
AI: I'll try to answer the question without saying too much so that you can still work on it. I can edit my answer to give a complete solution if need be.
Let $\mathcal{U}$ be an open cover of $E$. Then for each $x\in B$ there exist $p^{-1}(x)$ is finite. Thus we can choose $U^x_1,\ldots, U^x_{n_x}\in\mathcal{U}$ such that $p^{-1}(x)$ is in the union of these sets.
Hints: Look at the image of $U^x_1,\ldots,U^x_{n_x}$ under $p$. Can you get an open set of $B$ from this containing $x$? How can you use this to get an open cover of $B$? How do you extract an open cover of $E$ from this information? |
H: What did Cantor take to be the relationship between the countable ordinals and the power set of the naturals?
I've been told that Cantor sees a relationship between the countable ordinals (Cantor's second number class) and the powerset of the natural numbers.
I've read the "Grundlagen" a few times, but can't seem to locate what he takes this relation to be.
I'm suspecting this is related to the continuum hypothesis CH, though I do not know if my suspicion is correct. Here is why I suspect that this is related to CH. According to wikipedia,
There is no set whose cardinality is strictly between that of the integers and that of the real numbers.
The |powerset of integers| = |powerset of naturals| = |reals| = $2^{\aleph_0}$.
The cardinality of the countable ordinals (the second number class) is identified with $\aleph_1$.
Cantor took $2^{\aleph_0} = \aleph_1$. Hence my suspicion.
My questions:
Am I right that Cantor "conjectured" that the relationship between the countable ordinals and the power set of the naturals is that they are of the same cardinality?
Am I right in my suspicion that this is (or is related to) the continuum hypothesis?
A direct answer to these questions and some commentary would be most appreciated.
Thanks
AI: As Wikipedia says in article about Cantor:
Cantor was the first to formulate what later came to be known as the continuum hypothesis or CH: there exists no set whose power is greater than that of the naturals and less than that of the reals (or equivalently, the cardinality of the reals is exactly aleph-one, rather than just at least aleph-one).
Cantor believed the continuum hypothesis to be true and tried for many years to prove it, in vain.
I'll add two quotes from the book Georg Cantor: His Mathematics and Philosophy of the Infinite by Joseph Warren Dauben,
Cantor's second number class is what we would call today $\omega_1$, the smallest uncountable ordinal,
which might perhaps be interesting for you in connection with this question.
p.110
Despite all that the Grundlagen had accomplished, there was a serious
lacuna. Though Cantor had made it clear that his introduction of the transfinite
numbers, particularly those of the second number class, was essential to
sharpening the concept of power, the question of the power of the continuum
was still unanswered. He hoped a proof would be forthcoming, establishing his
continuum hypothesis that the power of the continuum was none other than that
of the second number class (II). The benefits of such a proof would be
numerous. It would immediately follow that all infinite point sets were either of
the power of the first or second number class, something Cantor had long
claimed. It would also establish that the set of all functions of one or more
variables represented by infinite series was necessarily equal in power to the
second number class. Likewise, the set of all analytic functions or that of all
functions represented in terms of trigonometric series would also be shown to
have the power of the class (II).
The Grundlagen went no further in settling any of these issues. Instead,
Cantor published a sequel in the following year as a sixth in the series of papers
on the Punktmannigfaltigkeitslehre. Though it did not bear the title of its
predecessor, its sections were continuously numbered, 15 through 19; it was
clearly meant to be taken as a continuation of the earlier 14 sections of the
Grundlagen itself. In searching for a still more comprehensive analysis of
continuity, and in the hope of establishing his continuum hypothesis, he
focused chiefly upon the properties of perfect sets and introduced as well an
accompanying theory of content.
p.137
In the fall of 1884 he again took up the intricate problem of
the continuum hypothesis. On August 26, 1884, little more than a week after his
letter of reconciliation to Kronecker, Cantor had written to Mittag-Leffler
announcing, at last, an extraordinarily simple proof that the continuum was
equal in power to the second number class (II). The proof attempted to show
that there were closed sets of the second power. Based upon straightforward
decompositions and the fact that every perfect set was of power equal to that of
the continuum, Cantor was certain that he had triumphed. He summarized the
heart of his proposed proof in a single sentence: "Thus you see that everything
comes down to defining a single closed set of the second power. When I've put
it all in order, I will send you the details."
But on October 20 Cantor sent a lengthy letter to Mittag-Leffler followed
three weeks later by another announcing the complete failure of the continuum
hypothesis. On November 14 he wrote saying he had found a rigorous proof
that the continuum did not have the power of the second number class or of any
number class. He consoled himself by saying that "the eventual elimination of
so fatal an error, which one has held for so long, ought to be all the greater an
advance." Perhaps he was thinking back to the similar difficulties he had
encountered in trying to decide whether or not the real numbers were denumerable,
or how lines and planes might be corresponded. Cantor had come to learn
that one should never be entirely surprised by the unexpected. Nevertheless,
within twenty-four hours he had decided that his latest proof was wrong and that
the continuum hypothesis was again an open question. It must have been
embarrassing for him to have been compelled to reverse himself so often within
such a short period of time in his correspondence with Mittag-Leffler. But even
more discouraging must have been the realization that the simplicity of the
continuum hypothesis concealed difficulties of a high order, ones that, despite
all his efforts and increasingly sophisticated methods, he seemed no better able
to resolve.
In attempting to find a solution for his continuum hypothesis, Cantor was led
to introduce a number of new concepts enabling more sophisticated decompositions
of point sets. These, he hoped, would eventually lead to a means of
determining the power of the continuum. His attempt to publish these new
methods and results marked the final and most devastating episode responsible
for his disillusionment with mathematics and his discontent with colleagues
both in Germany and abroad.
EDIT: You've added clarification to your question that you would like to know this:
Am I right that Cantor "conjectured" that the relationship between the countable ordinals and the power set of the naturals is that they are of the same cardinality?
Am I right in my suspicion that this is (or is related to) the continuum hypothesis?
(1) Yes, I believe that the excerpts I provided above give sufficient support for this claim.
(2) I am used to $\aleph_1=2^{\aleph_0}$ as the usual formulation of CH. And this is the same thing as you wrote in your Question 1. However, we should be careful if we want to avoid Axiom of Choice.
If we are working in ZF, i.e. without Axiom of Choice, this formulation is not equivalent to "There is no set whose cardinality is strictly between that of the integers and that of the real numbers." The reason is that $\aleph_1$ and $2^{\aleph_0}$ can be incomparable.
The relation between these two claims (which are in ZFC both equivalent formulations of CH) is explained in detail here. |
H: Center of mass of a semi-annular plane
I am trying to find the $y$-coordinate of the center of mass of a semi-annular plane region bounded by $1\leq x^2+y^2 \leq 4$ and $ y > 0$. I know that the sketch should look similar to half a donut above the y axis, but I am not sure how to set up the integral with the appropriate limits.
AI: Assuming that the density of the region is uniform, the $y$ coordinate of the center of mass is given as $$y_c = \dfrac{\int_R y dx dy}{\int_R dx dy}$$
The region, $R$, you are interested in i.e. the semi-annular plane as shown below.
Converting this to polar coordinates, note that the $r$ varies from $1$ to $2$ while $\theta$ varies from $0$ to $\pi$. Now you should be able to the integral.
Move your mouse over the gray area for a hint.
Converting to polar coordinates, we get that $y = r \sin(\theta)$ and the area element $dxdy$ gets converted to $r dr d \theta$. Hence, this gives us the center of mass as $$y_c = \dfrac{\int_{r=1}^{2} \int_{\theta=0}^{\pi} r \sin(\theta) r dr d \theta}{\int_{r=1}^{2} \int_{\theta=0}^{\pi} r dr d \theta}$$
Move your mouse over the gray area below for the complete solution.
Hence, we get that \begin{align}y_c & = \dfrac{\int_{r=1}^{2} \int_{\theta=0}^{\pi} r \sin(\theta) r dr d \theta}{\int_{r=1}^{2} \int_{\theta=0}^{\pi} r dr d \theta} = \dfrac{\left(\int_{r=1}^{2} r^2 dr \right) \left(\int_{\theta=0}^{\pi} \sin(\theta) d \theta \right)}{\left(\int_{r=1}^{2} r dr \right) \left(\int_{\theta=0}^{\pi} d \theta \right)}\\ & = \dfrac{\left(\dfrac{2^3-1^3}{3} \right) \times2}{\left(\dfrac{2^2-1^2}{2} \right)\times \pi} = \dfrac{14}{3} \times \dfrac{2}{3 \pi} = \dfrac{28}{9 \pi}\end{align} |
H: Liouville's theorem for Banach spaces without the Hahn-Banach theorem?
Let $B$ be a (complex) Banach space. A function $f : \mathbb{C} \to B$ is holomorphic if $\lim_{w \to z} \frac{f(w) - f(z)}{w - z}$ exists for all $z$, just as in the ordinary case where $B = \mathbb{C}$. Liouville's theorem for Banach spaces says that if $f$ is holomorphic and $|f|$ is bounded, then $f$ is constant.
The only way I know how to prove this uses the Hahn-Banach theorem: once we know that continuous linear functionals on $B$ separate points, we can apply the usual Liouville's theorem to $\lambda(f)$ for every such functional $\lambda : B \to \mathbb{C}$.
Can we avoid using Hahn-Banach? What if $B$ is in addition a Banach algebra?
Motivation: Liouville's theorem is useful in the elementary theory of Banach algebras, where it seems to me that we usually don't need the big theorems of Banach space theory (e.g. the closed graph theorem), and I would like to be able to develop this theory within ZF if possible. It would be very interesting if this were actually independent of ZF.
AI: The usual argument (from Ahlfors) is to use the estimate
$|f'(a)| \leq M/r$, where M is a bound for $|f|$ and $r$
is the radius of a large circle about $0$ containing $a$.
This follows from Cauchy's integral formula. I believe
there is no difficulty proving Cauchy's theorem and
integral formula for Banach space valued functions using
classical methods, since these just estimate absolute
values (replace by norms) and then use completeness.
First you need some integration, but the integral that
is the limit of the integral on step maps suffices. |
H: The boundary is a closed set
A point $p$ in a metric space $X$ is a boundary point of the set $A$, if any neighbourhood of $p$ has points of both $A$ and $X-A$.Prove that the set of all boundary points of $A$ is closed.
My attempt:
By definition of an open set this means that for every $x$ in the boundary there is an open ball centred at $x$ contained in the boundary. An open ball is a neighbourhood of $x$, which implies it contains points of $A$ and $X - A$, which in turn implies there are points in both $A$ and $X - A$ that are in the boundary of $A$.
If $A$ is open, then pick any such point in $A$ that is also in the boundary. This point cannot be in $X - A$ by definition of set subtraction. Further, because $A$ is open there exists an open ball centred around this point contained in $A$. Again, an open ball is a neighbourhood, which means a neighborhood of this point does not contain points of $X - A$, implying it cannot be in the boundary, a contradiction. If A is closed then $X - A$ is open and a symmetric argument holds. Hence the boundary is closed.
Is my work correct?
AI: The boundary of a set $A$ is defined as $\overline{A} \cap \overline{X - A}$. It is the intersection of two closed sets and hence is closed.
By the way your proof is not correct because you assumed that $A$ is either open or closed. There are sets like $(0,1] \subset \Bbb{R}$ in the usual topology that are neither open nor closed. |
H: Exhibiting a ring isomorphism between a ring and itself.
I recently proved to myself that if $R$ is a ring, and $R'$ a set in bijection with $R$, say by $f\colon R'\to R$, then one can turn $R'$ into a ring by defining $0'=f^{-1}(0)$, $1'=f^{-1}(1)$,
$$
r'+s'=f^{-1}(f(r')+f(s')),\qquad r's'=f^{-1}(f(r')f(s')),
$$
and then $f$ is a ring isomorphism.
Now suppose you put a new ring structure on $R$, say $(R,+,\cdot_u,0,u^{-1})$, where $a\cdot_u b=aub$. I want to use the above result as a shortcut to show $(R,+,\cdot, 0,1)$ is isomorphic to $(R,+,\cdot_u, 0,u^{-1})$ by exhibiting a bijection on $R$ which satisfies the four properties I listed above. I've had trouble thinking of what the map would look like. Does anyone see what the map would be?
AI: If $u$ is invertible, then $f: R \rightarrow R$, $r \mapsto ru$ is a bijection and satisfies the properties you want.
$f : R \rightarrow R$, $r \mapsto ur$ will also work. |
H: Matrix commutator question
Here's a nice question I heard on IRC, courtesy of "tmyklebu."
Let $A$, $B$, and $C$ be $2\times 2$ complex matrices. Define the commutator $[X,Y]=XY-YX$ for any matrices $X$ and $Y$. Prove
$$[[A,B]^2,C]=0.$$
AI: Since the trace is additive, and $\mathrm{trace}(XY)=\mathrm{trace}(YX)$, it follows that the trace of any commutator of matrices is zero.
Thus, the trace of $[A,B]$ is $0$, hence it has eigenvalues $\lambda$ and $-\lambda$.
Case 1. $\lambda\neq 0$. Then $[A,B]$ is diagonalizable, and hence so is $[A,B]^2$ (same conjugating matrix). But the eigenvalues of $[A,B]^2$ are $(\lambda)^2$ and $(-\lambda)^2$, so $[A,B]^2$ is diagonalizable with all eigenvalues equal; hence $[A,B]^2$ is actually a scalar multiple of the identity, and therefore commutes with every matrix. Therefore, $[[A,B]^2,C] = 0$.
Case 2. $\lambda=0$. If $[A,B]$ is diagonalizable, then it is the zero matrix, and therefore $[A,B]^2=0$. If $[A,B]$ is not diagonalizable, then its Jordan canonical form is
$$\left(\begin{array}{cc}0 &1\\
0 & 0
\end{array}\right),$$
and hence $[A,B]^2=0$. Either way, $[A,B]^2=0$, hence $[A,B]^2$ commutes with any matrix, so $[[A,B]^2,C]=0$. |
H: Prove equality of integrals
Let $f(x,y) = \text{sgn}(x-y)e^{-|x-y|}$ (Where $\text{sgn}(t)$ is the sign of $t$)
I want to prove the equation below.
$$\int^\infty_0dx \int^\infty_0 f(x,y)dy = -\int^\infty_0 dy \int^\infty_0 f(x,y)dx =-1$$
I don't know how can I start to prove this. Please give some outline for that.
AI: \begin{align}
\int_0^{\infty} f(x,y) dy & = \int_0^{\infty} \text{sgn}(x-y) \exp(-\lvert x - y\rvert) dy\\
& = \int_0^{x} \text{sgn}(x-y) \exp(-\lvert x - y\rvert) dy + \int_x^{\infty} \text{sgn}(x-y) \exp(-\lvert x - y\rvert) dy\\
& = \int_0^{x} \exp(-\lvert x - y\rvert) dy - \int_x^{\infty} \exp(-\lvert x - y\rvert) dy\\
& = \int_0^{x} \exp(-(x - y)) dy - \int_x^{\infty} \exp(x-y) dy\\
& = \int_0^{x} \exp(y-x) dy - \int_x^{\infty} \exp(x-y) dy\\
& = \left. \exp(y-x) \right \rvert_{0}^{x} + \left. \exp(x-y) \right \rvert_{x}^{\infty}\\
& = \left(1 - \exp(-x) \right) + \left( 0 - 1\right)\\
& = - \exp(-x)
\end{align}
Hence,
\begin{align}
\int_0^{\infty} \left(\int_0^{\infty} f(x,y) dy \right) dx & = \int_0^{\infty} - \exp(-x) dx = \left. \exp(-x) \right \rvert_{0}^{\infty} = 0 - 1 = -1
\end{align}
You can do a similar thing for $$\int_0^{\infty} \left(\int_0^{\infty} f(x,y) dx \right) dy$$
\begin{align}
\int_0^{\infty} f(x,y) dx & = \int_0^{\infty} \text{sgn}(x-y) \exp(-\lvert x - y\rvert) dx\\
& = \int_0^{y} \text{sgn}(x-y) \exp(-\lvert x - y\rvert) dx + \int_y^{\infty} \text{sgn}(x-y) \exp(-\lvert x - y\rvert) dx\\
& = \int_0^{y} -\exp(-\lvert x - y\rvert) dx + \int_y^{\infty} \exp(-\lvert x - y\rvert) dx\\
& = \int_0^{y} -\exp(x - y) dx + \int_y^{\infty} \exp(y-x) dx\\
& = -\left. \exp(x-y) \right \rvert_{0}^{y} - \left. \exp(y-x) \right \rvert_{y}^{\infty}\\
& = - \left(1 - \exp(-y)\right) - \left( 0 - 1\right)\\
& = \exp(-y)
\end{align}
Hence,
\begin{align}
\int_0^{\infty} \left(\int_0^{\infty} f(x,y) dx \right) dy & = \int_0^{\infty} \exp(-y) dy = - \left. \exp(-y) \right \rvert_{0}^{\infty} = - \left(0 - 1 \right) = 1
\end{align} |
H: How do I derive these roots
Let $z = \cos(\frac{\pi k}{5}) + i\sin(\frac{\pi k}{5})$
Consider the imaginary part of $z^5$, and deduce that $x^4 - 3x^2 + 1
= 0$ has solutions:
$$2\cos(\frac{\pi}{5}), ~2\cos(\frac{2\pi}{5}),
~2\cos(\frac{3\pi}{5}), ~2\cos(\frac{4\pi}{5})$$
So, 'considering' the imaginary part of $z$, I considered the following to be true:
$$5\cos^4\theta~\sin\theta - 10\cos^2\theta~\sin^3\theta + \sin^5\theta = 0$$
where $\theta = \frac{\pi k}{5}$ since $z^5$ is real $\forall_k \in \mathbb{Z}$
How do I apply this to the polynomial from the question? Normally I would have just used the quadratic formula to solve it, but if I do that I get:
$$x = \pm\sqrt{\frac{3 \pm \sqrt{5}}{2}}$$
but I couldn't seem to relate the two sets of answers easily.
AI: If you assume $\sin\theta \neq 0$ you can divide through your trigonometric equation to get an equation of degree 4
Note that the given roots are in terms of $\cos k\theta$ so use $\sin^2\theta=1-\cos^2\theta$ to find a quartic in $\cos \theta$, and compare with the equation you have been given. Note the factor 2 in the given roots and make the appropriate substitution. |
H: Birational morphism
I have a question on rational and birational maps:
Is the map $$\mathbb{P}^1\rightarrow \mathbb{P}^2, (x:y) \mapsto (x:y:1)$$ rational? Birational? If birational what is its inverse?
Same questions for map $$\mathbb{P}^1 \rightarrow \mathbb{P}^2, (x:y) \mapsto (x:y:0).$$
My guess is that both aren't birational and that both are rational, but would like to hear another opinion.
Thank you
AI: The first map you define is not even a map (except if you consider only the set structure and if the base field is $\Bbb F_2$). In $\Bbb P^1$, the points $[ x : y ]$ and $[ \lambda x : \lambda y ]$ are the same for all non zero $\lambda$ in the base field. So their image must be the same. But obviously all the $[\lambda x : \lambda y : 1 ]$ are not equal.
The second one is a morphism, defined everywhere, but it's not birational since its image is not dense in $\Bbb P^2$. |
H: Tensor product of $R$-algebras
Let $f: R \to S$ and $g: R \to T$ be two $R$-algebras. To show that $S \otimes_R T$ is an $R$-algebra I need to define a ring structure (multiplication) on it and a ring homomorphism $h : R \to S \otimes_R T$.
Using the universal property of the (multi-)tensor product, defining multiplication is clear to me. What I'm confused about is the map $h : R \to S \otimes_R T$. According to this answer here, $r \mapsto 1 \otimes g(r) = f(r) \otimes 1$.
How are they the same?
AI: The basic fact is that for two $R$-modules $M,N$ you have:
$$r\cdot m\otimes _R n=m\otimes_R r\cdot n \quad (\text {for all} \quad r\in R,\; m\in M,\; n\in N) \quad (*)$$
In particular, in your case, you have: $$r\cdot 1_S\otimes _R 1_T=1_S \otimes_R r\cdot 1_T \quad (**)$$
In order to conclude, you just have to remember that built into the notion of algebra is the equality $r\cdot s=f(r)s$ where on the right hand side of the equality $f(r)s$ means the product of the elements $f(r),s$ in the ring $S$.
In particular $r\cdot 1_S=f(r)1_S=f(r)$ and similarly $r\cdot 1_T=g(r)$.
Transporting these last equalities into $(**)$, you get the required equation
$$ f(r) \otimes_R 1_T= 1_S \otimes_R g(r) \quad (***) $$ |
H: Limit finding of an indeterminate form: $\lim\limits_{x\to0} \frac{x^3}{\tan^3(2x)}$
Here is the limit I'm trying to find out:
$$\lim_{x\rightarrow 0} \frac{x^3}{\tan^3(2x)}$$
Since it is an indeterminate form, I simply applied l'Hopital's Rule and I ended up with:
$$\lim_{x\rightarrow 0} \frac{x^3}{\tan^3(2x)} = \lim_{x\rightarrow 0}\frac{6\cos^3(2x)}{48\cos^3(2x)} = \frac{6}{48} = 0.125$$
Unfortuntely, as far as I've tried, I haven't been able to solve this limit without using l'Hopital's Rule. Is it possibile to algebrically manipulate the equation so to have a determinate form?
AI: Use $\lim_{x \to 0} \frac{\sin x}{x} = 1$:
$$\lim_{x \to 0} \frac{x^3}{\tan^3 2x} = \lim_{x\to 0} \left( \frac{(2x)^3}{\sin^3 2x} \cdot \frac{\cos^3 2x}{8} \right) \stackrel{[1 \cdot \frac{1}{8}]}{=} \frac{1}{8}$$ |
H: Does $x\cos(x)$ have oblique asymptotes?
Looking at the graph of $x\cos(x)$ or $x\sin(x)$ etc., it looks like the magnitude of the waves are following a line. Are they oblique asymptotes or something else?
I am familiar with finding the oblique asymptotes of a rational function like $\frac{P(x)}{Q(x)}$ by dividing $Q(x)$ into $P(x)$; however, it doesn't seem like I can do that with $x
\cos(x)$ et al. what is going on here? and how would you find the equation of the function that is 'controlling' the original function's behavior?
AI: For a line to count as an asymptote, the distance between that line and the graph of the function has to tend to zero. This doesn't hold in your examples, so the lines are not asymptotes. |
H: Analytic proof for Circles of Apollonius
I'm looking for an analytic proof the statement for a Circle of Apollonius (I found a geometrical one already): If $\overline{AC}:\overline{BC}=s$, then $P \in k_s$. $s \in (0,1)$.
$k_s$ is the circle. I made the following drawing:
WLOG I can set $A=(0/0)$ and $B=(0/1)$. I really don't know how to go on. Perhaps someone can give a hint?
Greetings
AI: Assume $s\in(0,1)$ and start with
$$
\begin{align}
s|A-C|&=|B-C|\\
s^2|A-C|^2&=|B-C|^2\\
s^2(|A|^2+|C|^2-2A\cdot C)&=|B|^2+|C|^2-2B\cdot C\\
0&=(1-s^2)|C|^2-2(B-s^2A)\cdot C+|B|^2-s^2|A|^2\\
0&=|C|^2-2\frac{B-s^2A}{1-s^2}\cdot C+\frac{|B|^2-s^2|A|^2}{1-s^2}\\
0&=\left|C-\frac{B-s^2A}{1-s^2}\right|^2+\frac{|B|^2-s^2|A|^2}{1-s^2}-\left|\frac{B-s^2A}{1-s^2}\right|^2\\
\left|C-\frac{B-s^2A}{1-s^2}\right|^2&=\left|\frac{B-s^2A}{1-s^2}\right|^2-\frac{|B|^2-s^2|A|^2}{1-s^2}\\
\left|C-\frac{B-s^2A}{1-s^2}\right|^2&=\left(\frac{s}{1-s^2}|B-A|\right)^2\tag{1}
\end{align}
$$
Equation $(1)$ says that $C$ is on the circle with center $\dfrac{B-s^2A}{1-s^2}$ and radius $\frac{s}{1-s^2}|B-A|$. |
H: Applications of graph colorings to discrete math
I'm looking for interesting examples of questions in discrete math that can be proved using graph colorings (as, for example existence of a particular division of people into $k$ distinct groups with given conditions, using, in the proof, that an associated graph is $k$-colorable). I need those example for a basic course in discrete math for undergraduates.
Thanks in advance for any help.
AI: Timothy Gowers discusses one such example 7:10 out in this clip and it continues in part 6.
Summary:
You have seven candidates and eight papers. Each candidate takes either two or three papers and cannot take two at the same time. Can this be done in three sittings? Form a graph with eight nodes, each representing a paper, and draw an edge between two nodes if they cannot be performed in the same sitting (i.e. at least one student is doing both papers). Then three sittings will be sufficient if and only if there exists a coloring of the graph with three colors such that no two nodes of the same color are connected by an edge. |
H: Examples of rings of fractions
I wanted to come up with a few examples of rings of fractions $S^{-1}R$. Can you tell me if these are correct:
1.Let $R = \mathbb Z$, $S = (2 \mathbb Z \setminus \{0\}) \cup \{1\}$. Then every $[x] = \frac{r}{s} \in S^{-1}R$ consists of the elements: $[x] = \{ 2x, \frac12 x\}$.
The ring homomorphism $f: R \to S^{-1}R$ is injective since if $f(r) = [\frac{r}{1}] = [\frac{r^\prime}{1}] = f(r^\prime)$ we have $2k r = 2k r^\prime$ for $2k \in \mathbb Z$ and since $\mathbb Z$ is an integral domain, $r=r^{\prime}$.
2.Now let $R=\mathbb Z / 12 \mathbb Z$ and $S = \{1,2,4,6,8,10\} = (2 R \setminus \{0\}) \cup \{1\}$. Then $f: R \to S^{-1}R$ is not injective since $f(6) = f(3)$.
Are there more interesting examples where $f: R \to S^{-1}R$ is not injective?
AI: It's difficult to find a place to begin, because there are so many examples.
For $R=\mathbb{Z}$,
there is of course the choice $S=R\setminus\{0\}$, producing $\mathbb{Q}$.
You could also just choose a prime ideal, say $P=2\mathbb{Z}$, and use $S=R\setminus P$. This yields the subring of $\mathbb{Q}$ consisting of all fractions of odd denominator.
You could fix a nonzero element, say $a\in R$ and use $S=\{a^i\mid i\in\mathbb{Z}^+\}$. This produces the subring of $\mathbb{Q}$ whose denominators are a power of $a$.
To break from $\mathbb{Z}$ for a second, I'd like to apply the last point to $R=F[x]$ and choose $a=x$. The ring of fractions based on the powers of $x$ then yield $F[x,x^{-1}]$, the Laurent polynomials.
In your second example you put $0$ in your multiplicative set... are you aware that will make $S^{-1}R=\{0\}$?
If the map is not injective, and you exclude $0$ from $S$, then there must be a nonzero zero divisor in $S$. If $a\neq b$ in $R$, but $(a,1)=(b,1)$ in $S^{-1}R$, then that says there exists $s\in S$ such that $as=bs$. This means $(a-b)s=0$.
Using that idea, let $as=0$ for some nonzero $a\in R$, nonzero $s\in S$. Let $b$ be anything nonzero in $R$. Note that $0=as=(a+b-b)s=(a+b)s-bs$. Then $(a+b,1)=(b,1)$.
A simple way to produce a nontrivial ring of fractions such that the canonical map is not an injection would be to find a nonzero zero divisor $x$ in a ring, such that $x$ is not nilpotent. Then localize at the nonnegative powers of $x$. |
H: some question of combination
we know that hilbert seris of n- variables polynomial ring is $\Sigma_{i} \binom{n-1+i}{i}t^{i}$
But, I don't know $\Sigma_{i} \binom{n-1+i}{i}t^{i}=(1-t)^{-n}$.
I wonder to prove in detail.
AI: After the modifications in my comment, we can write the right hand side as
$\frac{1}{(1 - t)^n} = (1 + t + t^2 + \ldots)^n$.
Now the coefficient of $t^i$ in $(1 + t + t^2 + \ldots)^n$ is the number of ways to distribute $i$ identical objects to $n$ distinct containers, which is the coefficient of $t^i$ on the left hand side. |
H: Limits of Subsequences
If $s=\{s_n\}$ and $t=\{t_n\}$ are two nonzero decreasing sequences converging to 0, such that $s_n ≤t_n$ for all $n$. Can we find subsequences $s ′$ of $s$ and $t ′$ of $t$ such that $\lim \frac{s'}{t'}=0$ , i.e., $s ′$ decreases more rapidly than $t ′$ ?
AI: Yes, we can. So we have two positive decreasing sequences with $s_n \leq t_n$, and $s_n \to 0, t_n \to 0$.
Then we can let $t' \equiv t$. As $\{t_n\}$ is positive, $t_1 > 0$. As $s_n \to 0$, there is some $k$ s.t. $s_k < t_1/1$. Similarly, there is some $l > k$ s.t. $s_l < t_2/2$. Continuing in this fashion, we see that we can find a sequence $s'$ so that $s'_n < t_n/n$, so that $\dfrac{s'_n}{t_n} < \dfrac{1}{n}$. |
H: Elementary results from Algebraic Number Theory
The purpose of this question is to motivate me to study algebraic number theory.
Let me explain. My motivation for studying number theory is to learn about beautiful results with simple, accessible statements. For example, the theorem that a prime can be written as the sum of two squares if and only if it is 1 mod 4. I have been reading the first few pages of both Neukirch's Algebraic Number Theory and Serre's Local Fields. I have looked ahead to see what I have to look forward to in terms of such results.
A prime is the sum of two squares if and only if it is 1 mod 4 (done in the first few pages by describing characterizing factorization in the Gaussian integers)
Pythagorean triples (an exercise in the first section)
Solving Pell's Equation
Quadratic Reciprocity (using a result about cyclotomic number fields)
Lagrange's four-square theorem
All of these appear relatively early in the book, and are provable with elementary number theory accessible to a motivated high school student. Looking at some of the later chapters, I wonder: What is the point of all this theory, besides being aesthetically pleasing?
I do not wish to denigrate the absolutely beautiful theory contained in these books. It is marvelous, and I want to learn it. However, I wonder how it can be applied to produce results on the standard integers, especially ones not provable by elementary methods. Fermat's Last Theorem is the obvious example, but surely there are others. A reference where I could find a collection of such results would be especially appreciated. In case it is not clear, I am looking for results provable with methods and theorems developed in the books I mentioned above (or other similar standard graduate texts).
Perhaps the problem is that I don't yet appreciate that results about algebraic integers and number fields can be as thrilling as those about integers. If this is the case, I'm interested in seeing a short list of remarkable theorems about such objects. Maybe I just need to refine my tastes.
To summarize: How can the powerful theorems of algebraic number theory (e.g. the version of Rieman-Roch found in Neukirch, Class Field Theory) be used to give interesting elementary results?
AI: Quite a lot of algebraic number theory was invented through trying to prove Fermat's last theorem and other Diophantine problems.
For example, if I asked you to solve the equation $x^2 - y^2 = 5$ in integers it is very simple, you can factorise $(x+y)(x-y) = 5$ and solve the problem by linking to divisors of $5$, in order to get the solutions $x = \pm 3, y=\pm 2$.
Now say I ask you to solve the equation $y^3 = x^2 + 2$ in integers. This is not so easy if we work entirely in $\mathbb{Z}$ and use elementary methods. However, if we shift focus into a bigger ring of numbers, $\mathbb{Z}[\sqrt{-2}] = \{x+y\sqrt{-2}\,|\,x,y\in\mathbb{Z}\}$ then the problem again turns into a multiplicative problem:
$(x + \sqrt{-2})(x- \sqrt{-2}) = y^3$
so that solving the original Diophantine is really the same as solving a "product" style equation in $\mathbb{Z}[\sqrt{-2}]$.
The point of (basic) algebraic number theory is to study rings like this. How do the elements in these rings factorise?
In our problem above it turns out that the ring $\mathbb{Z}[\sqrt{-2}]$ has properties that are strangely close to properties of $\mathbb{Z}$. In fact the elements in this ring "factorise uniquely" into irreducible elements (the analogue of prime numbers in $\mathbb{Z}$).
The phrase "factorise uniquely" does not have quite the meaning you might think, we have to allow for multiplication of units (things that "divide $1$"). It is the ordering of $\mathbb{Z}$ allows us to consider unique factorisations into "positive primes".
There is also a notion of coprimality. This allows us to solve our problem since for odd $x$ it can be shown that $x\pm\sqrt{-2}$ are coprime in $\mathbb{Z}[\sqrt{-2}]$. But their product is a cube so (as in $\mathbb{Z}$), we must have that $x + \sqrt{-2} = (a+b\sqrt{-2})^3$ for some $a+b\sqrt{-2}\in \mathbb{Z}[\sqrt{-2}]$. Comparing coefficients lets you find the possibilities for $a,b$, hence for $x$.
The ideas of Lame and Kummer were to study FLT in the same way by considering the factorisation (for $\zeta$ a primitive $p$-th root of unity):
$z^p = x^p + y^p = (x + y)(x + \zeta y) ... (x + \zeta^{p-1} y)$
forming yet another product equation, now in the ring $\mathbb{Z}[\zeta]$.
Now this is not the entire story, since some of the rings we study in algebraic number theory do not have unique factorisation. For example the ring $\mathbb{Z}[\sqrt{-5}]$ does not since:
$6 = 2\times 3 = (1+\sqrt{-5})(1 - \sqrt{-5})$
gives two totally different factorisations of $6$. Actually the ring $\mathbb{Z}[\zeta]$ does not have unique factorisation for $p=23$, so that FLT could not be solved entirely by the above method.
The thing that stopped factorisation being unique was the fact that the ring wasn't big enough to factorise everything further into the same things.
Fortunately we can restore unique factorisation without having to extend! Through the genius of Kummer and Dedekind, they realised that by considering the "multiples" of an element as an object in its own right, we can reform factorisation in a way that becomes unique upto ordering.
In modern language these objects are called ideals of a ring. There is a notion of a prime ideal, capturing the notion of prime number. The different factorisations of $6$ above can be explained as reordering of the prime ideals in the factorisation of the ideal generated by $6$. These prime ideals are NOT generated by one element, so they dont correspond to "multiples" of something in $\mathbb{Z}[\sqrt{-5}]$, they correspond more to "multiples" of something that doesn't exist in the ring, but would exist after making an extension.
Kummer was able to prove a huge number of cases of FLT by using the ideal theory. This is outlined in many books.
Focus in algebraic number theory now turns to studying these algebraic constructions. We see that in a given "nice" ring, certain prime numbers may factorise, whereas others don't.
For example, in $\mathbb{Z}[i]$ we find that a prime $p$ factorises further if and only if $p=2$ or $p \equiv 1$ mod $4$. The factorisation of $2$ is different to the others in that $2 = (1+i)^2$ is not "square-free". All the others factorise into two different factors. We say that $2$ ramifies, primes $p\equiv 1$ mod $4$ split and primes $p\equiv 3$ mod $4$ are inert.
This congruence relationship describing the factorisation of primes is in some sense really explained by the values of the Legendre symbol $\left(\frac{-1}{p}\right)$, which is also explaining sums of two squares! Working in similar rings gives you the entire quadratic reciprocity law.
The goal of class field theory is to explain the splitting of primes in ANY extension of "number fields" to get similar characterisations in terms of congruences. In fact I just told a lie, we cannot yet do this for ANY extension, class field theory does it for abelian extensions (ones with abelian Galois group) but never-the-less it is quite a strong theory that has many applications (for example it solves the question of which primes can be written as $x^2 + ny^2$.
In the case of abelian extensions of $\mathbb{Q}$ we find that there are simple congruence conditions mod some integer $N$ that completely describe splitting behaviour of primes!
Another side of class field theory is the Cebotarev density theorem, which states essentially that most splitting types happen infinitely often. This is a huge generalisation of Dirichlet's theorem on primes in arithmetic progressions...in fact it provides an infinite amount of Dirichlet theorems, one for each abelian extension.
These days the (mostly unsolved) Langlands program is supposed to be filling in the gaps for non-abelian extensions but this is very difficult to understand and is not yet completely understood. When this is fully understood it will prove to be the holy grail of number theory, it will characterise in a huge way the splitting of primes.
Anyway, I hope this somewhat rushed introduction will whet your appetite. The book I first started with was Stewart/Tall - Algebraic number theory and fermat's last theorem. This is a good book to start you off. Also Lang - Algebraic number theory, Cox - Primes of the form $x^2 + ny^2$ and Childress - Class field theory are good ones to start with for class field theory. |
H: Proving $\frac{\sin x}{x} =\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{2^2\pi^2}\right) \left(1-\frac{x^2}{3^2\pi^2}\right)\cdots$
How to prove the following product?
$$\frac{\sin(x)}{x}=
\left(1+\frac{x}{\pi}\right)
\left(1-\frac{x}{\pi}\right)
\left(1+\frac{x}{2\pi}\right)
\left(1-\frac{x}{2\pi}\right)
\left(1+\frac{x}{3\pi}\right)
\left(1-\frac{x}{3\pi}\right)\cdots$$
AI: Real analysis approach.
Let $\alpha\in(0,1)$, then define on the interval $[-\pi,\pi]$ the function $f(x)=\cos(\alpha x)$ and $2\pi$-periodically extended it the real line. It is straightforward to compute its Fourier series. Since $f$ is $2\pi$-periodic and continuous on $[-\pi,\pi]$, then its Fourier series converges pointwise to $f$ on $[-\pi,\pi]$:
$$
f(x)=\frac{2\alpha\sin\pi\alpha}{\pi}\left(\frac{1}{2\alpha^2}+\sum\limits_{n=1}^\infty\frac{(-1)^n}{\alpha^2-n^2}\cos nx\right),
\quad x\in[-\pi,\pi]\tag{1}
$$
Now take $x=\pi$, then we get
$$
\cot\pi\alpha-\frac{1}{\pi\alpha}=\frac{2\alpha}{\pi}\sum\limits_{n=1}^\infty\frac{1}{\alpha^2-n^2},
\quad\alpha\in(-1,1)\tag{2}
$$
Fix $t\in(0,1)$. Note that for each $\alpha\in(0,t)$ we have $|(\alpha^2-n^2)^{-1}|\leq(n^2-t^2)^{-1}$ and the series $\sum_{n=1}^\infty(n^2-t^2)^{-1}$ is convergent. By Weierstrass $M$-test the series in the right hand side of $(2)$ is uniformly convergent for $\alpha\in(0,t)$. Hence we can integrate $(2)$ over the interval $[0,t]$. And we get
$$
\ln\frac{\sin \pi t}{\pi t}=\sum\limits_{n=1}^\infty\ln\left(1-\frac{t^2}{n^2}\right),
\quad t\in(0,1)
$$
Finally, substitute $x=\pi t$, to obtain
$$
\frac{\sin x}{x}=\prod\limits_{n=1}^\infty\left(1-\frac{x^2}{\pi^2 n^2}\right),
\quad x\in(0,\pi)
$$
Complex analysis approach
We will need the following theorem (due to Weierstrass).
Let $f$ be an entire function with infinite number of zeros $\{a_n:n\in\mathbb{N}\}$. Assume that $a_0=0$ is zero of order $r$ and $\lim\limits_{n\to\infty}a_n=\infty$, then
$$
f(z)=
z^r\exp(h(z))\prod\limits_{n=1}^\infty\left(1-\frac{z}{a_n}\right)
\exp\left(\sum\limits_{k=1}^{p_n}\frac{1}{k}\left(\frac{z}{a_n}\right)^{k}\right)
$$
for some entire function $h$ and sequence of positive integers $\{p_n:n\in\mathbb{N}\}$. The sequence $\{p_n:n\in\mathbb{N}\}$ can be chosen arbitrary with only one requirement $-$ the series
$$
\sum\limits_{n=1}^\infty\left(\frac{z}{a_n}\right)^{p_n+1}
$$ is uniformly convergent on each compact $K\subset\mathbb{C}$.
Now we apply this theorem to the entire function $\sin z$. In this case we have $a_n=\pi n$ and $r=1$. Since the series
$$
\sum\limits_{n=1}^\infty\left(\frac{z}{\pi n}\right)^2
$$
is uniformly convergent on each compact $K\subset \mathbb{C}$, then we may choose $p_n=1$. In this case we have
$$
\sin z=z\exp(h(z))\prod\limits_{n\in\mathbb{Z}\setminus\{0\}}\left(1-\frac{z}{\pi n}\right)\exp\left(\frac{z}{\pi n}\right)
$$
Let $K\subset\mathbb{C}$ be a compact which doesn't contain zeros of $\sin z$. For all $z\in K$ we have
$$
\ln\sin z=h(z)+\ln(z)+\sum\limits_{n\in\mathbb{Z}\setminus\{0\}}\left(\ln\left(1-\frac{z}{\pi n}\right)+\frac{z}{\pi n}\right)
$$
$$
\cot z=\frac{d}{dz}\ln\sin z=h'(z)+\frac{1}{z}+\sum\limits_{n\in\mathbb{Z}\setminus\{0\}}\left(\frac{1}{z-\pi n}+\frac{1}{\pi n}\right)
$$
It is known that (here you can find the proof)
$$
\cot z=\frac{1}{z}+\sum\limits_{n\in\mathbb{Z}\setminus\{0\}}\left(\frac{1}{z-\pi n}+\frac{1}{\pi n}\right).
$$
hence $h'(z)=0$ for all $z\in K$. Since $K$ is arbitrary then $h(z)=\mathrm{const}$. This means that
$$
\sin z=Cz\prod\limits_{n\in\mathbb{Z}\setminus\{0\}}\left(1-\frac{z}{\pi n}\right)\exp\left(\frac{z}{\pi n}\right)
$$
Since $\lim\limits_{z\to 0}z^{-1}\sin z=1$, then $C=1$. Finally,
$$
\frac{\sin z}{z}=\prod\limits_{n\in\mathbb{Z}\setminus\{0\}}\left(1-\frac{z}{\pi n}\right)\exp\left(\frac{z}{\pi n}\right)=
\lim\limits_{N\to\infty}\prod\limits_{n=-N,n\neq 0}^N\left(1-\frac{z}{\pi n}\right)\exp\left(\frac{z}{\pi n}\right)=
$$
$$
\lim\limits_{N\to\infty}\prod\limits_{n=1}^N\left(1-\frac{z^2}{\pi^2 n^2}\right)=
\prod\limits_{n=1}^\infty\left(1-\frac{z^2}{\pi^2 n^2}\right)
$$
This result is much more stronger because it holds for all complex numbers. But in this proof I cheated because series representation for $\cot z$ given above require additional efforts and use of Mittag-Leffler's theorem. |
H: Inner product space over $\mathbb{R}$
Definition of the problem
I have to prove the following statement:
Let $\left(E,\left\langle \cdot,\cdot\right\rangle \right)$ be an
inner product space over $\mathbb{R}$. prove that for all $x,y\in E$
we have
$$
\left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)\left|\left\langle x,y\right\rangle \right|\leq\left\Vert x+y\right\Vert \cdot\left\Vert x\right\Vert \left\Vert y\right\Vert .
$$
My efforts
I tried two different ways to prove that, both unsuccessfull..
First:
First, by squaring the whole inequality:
$$
\left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)^{2}\left|\left\langle x,y\right\rangle \right|^{2}\leq\left\Vert x+y\right\Vert ^{2}\cdot\left\Vert x\right\Vert ^{2}\left\Vert y\right\Vert ^{2}.
$$
We have from Cauchy-Schwarz that
$$
\left|\left\langle x,y\right\rangle \right|\leq\left\Vert x\right\Vert \cdot\left\Vert y\right\Vert
$$
So we obtain
$$
\left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)^{2}\left|\left\langle x,y\right\rangle \right|^{2}\leq\left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)^{2}\cdot\left\Vert x\right\Vert ^{2}\left\Vert y\right\Vert ^{2}=\left(\left\Vert x\right\Vert ^{2}+\left\Vert y\right\Vert ^{2}+2\left\Vert x\right\Vert \left\Vert y\right\Vert \right)\cdot\left\Vert x\right\Vert ^{2}\left\Vert y\right\Vert ^{2}.
$$
By Pythagorean theorem, we obtain
$$
\left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)^{2}\left|\left\langle x,y\right\rangle \right|^{2}\leq\left(\left\Vert x+y\right\Vert ^{2}+2\left\Vert x\right\Vert \left\Vert y\right\Vert \right)\cdot\left\Vert x\right\Vert ^{2}\left\Vert y\right\Vert ^{2}.
$$
We're almost there, except an extra term very annoying:
$$
\left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)^{2}\left|\left\langle x,y\right\rangle \right|^{2}\leq\left\Vert x+y\right\Vert ^{2}\cdot\left\Vert x\right\Vert ^{2}\left\Vert y\right\Vert ^{2}+2\left\Vert x\right\Vert ^{3}\left\Vert y\right\Vert ^{3}.
$$
Second
I tried after to use only the Cauchy-Schwarz inequality, not squared:
$$
\left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)\left|\left\langle x,y\right\rangle \right|\leq\left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)\cdot\left\Vert x\right\Vert \left\Vert y\right\Vert .
$$
My question
Could you give me a hint/idea on how to solve this problem? which Lemma/Theorem should I use?
Thank you
Franck
AI: Then, let me remove this question from the dead list of "unanswered questions" by answering it.
The statement is false. A counter-example is as follows. Let $E$ be $\mathbb{R}$ itself, and the inner product be the ordinary multiplication of real numbers. Let $x = 1$ and $y = -1$. Then the left hand side is $(||x||+||y||) \cdot | \langle x, y \rangle| = (1 + 1) 1 = 2.$ The right hand side is $||x + y|| \cdot ||x|| \cdot ||y|| = ||0|| \cdot 1 \cdot 1 = 0$. |
H: is the approximation of the sum true?
Someone commented under my question Calculation of the moments using Hypergeometric distribution that
$$
\sum_{k=0}^l\frac{{l \choose k}{2n-l \choose n-k}(2k-l)^q}{{2n\choose n}}\sim \sum_{k=0}^l (2k-l)^q {l \choose k}.
$$
I've tried to use the Stirling's approximation formula, but I did not get the result. Maybe I have mistake... So is this approximation is true?
Thank you for your help.
AI: It's only true when $l = 0$ or $q$ is odd (since both sides are $0$), otherwise it is not true.
(I'll just do the case when $l$ is even, the case when $l$ is odd is pretty much identical.)
Since $\binom{a}{b}$ is at a maximum when $b = a/2$ (or $b = (a\pm 1)/2$ if $a$ is odd) and all the terms in the summation are non-negative ($q$ is even), we have
$$0 \le \sum_{k=0}^l \frac{\binom{2n-l}{n-k}}{\binom{2n}{n}} (2k-l)^q\binom{l}{k} \le \frac{\binom{2n-l}{n-l/2}}{\binom{2n}{n}} \sum_{k=0}^l (2k-l)^q\binom{l}{k}$$
$\binom{a}{a/2}$ is increasing, so it is maximised by taking $a$ as large as possible, that is, choosing $l$ as small as possible; so $l = 2$ ($0$ was eliminated above). Thus
\begin{align*}\frac{\binom{2n-l}{n-l/2}}{\binom{2n}{n}} \sum_{k=0}^l (2k-l)^q\binom{l}{k} &\le \frac{\binom{2n-2}{n-1}}{\binom{2n}{n}} \sum_{k=0}^l (2k-l)^q\binom{l}{k} \\
&= \frac{n}{2n(2n-1)}\sum_{k=0}^l (2k-l)^q\binom{l}{k} \\
&= \frac{1}{2(2n-1)}\sum_{k=0}^l (2k-l)^q\binom{l}{k}
\end{align*}
Thus, the approximation can't hold, since the left-hand side goes to $0$ as $n \to \infty$ while the right hand side doesn't. |
H: probability of passing an exam
I have an exam that I can pass with a probability p.
If I fail the exam, I can retry as many times as I want until I do (the chances to succeed at the 3rd try is still p).
What's the probability that I succeed at the n th retry (which is to say that I took n-1 exams, failed them, and then succeeded at the nth) ?
AI: You have to fail the first with probability $(1-p)$, in the second ..., in (n-1)th and pass in the nth. Then the probability is $(1-p)^{n-1}p$. |
H: Trigonometry in Simple Harmonic Motion
In one of my high school maths questions the example given to find the maximum displacement of a Simple Harmonic Motion where
$ x=2+4\cos \left (2t + \frac{\pi}{3} \right ) $
and the motion lies in the interval:
$-2 \leq x\leq 6$
is:
let $x=2+4\cos \left (2t + \frac{\pi}{3} \right )= 6$
then $\cos
\left (2t + \frac{\pi}{3} \right )= 1$
$2t + \frac{\pi}{3} = 2\pi$
$t = \frac{5\pi}{6}$
I am getting confused on the transition from line 2 to line 3. I'm confused as to how getting rid of the cos turns 1 into $2\pi$. Could someone please explain this at a high school level?
AI: Remember: $$\cos x=1\Longleftrightarrow x=2k\pi\,,\,k\,\,\text{an integer}$$so$$\cos\left(2t+\frac{\pi}{3}\right)=1\Longleftrightarrow 2t+\frac{\pi}{3}=2k\pi$$
Assuming, as surely is the case, that it must be $\,t\geq 0\,$ , we get that$$2k\pi>\frac{\pi}{3}\Longrightarrow k=1,2,3,...$$and we can choose $k=1\Longrightarrow 2t+\frac{\pi}{3}=2\pi$ |
H: Regular polygons that touching to a sphere surface
What is the possible number of n sided polygons(every face is the same regular polygon) that touching their corners to sphere surface and also touching each other ?
I would like to know the relation between n sides polygon and possible placing number.
And also I would like to find a (one side of polygon ) depends on r, n (number of sides), m (total number of polygons that placed on the sphere).
Examples:
Can I place 17 regular triangles on a sphere surface?
Can I place 7 squares on a sphere surface?
Can I place 10 regular hexagons on a sphere surface?
I do not know the related tag about my question. Please feel free to retag it. Thanks a lot for answers.
AI: There is a special case where any $r$, $n$ and $m$ will work for some $a$: if you place the polygons along a single strip, $a$ just needs to be small enough so that the strip doesn't intersect itself. More generally, for any planar arrangement of polygons, if no vertex lies in the interior of the figure, you can embed it on the sphere for any small enough $a$.
Given your figure I don't think this is what you're looking for, so we'll require that (hypothesis $H_1$) some vertex lies in the interior, or equivalently there should be a loop of $k\ge 3$ faces around some vertex. Then you must have $2\pi/k>\pi-2\pi/n$ because of the sphere's curvature. There are exactly 5 $(n,k)$ pairs satisfying those constraints (in particular $3\le n\le 5$, so hexagons won't work).
For each of these pairs there is a Platonic solid:
$$\begin{matrix}
(n,k)&\text{Platonic solid $S_{nk}$}&a/r&m_0\\
\hline
(3,3)&\text{tetrahedron}&\sqrt{8/3}&4\\
(3,4)&\text{octahedron}&\sqrt 2&8\\
(3,5)&\text{icosahedron}&\sqrt{2-2/\sqrt 5}&20\\
(4,3)&\text{cube}&2/\sqrt 3&6\\
(5,3)&\text{dodecahedron}&4/(\sqrt 3\cdot(1+\sqrt 5))&12
\end{matrix}$$
$(n,k)$ uniquely determines the angle between faces and thus the ratio $a/r$, in an injective way as can be seen above. So $k$ must be constant for all vertices lying in the interior.
Platonic solids are solutions, but is every solution a subset of a Platonic solid?
This is always true when $(n,k)=(3,3)$. In the other cases, we must require that the figure be connected, but this still leaves a degree of freedom in rotation around a vertex: so we must require that any two faces are path-connected through edges. Note that because the vertices lie on a sphere and edge length is constant, two edges from the figure are either equal or disjoint except for at most one point.
But given the sphere and a face $F$, the only faces sharing an edge with $F$ are the neighbors of $F$ in the unique Platonic solid containing $F$. So if we define the equivalence relation $F\sim G$ iff $F$ and $G$ are in the same Platonic solid, the connectivity hypothesis implies that all faces are in the same equivalence class, hence in the same Platonic solid. QED.
If we don't require the connectivity constraint, we must at least require that ($H_2$) the interior of the cones around the center of the sphere containing each face is disjoint (otherwise you can "stack" faces above each other, which is probably not what you want). The sum of the area of their intersections with the sphere is proportional to $m$ and maximal for the Platonic solid, proving that the Platonic solid maximizes $m$.
So, conditional to ($H_1$) and ($H_2$), the feasible $(a,r,n,m)$ are those where $a/r$ matches that of $S_{nk}$ for some $k$, and $k\le m\le m_0(S_{nk})$.
Thus:
you can place 17 triangles, by removing 3 triangles from an icosahedron
you can't place 7 squares
you can't place 10 hexagons |
H: proof that translation of a function converges to function in $L^1$
Let $f \in L^1(\mathbb{R})$, for $a\in \mathbb{R}$ let $f_a(x)=f(x-a)$, prove that:
$$\lim_{a\rightarrow 0}\|f_a -f \|_1=0$$
I know that there exists $g\in C(\mathbb{R})$ s.t $\|f-g\|_1 \leq \epsilon$, this is also true for $f_a$ and $g_a$.
Now I have the next estimation:
\begin{align*}
\|f_a-f\|_1 &= \int_{\mathbb{R}} |f(x)-f(x-a)| dx \\
&= \int |f(x)-g(x)+g(x)-g(x-a)+g(x-a)-f(x-a)| \\
&\leq \|f-g\|_1 + \|f_a-g_a\|_1 + \int |g(x)-g(x-a)|
\end{align*}
my question is: I can argue that $\lim_{a\to0} \int |g(x)-g(x-a)| = \lim_{a\to0} \lim_{T\to\infty} |g(x_0)-g(x_0-a)| 2T$, now I think that I can change the order of the limits here, but I am not sure why?
P.s
$x_0$ is some point in $[-T,T]$, and the above is valid from the intermediate integral theorem for continuous functions, right?
Thanks.
AI: What you need to do is to be a little more picky in your choices. You can choose $g$ not only continuous, but also in $L^1$ and compactly supported.
You have already reduced the problem to proving it for $g$. If you know that $g$ is compactly supported, then there exists a $T$ such that
$$
\int_{\mathbb{R}}|g(x)-g(x-a)|=\int_{-T}^T|g(x)-g(x-a)|
$$
and then you don't need to take a limit on $T$.
(By the way, I don't think you can exchange the limits at the end like you want) |
H: Recurrence relation $T_{k+1} = 2T_k + 2$
I have a series of number in binary system as following:
0, 10, 110, 1110, 11110, 111110, 1111110, 11111110, ...
I want to understand : Is there a general seri for my series?
I found this series has a formula as following:
(Number * 2) + 2
but i don't know this formula is correct or is there a general series (such as fibonacci) for my issue.
AI: The series is...
$T_k = 2^k - 2$ |
H: Do improper integrals like $\int_{-\infty}^{+\infty} f$ converge if $xf(x)\rightarrow 0$?
My teacher assumes without proof in his notes that, given a rational function $R(x)$, the improper integral $\int_{-\infty}^{+\infty} R(x)dx$ converges if $\lim_{|x|\rightarrow\infty} xR(x) = 0$. He then proceeds to explain the Estimation lemma, that is a very similar result.
However it is unclear for me whether this fact is true, and if it is bidirectional (i.e. if $\int_{-\infty}^{+\infty} R(x)dx$ converges then $\lim_{|x|\rightarrow\infty} xR(x) = 0$) and holds true even for non-rational functions
AI: One cannot extend the result to general functions. For example, let $f(x)=\frac{1}{|x|\log(|x|)}$ if $|x|\ge 2$, and let $f(x)=f(2)$ for $-2 \lt x\lt 2$.
Then $\lim_{x\to\infty} xf(x)=\lim_{x\to -\infty} xf(x)=0$, but the improper integral does not converge.
The result does hold, bidirectionally, for rational functions whose denominator vanishes nowhere. It is essentially obvious, though writing out the proof is typographically unpleasant. For a rational function $R(x)=\frac{P(x)}{Q(x)}$, where $P$ and $Q$ are polynomials with $Q(x)$ nowhere $0$ (and $P$ not identically $0$), there is convergence iff $\deg(Q)\ge 2+\deg(P)$. This is precisely the condition for $\lim_{|x|\to \infty} xR(x)$ to be $0$. The reason is basically that degrees can only take on integer values. |
H: On the $PSL(2, p)$, $p$ a Mersenne prime
We know $|PSL(2,p)|=p(p+1)(p-1)/2$. Let $p$ be Mersenne prime (that is $p+1=2^{n}$) and $r$ be prime divisor of $(p-1)/2$.
My question: What is the number of Sylow $r$-subgroups of $PSL(2,p)$?
AI: Let $G=\operatorname{PSL}(2,q)$ for q an odd prime power. If r is an odd prime not dividing q but dividing $|G|$, then r divides exactly one of $\{q+1,q-1\}$ and a Sylow r-subgroup of G is cyclic with normalizer dihedral of order $q-\epsilon$ ($\epsilon = \pm1$). In particular, the number of Sylow r-subgroups is $q(q+\epsilon)/2$.
The normalizer is a at least this big because it obviously normalizes the subgroup, and it is no bigger, since it is a maximal subgroup.
This does not require q to be a Mersenne prime, or even prime (just odd), and does not require r divide $q-1$ (though in your case this means $\epsilon=1$). |
H: A "prime-mapping" polynomial
Suppose that $f$ is a polynomial with integer coefficients with the property that for any prime $p$, $f(p)$ is a prime. Is there any such polynomial $f$ other than $f(x)=x$ of course?
My approach was that if the leading coefficient $a_{0}$ of $f$ is $0$, then $f(p)=p$ for any prime $p$, so $f(x)-x$ has infinite roots $\implies f(x)=x$. If $\deg{f}=n$ and if $a_{0}$ has $\gt n$ prime factors, then also, the same argument works - but I couldn't complete my argument.
Any help will be greatly appreciated!
AI: The constant functions $f(x)=p$, where $p$ is prime, have the desired property, as does the function $f(x)=x$. To show there are no others, suppose that $f$ has positive degree. Suppose also that for some prime $p$, we have $f(p)=q$, where $q$ is a prime different from $p$. (If $f(p)=p$ for all primes $p$, then $f(x)=x$, since a non-zero polynomial of degree $m$ cannot have more than $m$ zeros.)
Then $f(p+nq)\equiv 0 \pmod{q}$ for every integer $n$. But by Dirichlet's Theorem, there are infinitely many primes in the arithmetic progression $p, p+q, p+2q, \dots$. If $p^\ast$ is a large enough such prime, then $f(p^\ast)$ is larger than $q$, but divisible by $q$, and therefore not prime. |
H: Show that $\sum\nolimits_{d|n} \frac{1}{d} = \frac{\sigma (n)}{n}$ for every positive integer $n$.
Show that $\sum\nolimits_{d|n} \frac{1}{d} = \frac{\sigma (n)}{n}$ for every positive integer $n$.
where $\sigma (n)$ is the sum of all the divisors of $n$
and $\sum\nolimits_{d|n} f(d)$ is the summation of $f$ at each $d$ where $d$ is the divisor of $n$.
I have written $n=p_1^{\alpha_1}p_2^{\alpha_2}p_3^{\alpha_3}.......p_k^{\alpha_k}$ then:-
$$\begin {align*} \sum\nolimits_{d|n} \frac{1}{d}&=\frac{d_2.d_3......d_m+d_1.d_3......d_m+........+d_1.d_2.d_3......d_{m-1}}{d_1.d_2.d_3......d_m} \\&=\frac{d_2.d_3......d_m+d_1.d_3......d_m+........+d_1.d_2.d_3......d_{m-1}}{p_1^{1+2+...+\alpha_1}p_2^{1+2+...+\alpha_2}p_3^{1+2+....+\alpha_3}.......p_k^{1+2+....+\alpha_k}} \\ \end{align*}$$
where $d_i$ is some divisor among the $m$ divisors.
Then I cannot comprehend the numerator so that to get the desired result.
Also suggest some other approches to this question.
AI: $\displaystyle
n\sum_{d|n} \frac{1}{d} = \sum_{d|n} \frac{n}{d} = \sum_{d|n} {d} = \sigma (n)
$
or
$\displaystyle
\frac{\sigma (n)}{n} = \frac{1}{n} \sum_{d|n} {d} = \sum_{d|n} \frac{d}{n} = \sum_{d|n} \frac{1}{d}
$ |
H: How many numbers $\in [1 .. 10^9]$ that are not of the form $ x^2, x^3$ or $x^5$?
We've started with $x^2$, saying that there are $\sqrt{10^9}$ numbers that are not $\in 10^9$
i'm thinking that if we add $\sqrt[3]{10^9}$ and $\sqrt[5]{10^9}$ to $\sqrt{10^9}$ and subtract them from $10^9$, we would have gone too far and excluded too many numbers due to repeating numbers such as $2^2$,$2^3$ and $2^5$, for example.
does it even make sense to do so? aren't all numbers in $\sqrt[3]{10^9}$ and $\sqrt[5]{10^9}$ $< 10^9$ already in $\sqrt{10^9}$ ?
AI: Let $a=10^9$. There are $\lfloor a^{1/2} \rfloor$ squares, $\lfloor a^{1/3}\rfloor$ cubes and $\lfloor a^{1/5}\rfloor$ $5^{th}$ powers within $1$ to $10^9$. So, the P.I.E. gives the answer to be: $a-(\lfloor a^{1/2}\rfloor+\lfloor a^{1/3}\rfloor+\lfloor a^{1/5}\rfloor-\lfloor a^{1/6}\rfloor-\lfloor a^{1/10}\rfloor-\lfloor a^{1/15}\rfloor+\lfloor a^{1/30}\rfloor)$. |
H: The limit of integral
Let $1 \le p < \infty$ and assume $f \in L^p(\mathbb{R})$.
I'm trying to prove the limit of integral
$$\lim_{x \to \infty} \int^{x+1}_x f(t)dt =0.$$
Can I use Riesz Theorem for Banach spaces?
AI: Note that we need only prove this for real-valued $f\geq 0$, since
$$\left|\int_x^{x+1}f(x)dx\right|\leq \int_x^{x+1}|f(x)|dx.$$
Let $\epsilon>0$, and choose some simple function $s\leq f$ such that $\int fd\mu<\int sd\mu+\epsilon/2$. Since $s$ is simple and integrable, we have that
$$s(x)=\sum\limits_{n=0}^k \alpha_n\chi_{A_n}(x)$$
for some collection $\{A_n\}$ of measurable sets, and each $A_n$ has finite measure except $A_0$ (for which $\alpha_0=0$). Thus $A=\bigcup\limits_{n=1}^k A_n$ has finite measure, so we have some interval $I$ such that $\mu(A\setminus I)<\epsilon/2\max\{|\alpha_1|,\ldots,|\alpha_n|\}$. Thus for sufficiently large $x$ we have $(x,x+1)\cap I=\emptyset$ so
$$\int_x^{x+1} f(x)dx<\int_x^{x+1}s(x)dx+\epsilon/2< \max\{|\alpha_1|,\ldots,|\alpha_n|\}\cdot \epsilon/2\max\{|\alpha_1|,\ldots,|\alpha_n|\}+\epsilon/2=\epsilon$$
i.e. we have $\lim\limits_{x\to \infty} \left|\int_x^{x+1}f(x)dx\right|<\epsilon$. Since this is true for all $\epsilon>0$, the limit must be $0$. |
H: Measurable function on the interval $[0,1]$
Assume that $f$ is a measurable function on the interval $[0,1]$ such that $0<f(x)<\infty$ for $x \in [0,1]$. Then, how can I prove the inequality below?
$$\int^1_0 f(x)dx \int^1_0 {1 \over {f(x)}} dx \ge 1$$
AI: Here is another way. What you have written is nothing but the Arithmetic mean-Harmonic mean inequality for functions. In case, of a finite set of positive numbers $\{a_i\}$'s, the AM-HM inequality reads
$$\dfrac{\sum_{i=1}^{n} w_i a_i}{\sum_{i=1}^{n} w_i} \geq \dfrac{\sum_{i=1}^{n} w_i}{\sum_{i=1}^{n} \dfrac{w_i}{a_i}}$$where $w_i \geq 0$ are the weights. The weights can be normalized to $1$ i.e. if we enforce that $\displaystyle \sum_{i=1}^{n} w_i = 1$, then we get that $$\sum_{i=1}^{n} w_i a_i \geq \dfrac1{\displaystyle \sum_{i=1}^{n} \dfrac{w_i}{a_i}}$$
Hence, we get that $$\left(\sum_{i=1}^{n} w_i a_i \right) \left(\sum_{i=1}^{n} \dfrac{w_i}{a_i} \right)\geq 1$$
If you partition the interval $[0,1]$ into $n$ disjoint measurable sets, say $E_1^{(n)},E_2^{(n)},\ldots,E_n^{(n)}$, such that $\displaystyle \bigcup_{k=1}^{n} E_k^{(n)} = [0,1]$, choose $w_i^{(n)} = \mu^{(n)} \left(E_i^{(n)}\right)$. Note that $\sum_i w_i^{(n)} = 1$.
Approximate $f(x)$ from below using step functions on these intervals i.e. $f_{step}^{(n)} = f_i^{(n)}$ on the interval $E_i^{(n)}$ such that $\displaystyle \sum_{i=1}^{n} f_i^{(n)} \mu(E_i^{(n)}) = \int f_{step}^{(n)} \to \int f$.
Since $f>0$ on $[0,1]$, $\dfrac1f$ can be approximated from above using the step function $(1/f)_{step}^{(n)} = 1/f_i^{(n)}$ on the interval $E_i^{(n)}$ such that $\displaystyle \sum_{i=1}^{n} \dfrac1{f_i^{(n)}} \mu(E_i^{(n)}) = \int \dfrac1{f_{step}^{(n)}} \to \int \dfrac1{f}$.
From the AM-HM inequality we have that $\left(\displaystyle \sum_{i=1}^{n} \mu(E_i^{(n)}) f_i^{(n)} \right) \left( \displaystyle \sum_{i=1}^{n} \dfrac{\mu(E_i^{(n)}) }{f_i^{(n)}} \right) \geq 1$.
Take the limit to conclude that $$\left(\int f dx \right) \times \left(\int \dfrac1f dx \right)\geq 1$$
You may need to argue out and justify some parts of the above argument to make it into a rigorous proof but hope the idea is clear. |
H: Minimal polynomial of $\sqrt2+1$ in $\mathbb{Q}[\sqrt{2}+\sqrt{3}]$
I'm trying to find the minimal polynomial of $\sqrt2+1$ over $\mathbb{Q}[\sqrt{2}+\sqrt{3}]$. The minimal polynomial of $\sqrt2+1$ over $\mathbb{Q}$ is
$$ (X-1)^2-2.$$
So I look at $\alpha = \sqrt2 + \sqrt3$
$$ \alpha^0 = 1$$
$$ \alpha^1 = \sqrt2 + \sqrt3.$$
I cannot find $a,b$ such that $a\alpha + b\alpha = \sqrt2+1$. So the degree 2 is mimimal. Is that correct?
AI: Letting $t = \sqrt{2} + \sqrt{3}$, $\sqrt{2} + 1 = 1 + \frac{1}{2}(t^3 - 9t)$ so the minimal polynomial is $x - \sqrt{2} - 1$. |
H: probability of passing an exam (continued)
Following my previous question, which can be found here:
probability of passing an exam,
I found out that the probability of passing an exam at the nth try is $p(1-p)^{n-1}$.
If I now assume that taking an exam takes me one hour of work, how many hours on average will I have worked if the maximum number of retries is N (regardless of whether I end up passing the exam or not) ?
AI: The probability of taking the exam $n$ times is then $p(1-p)^{n-1}$ except for the last, which has probability $(1-p)^{N-1}$ as you stop in any case. The average or expected number of hours is then $$\sum_{i=1}^N P(i)i=\sum_{i=1}^{N-1} ip(1-p)^{i-1}+(1-p)^{N-1}N$$
To sum the series, let $q=1-p$. Then $$\sum_{i=1}^{N-1} ip(1-p)^{i-1}=p\sum_{i=1}^{N-1} iq^{i-1}=p\sum_{i=1}^{N-1} \frac d{dq} q^i=p \frac d{dq} \frac {q-q^N}{1-q}$$ |
H: Show that $\frac{n}{\sigma(n)} > (1-\frac{1}{p_1})(1-\frac{1}{p_2})\cdots(1-\frac{1}{p_r})$
If $n=p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r}$ is the prime factorization of $n>1$ then show that :
$$1>\frac{n}{ \sigma (n)} > \left(1-\frac{1}{p_1}\right)\left(1-\frac{1}{p_2}\right)\cdots\cdots\left(1-\frac{1}{p_r}\right)$$
I have solved the $1^\text{st}$ inequality($1>\frac{n}{ \sigma (n)}$) and tried some manipulations on the right hand side of the $2^\text{nd}$ inequality but can't get much further.Please help.
AI: Note that the function $\dfrac{n}{\sigma(n)}$ is multiplicative. Hence, if $n = p_1^{k_1}p_2^{k_2} \ldots p_m^{k_m}$, then we have that $$\dfrac{n}{\sigma(n)} = \dfrac{p_1^{k_1}}{\sigma \left(p_1^{k_1} \right)} \dfrac{p_2^{k_2}}{\sigma \left(p_2^{k_2} \right)} \ldots \dfrac{p_m^{k_m}}{\sigma \left(p_m^{k_m} \right)}$$ Hence, it suffices to prove it for $n = p^k$ where $p$ is a prime and $k \in \mathbb{Z}^+$.
Let $n=p^k$, then $\sigma(n) = p^{k+1}-1$. This gives us that $$\dfrac{n}{\sigma(n)} = p^k \times \dfrac{p-1}{p^{k+1}-1} = \dfrac{p^{k+1} - p^k}{p^{k+1} - 1} = 1 - \dfrac{p^k-1}{p^{k+1}-1}.$$ Since $p > 1$, we have that $p(p^k-1) < p^{k+1} - 1 \implies \dfrac{p^k-1}{p^{k+1}-1} < \dfrac1p \implies 1 - \dfrac{p^k-1}{p^{k+1}-1} > 1 - \dfrac1p$. Hence, if $n=p^k$, then $$\dfrac{n}{\sigma(n)} > \left( 1 - \dfrac1p\right)$$
Since, $\dfrac{n}{\sigma(n)}$ is multiplicative, we have the desired result. |
H: Is there any permutation $x≠1$ leaving at least $n-2k$ letters fixed at this group?
This question has an answer which I am noting both here.
Q: Suppose that $G$ is permutation group of degree $n$. If for an integer $k$ where $4≤2k≤n$ we have $|G|≥(n-k)!k$ then $G$ contains a permutation $x≠1$ that leaves at least $n-2k$ letters fixed.
A: The number of $k-$cycles in $S_n$ is equal to $\binom{n}{k}(k-1)!$. Since $|G|≥(n-k)!k$ so, $|S_n:G|≤\frac{n!}{(n-k)!k}=\binom{n}{k}(k-1)!$. Hence either $G$ contains a $k-$cycle like $x$ or some right coset of $G$ contains at least two $k-$cycles $y$ and $z$. In the latter case we take $x=yz^{-1}$. (This result is due to Netto).
Why does he insist working with a $k-$cycle say $x$ rather than a $2k-$cycle permutaion? Doesn't a $2k-$cycle leave $n-2k$ letters unmoved? Unfortunately, I am baffling of his second result. Why $yz^{-1}$? Wasn't $yz$ useful? Thanks so much for any help.
AI: Here is his argument in the other order, so that maybe it makes more sense.
There are N different k-cycles and at most N different cosets of G in $S_n$. That means that if any coset lacks a k-cycle, then some coset Ga has two k-cycles, say y and z. Since cosets are equal if they are not disjoint, we get $Ga = Gy = Gz$. More usefully we get that $yz^{-1} \in G$. Now $yz^{-1}$ is a product of two k-cycles so moves at most $2k$ points, and leaves the other (at least) $n-2k$ points alone.
This means we are done if some coset lacks a k-cycle. The only alternative is that every coset has a k-cycle. Well G is a coset of G, and so it must have a k-cycle and we are definitely done. |
H: Is $\mathbb{Q}[\sqrt2]$ = $\mathbb{Q}[\sqrt2+1]$?
Is $\mathbb{Q}[\sqrt2]$ = $\mathbb{Q}[\sqrt2+1]$?
I think so because
$$\mathbb{Q}[\sqrt{2}+1] = \{\sum_{i=0}^{n}c_i(\sqrt{2}+1)^i\mid n\in\mathbb{N}, c_i\in\mathbb{Q}\}$$
$$= \{\sum_{i=0}^{n}c_i(\sqrt{2})^i\mid n\in\mathbb{N}, c_i\in\mathbb{Q}\} = \mathbb{Q}[\sqrt{2}].$$
This could be worked out with Binomial theorem right?
AI: Actually show that: $$\mathbb Q[\sqrt2]=\left\{a+b\sqrt2\mid a,b\in\mathbb Q\right\}\\ \mathbb Q[\sqrt2+1]=\left\{a+b(\sqrt2+1)\mid a,b\in\mathbb Q\right\}\\$$
So you do not need the "entire" binomial argument, not to take arbitrarily long combinations. Two is enough.
Now you can either use the argument Alex gave, or we can show two-sided inclusions directly:
Suppose that $a+b(\sqrt2+1)$ is in $\mathbb Q[\sqrt2+1]$, take $c=a+b$ (which is a rational number) and then $a+b(\sqrt2+1)=a+b\sqrt2+b=c+b\sqrt2\in\mathbb Q[\sqrt2]$. Therefore $\mathbb Q[\sqrt2+1]\subseteq\mathbb Q[\sqrt2]$.
Take now $a+b\sqrt2\in\mathbb Q[\sqrt2]$ then $a+b\sqrt2=a+(-b+b)+b\sqrt2=(a-b)+b(\sqrt2+1)\in\mathbb Q[\sqrt2+1]$. Therefore $\mathbb Q[\sqrt2]\subseteq\mathbb Q[\sqrt2+1]$.
Therefore we have equality. |
H: Bounded operator that does not attain its norm
What is a bounded operator on a Hilbert space that does not attain its norm? An example in $L^2$ or $l^2$ would be preferred.
All of the simple examples I have looked at (the identity operator, the shift operator) attain their respective norms.
AI: For an example in $L^2[0,1]$, consider the operator of multiplication by $x$, i.e. $(Tf)(x) = x f(x)$. |
H: Evaluating $ \int_0^{\infty } \exp\left(-g x-\frac{x^2}{2}-\frac{x^2 z}{1-z}\right) x^k \sin(hx) \, dx $
I'm attempting to evaluate the following integral, so far, with little success. Any help would be appreciated:
$$
\ \int_0^{\infty } \exp\left(-g x-\frac{x^2}{2}-\frac{x^2 z}{1-z}\right) x^k \sin(hx) \, dx
$$
All paramaters are real.
Mx
AI: Combining summand is the exponents one has that the integral is a Laplace transform $x\to g$ of the function
$f(x,a,h)=e^{-a x^2} x^k \sin hx\;$.
For $k=0$, $a,g>0\;$ Mma gives
$$
L[f]=
-\frac{i \sqrt{\pi } e^{\frac{(g-i h)^2}{4 a}} \left(\text{erfc}\left(\frac{g-i h}{2
\sqrt{a}}\right)-e^{\frac{i g h}{a}} \text{erfc}\left(\frac{g+i h}{2
\sqrt{a}}\right)\right)}{4 \sqrt{a}}.
$$
Now note that for natural $k$ the result is equal to
$(-1)^k\frac{\partial^k}{\partial g^k}L[f]$, so it seems where is no good formula for arbitrary $k$. |
H: L-measurable function and integral
Assume that $f:E \to [0,\infty]$ where $E \subseteq \mathbb{R}^n$ is a measurable set, and $f$ is $\mathbb{L}$-measurable. And use $x \in \mathbb{R}^n$ and $y \in \mathbb{R}$.
First I'm wondering why the subsets A and B stated below are measurable.
$$A=\{(x,y) \in \mathbb{R}^{n+1} | 0 \le y < f(x), x \in E\}$$
$$B=\{(x,y) \in \mathbb{R}^{n+1} | 0 \le y \le f(x), x \in E\}$$
And then how can I conclude following equation for measure value?
\begin{align}
\lambda(A)=\lambda(B)=\int_E f(x)dx & = \int_0^\infty \lambda(\{x \in E | f(x) >y\})dy \\
&= \int_0^\infty \lambda(\{x \in E | f(x) \ge y\})dy
\end{align}
AI: If $f$ is Lebesgue measurable, then the function $\tilde{f}(x,y)=y-f(x)$ is also Lebesgue measurable. Then $A=\{y \geq 0\} \cap \{ \tilde{f}(x,y)<0\}$ and $B=\{y \geq 0\} \cap \{\tilde{f}(x,y) \leq 0\}$ are both measurable.
For each $x \in E$, the slices $A_x=\{y \in \mathbb{R} \, | \, (x,y) \in A\}$ and $B_x=\{y \in \mathbb{R} \, | \, (x,y) \in B\}$ both have measure $f(x)$ (since they are, resp., the segments $[0,f(x))$ and $[0,f(x)]$). Thus,
$m(A)=\int \chi_A(x,y) \, dx \, dy=\int_E m(A_x)\,dx=\int_E f(x) \, dx$.
And similarly for $m(B)$. |
H: Evaluate the sum: $\sum\limits_{n=0}^{\infty} \frac1{F_{(2^n)}}$
Evaluate the sum:
$$\sum_{n=0}^{\infty} \frac{1}{F_{(2^n)}}$$
where $F_{m}$ is the $m$-th term of the Fibonacci sequence. I need some support here. Thanks.
AI: As wikipedia claims the result follows from the identity
$$
\sum\limits_{n=0}^N\frac{1}{F_{2^n}}=3-\frac{F_{2^N-1}}{F_{2^N}}
$$
You can try to prove it by induction. |
H: Multi variable integral : $\int_0^1 \int_\sqrt{y}^1 \sqrt{x^3+1} \, dx \, dy$
$$\int_0^1 \int_\sqrt{y}^1 \sqrt{x^3+1} \, dx \, dy$$
Here is my problem in my workbook. If I solve this problem by definition, that find integral for $x$, after that solve for $y$. so $\int_\sqrt{y}^1 \sqrt{x^3+1} \, dx$ is so complicate. I have used Maple, but the result still long and complicate that I cannot use it to find integral for $y$.
Thanks :)
AI: The way you have the integration setup, you integrate along $x$ first for a fixed $y$. The figure below indicates how you would go about with the integration. You integrate over the horizontal red strip first and then move the horizontal strip from $y=0$ to $y=1$.
Now for the ease of integration, change the order of integration and integrate along $y$ first for a fixed $x$. The figure indicates how you would go about with the integration. You integrate over the vertical red strip first and then move the vertical strip from $x=0$ to $x=1$.
Hence, if you swap the integrals the limits become $y$ going from $0$ to $x^2$ and $x$ goes from $0$ to $1$.
$$I = \int_0^1 \int_\sqrt{y}^1 \sqrt{x^3+1} dx dy = \int_0^1 \int_0^{x^2} \sqrt{x^3+1} dy dx = \int_0^1 x^2 \sqrt{x^3+1} dx$$
Now call $x^3+1 = t^2$. Then we have that $3x^2 dx = 2t dt \implies x^2 dx = \dfrac{2}{3}tdt$. As $x$ varies from $0$ to $1$, we have that $t$ varies from $1$ to $\sqrt{2}$.
Hence, we get that
$$I = \int_1^\sqrt{2}\dfrac23 t \times t dt = \dfrac23 \int_1^\sqrt{2}t^2 dt = \dfrac23 \times \left. \dfrac{t^3}3\right \vert_{t=1}^{t=\sqrt{2}} = \dfrac29 \left( (\sqrt{2})^3 - 1^3\right) = \dfrac29 (2 \sqrt{2} - 1).$$ |
H: Proof that $A \otimes B \cong B \otimes A$
Possible Duplicate:
There exists a unique isomorphism $M \otimes N \to N \otimes M$
I want to show that for Abelian groups $A$ and $B$ that the tensor product $A \otimes B$ is isomorphic to $B \otimes A$. I believe that I have accomplished this and have posted my attempt as an answer to my own question. I would appreciate any constructive feedback.
AI: Let's use the universal definition of $A\otimes B$: $(A\otimes B,\iota)$, where $\iota\colon A\times B\to A\otimes B$ is a bilinear (resp. $R$-bilinear) map, is the unique group (resp. $R$-module) with the property that for any bilinear map $\phi\colon A\times B\to C$ (resp. $R$-bilinear), where $C$ is any abelian group (rep. $R$-module), there exists a unique homomorphism $\Phi\colon A\otimes B\to C$ such that $\phi=\Phi\circ\iota$.
Let $s\colon A\times B\to B\times A$ be the map $s(a,b) = (b,a)$, and let $(B\otimes A,j)$ be the tensor product of $B$ and $A$. Note that $B\otimes A$ has the corresponding universal property relative to $B\times A$ and $j$.
Now, the map $js\colon A\times B\to B\otimes A$ is a bilinear map (composition of a homomorphism and a bilinear map). Therefore, there exists a unique $\mathcal{F}\colon A\otimes B\to B\otimes A$ such that $js = \mathcal{F}\iota$. Likewise, the map $\iota s^{-1}\colon B\times A\to A\otimes B$ is bilinear, so there exists a unique $\mathcal{G}\colon B\otimes A\to A\otimes B$ such that $\iota s^{-1} = \mathcal{G}j$.
Now consider $\mathcal{GF}\colon A\otimes B\to A\otimes B$. We have that
$$\mathcal{GF}\iota = \mathcal{G}js = \iota s^{-1}s = \iota.$$
But there is supposed to be a unique map $f\colon A\otimes B\to A\otimes B$ such that $f\circ\iota = \iota$ (since $\iota$ is bilinear), and clearly $f=\mathrm{id}_{A\otimes B}$ works. Therefore, $\mathcal{GF}=\mathrm{id}_{A\otimes B}$.
Symmetrically, $\mathcal{FG}j = \mathcal{F}\iota s^{-1} = jss^{-1} = j$. But $j$ is a bilinear map $B\times A\to B\otimes A$, so there is supposed to be a unique map $g\colon B\otimes A\to B\otimes A$ such that $gj=j$. Since $\mathrm{id}_{B\otimes A}$ works, that is the unique function with the desired property. Since $\mathcal{FG}$ also works, $\mathcal{FG}=\mathrm{id}_{B\otimes A}$.
Thus, $\mathcal{FG}=\mathrm{id}_{B\otimes A}$ and $\mathcal{GF}=\mathrm{id}_{A\otimes B}$. Therefore, $\mathcal{F}\colon A\otimes B\to B\otimes A$ is an isomorphism, as desired.
Note that we don't need to know how we represent $A\otimes B$; we just need the universal property (and that a tensor product exists for any [ordered] pair of groups). Though one can likewise use the universal property to show that if $(M,\iota)$ is a tensor product for $A\times B$, then $(M,\iota s^{-1})$ is a tensor product for $B\times A$, so you just need to know $A\otimes B$ exists. |
H: A continuity condition for a bilinear form on a Hilbert space
Let $H$ be a real Hilbert space, and let $B : H \times H \to \mathbb{R}$ be bilinear and symmetric. Suppose there is a constant $C$ such that for all $x \in H$, $|B(x,x)| \le C \|x\|^2$. Must $B$ be continuous?
This seems like it should just be a simple polarization argument, but for some reason I can't see it.
AI: For $t > 0$,
$$4 |B(x,y)| = |B(x/t+ty,x/t+ty) - B(x/t-ty,x/t-ty)| \le C (\|x/t+ty\|^2 + \|x/t-ty\|^2) \le 2 C (\|x\|/t+t\|y\|)^2$$
Take $t = \sqrt{\|x\|/\|y\|}$. |
H: Cyclic refinements of abelian towers
I was looking through Lang's Algebra and found the following statement,
Let $G$ be a finite group. An abelian tower of $G$ admits a cyclic refinement.
After some work, I understand the proof, and now I want to show that we cannot drop the hypothesis that $G$ is finite. Its enough to find an infinite abelian group which does not have a cyclic tower. Does anyone know of any such groups? Thanks!
AI: $$\large(\mathbb{Q},+){}{}{}{}{}{}{}{}{}$$
Also:
$$\large(\mathbb{Z}_{p^{\infty}},+).$$ |
H: Very Important question. Limiting distribution
I have an exam in the morning and there is still one question I cannot do.
$X_1, \ldots, X_n$ are iid random variables each having distribution with density $f_{X_i}(x;\theta)= 1/\theta$, for $x \in [0,\theta]$ where $\theta>0$ compute the CDF of the random variable $\max(X_1,\ldots X_n)$ and prove that
$n(\theta-\max(X_1,\ldots,X_n)) \to W$ in distribution and state the CDF of $W$.
How can I do this? I have worked out that the CDF of $\max(X_1,\ldots,X_n)$ is $ (x/\theta)^n$ but that is all :(
Thanks.
AI: You've already worked out the CDF of $\max(X_1,...,X_n)$. Now just find the CDF of $n(\theta-\max(X_1,\ldots,X_n))$ (use the definition of CDF, it's not hard) and take the limit as $n\to\infty$ (use $\lim_{n\to\infty} (1+x/n)^n = e^x$). |
H: $\wedge^k(V)^* \cong \mathrm{Alt}^k(V)$
Let $V$ be a finite dimensional real vector space, let $\mathrm{Alt}^k(V)$ denote the space of
alternating $k$-linear forms on $V$ and let $\wedge^k(V)$ denote the $k^{th}$ exterior power of $V$.
I am trying to see why the algebraic dual $\wedge^k(V)^* := (\wedge^k(V))^*$ is isomorphic to $\mathrm{Alt}^k(V)$. Here are
my thoughts:
By the universal property of the exterior power, for any alternating $k$-linear form $f$
with domain $V^k$ there exists a unique linear form $\phi$ with domain $\wedge^k(V)$ such that
$$
\phi(v_1 \wedge \cdots \wedge v_k) = f(v_1, \dots, v_k).
$$
The universal property thus provides a mechanism to produce elements in $\wedge^k(V)^*$
from elements in $\mathrm{Alt}^k(V)$ and this mechanism of production is unique.
Thus, we have an injection
$$
\Phi: \mathrm{Alt}^k(V) \longrightarrow \wedge^k(V)^*
$$
What I'm not sure about is how to argue surjectivity; what is the best way to approach this?
AI: The canonical map $\psi\colon V^k \to \bigwedge^k V$ given by $\psi(v_1, \ldots, v_k) = v_1 \wedge \cdots \wedge v_k$ is alternating. If you have an element $g \in (\bigwedge^k V)^*$ then $g \circ \psi\colon V^k \to \mathbf R$ is also alternating, and I believe that this assignment $g \mapsto g \circ \psi$ gives you an inverse to $\Phi$.
This is simpler than you might fear, and has little to do with the ground ring or the finiteness of $V$. And maybe we should expect that, since the universal property of $\bigwedge^k V$ more or less says that it represents the functor $W \mapsto L^k_a(V, W)$.
Extra trouble and hypotheses seem to enter once you try to write things such as an isomorphism $(\bigwedge^k V)^* \approx \bigwedge^k (V^*)$, an embedding $\bigwedge^k V \hookrightarrow V^{\otimes k}$, or the wedge product induced on alternating forms. Some examples of this are discussed in the back of Fulton and Harris. |
H: Collection of converging sequences determines the topology?
Is it the case that the set of converging sequences uniquely determines the open sets in a topological space?
In other words: Given a space $X$ and two topologies $T_{1}$, $T_{2}$ on $X$. such that the set of converging sequences under $T_{1}$ equals the set of converging sequences under $T_{2}$. Does it imply that $T_{1}=T_{2}$?
AI: No. A simple counterexample can be produced as follows. Let $D$ be an uncountable set, and fix a point $p\in D$. Let $\tau_1$ be the discrete topology on $D$. Let $\tau_2$ be the topology that makes each point of $D\setminus\{p\}$ isolated and gives $p$ nbhds of the form $D\setminus C$, where $C$ is any countable subset of $D\setminus\{p\}$. In other words, $$\tau_2=\Big\{\{x\}:x\in D\setminus\{p\}\Big\}\cup\{D\setminus C:C\subseteq D\setminus\{p\}\text{ and }C\text{ is countable}\}\;.$$
Then $\langle D,\tau_1\rangle$ and $\langle D,\tau_2\rangle$ have the same convergent sequences: the only sequences that converge in either topology are those that are eventually constant. However, the two topologies are clearly not homeomorphic.
Spaces whose topologies are completely determined by their convergent sequences are called sequential spaces. |
H: Probability of an observation message
I want to do inference in a Hidden Markov Modell (Gaussian Mixture), given observed continoues variables $Y$ and latent discrete variables $X$. For this I need to compute the probability of an observation message $\mu_{Y \rightarrow X}(x_t) =: \varrho_t)x_t = P(y_t|x_t)$.
But how can I obtain the probability $P(y_t|x_t)$ for some given time $t$? I only have the values of $Y$, the (actually hidden) corresponding values of $X$, and a transition probability table $P(x'|x)$ but no probabilities what so ever for $y_t$.
How can I approach this?
AI: Since X is discrete and your Y is a Gaussian mixture, the most simple thing to assume is that $$y_t | x_t = a \sim N(\mu_a, \sigma_a^2).$$
Now, given a sequence of observations and hidden states $(x_t, y_t)_{t=1}^T$, you can estimate $P(y_t|x_t = a)$ using the maximum likelihood principle. That is, take all times t such that $x_t = a$, ${\cal T}_a = \{t : x_t = a\}$, and estimate $(\mu_a, \sigma_a)$ from corresponding $\{y_t\}_{{\cal T}_a}$. Perform this for all possible values that X can take on. |
H: Probability of observed data in a HMM
Possible Duplicate:
Probability of an observation message
In a given Gaussian mixture model with observed continues variables $Y$ and latent discrete variables $X$ I want to apply the forward-backward algorithm in order to compute the marginal posteriors $P(x_t|y_{1:T})$.
Since this is computed as $$\frac{\alpha_t(x_t) \beta_t(x_t)}{P(Y)}$$
I was wondering how do I obtain the value of $P(Y)$? The only probabilities I have given is a transition probability $P(x'|x)$.
AI: If you observe the sequence of latent variables $\{x_t\}_{t=1}^T$, then you can estimate $P(y_t|x_t)$. Computing $P(Y)$ after that is straight forward (assuming that you know the initial state probability $P(x_0)$). |
H: What other substitutions could I use to evaluate this integral?
Consider the integral
$$ \int x^2\sqrt{2 + x} \, dx$$
I need to find the value of this integral, yet all its (seemingly) possible substitutions don't allow me to cancel appropriate terms. Here are three substitutions and their outcomes, all of which cover both terms in the integrand, as well as the composed function inside the root.
$\boxed{\text{Let }u = (2 + x)}$
$$u = 2 + x$$
$$du = 1 \space dx$$
$$dx = du$$
$$\int x^2 \sqrt{u} \space du$$
$\boxed{\text{Let } u = \sqrt{2+x}}$
$$ u = \sqrt{2 + x}$$
\begin{align*} du &= \frac{1}{2}(2 + x)^{-\frac{1}{2}} \space dx \\
&= \frac{1}{2\sqrt{2 + x}} \space dx
\end{align*}
$$ dx = 2\sqrt{2 + x} \space du$$
$$ 2\int ux^2\sqrt{2 + x} \space du $$
$\boxed{\text{Let }u = x^2}$
$$ u = x^2$$
$$ du = 2x \space dx$$
$$ dx = \frac{1}{2x} \space du$$
$$ \frac{1}{2} \int \frac{u\sqrt{2 + x}}{x} \space du$$
As you can see, none of these allow me to move forward into integrating with respect to $u$. What other substitutions can I use that would help me move on with this integral?
AI: Your first substitution should work. That is let $u=2+x$, then $ x= u-2$ and thus $$x^2 = (u-2)^2.$$ You should get $$ \int \left(u^{5/2}-4u^{3/2}+4u^{1/2}\right)~du. $$
The second substitution also works. Let $u =\sqrt{2+x}$. So $u^2 = 2+x$. Then $x^2 = (u^2-2)^2$. Putting everything together gives you $$\int 2\left(u^6-4u^4+4u^2\right)~du.$$ |
H: Centroid of a region
$$y = x^3, x + y = 2, y = 0$$
I am suppose to find the centroid bounded by those curves. I have no idea how to do this, it isn't really explained well in my book and the places I have looked online do not help either.
AI: Say $f(x)$ and $g(x)$ are the two bounding functions over $[a, b]$
The mass is
$$M=\int_a^b f(x)-g(x) \, dx$$
We find the moments:
$$M_x=\frac{1}{2}\int_{a}^b \left(\left[f(x)\right]^2-\left[g(x)\right]^2\right)\, dx$$
$$M_y=\int_{a}^b x\left(f(x)-g(x)\right)\, dx$$
And the center of mass, $(\bar{x}, \bar{y})$, is
$$\bar{x}=\frac{M_y}{M}$$
$$\bar{y}=\frac{M_x}{M}$$ |
H: hausdorff, intersection of all closed sets
Can you please help me with this question?
Let's $X$ be a topological space.
Show that these two following conditions are equivalent :
$X$ is hausdorff
for all $x\in X$ intersection of all closed sets containing the neighborhoods of $x$ it's $\{x\}$.
Thanks a lot!
AI: Recall the definition of Hausdorff:
$X$ is a Hausdorff space if for every two distinct $x,y\in X$ there are disjoint open sets $U,V$ such that $x\in U$ and $y\in V$.
Suppose that $X$ is Hausdorff and $x\in X$. Suppose $y\neq x$. We have $U,V$ as in the definition. So $x\in U$ and $y\in V$ and $U\cap V=\varnothing$. Suppose that $F$ is a closed set containing an open neighborhood of $x$, intersecting this open set with $U$ yields an open neighborhood of $x$ which is a subset of $F$ so without loss of generality $U\subseteq F$. $F'=F\cap(X\setminus V)$ is closed and does not contain $y$. Furthermore $U$ itself is a subset of this closed set and $y\notin F'$. Therefore when intersecting all closed subsets which contain an open environment of $x$ we remove every other $y$, so the result is $\{x\}$.
On the other hand, suppose that for every $x\in X$ this intersection is $\{x\}$, by a similar process as above deduce that $X$ is Hausdorff as in the definition above. |
H: Minimum sphere containing a tetrahedron
Is there an equation which would give me the radius of the smallest sphere containing a certain tetrahedron (no need to touch all vertices); given that I know the insphere, circumsphere radii and the longest edge of the tetrahedron.
For 2D example of a triangle:
http://demonstrations.wolfram.com/CircumcircleAndIncircleOfATriangle/
AI: Let $r_\min$ be the radius of the minimal enclosing sphere.
The circumradius is an upper bound on $r_\min$, because the circumsphere encloses the tetrahedron.
$\sqrt{3/8}$ times the length of the longest edge is an upper bound on $r_\min$, because for a given length $\ell$, the tetrahedron with the largest minimal enclosing sphere and no edge longer than $\ell$ is the regular tetrahedron of edge length $\ell$, and its circumradius is $\sqrt{3/8}\ell$.
Putting that together, the value $\min(r_{\text{circ}},\sqrt{3/8}\ell_\max)$ is an upper bound on $r_\min$.
I don't know how the inradius can be used to improve the bound. But I'm pretty sure all you can get are bounds; specifying the circumradius, length of longest edge, and inradius is not sufficient to uniquely specify a tetrahedron. |
H: Hydrostatic pressure on a triangle
I am attempting to follow Paul's calculus notes, but am having trouble, in particular at this page: http://tutorial.math.lamar.edu/Classes/CalcII/HydrostaticPressure.aspx
I get to the part with the "The height of this strip is $\Delta x$ and the width is $2a$. We can use similar triangles to determine a as follows,"
and I am completely lost, I do not follow at all what is happening. Is there a mistake? The calculation seems impossible. Is he writing things backwards? $\frac{3}{4}$ is a constant how is it equal to that weird statement? I don't get it, is that solving for $a$?
AI: $\hskip 1in$
In the above, the base and height of the larger right triangle are $A$ and $B$ respectively, and the base and height of the inside, smaller triangle is $a$ and $b$ respectively. The two triangles are similar, which is a term from geometry meaning the ratios between the sides of one triangle are equal to the ratios between the sides of the other triangle. That is, in the above, we have that
$$\frac{a}{b}=\frac{A}{B}.$$
(Equivalently, the inner triangle is the bigger one scaled down in size.) Hence we have:
$\hskip 1in$ $\huge\displaystyle \frac{3}{4}=\frac{a}{4-x_i^*}$
Keep in mind we've approximated the triangle by covering it with thin rectangular strips, and intend to calculate the hydrostatic force associated to each strip and then add them up. As we make the strips smaller and smaller our calculation will be a Riemann sum for an integral, and hence this integral will be our desired answer. In order to calculate the force for a strip, we first denote its dimensions as $\Delta x\times 2a$. Note that $a$ is not really a constant: it varies with choice $x_i^*$. In order to do the Riemann sum though we need to write the hydrostatic force (of each strip) purely in terms of $x_i^*$, which means we need to solve for $a$ in terms of $x_i^*$, and similarity is what allows us to do that. |
H: The comparison theorem for matrices
Let matrix $X$ satisfy a differential equation
$$
\dot X = f(t,X)
$$
where right side is real and symmetric. Let $X(0) = M = M^{T} \succeq 0$ and matrix $Y$ satisfy differential inequality
$$
\dot Y \succeq f(t,Y), \;\;\; Y(0) = M
$$
where $A \succeq B$ means that for any vector $v$ we have $v^{T}Av \geq v^{T}Bv$. Is it true that
$$
Y(t) \succeq X(t)
$$
for $t> 0$? Function $f$ is sufficiently smooth.
AI: I think this is a counterexample, even with a time-independent equation in which the matrices are diagonal. Let $f((a,b))=((0,-2a))$ where $((a,b))$ means the 2 by 2 diagonal matrix with diagonal entries $a,b$. Starting with $M=((0,0))$, we have $X=((0,0))$ for all times. On the other hand, $Y(t)=((t,-t^2))$ satisfies $\dot{Y}(t)=((1,-2t))\ge ((0,-2t))=f(Y)$. |
H: How to prove that $L=\{w \mid \#a(w)=\#b(w)=\#c(w)\}$ is not context free using closure
How can I prove that the language $L = \{w \mid \#a(w)=\#b(w)=\#c(w)\}$ is not context free using closure?
EDIT :
I know that the language $L_1 = \{a^i b^i c^i \mid i\geq 0\}$ is not a context free language.
Now I'm trying to find another language $L_2$, where $L_2$ would be a regular language, in order to make a contradiction, since if $L_1$ is context free and $L_2$ is a regular language, then $L_1 \cap L_2$ is also context free.
AI: $L_1$ is included in $L$: can you find a regular language $R$ so that $L_1=L\cap R$?
Hints:
$R$ needs to reject symbols other than $a,b,c$.
$R$ needs to enforce the order between appearances of $a$, $b$ and $c$. |
H: Terminology for a function computed by a finite-state transducer?
A finite-state transducer is a generalization of a finite state machine that accepts an input string and produces an output string (instead of just accepting or rejecting). Is there a name for a function $f : \Sigma_1^* \rightarrow \Sigma_2^*$ that can be computed by a finite-state transducer? Functions of this form computable by Turing machines are typically called computable functions, and I was curious if there was an analogous term for FSTs.
Thanks!
AI: It's called a Finite state transduc$tion$. |
H: Why is it undecidable whether two finite-state transducers are equivalent?
According to the Wikipedia page on finite-state transducers, it is undecidable whether two finite-state transducers are equivalent. I find this result striking, since it is decidable whether two finite-state automata are equivalent to one another.
Unfortunately, Wikipedia doesn't provide any citations that provide a justification for this result. Does anyone know of a proof of this result? Alternatively, if the article is incorrect, does someone know an algorithm that could be used to show equivalence?
Thanks!
AI: This web page from the on-line version of An Introduction to the Theory of Computation, by Eitan Gurari, shows that the equivalence problem for finite-state transducers is undecidable by reduction from the Post Correspondence Problem, whose undecidability is shown on the same page; he cites
Griffiths, T. (1968). "The unsolvability of the equivalence problem for $\Lambda$-free nondeterministic generalized machines," Journal of the Association for Computing Machinery 15, 409-413,
for the result.
Gurari himself proved that the equivalence problem is decidable for deterministic transducers:
The equivalence problem for deterministic two-way sequential
transducers is decidable. SIAM J. Comput. , 11(3):448–452, 1982.
Apparently the difficulty lies in the fact that some non-deterministic finite-state transducers cannot be reduced to deterministic ones. |
H: Cartan Theorem.
Cartan Theorem: Let $M$ be a compact riemannian manifold. Let $\pi_1(M)$ be the set of all the classes of free homotopy of $M.$ Then in each non trival class there is a closed geodesic. (i.e a closed curve which is geodesic in all of its points.)
My question: Why free classes? Why the theorem does not apply if we exchange free classes for classes with a fixed base point?
AI: Consider the Klein bottle $K$ with flat metric. I'm thinking of $K$ as a square where the left and right sides are identified in the "right" way (like on the torus), while the top and bottom are identified in the "wrong" way (like on $\mathbb{R}P^2$). In this picture, geodesics are straight lines that wrap around depending on the identifications on the edges.
Take your basepoint $p$ to be the center of the square. Consider the geodesic $\gamma$ emanating from $p$ with slope 1. So, it starts in the middle of the square moving towards the top right corner. After it gets to the top right corner, due to the identifications we're making, it becomes a straight line emanating from the bottom right corner with slope -1 until it eventually hits $p$ again, i.e., it closes up. However, it is not a closed geodesic because it makes a corner at $p$.
Further, I claim no other geodesic emanating from $p$ is in the same homotopy class as $\gamma$. To see this, work in the univeral cover, $\mathbb{R}^2$ (thought of as being tiled by squares with identification arrows as approrpriate, with corners on integer lattice points). Geodesics are still straight lines, but now there is a unique straight line from $(\frac{1}{2},\frac{1}{2})$ to $(\frac{3}{2},\frac{3}{2})$, given by lifting $\gamma$.
This means the Cartan's theorem fails on based loops, at least in this particular case. |
H: Union of compact sets in a convergence space
Let $X$ be a convergence space and let $K_1, K_2, \ldots, K_n$ be compact subsets of $X$. I'm trying to prove for myself that the union $K$ of the $K_i$ is compact. By definition, $K$ is compact if every ultrafilter on it converges, so given an ultrafilter $\mathcal U$ on $K$, I have to show that it converges. Somehow I need to produce ultrafilters on each of the $K_i$ and relate them to $\mathcal U$ but I have no idea how to do this. Any tips?
AI: If $\mathscr{U}$ is an ultrafilter on $K$, there must be an $i\in\{1,\dots,n\}$ such that $K_i\in\mathscr{U}$. Then the trace of $\mathscr{U}$ on $K_i$, $\{U\cap K_i:U\in\mathscr{U}\}$, is an ultrafilter on $K_i$ that converges iff $\mathscr{U}$ does. |
H: Sequence in $L^p(X,M,\mu)$
I have two question.
Suppose that {$f_k$} is a sequence in $L^p(X,M,\mu)$ such that
$f(x) = \lim_{k \to \infty} f_k(x)$
exists for $\mu$ -a.e. $x \in X$.
Assume $1\le p<\infty$,
$\liminf_{k\to \infty} ||f_k||_p = a$
is finite.
First one is proving that $f \in L^p$ and $||f||_p \le a$.
And if additionally assume that $||f||_p = \lim_{k \to \infty} ||f_k||_p $.
Second one is to prove $$\lim_{k \to \infty} ||f-f_k||_p =0 $$
Those are very natural fact, but I want have strict proof of them. How can I approach?
AI: 1) is easy using Fatou's lemma |
H: $L^p$ measurable functions equality
Suppose $1<p<\infty$ and $f,g \in L^p (X,M,\mu)$. Where $||f||_P$ and $||g||_p$ are non zero, and
$||f+g||_p = ||f||_p +||g||_p$ .
Proving that equality:
$${f \over ||f||_p} = {g \over ||g||_p} \text{ }\mu -a.e.$$
What theorem is available for that prove? I can't find start point.
AI: The triangle inequality (Minkowski) is proven using Hölder. To show what you want, you can repeat that argument:
(note that $q(p-1)=p$)
$$
\|f\|_p+\|g\|_p=\|f+g\|_p=\left(\int|f+g|^p\right)^{1/p}=\left(\int|f+g|\,|f+g|^{p-1}\right)^{1/p}\leq\left(\int|f|\,|f+g|^{p-1}+\int|g|\,|f+g|^{p-1}\right)^{1/p}
\leq\left(\|f\|_p \left(\int|f+g|^p\right)^{1/q}+\|g\|_p \left(\int|f+g|^p\right)^{1/q}\right)^{1/p}
=\left((\|f\|_p+\|g\|_p)\left(\int|f+g|^p\right)^{1/q} \right)^{1/p}
=\left(\|f+g\|_p \|f+g\|_p^{p/q} \right)^{1/p}
=\left(\|f+g\|_p \|f+g\|_p^{p-1} \right)^{1/p}
=\left( \|f+g\|_p^{p} \right)^{1/p}
=\|f+g\|_p=\|f\|_p+\|g\|_p
$$
In particular we have equality in the two Hölder inequalities we used in the middle (the second "$\leq$"). Equality in Hölder occurs only when the $p$ and $q$ powers of the two factors are linearly dependent. This means that there exist constants $\alpha,\beta$ such that
$$
|f|^p=\alpha|f+g|^{q(p-1)},\ \ |g|^p=\beta|f+g|^{q(p-1)}
$$
almost everywhere.
So $|g|^p=\gamma|f|^p$ a.e. for some constant $\gamma$, and then $|g|=\delta|f|$ a.e. for yet another constant. Then
$$
\frac{|f|}{\|f\|_p}=\frac{|g|}{\|g\|_p}\ \ \ \text{a.e.}
$$ |
H: Measure space inequality
$X,M,\mu$ is a measure space for which
$$\mu(A)>0 \to \mu(A) \ge 1.$$
If $1 \le p<q\le\infty $, then $L^p \subset L^q$ , and then
$$||f||_\infty \le ||f||_q \le ||f||_p \le ||f||_1$$
This inequality is introduced on my book, and very useful. So I'm trying to prove that inequality. How can I do that?
AI: For any measurable $f\ne0$, and for any $\varepsilon$, there exists a measurable subset $A$ with positive measure (and so $\mu(A)\geq1$) such that $|f|\geq\|f\|_\infty-\varepsilon$ on $A$. Then
$$
\|f\|_q=\left(\int|f|^q\right)^{1/q}\geq\left(\int_A|f|^q\right)^{1/q}
\geq\left(\int_A(|f|_\infty-\varepsilon)^q\right)^{1/q}=(\|f\|_\infty-\varepsilon)\,\mu(A)\geq\|f\|_\infty-\varepsilon.
$$
As $\varepsilon$ was arbitrary, the first inequality is proven.
For the second inequality, assume first that $\|f\|_q=1$. This implies, by the first inequality, that $|f|\leq1$ a.e. Then, using $p\leq q$ (and so $|f|^p\geq|f|^q$),
$$
\|f\|_p=\left(\int|f|^p\right)^{1/p}\geq\left(\int|f|^q\right)^{1/p}=1^{1/p}=1=\|f\|_q
$$
The case $\|f\|_q\ne1$ follows easily.
The third inequality is a particular case of the second one. |
H: Is there a differential limit?
I'm wondering if there's such a concept as a "differential limit". Let me give an example because my nomenclature is my own and unofficial, but hopefully indicative of the concept.
For some function f(x), there may exist a derivative of that function f'(x) which we call a first order differentiation of f(x). Likewise from f'(x) we could also derive a second order function f''(x). Is there a mathematical concept of the limit of a function as the order of differentiation goes to infinity?
I can think of a function that I can take the derivative of an infinite number of times ( albeit without getting a constant value): f(x) = sin(x)
The function g(x) = x^2 might have a differential limit of 0. (g'(x) = 2x; g''(x) = 2; g'''(x) = 0 ...)
Is there any usefulness to this concept? A topic of study or keyword that discusses this? Any more examples of infinitely derivable functions?
Most importantly if this is interesting to anyone other than myself: what is a good resource for learning more?
AI: First: yes, we care about how many times a function can be differentiated. In fact, we have names for such classes: $\mathcal{C}_0$ is the collection of all continuous functions, $\mathcal{C}_1$ the class of functions that have continuous derivative; $\mathcal{C}_2$ the class of functions that have continuous second derivative; and so on. We have that
$$\mathcal{C}_0 \supsetneq \mathcal{C}_1 \supsetneq\mathcal{C}_2\supsetneq\mathcal{C}_3\supsetneq \cdots \supsetneq \mathcal{C}_n\supsetneq \cdots$$
Then we have "infinitely differentiable functions", functions that have continuous derivatives of all orders; this is denoted $\mathcal{C}_{\infty}$. There are functions that have derivatives of all orders that don't cycle through values, such as $e^{x^2}$.
Even beyond the infinitely differentiable functions we have the notion of "analytic function", which are functions that can be defined at every point by a power series that converges on an open interval containing the point.
Now, I think you are talking about trying to make sense of a "limit" of the functions as you take derivatives. So that for a polynomial the "limit" should be $0$, a function like $\sin x$ should not have a limit, and so on.
There are in fact several different ways of making the idea of "differential limit" precise. I'm going to talk more generally about what "limit" and "convergence" can mean here, before addressing the specific construction you are talking about. I'll spill the beans and say that I haven't really seen this particular construction before (there's too much to wade through below, and I don't want you to be annoyed when you reach the end and find that I say "I haven't seen it...") But that does not mean you can't ask interesting questions about it.
What you describe is a sequence of functions. The main issue is that one has to define what one means by saying that two functions are "close to one another". Then you get the notion of convergence of functions.
Turns out that there are many different ways of saying this, and which one is useful depends on the context.
Let's take the case of a sequence of real valued functions of real variable, with domain all of $\mathbb{R}$. Let's call the sequence $f_n$ (in your case, $f_n = \frac{d^nf}{dx^n}$ for a given $f$).
The most natural way of saying that the sequence $f_n$ converges to a function $f$ is:
$f_n\to f$ if for every $x$, $\lim\limits_{n\to\infty}f_n(x) = f(x)$.
This is called pointwise convergence. For example, the sequence $f_n(x) = x^n$ defined on $[0,1]$ converges pointwise to the function
$$f(x) = \left\{\begin{array}{ll}
0 &\text{if }0\leq x\lt 1\\
1 & \text{if }x=1.
\end{array}\right.$$
It turns out that pointwise convergence is natural, but is not very "good", in the sense that a lot of properties we find interesting about functions are not preserved by pointwise convergence. For example, above, all functions $f_n$ are continuous, but their limit is not.
The problem arises because we are forcing no connection between how $f_n(x)$ converges to $f(x)$ and how $f_n(y)$ converges to $f(y)$ with $x\neq y$; that means that convergence on some points can "lag behind" convergence in points very close to them. This can be remedied by making things a bit tighter. If we write out the definition of the limits above, we obtain the following description:
$\{f_n\}$ converge pointwise to $f$ if and only if for every $x$, for every $\epsilon\gt 0$, there exists $N$, which may depend on both $x$ and $\epsilon$, such that for all $n\geq N$ we have $|f_n(x)-f(x)|\lt \epsilon$.
The way to make this tighter, to make sure the the $f_n(x)$ converge to $f$ more or less "the same way", is to force the $N$ to depend only on $\epsilon$, and not on $x$ as well. This is accomplished with:
We say that $\{f_n\}$ converge uniformly to $f$ if and only if for every $\epsilon\gt 0$ there exists $N$, which depends only on $\epsilon$, such that for all $x$ and all $n\geq N$ we have $|f_n(x)-f(x)|\lt\epsilon$.
Uniform convergence is better than pointwise convergence: if every element of the sequence is continuous, and the sequence converges uniformly, then the limit is continuous, for example.
There are yet other ways of talking about functions converging. Here are a few:
Fix a real number $p$, $1\leq p$. We say that $\{f_n\}$ converges to $f$ in $p$-norm if and only if
$$\lim_{n\to\infty}\int_{\infty}^{\infty}|f_n(x)-f(x)|^p\,dx = 0.$$
Closely related: we say that $\{f_n\}$ converges to $f$ in the sup-norm if and only if
$$\lim_{n\to\infty}\sup\{|f_n(x)-f(x)|\} = 0.$$
There is a way of measuring sets of real numbers, called the Lebesgue measure, $\lambda$, the details of which are perhaps too much to go over now. But intuitively, it is a way of assigning a "size" to a lot of sets of real numbers, including all intervals and much more complicated sets, in a nice and consistent way. We say that $\{f_n\}$ converges in measure to $f$ if and only if for every $\epsilon\gt 0$,
$$\lim_{n\to\infty}\lambda(\{x\mid |f_n(x)-f(x)|\geq\epsilon\}) = 0.$$
We say that $\{f_n\}$ converges almost uniformly to $f$ if and only if for every $\epsilon\gt 0$ there is a set $X\subseteq \mathbb{R}$, with $\lambda(X)\lt\epsilon$, such that $\{f_n\}$ converges uniformly to $f$ on $\mathbb{R}-X$.
There are small variations to each of those; depending on context, one may be able to define other ways of convergence, or some of the above may no longer make sense. A typical variation is to take any of the above types of convergence (except almost uniform, for which the concept doesn't add anything), and add the condition "almost everywhere" (or "almost surely" if you are doing probability). That means that there exists a set $Y$ contained in the domain, of measure $0$, such that the type of convergence you are interested in occurs in $\mathbb{R}-Y$. (Sets of measure $0$ need not be empty, and they can be quite complicated: for example, any countable set has measure $0$, and there are uncountable sets that do as well, like the Cantor ternary set.
Another variation is to define the concept of "Cauchy sequence" for each of the types, which means making the analog of the definition of "Cauchy sequence" of real numbers: instead of comparing to a limit, we ask that for all $n$ and $m$ greater than $N$, $f_n$ and $f_m$ have the relevant property. For instance, "pointwise Cauchy" means that for all $x$ and all $\epsilon\gt 0$ there exists $N$ such that for all $n,m\gt N$ we have $|f_n(x)-f_m(x)|\lt \delta$. "Cauchy in measure" means that for all $\epsilon\gt 0$ and all $\delta\gt 0$ there exists $N$ such that if $n,m\gt N$, then $\lambda(\{x\mid |f_n(x)-f_m(x)|\geq\epsilon\}\lt\delta$; and so on.
All of these concepts are extremely important and useful. They show up in probability, statistics, functional analysis, analysis, and other areas. There are known implications (e.g., uniform convergence implies pointwise convergence, and so on).
Now, what about the specific case in which we start with a function $f_0$ (necessarily in $\mathcal{C}_{\infty}$) and we define $f_n$ to be the $n$th derivative of $f_0$? I confess I haven't seen this particular sequence addressed, but you can ask all the questions related to the convergence types above: is the sequence pointwise Cauchy? $p$-norm Cauchy? Cauchy in measure? Does it converge uniformly? Almost uniformly? Pointwise? In measure? And so on. Seems like an interesting question, if nothing else. |
H: Generalization of Hölder's inequality
Assume $1<p_k< \infty$ for $k=1,\ldots,N$ , and $\displaystyle\sum^N_{k=1}\frac{1}{p_k} =1$.
I want to prove that
$$\left|\int_X f_1 f_2\cdots f_N\; d\mu \right| \le \lVert f_1\rVert_{p_1} \lVert f_2\rVert_{p_2} \cdots \lVert f_N\rVert_{p_N}.$$
How can I directly adjust Hölder's inequality for it?
AI: Hint
Start with Hölder on the function $f_1$ and $g_1=f_2f_3\dots f_n$ using $p_1$ and $p_1'$. That is
$$\left|\int f_1 g_1 dx\right|\leq\left(\int |f_1|^{p_1}dx\right)^{1/p_1}\left(\int |g_1|^{p_1'}dx\right)^{1/p_1'}$$
where $p_1'=p_1/(p_1-1)$.
Apply Hölder on $|f_2|^{p_1'}$ and $g_2 = |f_3f_4\dots f_n|^{p_1}$ using $p_2/p_1'$ and $(p_2/p_1')'$... |
H: Simple Harmonic Motion with trigonometry
I need some help with my high school maths question:
A particle is moving in simple harmonic motion has speed 12m/s at the origin. Find the displacement-time equation if it is known that for positive constants a and n:
$x=a\cos 8t$
$x=16\cos nt$
So far this is what I know:
The particle has a speed of 12m/s at the origin so when t=0, v=12. The origin is where the maximum speed occurs (Or is it at the centre of the motion?).
AI: Problem $1.$ "At the origin" presumably refers to $x=0$. At the time $t_1$ when the particle is at the origin, $8t_1=\pi/2$, or $8t_1$ has some other value whose cosine is $0$. The sine of $8t_1$ is therefore $\pm 1$.
The velocity at any time $t$ is $\frac{dx}{dt}$, which is $-8a\sin(8t)$. The speed is the absolute value of this, and is $12$ when we are at the origin. Thus at the origin $-8a\sin(8t)$ has absolute value $12$. Since the sine of $8t_1$ has absolute value $1$, it follows that $8a$ has absolute value $12$. Since $a$ is positive, $a=\frac{12}{8}$. Thus $x=\frac{12}{8}\cos(8t)$.
The reasoning for Problem $2$ is similar. |
H: How can the jth level of a binary tree with n nodes has problems of size $({\frac{n}{2}})^j$?
I read from a book that the jth level (starting from j=0 or the root) of a binary tree with n nodes divides a problem into $2^j$ subproblems, each of size $\frac{n}{2^j}$. I understand where $2^j$ comes from, but where does $\frac{n}{2^j}$ come from?
$n$ includes nodes above the level, right? Hence, how come the sum of the sizes of the subproblems of the subtrees of the level is $n$? Shouldn't it be less than $n$?
AI: The idea is that if one node of a binary tree representing a divide-and-conquer algorithm has associated 'problem size' $n$, then the two nodes beneath it have problem sizes $x$ and $y$ with $x+y=n$; you split the problem of size $n$ into two sub-problems whose sizes add to $n$. In the best-case scenario, these two subproblems are the same size, which means that each of them is (roughly) $n/2$. By iterating this down you can see that each of the problems two levels beneath are of size roughly $n/4$, the problems on the level below that are of size roughly $n/8$, etc; since you have $2^j$ nodes at level $j$ but the overall size of the problem (the sum of the sub-problem sizes in those nodes) is still $n$, each of those sub-problems must be of size $n/2^j$ so that their sum can be $2^j\cdot(n/2^j)=n$. |
H: Cardinality, $|x|=|y|$ implies $|A^x|=|A^y|$
I am trying to prove, without using the Schroder-Bernstein theorem, (where a modulus defines cardinality) that
i.) $|x|=|y|$ implies $|A^x|=|A^y|$
and
ii.) $|x|=|y|$ implies $|x^A|=|y^A|$.
Thank you!
AI: To say that $|x|=|y|$ is to say that the difference between $x$ and $y$ is just in the labels of the names. An element of $A^x$ is a function whose domain is $x$ and its co-domain is $A$.
If we agreed that simply by changing the name of every element in $x$ we can get $y$, fix a renaming method (that is, pick a particular way to rename from $x$ to $y$) and take some $f\in A^y$, define $\hat f\in A^x$ to be the function which sends $x\in X$ to $f(y)$ such that $y$ is the element we get by renaming $x$ in the renaming method.
Rigorously a renaming method is simply a bijection $\varphi\colon y\to x$, and we simply define $\hat f = \{\langle\varphi(y),f(y)\rangle\mid \langle y,f(y)\rangle\in f\}$. We can show that this is a function from $x$ into $A$ and if $f\neq g$ then $\hat f\neq\hat g$.
Therefore $\hat\bullet\colon A^y\to A^x$ is a bijection and $|A^x|=|A^y|$.
The second question is essentially the same, this time we have a function from $A$ to $x$, so we simply compose it with $\varphi$, our bijection.
The Cantor-Bernstein theorem simply tells us that if we have two injections then we have a bijection. To prove something like $|x|=|y|\implies |A^x|=|A^y|$ without the use of Cantor-Bernstein would depend on how many laws you have already proved using bijections and injections.
For example, if you prove that $|x|\leq |y|\implies |A^x|\leq|A^y|$ then you can simply write:
Since $|x|=|y|$ we have that $|x|\leq |y|$ and therefore $|A^x|\leq|A^y|$; similarly $|y|\leq|x|$ and therefore $|A^y|\leq|A^x|$.
By Cantor-Bernstein we have that $|A^x|=|A^y|$.
However, after quite some time you realize that proving that aforementioned property is not easier than constructing explicit bijections. |
H: Understanding the $L^\infty$ norm
I'm very confused about $L^\infty$. So I'm trying to prove this:
Is $\|f\|_{\infty}$ the smallest of all numbers of the form $\sup\{|g(x)| \,:\,x \in X\}$, where $f=g $ $\mu$-a.e.?
AI: The answer to your question is yes. Recall that the definition of $\|f\|_\infty$ is
$$\|f\|_\infty=\operatorname{ess sup}(f)=\inf\{a\in\mathbb{R}\mid \mu(\{x\in X\mid |f(x)|>a\})=0\}.$$
I'll interpret the phrase "the smallest of all numbers of the form $\sup\{|g(x)|\,:\,x\in X\}$, where $f=g$ $\mu$-a.e." to mean $$M=\inf_{g=f\text{ a.e.}}\left\{\sup_{x\in X}|g(x)|\right\}.$$
For any $\epsilon>0$, we can define $g:X\to\mathbb{R}$ by
$$g(x)=\begin{cases}f(x) & \text{if }|f(x)|\leq \|f\|_\infty+\epsilon,\\0 & \text{otherwise}.\end{cases}$$
Note that $$\mu(\{x\in X\mid f(x)\neq g(x)\})=\mu(\{x\in X\mid |f(x)|>\|f\|_\infty+\epsilon\})=0$$
by the definition of $\|f\|_\infty$, so that $f=g$ almost everywhere, and that $\sup_{x\in X}|g(x)|\leq \|f\|_\infty+\epsilon$.
Thus,
$$M=\inf_{g=f\text{ a.e.}}\left\{\sup_{x\in X}|g(x)|\right\}\leq \|f\|_\infty+\epsilon$$
for all $\epsilon>0$, and thus $M\leq \|f\|_\infty$.
For the other direction, note that for any $\epsilon>0$, $$\mu(\{x\in X\mid |f(x)|>(\|f\|_\infty-\epsilon)\})>0$$ by the definition of $\|f\|_\infty$. If $g=f$ a.e., then certainly $|g|=|f|$ a.e., so we must have that $$\mu(\{x\in X\mid |g(x)|>(\|f\|_\infty-\epsilon)\})>0$$ too, so that $g(x)>\|f\|_\infty-\epsilon$ for some $x\in X$. Thus, for any $g$ such that $g=f$ a.e., we have that $$\sup_{x\in X}|g(x)|\geq \|f\|_\infty-\epsilon$$
so that
$$M=\inf_{g=f\text{ a.e.}}\left\{\sup_{x\in X}|g(x)|\right\}\geq \|f\|_\infty-\epsilon$$
for all $\epsilon>0$, and thus $M\geq \|f\|_\infty$.
This shows that $M=\|f\|_\infty$. |
H: An inequality problem.
Possible Duplicate:
Showing the inequality $|\alpha + \beta|^p \leq 2^{p-1}(|\alpha|^p + |\beta|^p)$
In the condition $a,b \in[0,\infty)$, $1\le p<\infty$,
How can I conclude this inequality?
$$(a+b)^p \le 2^{p-1} (a^p + b^p)$$
AI: you could solve it using Hölder inequality.
$ (a+b)^p \leq 2^{p-1}(a^p+b^p) \Leftrightarrow
(a+b) \leq (1+1)^{1- \frac{1}{p}} (a^p +b^p)^{\frac{1}{p}} $
and then apply Hölder. |
H: On the gaps between consecutive primes
I have observed something, that either:
Given any natural number $n$, there exists some natural number $k$, such that above $k$, the difference between any two consecutive primes is $> n$ ($\implies$ the prime gap increases steadily, having limit infinity), or
For some $n$, there are infinitely many consecutive prime pairs of the form $(p,p+n)$.
Both (1) and (2) look remarkable results to me and one of them must be true! Is there any information as to which one is true? (can both be true?) And then, why is the classical twin prime conjecture important and why can't the conjecture be put in this way:
There exists infinitely many consecutive prime pairs of the form $(p,p+n)$, for some natural number $n$?
AI: (2) is universally believed to be true, but we are far from having a proof. It is a recent breakthrough of Goldston, Pintz and Yıldırım (arXiv) that (2) is true if we know more about a certain constant $\theta$ called the level of distribution for primes in arithmetic progressions.
The Bombieri-Vinogradov theorem is equivalent to $\theta \ge 1/2$. Goldston-Pintz-Yıldırım shows that if one can obtain any slight improvement $\theta > 1/2$, then (2) provably holds for some $n$. (This does not contradict @GerryMyerson's answer, since no one has any idea how to get any improvement. Even GRH probably isn't enough to imply this.)
Elliot and Halberstam have conjectured that $\theta = 1$. If this is true, or even if just $\theta > 0.971$, then (2) holds for some $2 \le n \le 16$. So far no one has succeeded in tightening this bound to get twin primes conditionally under Elliot-Halberstam. |
H: Why do $\mathbb{C}$ and $\mathbb{H}$ generate all of $M_2(\mathbb{C})$?
For this question, I'm identifying the quaternions $\mathbb{H}$ as a subring of $M_2(\mathbb{C})$, so I view them as the set of matrices of form
$$
\begin{pmatrix}
a & b \\ -\bar{b} & \bar{a}
\end{pmatrix}.
$$
I'm also viewing $\mathbb{C}$ as the subfield of scalar matrices in $M_2(\mathbb{C})$, identifying $z\in\mathbb{C}$ with the diagonal matrix with $z$ along the main diagonal.
Since $\mathbb{H}$ contains $j=\begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}$ and $k=\begin{pmatrix} 0 & i \\ i & 0\end{pmatrix}$, I know that
$$
ij+k=\begin{pmatrix} 0 & 2i\\ 0 & 0 \end{pmatrix}
$$
and
$$
-ij+k= \begin{pmatrix} 0 & 0\\ 2i & 0 \end{pmatrix}
$$
are in the generated subring. I'm just trying to find matrices of form $\begin{pmatrix} a & 0 \\ 0 & 0 \end{pmatrix}$ and $\begin{pmatrix} 0 & 0\\ 0 & d \end{pmatrix}$ for $a,d\neq 0$ to conclude the generated subring is the whole ring. How can I get these remaining two pieces? Thanks.
AI: Hint: Use linear combinations of
$$
jk=\pmatrix{i&0\cr0&-i\cr}\qquad\text{and the scalar}\qquad \pmatrix{i&0\cr 0&i\cr}.
$$ |
H: Quick question about lim and sup
I've got a question regarding a step in a proof, the situation is following:
Let $X_{1},\dots,X_{n}$ be independent, symmetric stochastic variables so that $\sum\limits_{n=1}^{\infty}X_{n}$ exists in probability.
If I use the fact that convergent in probability implies the same convergence of the corresponding Cauchy sequence i can write: $$\lim_{n,m\to\infty}P(|S_{m}-S_{n}|>t)=0$$ where $S_{n}=\sum\limits_{i=1}^{n}X_{i}$.
My textbook then says that from the triangle inequality we can write that as: $$\lim_{n\to\infty}\sup_{m\geq n+1}P(|S_{m}-S_{n}|>t)=0$$
Could someone please help me fill in the blanks?
I hope I didn't leave anything relevant out, let me know if you feel i did.
AI: This uses the property of Cauchy sequences. One has that for every $\epsilon>0$, there exists an $N$ such that for all $n,m\geq N$, $P(|S_n-S_m|>t)<\epsilon$. So if we temporarly anchor $n>N$, the inequality holds even after taking the supremum over $m\geq n+1$, so long as $n\geq N$.
As far as triangle inequality goes, likely the book is referring to the fact that $|S_n-S_m|\leq |S_n-S_N|+|S_m-S_N|$ which implies that if $|S_n-S_m|>t$ then either $|S_n-S_N|>t/2$ or $|S_m-S_N|>t/2$, so that one can write:
$P(|S_n-S_m|>t) \leq P(|S_n-S_N|>t/2) + P(|S_m-S_N|>t/2)$
and so if we again think of $N$ as any large anchored quantity, then the two terms on the right can be made uniformly small for $n,m\geq N$. In other words if you take the supremum in $m$ over $m\geq n+1$, then the right hand side can be made arbitrarely small by picking a large enough $n$ and $N$. Essentially this "trick" decouples $n$ and $m$ from eachother. |
H: Basic fact in $L^p$ space
I'm studying $L^p$ space.
$1 \le p < r <q < \infty$ then $L^p \cap L^q \subset L^r$.
More over $L^p \cap L^\infty \subset L^r$
I'm trying to prove that fact. Which theorem is useful for proving that?
AI: For $f(x)\ge0$, Jensen's Inequality yields
$$
\left(\frac{1}{\int_X f^p(x)\,\mathrm{d}x}\int_X f^{r-p}(x)f^p(x)\,\mathrm{d}x\right)^{\Large\frac{q-p}{r-p}}\le\frac{1}{\int_X f^p(x)\,\mathrm{d}x}\int_X f^{q-p}(x)f^p(x)\,\mathrm{d}x
$$
which becomes
$$
\left(\int_Xf^r(x)\,\mathrm{d}x\right)^{\Large\frac1r}\le\left(\int_Xf^p(x)\,\mathrm{d}x\right)^{\Large\frac1p\left(\frac{p}{r}\frac{q-r}{q-p}\right)}\left(\int_X f^q(x)\,\mathrm{d}x\right)^{\Large\frac1q\left(\frac{q}{r}\frac{r-p}{q-p}\right)}
$$
Thus for $f\in L^p\cap L^q$,
$$
\|f\|_r\le\|f\|_p^{\Large\frac{p}{r}\frac{q-r}{q-p}}\;\|f\|_q^{\Large\frac{q}{r}\frac{r-p}{q-p}}\tag{1}
$$
where
$$
\frac{p}{r}\frac{q-r}{q-p}+\frac{q}{r}\frac{r-p}{q-p}=1\tag{2}
$$
Note that when $q\to\infty$, $(1)$ becomes
$$
\|f\|_r\le\|f\|_p^{\Large\frac{p}{r}}\;\|f\|_\infty^{\Large1-\frac{p}{r}}\tag{3}
$$ |
H: How are the limits of this integral transformed?
Using $$\ln(x) = \int_1^x \frac{1}{t} dt$$
Show that for $x > 0$, $\ln\left(\frac{1}{x}\right) = -\ln(x)$
I am following a provided answer and didn't quite understand the following transformation and why/how it is done:
$$\ln\left(\frac{1}{x}\right) = \int_1^{\frac{1}{x}} \frac{1}{t}dt$$
$$~u = \frac{1}{t}, du = \frac{-1}{t^2}dt$$
$$= -\int_1^{\frac{1}{x}} t\frac{dt}{t^2}$$
$$=-\int_1^x \frac{1}{u}du$$
$$= -\ln(x)$$
I don't understand how the limits of integration were changed in the $2nd$ and $3rd$ term (from $\frac{1}{x}$ to $x$). Is there a property that makes this correct or is there some other reasoning behind the change?
AI: \begin{align}
\ln\left(\frac{1}{x}\right) & = \int_1^{\frac{1}{x}} \frac{1}{t}dt & & {\text{Replace $x$ by $\dfrac1x$ in the definition}}\\
u & = \frac{1}{t} & &{\text{This is the substitution you are making}}\\
du & = \frac{-1}{t^2}dt & &{\text{This is because }\dfrac{du}{dt} = - \dfrac1{t^2}}\\
\ln\left(\frac{1}{x}\right) & = \int_1^{\frac{1}{x}} \frac{1}{t}dt & = \int_1^{\frac{1}{x}} -t \times \left(-\frac{1}{t^2}dt \right) & \text{Multiply and divide the integrand by $-t$}
\end{align}
In the above integrand, we can replace $-\dfrac{1}{t^2}dt$ by $du$ and $-t$ by $- \dfrac1u$.
Note that we are making the change of variable, the integrand is in terms of $u$ now.
Hence, we need to look at the limits for the variable $u$ in the integral.
We have the transformation that $u = \dfrac1t$. Hence, if $t$ goes from $1$ to $1/x$, then $u$ goes from $1$ to $x$. This is because when $t=1$, $u = \dfrac11 = 1$. Similarly, if $t = 1/x$, then $u = \dfrac1{1/x} = x$.
(For example, if $t$ goes from $1$ to $2$, then $1/t$ goes from $1$ to $1/2$.)
Hence, we get that
\begin{align}
\ln \left( \dfrac1x\right) & = \int_1^{\frac{1}{x}} -t \times \left(-\frac{1}{t^2}dt \right) & = \int_1^x \left(-\dfrac1u \right) \times du = - \int_1^x \dfrac{du}{u} = - \ln(x)
\end{align}
where the last equality is obtained since we have defined $\ln(x)$ as $\displaystyle \int_1^x \dfrac{dt}{t}$. |
H: Show that $f'$ is not continuous at 0 for the following function:
$$ f(x) = \begin{cases} x + 2x^2\sin(1/x) & \text{ for }x \neq 0 \\ 0 & \text{ for } x = 0\end{cases} $$
This is another exam practice question I am working on.
I simply took the derivative:
$$f'(x) = 1 + 4x\sin(1/x) - 2x^4\cos(1/x) $$
Now we see that $f'$ is undefined at $x = 0$, therefore $f'$ cannot be continuous there.
Is this sufficient to answer this question? The previous part of the question required me to use the limit definition of the derivative to show that $f'(0) = 1$. Would I need to use the limit definition here as well?
Thanks for any feedback!
AI: No. What you have is incorrect.
$f'(x)$ is in-fact well defined at all $x$ including $0$. Just that $f'$ is not continuous at $0$.
When you compute the derivative $f'(x)$ as $$f'(x) = 1 + 4x \sin(1/x) - 2 \cos(1/x),$$ this computation is valid only when $x \neq 0$ (There is an error in your derivative computation for $x \neq 0$. There is no $x^4$ infront off $\cos(1/x)$).
To compute the derivative at $x=0$, lets go back to the definition of the derivative at $0$. We have that
$$f'(0) = \lim_{h \to 0} \dfrac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \dfrac{h + 2h^2 \sin(1/h) - 0}{h} = \lim_{h \to 0} \left(1 + 2h \sin(1/h) \right) = 1$$
In the above derivation, we have used the fact that $\displaystyle \lim_{h \to 0} h \sin(1/h) =0$.
This is because of the fact that $-1 \leq \sin(1/h) \leq 1$ and hence $$-\lvert h \rvert \leq h \sin(1/h) \leq \lvert h \rvert \implies -\lim_{h \to 0} \lvert h \rvert \leq \lim_{h \to 0} h \sin(1/h) \leq \lim_{h \to 0} \lvert h \rvert$$
Whereas for $x \neq 0$, we have that $f'(x) = 1 + 4x \sin(1/x) - 2 \cos(1/x)$. Putting these together, we get that
$$ f'(x) = \begin{cases} 1 + 4x \sin(1/x) - 2 \cos(1/x) & x \neq 0\\ 1 & x = 0\end{cases}$$Now you can see that $f'(x)$ is discontinuous at $0$ even though $f'(x)$ is well-defined at all $x$ including $x=0$ where $f'(0) = 1$. However, \begin{align}
\lim_{x \to 0} \left(1 + 4x \sin(1/x) - 2 \cos(1/x) \right) & = 1 + \lim_{x \to 0} 4x \sin(1/x) - \lim_{x \to 0} 2 \cos(1/x)\\
& = 1 - \lim_{x \to 0} 2 \cos(1/x)\\
\text{ And the above limit doesn't exists}
\end{align}
Hence, $\lim_{x \to 0} f'(x)$ doesn't exist. This means that even though $f'(0)$ is well-defined and in-fact $f'(x)$ is well-defined for all $x$, $f'(x)$ is not continuous at $x=0$. |
H: How large need $n$ be taken to ensure that $T_n(x)$ gives a value of $\ln(1.3)$ which has an error of less than $0.0002$?
$$ f(x) = \ln(1+x)$$
The previous part of this question required me to write down the remainder term for the taylor polynomial of order n.
My remainder term worked out to be:
$$R_n(x) = (-1)^n \frac{x^{n+1}}{n+1}$$
I checked this manually, and it seemed to be correct.
For this question I am quite sure I need to solve for $n$, such that $R_n(x) < 0.0002$.
I have:
$$(-1)^n \frac{(0.3)^{n+1}}{n+1} < 0.0002 $$
(I have $0.3$ for $x$, since $\ln(1+0.3) = \ln(1.3)$.) Am I on the right track here? I am a little stuck with the algebra. I have:
$$ (-1)^n(0.3)^{(n+1)} < 0.0002(n+1) $$
I have tried to take the natural log of both sides, but it seems like I can't isolate the $n$.
Any help would be greatly appreciated!
AI: You actually want $|R_n(0.3)|<0.0002$, so you can ignore the factor of $(-1)^n$.
Here I think that your easiest bet is a little intelligently directed trial and error. Note that after you take logs, you have $(n+1)\ln 0.3<\ln 0.0002+\ln(n+1)$. Now $\ln 0.3$ is roughly $-1.2$, $\ln 0.0002$ is roughly $-8.5$, and $7\cdot12=84$, so a very rough first approximation would be to take $n+1=7$. Since $\ln 7$ is roughly $2$, in round numbers we have $-8.4$ on the left and $-6.5$ on the right, so $n=6$ is big enough: $-8.4<-6.5$. Now it only remains to be seen how much smaller you can go. By actual computation I get
$$\frac{0.3^6}6=0.0001215<0.0002$$ and
$$\frac{0.3^5}5=0.000486>0.0002\;,$$
so it appears that $n=5$ is the best we can do. |
H: Parametric equation involving exponents and e
How do you go about solving this problem?
For each plane curve given below, find a rectangular equation. State the appropritate interval for $x$ and $y$.
$x(t) = e^{5t}$, $y(t) = e^t$, $t \in (-\infty, \infty)$.
Which is the correct rectangular equation?
(a) $x = \frac 1{y^5}$
(b) $y = x^5$
(c) $y = \frac 1{x^5}$
(d) $x = y^5$.
AI: Since you are given alternatives, you can simply plug in and check:
(A) $x = 1/y^5$ gives $e^{5t} = 1/(e^t)^5$, which is not true. So (A) is wrong.
Etc. |
H: Characterization of ideals in rings of fractions
Let $R$ be a commutative unital ring. Let $S$ be a multiplicative subset.
Is there a characterisation of the ideals in the ring of fractions $S^{-1}R$ in terms of ideals $I$ in $R$ and $R$?
AI: I don't believe in general you can say much about how ideals in $R$ are related to those in the localization. I will state two instances where we do have some correspondence:
Prime ideals:
The prime ideals of $S^{-1}R$ are in one to one correspondence with prime ideals of $R$ that don't meet $S.$ That is to say prime ideals $\mathfrak{p} \subset R$ such that $\mathfrak{p} \cap S = \emptyset$. The correspondence is given by $\mathfrak{p} \leftrightarrow S^{-1}\mathfrak{p}$. The proof of this fact is not hard, one direction is easy and for the other you just need to use the isomorphism $\overline{S}^{-1}(R/I) \cong S^{-1}R/S^{-1}I$. This isomorphism comes from applying $S^{-1}$ to the exact sequence
$$0 \longrightarrow I \longrightarrow R \longrightarrow R/I \longrightarrow 0$$
of $R$ - modules. By $\overline{S}$ I mean the image of $S$ in the quotient, which is still a multiplicative set.
Ideals $I\subset R$ such that $xs \in I$ implies that $x \in I$ for all $s \in S$:
It is not hard to show that for any ideal $J$ of $S^{-1}R$, we have that $(J^c)^e = J$ where $()^c$ denotes contraction, $()^e$ denotes extension. Now if we are given an ideal $I$ of $R$ instead, it is not true that $(I^e)^c = I$. What is true though (which is not hard to prove) is that
$$(I^e)^c = \{r \in R: rs \in I \hspace{2mm} \text{for some} \hspace{2mm} s \in S\}.$$
It is easy to see from here that the processes of extension and contraction define the following bijective correspondence:
$$\{ \text{ ideals $I$ of $R$ | $xs \in I \implies x \in I$ for all $s \in S$}\} \leftrightarrow \{ \text{ideals of $S^{-1}R$}\}.$$ |
H: Prove that $\frac{{a}^{2}}{b-1}+\frac{{b}^{2}}{a-1}\geq8$
I need to prove that for any real number $a>1$ and $b>1$ the following inequality is true:
$$\frac{{a}^{2}}{b-1}+\frac{{b}^{2}}{a-1}\geq8$$
AI: Let $a = 1+x$ and $b = 1+y$. Then we need to prove that $$\dfrac{(x+1)^2}{y} + \dfrac{(y+1)^2}{x} \geq 8$$ i.e. $$\dfrac{x^2}{y} + 2 \dfrac{x}{y} + \dfrac1y + \dfrac{y^2}{x} + 2 \dfrac{y}{x} + \dfrac1x \geq 8$$
for $x,y \geq 0$.
Now apply AM-GM as shown below.
$$\dfrac{\dfrac{x^2}{y} + \dfrac{x}{y} + \dfrac{x}{y} + \dfrac1y + \dfrac{y^2}{x} + \dfrac{y}{x} + \dfrac{y}{x} + \dfrac1x}{8} \geq \sqrt[8]{\dfrac{x^2}{y} \times \dfrac{x}{y} \times \dfrac{x}{y} \times \dfrac1y \times \dfrac{y^2}{x} \times \dfrac{y}{x} \times \dfrac{y}{x} \times \dfrac1x} = 1$$
Hence, we get that $$\dfrac{x^2}{y} + 2 \dfrac{x}{y} + \dfrac1y + \dfrac{y^2}{x} + 2 \dfrac{y}{x} + \dfrac1x \geq 8$$ which is what we wanted to show. |
H: Bijective holomorphic map
A bijective holomorphic map from unit disk to itself will be rotation? That mean $f(z)=e^{i\alpha}z$? How do I approach to solve this problem?In addition I want to know how one can remember the conformal maps which sends unit disk to upper half plane or conversely,and all possible known place to other places like that?
AI: For $w\in\mathbb{D}$, consider the following function defined on $\mathbb{D}$:
$$B_w(z) = \frac{z-w}{1-\overline{w}z}.$$
This function is called a "Blaschke factor." It defines a bijective holomorphism from $\mathbb{D}$ to $\mathbb{D}$, but it is not a rotation if $w\neq 0$. This demonstrates there are many such mappings that are not rotations.
To see this, first note that that by an easy computation, $B_{w}$ is its own inverse, so it is biholomorphic. To see it maps $\mathbb{D}\rightarrow \mathbb{D}$, just take the modulus and do the standard manipulations, noting that $|z|<1$ and $|w|<1$.
However, if you make the assumption that $f(0)=0$, your guess is correct. All biholomorphic maps of the unit disk that fix the origin are rotations. We use the Schwartz lemma to show this. If you are not familiar with this, leave a comment below and I will edit this answer to include a proof and discussion.
By the lemma, we have $|f'(0)|\le 1$. Because $f^{-1}$ is another origin preserving biholomorphism, we also have $|f^{-1}$'$(0)|\le 1$. Because we know by the differentiation rule for inverse functions that $f'(0)=\frac{1}{f^{-1}\text{'}(0)}$, we must have $|f'(0)|=|f^{-1}\text{'}(0)|=1$, and considering the equality case in the Schwartz lemma, we see $f$ must be a rotation.
We know that $f^{-1}$ is differentiable. Then the chain rule gives
$$f(f^{-1}(x))=x\rightarrow f'(f^{-1}(x))=\frac{1}{f^{-1}\text{'}(x)}$$
and noting here that $f^{-1}(0)=0$ gives the result.
To answer the second part of your question, I suggest you use Google and the literature available to you. The magic search terms are "automorphism group of the disk" and "automorphism group of the upper half plane." This will tell you about all possible biholomorphisms of each space. Now, because the half-plane is conformally equivalent to the unit disk, knowing the automorphism group of the unit disk is enough to know of all the the maps from the unit disk to the upper half-plane. |
H: direct product of center of group
Let $Z(G)$ denote the center of a group $G$, let $J_n=Z(G)\times\dots \times Z(G)$, is it true that as a subset of external direct product $G\times\dots\times G$, $J_n$ is a subgroup?normal sybgroup?is it isomorphic to $Z(G)\times\dots \times Z(G)$ ($(n-1))$ times ?
I know $Z(G)$ is a subgroup, so I hope $J_n$ will be so, but I am not sure about the other options.thank you for your help. Well, about the isomorphism I think projection map will work?
AI: Let $G$ and $H$ be groups, let $A<G$ and $B<H$. The group operation in $G\times H$ is defined to be
$$(g_1,h_1)(g_2,h_2)=(g_1g_2,h_1h_2).$$
Clearly, the identity of the operation is $(e_G,e_H)$ where $e_G$ is the identity of $G$ and $e_H$ is the identity of $H$, and for any $(g,h)\in G\times H$, its inverse is $(g^{-1},h^{-1})$.
Because $A$ is a subgroup of $G$ and $B$ is a subgroup of $H$, we have that $e_G\in A$ and $e_H\in B$, so that $(e_G,e_H)\in A\times B$.
For any $(a_1,b_1),(a_2,b_2)\in A\times B$, we have that $a_1,a_2\in A$ and $b_1,b_2\in B$, so that $a_1a_2\in A$ and $b_1b_2\in B$ because $A$ and $B$ are subgroups of $G$ and $H$ respectively, so that $(a_1,b_1)(a_2,b_2)=(a_1a_2,b_1b_2)\in A\times B$.
For any $(a,b)\in A\times B$, we have that $a\in A$ and $b\in B$, so that $a^{-1}\in A$ and $b^{-1}\in B$ because $A$ and $B$ are subgroups of $G$ and $H$ respectively, so that $(a^{-1},b^{-1})\in A\times B$.
Thus, $A\times B$ is a subgroup of $G\times H$ for any subgroups $A<G$ and $B< H$. In fact, for any family of groups and subgroups (not just finite ones), the direct product of the subgroups is a subgroup of the direct product of the groups - just modify the above argument.
Now, we consider the case of normal subgroups. Let $N\triangleleft G$, $K\triangleleft H$, so that for any $g\in G$ and $h\in H$, we have $gNg^{-1}=N$ and $hKh^{-1}=K$. For any $(g,h)\in G\times H$, we have that
$$(g,h)(N\times K)(g,h)^{-1}=(g,h)(N\times K)(g^{-1},h^{-1})=\{(g,h)(a,b)(g^{-1},h^{-1})\mid (a,b)\in N\times K\}=\{(gag^{-1},hbh^{-1})\mid (a,b)\in N\times K\}\subseteq N\times K$$
and by symmetry we get $(g,h)(N\times K)(g,h)^{-1}=N\times K$. Thus $(N\times K)\triangleleft (G\times H)$.
In fact, for any family of groups and normal subgroups (not just finite ones), the direct product of the normal subgroups is a normal subgroup of the direct product of the groups - just modify the above argument. Or, you can use induction to show that this is true for a finite collection of groups $G_1,\ldots,G_n$ and normal subgroups $N_1\triangleleft G_1,\ldots,N_n\triangleleft G_n$. Letting $G_1=G_2=\cdots=G_n=G$ and $N_1=N_2=\cdots=N_n=Z(G)$, you have that $J_n$ is a normal subgroup of $G\times\cdots\times G$.
It certainly need not be the case that $J_n=\underbrace{Z(G)\times\cdots\times Z(G)}_{n\text{ times}}$ is isomorphic to $\underbrace{Z(G)\times\cdots\times Z(G)}_{n-1\text{ times}}$. They don't even necessarily have the same cardinality. For example, letting $G=C_2$ be the cyclic group of order 2, so that $|G|=|Z(G)|=2$, then the former has $2^n$ elements and the latter has $2^{n-1}$ elements. |
H: Loopspace adjunction: when are unit or counit equivalences?
For (nice?) pointed spaces, the reduced suspension $\Sigma$ is left adjoint to the loop space $\Omega$. This adjunction is given by the unit maps
$\eta_X : X \to \Omega \Sigma X$, $x \mapsto (t \mapsto [x,t])$
and the counit maps
$\varepsilon_X : \Sigma \Omega X \to X , [\omega,t] \mapsto \omega(t).$
Question. For which $X$ is $\eta_X$ a homotopy equivalence? For which $X$ is $\varepsilon_X$ a homotopy equivalence?
If this is useful, let's assume that $X$ is sufficiently nice (for example a CW complex). You may also replace "homotopy equivalence" by "homology equivalence" etc., if this yields to interesting statements. If there is no characterization: What are interesting classes of examples? And is there any source in the literature where this sort of question is studied?
AI: Hatcher Theorem 4J.1:
The map $J(X)\to \Omega \Sigma X$ is a weak homotopy equivalence for every
connected CW complex $X$.
Here $J(X)$ is James' reduced product of $X.
Hatcher Proposition 3.22:
For $n > 0$, $H^*(J(S^n);\mathbb Z)$ consists of a $\mathbb Z$ in each dimension a
multiple of $n$.
So your question fails already for spheres.
Edit: Corollary 4.J.3 also tells you how close you can get. The map $X\to \Omega\Sigma X $ factors through $J(X)$ and $(J(X),X)$ is $2n+1$ connected.
Edit2: And in this paper the authors show that the homotopy fibre of the map $S\Omega X\to X$ has the homotopy type of $\Omega X\ast\Omega X$, the join of $\Omega X$ with itself. |
H: Does a solved sudoku game always have same sum? Is this sum unique to solved game?
Fundamentally, I'm looking for help on two things:
Verification that my math is correct for the assumption that all Xs are Y.
Proof that are the inverse is true, that all Y's are X, or, if it's not true, example of X that is outside the Y set.
More specifically, the assumption is :
All solved Sudoku puzzles, where "solved" is defined as a 9x9 grid having the set 1-9 in each column, row, and 3x3 quadrant, can be verified as solved by taking the sum of each row, column, and quadrant, as the sum will always be 1215. This is based on taking the sum of each digit in the set (1+9 + 8+2 + 7+3 + 4+6 + 5 = 45) and multiplying by the total number of sets (9 rows + 9 columns + 9 quadrants = 27), so 45 * 27 = 1215.
If the assumption above is correct, is this sum unique to solved puzzles, or are there permutations of a completed (but not solved) board that would give 1215 using the same method?
After a general search and skimming 2 wikipedia articles on Sudoku mechanics (Mathematics of Sudoku and Sudoku Algorithms), I'm still not seeing this simple approach to verifying a board as solved. All math and logic seem dedicated to solving puzzles (as in finding the correct digit for each cell) or to generating solvable puzzles. This has me thinking I am overlooking something with my math and logic, but I can't think of a board where the sum of the sets wouldn't be 1215 or a board where the sum would be 1215 and it would be an invalid board.
I am ready to be shown the err of my ways, but it would be cool to confirm that a board can be solved without confirmation that each cell value is valid.
AI: Apropos your comment,
I'm guessing... that there is not an arithmetic way to check for set uniqueness?
there actually is a way, using only arithmetic, to check whether a sequence of nine numbers each between $1$ and $9$ contains all unique elements: for each number $n$ in the sequence, replace it with $2^{n-1}$, then add them all up and check that the sum is $511$. For example, the row $[5 | 3 | 4 | 6 | 7 | 8 | 9 | 1 | 2]$ is valid because
$$2^{5-1} + 2^{3-1} + 2^{4-1} + 2^{6-1} + 2^{7-1} + 2^{8-1} + 2^{9-1} + 2^{1-1} + 2^{2-1} \\ = 16 + 4 + 8 + 32 + 64 + 128 + 256 + 1 + 2 \\ = 511,$$
but $[5 | 5 | 5 | 5 | 5 | 5 | 5 | 5 | 5]$ is not because
$$2^{5-1} + 2^{5-1} + 2^{5-1} + 2^{5-1} + 2^{5-1} + 2^{5-1} + 2^{5-1} + 2^{5-1} + 2^{5-1} \\ = 16 + 16 + 16 + 16 + 16 + 16 + 16 + 16 + 16 \\ = 144 \ne 511.$$
It should be clear that any permutation of the same set of numbers should give the same sum, so any valid solution should have the sum $511$. What is not immediately obvious is that no other sequence of nine numbers can give this sum. This is true because the Hamming weight of $511$ is $9$. The sum of $N$ numbers that are powers of $2$ can have Hamming weight at most $N$, and if there are duplicates among them, the weight of the sum is strictly less than $N$. So the only way to get a Hamming weight of $9$ by adding nine powers of $2$ is if they are all distinct. |
H: Inequality between volume and its projections
Let $A \subset \mathbb{R}^3$ be connected and let's define $A_1, A_2, A_3 \subset \mathbb{R}^2$ as projections of $A$ onto three perpendicular (to each other) planes. Show that:
$$|A| \le \sqrt{|A_1| |A_2| |A_3|}\;,$$
where $|\cdot|$ is volume when applied to $A$ and area whenas $A_{1,2,3}$.
AI: This may be overkill, but this inequality is a simple consequence of the inequality on the first page of this document. It says that, for any three sigma-finite measures spaces $(X,\mu)$, $(Y,\nu)$ and $(Z,\xi)$ and any three square-integrable functions $f:X \times Y \to \mathbb{R}$, $g:Y \times Z \to \mathbb{R}$ and $h:Z \times X \to \mathbb{R}$,
$$\int_{X \times Y \times Z} f(x,y) g(y,z) h(z,x) d\mu (x) d\nu (y) d\xi (z)
\leq \sqrt{\int_{X \times Y} f^2(x,y) d\mu (x) d\nu (y) \int_{Y \times Z} g^2(y,z) d\nu (y) d\xi (z) \int_{Z \times X} h^2(z,x) d\xi (z) d\mu (x)}.$$
We can assume without loss of generality that the three perpendicular planes are the canonical planes $(x,y,0)$, $(0,y,z)$ and $(x,0,z)$. Then, taking $f(x_0,y_0)=1$ if there exists a $z$ suchat that $(x_0,y_0,z)$ belongs to $A$ (in other words, $f$ is conjugated to $1_{A_1}$ via the trivial morphism between $\mathbb{R}^2$ and the first canonical plane), an the same for $g$ and $h$, one gets the result.
A few remarks:
Connectedness is not necessary.
The article I linked works with probability measures. However, since the inequality is homogenous in the measures, one can extend it for free to finite measures (which solves the case of bounded $A$), and with some work the sigma-finite case is obvious - at least for positive functions.
There are simpler proofs, if only because we work with $\{0,1\}$-valued functions, but the trick is nice to know anyway. |
H: Definition of neighborhood and open set in topology
I am a Physics undergrad, and just started studying Topology. How do you define neighborhood and open set in Topology.Wikipedia gives a circular definition.
An open set is defined as follows.
In topology, a set is called an open set if it is a neighborhood of every point
While a neighborhood is defined as follows:
If $X$ is a topological space and $p$ is a point in $X$, a neighbourhood of $p$ is a subset $V$ of $X$, which includes an open set $U$ containing $p$
which itself contains the term open set.
How do you define it exactly?
AI: One defines a topology on a set by specifying the open sets.
Let $X$ be a set. If $\tau$ is a family of sets with the following properties, it is called a topology.
$X$ and $\varnothing$ are in $\tau$
Any (possibly infinite, even uncountably infinite) union of sets in $\tau$ is in $\tau$.
The intersection of any finite number of elements of $\tau$ is in $\tau$.
We call the sets in $\tau$ the open sets. You can see that the collection of open sets in, for example, $\mathbb{R}^2$ has exactly this set of properties.
A neighborhood of a set $S$ is a set $P$ that contains an open set $U$ so $S\subset U\subset P$.
For more information, see Topology by Munkres. |
H: How to check that whather a Polygon is completly inside of another Polygon?
Let's say I have two polygons. I know the co-ordinates of both polygons. Now, I need to check whether the first Polygon is completely inside of second polygon? IN this figure only 1 polygon is completely inside of red polygon.
AI: One way would be to compile a list of all edge pairs, and check whether any edges cross each other. If none do, you can choose any point inside one of the polygons, and check weather it is is the second polygon by using the following theorem:
A point is inside a polygon if a ray from this point crosses the polygon's edges
an odd number of times.
If the point is inside both, then one of the polygons is inside the other. |
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