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H: Uniqueness of meets and joins in posets
Exercise 1.2.8 (Part 2), p.8, from Categories for Types by Roy L. Crole.
Definition: Let $X$ be a preordered set and $A \subseteq X$. A join of $A$, if such exists, is a least element in the set of upper bounds for $A$. A meet of $A$, if such exists, is a greatest element in the set of lower bounds for $A$.
Exercise: Suppose that $X$ is a poset (and thus also a preorder). Show that meets and joins in a poset are unique if they exist.
AI: Let $A \subseteq X$ and suppose $x$ and $y$ are both joins of $A$. Then $x$ and $y$ are both upper bounds of $A$. So $A \le x$ and $A \le y$. But since $x$ and $y$ are joins of $A$ and $x, y \in X$, $y \le x$ and $x \le y$. Therefore, $x = y$ because because $\le$ is anti-symmetric. Meets are unique by a similar argument. |
H: A simple algebraic question
Show, as elegantly as possible, that
$2p_0(1-p_0^r) > (p_0+\Delta)[1-(p_0+\Delta)^r] + (p_0 - \Delta)[1-(p_0-\Delta)^r]$
for $r \in \mathbb{R^+}$, $0 < p_0 - \Delta < p_0 + \Delta < (r+1)^{-\frac{1}{r}}$.
This is trivial for $r=1$, but I believe it should also hold for general $r > 0$. I would be interested in elementary proofs for that inequality -- if it is indeed true in general. Many thanks!
AI: Your goal is to show that $u_r(p)\gt\frac12(u_r(p-\Delta)+u_r(p+\Delta))$ for every $0\lt\Delta\lt p$, where $u_r:p\mapsto p(1-p^r)$. Thus, if $u$ is concave, you are done. But $u_r''(p)=-(r+1)rp^{r-1}$ hence $u_r''(p)\lt0$ for every $p\gt0$, as soon as $r\gt0$. This shows that $u_r$ is indeed strictly concave.
Note: The upper bound $(r+1)^{-1/r}$ is irrelevant. |
H: Integral of exponential function
Consider $f$ being a measurable function on $R^n$ such that $$\int_{E} e^{|f|}=1$$ ($E$ measurable) and $f$ vanishes outside $E$ . Then $f\in L^p(R^n)$ for all $p\in (0,\infty)$.
I tried using that measure of $E$ cannot be bigger than $1$ and the formulae
$$\int|f|^p=p\int_0^\infty \alpha^{p-1}\omega(\alpha)d\alpha$$
where
$\omega(\alpha)=\{x \in R^n: |f(x)|>\alpha\}$.
AI: For all $x\ge0$, $x<e^x$. Therefore, substituting $x\mapsto x/p$ and raising both sides to the $p^{\text{th}}$ power yields
$$
\left(\frac{x}{p}\right)^p<e^x
$$
Therefore,
$$
\left(\int_E|f(x)|^p\,\mathrm{d}x\right)^{1/p}\le p\left(\int_E e^{|f(x)|}\,\mathrm{d}x\right)^{1/p}
$$ |
H: Derivative of an indicator inside an integral
I have a fairly basic question that relates to understanding a particular derivation.
I have the following function $Q(x) = E\left[I(F(x+\varepsilon)>c)\right]$, where $x \in R$, $\varepsilon \sim N(0,\sigma^2)$ for a known $\sigma>0$, $c$ is a scalar constant, $F(\cdot)$ is a known, bounded function, and $I(\cdot)$ is an indicator function.
This can be written as:
$E[I(F(x+\varepsilon)>c)] = \frac{1}{{\sqrt{2\pi}}\sigma} \int{I(F(x+\varepsilon)>c) \exp\left(-\frac{\varepsilon^2}{2\sigma^2} \right)}d\varepsilon $
Let $y = x + \varepsilon$ and rewrite the above as:
$E[I(F(x+\varepsilon)>c)] = \frac{1}{{\sqrt{2\pi}}\sigma} \int{I(F(y)>c) \exp\left(-\frac{(y-x)^2}{2\sigma^2} \right)}dy$
I am interested in calculating $\frac{\partial Q(x)}{\partial x}$. The derivative is:
$\frac{\partial E[I(F(x+\varepsilon)>c)]}{\partial x}= \frac{1}{{\sqrt{2\pi}}\sigma} \int{I(F(y)>c) \exp\left(-\frac{(y-x)^2}{2\sigma^2} \right)} \frac{(y-x)}{\sigma^2}dy \quad \quad $ (1)
Substituting back in again and rearranging slightly:
$\frac{\partial E[I(F(x+\varepsilon)>c)]}{\partial x} = \frac{1}{\sigma^2} E[I(F(x+\varepsilon)>c)\varepsilon]$
Finally, here are my two questions: (1) is the above derivation correct? (2) If so, I am trying to understand why in eq(1) I do not need to take the derivative of $I(F(y)>c)$ with respect to $x$. I know there has been a change of variables, but $y$ is obviously a function of $x$. Yet I was told that "once you do the substitution, $y=x+\varepsilon$, then y is just a dummy variable (i.e. an index) which runs from -infinity to infinity, and one cannot take the derivative in a dummy variable." I know you can't take the derivative of an indicator function, but somehow I can't get this clear in my head.
Thanks.
More broadly, in my actual problem, $x$ and $\varepsilon$ are high dimensional and I am trying to compute the derivative via Monte Carlo integration....
AI: This is true and might be recognized as the derivative of the function $x\mapsto\mathrm P(x+\varepsilon\in A)$, where $A=\{F\gt c\}$. Thus, one is considering $U(x)=\mathrm E(u(x+\varepsilon))$, for $u=\mathbf 1_A$ and the assertion is that $U'(x)=\sigma^{-2}\mathrm E(\varepsilon u(x+\varepsilon))$. This holds in full generality and the proof is as you explain.
Re your (2), the derivative of $\mathbf 1_{F(y)\gt c}$ with respect to $x$ exists and it is zero, simply because the function does not depend on $x$. Recall that, under suitable hypotheses,
$$
\frac{\mathrm d}{\mathrm dx}\int f(x,y)\,\mathrm dy=\int \frac{\partial f(x,y)}{\partial x}\,\mathrm dy.
$$ |
H: Given $\lambda$ and $A$, find $v$ such that $\lambda = v^{\intercal}Av$
If I know the values of $\lambda$ and $A$, how do I find a vector $v$ such that $\lambda = v^{\intercal}Av $?
This isn't a homework question; I just ran into this problem in Real Life and realized I couldn't solve it!
AI: If $A$ is positive definite then for any $\lambda>0$ pick any vector $v\not=0$ and compute
$$
c=v^TAv
$$
Then
$$
v\sqrt{\frac{\lambda}{c}}
$$
satisfies your equation. |
H: Integral over circle
Consider the following integral:
$\displaystyle \int_{|z-3i|=1} \frac{e^{z^2+z}}{z} dz$.
Why is my approach wrong?
Set $f(z)=e^{z^2+z}$, then from the Cauchy integral formula for circles, we have:
$\displaystyle f(0)=\frac{1}{2\pi i} \int_{|z-3i|=1} \frac{f(z)}{z} dz$.
So we have: $\displaystyle \int_{|z-3i|=1} \frac{e^{z^2+z}}{z} dz = 2 \pi i$.
Where is the mistake?
Thank you for your time!
AI: The problem is that $0$ is outside your circle $|z-3i|=1$, not inside. |
H: $<$ on a preorder is a strict partial order
Definition: Suppose $X$ is a preorder. Define $x < y$ as $x \le y$ and $y \not\le x$ for each $x, y \in X$.
Question: Show that this gives a strict partial order on $X$.
AI: We must show that the relation $<$ is irreflexive and transitive.
Irreflexive: Suppose $x \in X$. Then $x \le x$, so $x \not< x$.
Transitive: Suppose $x, y, z \in X$ such that $x < y$ and $y < z$. Then $x \le y$, $y \not\le x$, $y \le z$, and $z \not\le y$. By the transitivity of $\le$, we immediately have $x \le z$. Now assume that $z \le x$. Then $z \le y$. This is a contradition. |
H: Determining if sets are nonempty
How to show that these sets are nonempty (here $\mid $ means "divides")?
Here N is an arbitrary large integer and q is some fixed integer.
$R = \lbrace k \in {\mathbb N}:(kN\mid k!) \wedge ((k - 1)N\mid k!) \wedge \cdots \wedge (N\mid k!) \wedge (k > Nq)\rbrace$
$S = \lbrace k \in {\mathbb N}:({(2k - 1)^2}N\mid k!) \wedge ({(2k - 3)^2}N\mid k!) \wedge \ldots \wedge (N\mid k!) \wedge (k > Nq)\rbrace$
$T = \lbrace k \in {\mathbb N}:({k^5}N\mid k!) \wedge ({(k - 1)^5}N\mid k!) \wedge \ldots \wedge (N\mid k!) \wedge (k > Nq)\rbrace$
They exist by the axiom schema of separation, but how do I determine which $k$ to choose so that it satisfies all the properties? Is there a general approach?
AI: For example, for $R$, you want $k!/(k-j)$ to be a multiple of $N$ for each $j$ from $0$ to $k-1$. That
will certainly be true if $k \ge 2N$. |
H: What could this sum possibly converge to?
Consider
$$\sum_{i=0}^{\infty} \dfrac{n-i}{n!}$$
For $n \geq i$
Consider $n$ to be any natural number. I know for sure it's going to converge, but how do I write a formula for the sum?
Possible interpretation:
Find:
$$\lim_{n\to \infty}\sum_{i=0}^{n} \dfrac{n-i}{n!}$$
AI: $$u_n= \sum_{i=0}^n \frac{n-i}{n!}$$
Then
$$ u_n = \frac{1}{n!} \sum_{i=0}^n i = \frac{1}{n!} \frac{n(n+1)}{2} = \frac{n+1}{2(n-1)!} \rightarrow 0$$ |
H: Generating Function of Even Fibonacci
I was posed the following question recently on an exam:
Determine the generating function of the even-indexed Fibonacci numbers $F_{2n}$ given that the generating function of Fibonacci numbers is $\frac{x}{1-x-x^2}$.
I could not think of how apply knowledge of the generating function of the Fibonacci numbers to determine the generating function of the even Fibonacci, so I tried to to determine the generating function using the identity $F_{2n}=F_{n+1}^2-F_{n-1}^2$, but to no avail.
Any thoughts about how to find its generating function?
AI: HINT: For $f(x) = \displaystyle\sum_{n \geqslant 0} c_n x^n$, what is the series for $\frac{1}{2} \left( f(x) + f(-x)\right)$? |
H: Qubits and vector projections
In $\Bbb C^2$, how many real unit vectors are there whose projection onto $|1\rangle$ has length $\sqrt{3}/2$?
I would think zero as $\bigl(\frac{\sqrt{3}}{2}\bigr)^2 + x^2 = 1$, therefore there are no real values of $x$ to satisfy the equation. Please help.
AI: Write $\Bbb C^2$ as $\Bbb C|1\rangle\oplus \Bbb C|2\rangle$ and the canonical projection $P$ down to the subspace spanned by $|1\rangle$,
$$P:\alpha|1\rangle+\beta|2\rangle\mapsto\alpha|1\rangle.$$
Then $P(\cdot)=\frac{\sqrt{3}}{2}|1\rangle$ means $\alpha=\frac{\sqrt{3}}{2}$. Furthermore, the condition of being a unit vector means
$$\left\|\frac{\sqrt{3}}{2}|1\rangle+\beta|2\rangle\right\|=1\iff \left|\frac{\sqrt{3}}{2}\right|^2+|\beta|^2=1\iff |\,\beta|=\frac{1}{2}.$$
The only two real solutions for $\beta$ are $\pm1/2$, so the answer is two. The two real-coordinate unit vectors are given by $\frac{\sqrt{3}}{2}|1\rangle\pm\frac{1}{2}|2\rangle$. Note that losing the restriction of real-coordinates means that $\beta$ could be chosen anywhere on the circle of radius $1/2$ about the origin in the complex plane $\Bbb C$.
As an anonymous user points out in a suggested edit to my answer, I did not really pay attention to the first comment below. The question doesn't say the projection has coordinate $\sqrt{3}/2$, it says this is the length of the projection, so $\alpha$ may be $\pm\sqrt{3}/2$. This yields four possible solutions total. |
H: Evaluating $ \int_{|z|=1} \frac{e^z}{(z+3)\sin(2z)} \ dz$
Consider the following integral:
$\displaystyle \int_{|z|=1} \frac{e^z}{(z+3)\sin(2z)}dz$.
To apply the Cauchy integral formula, I rewrite it as:
$\displaystyle \int_{|z|=1} \frac{ze^z}{z(z+3)\sin(2z)}dz$ and take
$\displaystyle f(z)=\frac{ze^z}{(z+3)\sin(2z)}$.
The problem now is that I would compute $f(0)$ in the next step, but $\sin(2\cdot 0)=0$, so the denominator of $f$ is undefined. How would I deal with this?
AI: Hint: What type of singularity is $z = 0$ for the function $$f(z) = \frac{ze^z}{(z+3)\sin(2z)} = \frac{e^z}{z+3}\frac{z}{\sin(2z)}?$$
In other words: is there any reasonable way of defining $f(0)$, perhaps by some sort of limit? |
H: Is it possible to solve a linear inequality system using SVD?
I have a large linear inequality system of the form $Ax ≤ 0$.
Is there a way to solve this system using linear algebra tools like SVD?
AI: No; loosely speaking this problem is more related to linear programming, which is not equal to linear algebra.
(Plugging in an SVD of $A$ will be too helpful as you probably have seen.)
The set $Ax\leq 0$ is a convex polytope and the representation $Ax\leq 0$ is called its H-representation because it is a description as intersection of half-spaces. Another representation is its V-representation which consists of a list of the vertices. The Wikipedia entry on convex polytopes has more pointers. |
H: Evaluating $\int_{-\pi}^{\pi} \cos(e^{it})dt$
We have to calculate:
$$\displaystyle \int_{-\pi}^{\pi} \cos(e^{it})dt.$$
Is there something more promising one could try instead of a subsitution $u=e^{it}$?
AI: Actually the Taylor series approach is fairly straightforward too...
$\int_{-\pi}^{\pi} \cos(e^{it}) \,dt = \int_{-\pi}^{\pi} \sum_{n=0}^{\infty} (-1)^n \frac{e^{2int}}{(2n)!} = \sum_{n=0}^{\infty} (-1)^n \int_{-\pi}^{\pi} \frac{e^{2int}}{(2n)!} \, dt = 2 \pi$.
The exchange of summation and integration is justified by uniform convergence on the unit circle. All terms except the constant disappear. |
H: what can we say about the SVD of a matrix with respect to another SVD?
I have two matrices, both are $n \times m$ where $m < n$. These two matrices are $A$ and $B$.
I also know the singular value decomposition of $AB^{\top} = U \Sigma V^{\top}$.
Is there anything I can say about the SVD of $A$ or $B$? (for example, can I say that the left singular vectors of $A$ are $U$?) Are there any conditions under which I could say something like that?
Thanks.
AI: Let $V_A$ be the right singular vectors of $A$, and $U_B$ be the left singular vectors of $B^T$, we have $$AV_A=U_A\Sigma_A, \quad B^TV_B=U_B\Sigma_B$$
If $V_A= U_B$, then $$AB^TV_B=AU_B\Sigma_B=AV_A\Sigma_B=U_A\Sigma_A\Sigma_B$$
Also, because the the left singular vectors of $B^T$ are right singular vectors of $B$,
we can see that, if the right singular vectors of $B$ are right singular vectors of $A$, then we have that the right singular vectors of $B^T$ is the right singular vectors of $AB^T$, and the left singular vectors of $A$ is the left singular vectors of $AB^T$. |
H: how to solve an ODE with boundary conditions $ y(0)=y(\infty) $ by shooting method
how could I solve the linear differential equation
$$ -y''(x)+x^{3}y(x)=0$$ with the boundary conditions $ y(0)=y(\infty) $ by the linear shooting method ?
If we had $ y(0)=y(1)=0$ then the interval is finite however how can i deal with $ y(\infty) =0 $ ? i guess that we can get a large number $ L \rightarrow 10^{6} $ and solve it $ y(0)=y(L)=0$ however I do not know what is the best method to solve numerically my eigenvalue problem.
AI: First, it is not necessarily sufficient to choose a large number because it is large. Rather, you will want to choose a number $L$ such that $y''(L) \approx y'(L) \approx 0$. In some cases, it is sufficient to replace $\infty$ with something as small as $5$ (such as computing the boundary layer velocity profile over a flat plate). Other cases, you may need a number quite large. To estimate a sufficient value, make a guess for $y'(0)$, and integrate. Find an acceptable $L$ such that you achieve the desired error tolerance.
Second, assume you can find such an $L$. There are a number of ways to implement a shooting method. One way is to re-frame the problem as a "root finding" algorithm. First, you guess a value $c$ and let $y'(0)=c$. Then, integrate the ODE using whatever method suits your needs. Then, you compute the error $||y(L)-0|| = e$ and compute a new value of $c$. You can do this using Newton's method, for example, or via bisection. |
H: Constructing $\mathbb N$ from the set of factorials
Let S be the set $\{0!, 1!, 2!, \ldots\}$. Is it possible to construct any positive integer using only addition, subtraction and multiplication, and using any element in S at most once? For example:
$$ 3 = 2! + 1!$$
$$ 4 = 3! - 2! = 2! + 1! + 0!$$
$$ 146 = 4!\cdot3! + 2!$$
etc. My gut instinct says that this isn't true, but I can't see why. Something like 8076 doesn't have an obvious solution, but maybe you can get it by subtraction a huge factorial from the product of two smaller factorials or something. Or maybe there's a way of finding sets of factorials that add/subtract/multiply to 1, in which case any number can be constructed this way. I've tried finding something but haven't had much luck.
EDIT: Oops, positive integer, not positive number.
AI: Let me assume you're only allowed to use $0! = 1!$ once. In that case, all factorials past $4!$ are divisible by $24$, so working $\bmod 24$ the only numbers you're allowed to use are $1, 2, 6$, each at most once, and I am reasonably certain you cannot get any numbers congruent to $10 \bmod 24$ this way.
Edit: If you want to use both $0!$ and $1$, then all factorials past $5!$ are divisible by $120$, so working $\bmod 120$ the only numbers you're allowed to use are $1, 1, 2, 6, 24$. This time I am reasonably certain you cannot get any numbers congruent to $57 \bmod 120$. Just kidding! Every conjugacy class $\bmod 120$ is reachable.
Okay, working $\bmod 720$ the only numbers you're allowed to use are $1, 1, 2, 6, 24, 120$... |
H: If $g(x) := \int_1^2 f(xt)dt \equiv 0$ then $f \equiv 0$
Let $f \colon \mathbb R \to \mathbb R$ be a continuous function. Let's
define $$ g(x) := \int_1^2 f(xt)dt. $$
Prove that $g \equiv 0 \Rightarrow f \equiv 0$.
Well, I show you what I have done. First of all, I've noted that $g$ is differentiable for every $x \in \mathbb R \setminus\{0\}$. Indeed,
$$
g(x) = \frac{1}{x}\int_x^{2x} f(y)dy
$$
hence (by the fundamental theorem of calculus)
$$
\frac{dg}{dx} = -\frac{1}{x^2}\int_x^{2x} f(y)dy + \frac{1}{x}\left[2f(2x)-f(x)\right] = -\frac{1}{x}g(x) + \frac{1}{x}\left[2f(2x)-f(x)\right]
$$
So if $g(x)=0$ for every $x \in \mathbb R$ we must have
$$
\frac{dg}{dx} = 0 \Leftrightarrow 2f(2x)-f(x) = 0
$$
i.e.
$$
f(x)=\frac{1}{2}f\left(\frac{1}{2}x\right)
$$
for every $x \in \mathbb R$. Is it correct what I've done so far? How would you conclude? I don't manage to prove $f \equiv 0$: I've just noted that $f(0)$ must be $0$, but no more...
Thanks in advance.
AI: Yes, it's correct so far. Now show by induction that for each integer $n$,
$$f(x)=\frac 1{2^n}f\left(\frac x{2^n}\right).$$
Since $f$ is continuous (it has already been used when you showed that $g$ was differentiable) on $[0,1]$, we have that $\frac x{2^n}\to 0$ when $n\to +\infty$ for a fixed $x$, hence
$$f(x)=\lim_{n\to +\infty}\frac 1{2^n}f\left(\frac x{2^n}\right)=\lim_{n\to +\infty}\frac 1{2^n}f\left(0\right)=0.$$ |
H: conformally equivalent riemann surfaces
Two riemann surfaces $S$ and $R$ are said to be conformally equivalent if there exist a holomorphic map $f:S\rightarrow R$ which is one-one and onto, and inverse is also holomorphic.
I have to show No two of them are conformaly equivalent:
$a$) $\hat{\mathbb{C}}$, $b$) $\mathbb{C}$, $c$) $D$(Open Unit Disc)
Ok let $f:a\rightarrow b$ be a bijective holomorphic map and $f^{-1}: b\rightarrow a$ is also holomorphic, well,the contracdiction is $f(\hat{\mathbb{C}})=\mathbb{C}$ as $\mathbb{C}$ is non-compact? for $b\rightarrow c$ riemann mapping theorem can be applied to show the contradiction? and $a\rightarrow c$ is same argument as $a\rightarrow b$?
please help me.
AI: The sphere is the only compact one, so neither of the other two can be a continuous image of it, so cannot be conformally equivalent to it.
$\mathbb{D}$ cannot be the holomorphic image of $\mathbb{C}$ by Liouville's Theorem. (If it were we would have a bounded non constant entire analytic function).
And we're done! |
H: Re-writing in sign basis.
$\newcommand\ket[1]{\left\vert #1\right\rangle}$ Let $\ket\phi = 12 \ket{0} + 1 + 2\sqrt{i2}\ket{1}$. Write $\ket\phi$ in the form $\alpha_0\ket{+} + \alpha_1\ket{-}$. What is $\alpha_0$?
I came across this problem in a course i am doing, i have been struggling writing things in sign basis, much appreciated.
AI: The "sign basis" is defined as
$$\begin{aligned}
\lvert+\rangle &= \frac{1}{\sqrt{2}}\left(\lvert0\rangle+\lvert1\rangle\right)\\
\lvert-\rangle &= \frac{1}{\sqrt{2}}\left(\lvert0\rangle-\lvert1\rangle\right)
\end{aligned}$$
Those equations can be solved for $\lvert0\rangle$ and $\lvert1\rangle$. Then it's just a matter of inserting in the given state and comparing the coefficients. |
H: Compute $\sum \frac{1}{k^2}$ using Euler-Maclaurin formula
I read that Euler used the summation formula to calculate the value of the series $\sum_{k =1}^{\infty} \frac{1}{k^2}$ to high precision without too much hassle. The article Dances between continuous and discrete: Euler’s summation formula goes into the calculation, however without too much justification of why it works (especially since the series used to calculate the limit does not converge and one has to truncate it at a certain point). I would be glad if someone could elaborate from a more modern viewpoint on how and why it works.
AI: First, it can be shown, in many ways, that
$$
\sum_{k=1}^\infty\frac1{k^2}=\frac{\pi^2}{6}\tag{1}
$$
However, the Euler-Maclaurin Summation Formula can be used to numerically sum this series to high precision. Before we attempt to do this, let's discuss a bit about asymptotic series.
In general, asymptotic series, like those arising from the Euler-Maclaurin Summation Formula (EMSF), are divergent. This means that if you tried to sum all the terms arising from the formula, the sum would not converge. For example, the EMSF gives the following asymptotic expansion:
$$
\sum_{k=1}^n\frac1k\sim\gamma+\log(n)+\frac1{2n}-\frac1{12n^2}+\frac1{120n^4}-\frac1{252n^6}+\frac1{240n^8}-\frac1{132n^{10}}+\dots\tag{2}
$$
where $\gamma=0.5772156649015328606065121$ is the Euler-Mascheroni constant.
This expansion looks well behaved, and it is, up to a point. However, the coefficients grow on the order of $\frac{n!}{(2\pi)^n}$; by the time that we get to the term for $n^{50}$, we have
$$
\frac{19802288209643185928499101}{132n^{50}}\tag{3}
$$
Due to the growth rate of the coefficients, no matter how large $n$ is, this series cannot converge.
However, if we only use a finite number of terms, the series is a very good approximation for large $n$. As mentioned earlier, the expansion behaves well, up to a point. What this means is that for a given $n$, the terms get smaller, and then, at some point, start blowing up. The point at which the term start blowing up is further along the larger $n$ is. The good part is that if we terminate the sum while the terms are still getting smaller, the approximation is usually as good as the next term.
For example, let's approximate
$$
\sum_{k=1}^{1000}\frac1k=7.4854708605503449126565182\tag{4}
$$
using the first $4$ terms of $(2)$:
$$
\gamma+\log(1000)+\frac1{2000}-\frac1{12000000}=7.4854708605503365793271531\tag{5}
$$
The result in $(5)$ is $8.333329365\times10^{-15}$ smaller than the actual value in $(4)$. The next term is
$$
\frac1{120000000000000}=8.333333333\times10^{-15}\tag{6}
$$
Now let's see how the EMSF can be used to approximate $(1)$
The EMSF applied to $\dfrac1{k^2}$ yields
$$
\sum_{k=1}^n\frac1{k^2}\sim C-\frac1n+\frac1{2n^2}-\frac1{6n^3}+\frac1{30n^5}-\frac1{42n^7}+\frac1{30n^9}-\frac5{66n^{11}}\tag{7}
$$
Note that the EMSF always has a constant that needs to be determined in some other manner. In $(2)$ it was $\gamma$, the Euler-Mascheroni constant. Here, $C$ is the sum of the series; that is, as $n\to\infty$, the left side of $(7)$ tends to the sum, and everything on the right side of $(7)$, except $C$, tends to $0$.
To compute $C$, we will use $n=100$ and truncate the series at the $n^9$ term. The error we get should be less than $\dfrac5{66n^{11}}$, which would give us almost $23$ decimals places.
For $n=100$, the sum on the left of $(7)$ is
$$
\sum_{k=1}^{100}\frac1{k^2}=1.6349839001848928650771695\tag{8}
$$
For $n=100$, the sum of the terms on the right of $(7)$ other than $C$ is
$$
-\frac1n+\frac1{2n^2}-\frac1{6n^3}+\frac1{30n^5}-\frac1{42n^7}+\frac1{30n^9}
=-0.0099501666633335713952381\tag{9}
$$
Subtracting $(9)$ from $(8)$ gives
$$
C\stackrel{.}{=}1.6449340668482264364724076\tag{10}
$$
whereas
$$
\frac{\pi^2}{6}=1.6449340668482264364724152\tag{11}
$$
The value of $(10)$ is $7.6\times10^{-24}$ short of $(11)$ and $\dfrac5{66n^{11}}=7.6\times10^{-24}$.
Using a larger $n$, and possibly more terms of the series, would give more precision. |
H: Evaluating $\int_{|z|=1} \frac{\sin(z^2)}{ \left( \sin(z) \right)^2} dz.$
Consider $z \in \mathbb{C}$ and
$$\int_{|z|=1} \frac{\sin(z^2)}{ \left( \sin(z) \right)^2} dz.$$
How would we integrate this?
AI: The sine is an entire function, i.e. has no sigularities at finite points. So the only thing that could cause bad behavior is zeros in the denominator.
The only place inside the circle $|z|=1$ where the denominator is $0$ is $z=0$. Now notice that
$$
\lim_{z\to0} \frac{\sin(z^2)}{(\sin z)^2} = 1,
$$
so there's no pole there. It's a removable singularity. So you're integrating around a circle a function that has no singularities inside the circle, so you get $0$. |
H: Inverse image of a union equals the union of the inverse images
I just wonder if my following solution is true.
Let $X,Y$ be sets, let $f:X\to Y$ be a function, let $\{Y_i\}_{i\in I}$ be a family of subsets of $Y$. (Note: I use equalities instead of mutual containment)
$$\begin{align}f^{-1}\left[\bigcup_{i\in I} Y_i\right]
&= \{x \in X: \mbox{there exists an}\quad i \in I\mbox{ such that } y \in Y_i,f(x)=y\}
\\&=\bigcup_{i \in I} f^{-1}\left[Y_i\right]
\end{align}$$
I initially do not know how to get from the left to right, but when I put both sets in set notation, they turn out to be the same, hence the one line proof. Something go ultimately wrong?
AI: The statement is true, and your argument is essentially right, but I would say that you are skipping steps to achieve that identification. (Also, you should not have those $\infty$'s on top of the union symbol). I would instead add:
$$\begin{align*}
f^{-1}\left[\bigcup_{i\in I}Y_i\right] &= \left\{ x\in X\;\left|\; f(x)\in\bigcup_{i\in I}Y_i\right\}\right.\\
&=\Biggl\{x\in X\;\Biggm|\; \exists i\in I\text{ such that }f(x)\in Y_i\Biggr\}\\
&= \bigcup_{i\in I}\{x\in X\mid f(x)\in Y_i\}\\
&= \bigcup_{i\in I}f^{-1}[Y_i].
\end{align*}$$
The first equality is by definition of inverse image; the second by definition of the union; the third is by definition of union; and the fourth by definition of inverse image. |
H: Evaluating $\int_{|z|=1} \sin\left(e^{\frac{1}{z}}\right) \ dz$
Let
$$\int_{|z|=1} \sin\left(e^{\frac{1}{z}}\right) dz.$$
Is there an alternative to the residue theorem if we want to calculate the above integral?
AI: Sasha posted a terse comment suggesting the same substitution I was about to suggest. Here's why that's a good idea: the exponential function behaves in a messy way at $\infty$, and $1/z=\infty$ at a point that's inside the circle $|z|=1$. "Messy" means an essential singularity rather than a pole. After this substitution, the zero in the denominator is no longer in the exponential function. The singularity at $z=0$ is at $w=\infty$, and that point is not inside the curve you'll integrate along.
So $z=1/w$, $dz=\text{what?}$.
Then as $z$ moves along the curve $|z|=1$, what path does $w$ follow? |
H: Relating sum of a sequence to sums in subsequences.
I have been wondering about this question for some time now and would greatly appreciate if anyone had any leads on how to proceed.
Let $\{A_{i}\}_{i \in N}$ such that for, $i \neq j$, $A_{i} \cap A_{j} = \emptyset$ and $\bigcup{A_{i}} = N$. Let $(a_{i})_{i \in N}$ be a sequence in $R$. If, for all $A_{i}$, $\sum_{j \in A_{i}}{a_{j}} \geq 0$ (taking the usual ordering in $A_{i}$), is it true that $\sum_{i \leq n}{a_{i}} \geq 0$ infinitely often?
AI: Let $a_0=-1$, $a_1=-1$, $a_2=1$, $a_3=-1$, $a_4=1$, $a_5=-1$, $a_6=1$ and so on. Then all partial sums are negative, alternating between $-1$ and $-2$.
Let $A_0=\{0,2\}$, $A_1=\{1,4\}$, $A_2=\{3,6\}$, $A_3=\{5,8\}$, and so on. Then for any $j$ the sum of the elements of $A_j$ is $0$.
Thus there are quite simple partitions of the natural numbers for which the desired result fails.
Remark: We can produce a convergent series that converges to $0$ with the same property. Let $a_0=-1$. Let $a_1=1/2$, $a_2=1/3$. Adding $1/4$ would make the partial sum $\ge 0$, so $a_3=-1/4$. Now we can go forward again, twice. Let $a_4=1/5$ and $a_5=1/6$. Adding $1/7$ would make the partial sum positive, so $a_6=-1/7$. Continue. Again, we can partition the natural numbers into (finite) parts so that the sum over any part is $\ge 0$. |
H: Joint continuous random variables pdf
I have the following problem that I think I know how to solve, but I don't see why the given choices are as they are:
Suppose X and Y are jointly continuous random variables with joint
probability density function given by
f(x, y) = 1/c, x > 0, y > 0, x^2 + y^2 ≤ 2; or 0 otherwise
where c is a normalising constant which does not depend on x and y.
What's the value of c?
What's the upper limit of the integrals? We integrate from 0 to what? 2?
Thanks!
AI: Integrate over the part of the circle with radius $\sqrt{2}$ that lies in the first quadrant. If you really want to integrate, $y$ can travel from $0$ to $\sqrt{2-x^2}$, and then $x$ can travel from $0$ to $\sqrt{2}$. Or else we can change to polar coordinates.
There is no need to integrate, since the area is $\frac{\pi}{2}$. |
H: Complex contour integral
Let
$$\int_\gamma z(e^{z^2}+1)dz,$$
where $\displaystyle \gamma(t)=e^{it},t \in \left[0,\frac{\pi}{2}\right]$.
In order to apply Cauchy's integral formula, I'll set $f(z)=z^2\left(e^{z^2}+1\right)$ and rewrite
$$\int_\gamma z(e^{z^2}+1)dz=\int_\gamma \frac{f(z)}{z} dz.$$
$f(0)=0$, so we would have $$\int_\gamma z(e^{z^2}+1)dz=0$$
Unfortunately, this seems to be wrong, because another evaluation from the source, where I've taken this integral yields $\displaystyle \frac{e^{-1}-e}{2}-1$. Where did I make a mistake?
AI: The only thing you need Cauchy for in this problem is to tell you that you're allowed to integrate over a different path with the same endpoints. For instance, consider the change of variable $u=z^2$. Then
$$
\int_{\gamma}z\left(e^{z^2}+1\right)dz=\frac{1}{2}\int_{\gamma'}\left(e^u+1\right)du,
$$
where $\gamma'(t)=e^{it}$, $t\in[0,\pi]$. The integrand has no poles, so you can integrate over any contour with the same endpoints ($+1$ and $-1$) without changing the value; in particular you can just evaluate the real integral
$$
\frac{1}{2}\int_{+1}^{-1}\left(e^u+1\right)du = -\frac{1}{2}\left(e^u + u\right)\Big\vert_{-1}^{+1}=\frac{e^{-1}-e}{2}-1.
$$ |
H: Projective representations of loop groups
If $G$ is a Lie group and we take its loop group $LG$ why do we deal with projective representations of $LG$ and central extensions thereof? Where does the extra complexity come in to require us to consider this extra, very complicated, step?
AI: Projective representations are needed in quantum mechanics, and I think that is the main motivation behind studying projective representations. The point is that in QM, it is rays in a Hilbert space which have physical meaning. So the symmetry group is required to act on and transform these rays rather than vectors. |
H: The number $\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^n\right]$ is always an integer
For each $n$ consider the expression $$\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^n\right]$$
I am trying to prove by induction that this is an integer for all $n$.
In the base case $n=1$, it ends up being $1$.
I am trying to prove the induction step:
if $\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^n\right]$ is an integer, then so is
$\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right]$.
I have tried expanding it, but didn't get anywhere.
AI: Try writing
$$
\left(\frac{1+\sqrt{5}}{2}\right)^n = \frac{a_n+b_n\sqrt{5}}{2}
$$
with $a_1=b_1=1$ and
$$
\frac{a_{n+1}+b_{n+1}\sqrt{5}}{2} = \left(\frac{1+\sqrt{5}}{2}\right)\left(\frac{a_n+b_n\sqrt{5}}{2}\right)
$$
see what you get, then repeat as much as you need to with $\left(\frac{1-\sqrt{5}}{2}\right)^n$. |
H: why this map is injective?
let $U_0=\{[z:w]:z\neq 0\}$ and $U_1=\{[z:w]:w\neq 0\}$, $(z,w)\in \mathbb{C}^2$,and $[z:w]=[\lambda z:\lambda w],\lambda\in\mathbb{C}^{*}$ is a point in $\mathbb{CP}^1$, the map is $\phi:U_0\rightarrow\mathbb{C}$ defined by $$\phi([z:w])=w/z$$
$$\phi([z_1:w_1])=\phi([z_2:w_2])$$ which implies $w_1/z_1=w_2/z_2$, but how can I conclude $w_1=w_2$ and $z_1=z_2$ to show $\phi$ injectuve?
AI: The part in your question
$[z:w]=[\lambda z:\lambda w],\lambda\in\mathbb{C}^{*}$
is what you need to think about. Try multiplying by 1. I'm being vague like Dylan because its hard to give a hint that doesn't totally answer the question.
The one set $U_0$ is pairs modulo some equivalence (with the restriction that $z\neq 0$). So you need to show that if two representatives map to the same point, then they are representatives of the same equivalence class. |
H: Laurent series of $f(z)=\frac{1}{z(z-1)(z-2)}.$
Consider $$f(z)=\frac{1}{z(z-1)(z-2)}.$$
I want to determine the Laurent series in the point $z_0=0$ on $0<|z|<1$.
Partial decomposition yields:
$$f(z)=\frac{1}{z(z-1)(z-2)}=(1/2)\cdot (1/z) - (1/(z-1)) + (1/2)(1/(z-2)).$$
Is the general strategy now, to try to use the geometric series?
$(1/(z-1))=-(1/(1-z))=-\sum_{k=0}^\infty z^k$
$\displaystyle 1/(z-2)=-(1/2)\frac{1}{1-\frac{z}{2}}=-(1/2)\cdot\sum_{k=0}^\infty (z/2)^k$
So $f(z)=(1/2)\cdot (1/z)+(1/2)\cdot\sum_{k=0}^\infty z^k-(1/4)\cdot\sum_{k=0}^\infty (z/2)^k$ (*)
Some questions:
1) What is the difference between a Laurent and a Taylor series? I don't get it. It seems you calculate them the same.
2) Why didn't we write (1/z) as a series expansion too?
3) What makes the end result (*) to be a Laurent series?
AI: Partial fractions and geometric series give
$$
\begin{align}
\frac1{(1-x)(2-x)}
&=\frac1{1-x}-\frac1{2-x}\\
&=\frac1{1-x}-\frac12\frac1{1-x/2}\\
&=(1+x+x^2+x^3+x^4+\dots)\\
&-\left(\frac12+\frac14x+\frac18x^2+\frac1{16}x^3+\frac1{32x^4}+\dots\right)\\
&=\frac12+\frac34x+\frac78x^2+\frac{15}{16}x^3+\frac{31}{32}x^4+\dots
\end{align}
$$
Thus, the Laurent series for $\frac1{x(x-1)(x-2)}$ at $x=0$ is
$$
\frac1{x(x-1)(x-2)}=\frac1{2x}+\frac34+\frac78x+\frac{15}{16}x^2+\frac{31}{32}x^3+\dots
$$
We could also expand the series at $x=1$. Let $y=x-1$ and then
$$
\begin{align}
\frac1{x(x-1)(x-2)}
&=\frac1{(y+1)y(y-1)}\\
&=-\frac1y\frac1{1-y^2}\\
&=-\frac1y-y-y^3-y^5-y^7-\dots\\
&=-\frac1{x-1}-(x-1)-(x-1)^3-(x-1)^5-(x-1)^7-\dots
\end{align}
$$
The Laurent series is much like the Taylor series except terms of negative degree are allowed.
We don't expand $\frac1x$ since there is no power series for $\frac1x$ at $0$ other than $\frac1x$.
Definition. It is a power series at a point, $x_0$ which can have both positive and negative powers of $x-x_0$. |
H: map from $\mathbb{C}$ to $\mathbb{C}/L$ is open map?
Let $w_1,w_2\in\mathbb{C}$ be linearly independent vectors and let$$L=\{m_1w_1+m_2w_2:m_1,m_2\in\mathbb{Z}.\}$$
How does one show that the projection map $\pi:\mathbb{C}\rightarrow\mathbb{C}/L$ is open map? Well, we define $U \subset \mathbb{C}/L$ is open iff $\pi^{-1}(U)$ is open in $\mathbb{C}$ and let hence $\pi$ is continuous, let $V$ be open in $\mathbb{C}$.
Then to show $\pi(V)$ is open, enough to show $\pi^{-1}(\pi(V))$ is open in $\mathbb{C}$, but I am not able to visualize the situation or fact here, shall be happy if some one formally and informally write how to prove this. Thank you.
AI: Here are two "general" explanations for why $\pi: \mathbb{C} \rightarrow \mathbb{C}/\Lambda$ is an open map.
1) $\pi$ is a covering map, hence a local homeomorphism, hence an open map.
I think this is quite intuitive and easy to see: about any point on the torus $\mathbb{C}/\Lambda$, the preimage of a sufficiently small disk-shaped neighborhood will be a "lattice" of small disks in $\mathbb{C}$. In fact $\pi$ is a regular covering map, hence the quotient by an action of the group $\Lambda$.
This leads to a more general answer.
2) For a group $G$ acting on a topological space $X$, the quotient map $\pi: X \rightarrow X/G$ is open.
Proof: By definition of the quotient topology on $X/G$, we must show that if $U \subset X$ is open, then $\pi^{-1} \pi U$ is open. But $\pi^{-1} \pi U = \bigcup_{g \in G} g U$. Since each $g \bullet$ is a homeomorphism (that is part of the definition of a group action on a topological space) and $U$ is open, $g U$ is open, so $\pi^{-1} \pi U$ is a union of open sets, hence open.
Note that Paul Garrett remarks that most reasonable quotient maps are open. While I certainly agree with this as a principle, in any particular reasonable situation one still needs to summon a proof. Studying in general the problem of openness of quotient maps seems (to me, of course) to be unrewarding and technical -- c.f. Bourbaki's General Topology, which does entirely too much of this for my taste -- so it is worthwhile to collect "easy" explanations like those above. In fact 2) above was taken directly from lecture notes for a course on modular curves I taught recently: and it is from page 1 of those notes! |
H: Conditional probablity with tree diagram
Could someone tell me if I got this right?
So I drew a tree diagram
Sorry if it is too messy, but i had to do this on PaintBrush.
(a) I just added all the branches $\mathbb{P}(S_4 = 17) = 0.6 \times 0.4+0.4\times0.4=0.4$
(b)This one is a bit tricky, so basically I have $\mathbb{P}(S_4=17 | S_5=18)$ and forget about the equal sign and the numbers for now. So $\mathbb{P}(S_4 | S_5)= \dfrac{\mathbb{P}(S_4 \cap S_5)}{\mathbb{P}(S_5)} = \dfrac{\mathbb{P}(S_5|S_4)\mathbb{P}(S_4)}{\mathbb{P}(S_5)} = \dfrac{(0.6 + 0.4)\times 0.4}{0.336} = 1.19$
Which makes no sense for (b) because it is over 1...
Thank you
AI: You need to compute $\Pr((S_5=18)\cap (S_4=17))$ correctly, compute $\Pr(S_5=18)$, and divide.
For the numerator, multiply probabilities along all paths to $18$ that have $17$ the previous day, and add.
For the denominator, multiply probabilities along all paths that end up ay $18$, and add. After all, your conditional probability tells you that your sample space is being restricted to paths that end at $18$.
Renmark: It is not clear from the diagram what transition probabilities you used. But at least one entry is wrong. Look at the path that goes $17$, $16$, $17$, $18$. For the $17$ to $18$ part, you put what looks like $0.4$ for that transition. But the day before we went from $16$ to $17$, so the probability of $17$ to $18$ the next day is $0.6$. |
H: Laurent series - $f(z)=\frac{1}{z^2-4}+\frac{1}{6-z}$
We have $f(z)=\frac{1}{z^2-4}+\frac{1}{6-z}$. I want to expand this as a Laurent series in $z_0=2$ on $\{4<|z-2|<\infty\}$.
The partial decomposition is:
$$f(z)=\frac{1}{4}\frac{1}{z-2}-\frac{1}{4}\frac{1}{z+2}+\frac{1}{6-z}$$
In my reference, they expand $\frac{1}{z+2}$ and $\frac{1}{6-z}$ with the help of the geometric series, but they leave $\frac{1}{z-2}$ as it is. Why?
AI: $$\frac{1}{z+2}=\frac{1}{(z-2)+4}=\frac{1}{4}\frac{1}{1+\left(\frac{z-2}{4}\right)}=\frac{1}{4}\left(1-\frac{z-2}{4}+\frac{(z-2)^2}{16}-\frac{(z-2)^3}{64}+\cdots\right)$$
$$\frac{1}{6-z}=-\frac{1}{(z-2)-4}=\cdots$$
and now just mimic the first development above of the given fraction in powers of $\,z-2\,$
Pay attention to the fact that the first (and also the second one, btw) development is valid for
$$\left|\frac{z-2}{4}\right|<1\Longleftrightarrow|z-2|<4$$
Once you've done all the above just add the different fractions which form part of $\,f(z)\,$ |
H: Find the particular solution of the given differential equation that satisfies that indicated side condition
$$\frac{dy}{dx} = 3\sin(x/2)$$
$$y = 1 ,x= \frac{\pi}{3}$$
I just get stuck on the integration of the $3\sin(x/2)$.
AI: $$\int\sin kx\,dx=-\frac{1}{k}\cos kx+C\,\,,\,k\neq 0\,$$ |
H: map from $D$ to $\pi(D)$ is injective?
please follow this map from $\mathbb{C}$ to $\mathbb{C}/L$ is open map? ,let w be a non zero element of the lattice L so that |w|>2ϵ, fix such ϵ>0 and any $z_0$∈C and take an open disk of radius ϵ with centre at $z_0$, could you please tell me why $π:D\rightarrow π(D)$ is injective? what does it mean by a "lattice" of small disk in $\mathbb{C}$? what is π(D) pictorically? let my $L=\{m_1(1,0)+m_2(0,1):m_1,m_2\in\mathbb{Z}$}
AI: The statement you gave cannot be correct because there are points with arbitrarily high magnitude in $L$, so that would imply that $\pi$ is injective over $\mathbb C$.
You need $|w|>2\varepsilon$ for every non-zero $w\in L$, not just a single one.
Once you make that correction, $\pi(x)=\pi(y) \iff y-x\in L$, but since $|y-x|\le 2\varepsilon$ this can only be true if $x=y$. |
H: Comparing money earned by two social games
Which game makes more money on a daily basis?
game A, which has 175,000 DAU, a 32% second-day retention rate, a $0.05 ARPDAU, and a 30-day lifetime
or
game B, which has 150,000 DAU, a 22% second-day retention rate, a $0.08 ARPDAU, and a 15-day lifetime
AI: Lets label Dau $= D$, retention rate$=R$ and arpau $=P$ and Lifetime$=L$. Im guessing your DAU is for day 1. for any day n, The amount of money you make each day is$ D*R^{(n-1)}*P$. so the sum of the cash you make with a game is $$ \sum\limits_{n=1}^L D*R^{(n-1})*P$$ using wolfram alpha.
Using this formula for game A you get: $$ \sum\limits_{n=1}^{30} 175000*.32^{(n-1)}*.05\approx 12867.6$$
Using it for game B you get $$\sum\limits_{n=1}^{15} 150000*.22^{(n-1)}*.08\approx 15384$$
(I did these computations using wolfram alpha. I would appreciate it if someone double checkes using mathematica.) Also, this is based on what I understood from the question. |
H: Diagonalizing Matrices over UFDs
Suppose $R$ is a UFD, $K$ its field of fractions. If a matrix $M$ with entries in $R$ has distinct eigenvalues, then it is certainly diagonalizable over $K$. If those eigenvalues are actually in $R$, must it be diagonalizable over $R$?
If it helps: the specific case I'm interested in is when $K$ is a local field of characteristic $0$ and $R$ its integers.
AI: No. Consider the matrix
$$A=\left(\begin{array}{rr}
3&-1\\-1 & 3\end{array}\right)$$
over $\mathbb{Z}$. It is certainly diagonalizable over $\mathbb{Q}$, and the eigenvalues are $2$ and $4$, in $\mathbb{Z}$; the eigenspaces are $(a,-a)$ and $(b,b)$. Any diagonalizing matrix over $\mathbb{Q}$ would then be of the form
$$P = \left(\begin{array}{rr}a&b\\-a&b\end{array}\right)\qquad\text{or}\qquad \left(\begin{array}{rr}a & b\\a&-b
\end{array}\right),$$
so that $P^{-1}AP$ is diagonal. But the determinant of $P$ is either $2ab$ or $-2ab$. If $a$ and $b$ are nonzero integers, then the inverse of $P$ cannot be a matrix with integer coefficients, since the determinant is not $\pm 1$.
Added. By the way: the assumption that the eigenvalues lie in $R$ is unnecessary: if $M$ has coefficients in $R$, then the characteristic polynomial is monic and has coefficients in $R$. In particular, if it splits over $K$, then it splits over $R$ (by the Rational Root Theorem, or by Gauss's Lemma). So all you need to assume is that the minimal polynomial splits over $K$ and is squarefree; the polynomial will then necessarily split over $R$ as well. |
H: Integration by substitution in $n$ dimension
I just read the proof of Hardy-Littlewood-Sobolev inequality abaout fractional integral operator. Then I found the following identity (but don't understand it)
\begin{equation}
\int_{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{n}\backslash B\left( x,R\right) }\frac{1}{\left\vert x-y\right\vert
^{c(n-1)}}dy=\int_{S^{n-1}}\int_{R}^{\infty }\frac{1}{r^{^{c(n-1)}}}%
r^{n-1}drd\sigma
\end{equation}
where $B(x,R)$ denote $n$ dimensional ball, centered on $x$ and radius $R$. I guess that identity is generalization of substitution formula. Could you help me to understand that identity?
AI: The left integral is over all of space except the ball. $y$ is a point outside the ball and ranges over the volume of integration. $|x-y|$ is the distance between the points. The right integral is changing to spherical coordinates. $r$ is the distance between $x$ and $y$, so $r=|x-y|$. The inner integral is over the radius, and ranging from $R$ to $\infty$ avoids the ball nicely. $d\sigma$ represents all the angular variables ($n-$1 of them). The $r^{n-1}$ factor represents the radial variation of the surface area of an $n-1$ sphere. Presumably the next thing to happen is to do the $d\sigma$ integral getting the surface of the $n-1$ sphere ($4\pi$, for example, if $n=3$). |
H: Specifying plane waves
I'm having trouble understanding how to specify a transverse wave (including it's longitudinal axis and transverse direction) in 3d space.
I know this is called a "plane wave", and I know the formula for a plane wave along a unit vector $\hat{k}$ is:
$$
A( \vec{r}, t ) = A_0 cos( \vec{k} \cdot \vec{r} - \omega t + \phi )
$$
Which is from here
Where $\vec{k}$ is the longitudinal axis and $\vec{r}$ is the space position. The output of this function should be the magnitude of the wave at that point in space.
But I don't understand where you specify the transverse axis. From this picture it appears that the wave described must be a longitudinal wave with only one axis, modelling something like compressions and rarefactions in air.
What is the transverse axis here? How do you specify a transverse axis?
I want a single wave that looks like just either the red one or the blue one from here:
I can't see where the transverse axis comes in or how you specify it's direction.
AI: Longitudinal waves are along the direction of propagation, e.g., along the $y$-direction in your second figure.
Transverse waves are not longitudinal---the displacement is perpendicular to the direction of propagation, e.g., in the $x$- or $z$-direction in your figure.
These two types of transverse waves are in independent states of polarization.
They can be combined to form linearly, circularly, or elliptically polarized waves.
A transverse wave in the $x$-$y$ plane (blue in the figure) is described by
$A_x \cos(k y - \omega t) \hat x$.
A general transverse wave propagating in the $y$-direction would look like
$$A_x \cos(k y - \omega t)\hat x
+ A_z \cos(k y - \omega t + \phi)\hat z,$$
where $A_x,A_z \ge0$ and $-\pi < \phi \le \pi$.
If $\phi = 0$ or $\pi$ the wave is said to be linearly polarized.
If $A_x = A_z$ and $\phi = \pm \pi/2$ the wave is circularly polarized.
(The circular polarization is left- or right-handed depending on the sign of $\phi$ and your convention for handedness.)
Otherwise the wave is elliptically polarized.
Addendum:
To describe a transverse wave propagating with wave vector ${\bf k}$ we must first find a unit vector orthogonal to ${\bf k}$, call it $\hat n$.
This is one of the transverse directions.
The other can be chosen to be $\hat k\times \hat n$.
(It's a good exercise to show this is a unit vector orthogonal to $\hat k$ and $\hat n$.)
Then a general transverse wave will be represented by
$$A \cos({\bf k}\cdot {\bf r} - \omega t)\hat n
+ B \cos({\bf k}\cdot {\bf r} - \omega t + \phi)\hat k\times \hat n,$$
where $A,B \ge0$ and $-\pi < \phi \le \pi$.
Replace $A_x,A_y$ with $A,B$ above to determine the type of polarization.
For example, if we want a linearly polarized wave propagating along the $\frac{1}{\sqrt{2}}(1,0,1)$ direction with polarization vector $\hat y = (0,1,0)$ we will find the wave described by
$$A \cos\left(\frac{1}{\sqrt{2}}k(x+z) - \omega t\right) \hat y.$$
(Note that
${\bf k}\cdot {\bf r} = \frac{1}{\sqrt{2}}k(x+z)$.) |
H: Integrating $3\sin(x/2)$.
$$\int 3\sin\left(\frac{x}{2}\right)dx$$
can't figure this one out! I'm not sure if I'm supposed to substitute or not?
Here's where I'm at...
$$3\int\sin\left(\frac{x}{2}\right)dx$$
$$u = \frac{x}{2}$$
$$du = \frac{1}{2}dx$$
$$dx = 2du$$
$$3\int\sin\left(u\right)2du$$
and then...
$$-6cos\left(\frac{u^2}{2}\right)$$
thats not right is it..?
AI: Let $x=2t\implies dx=2dt$. Hence your problem becomes $2\int 3\sin(t)dt=-6\cos(t)+c=-6\cos(x/2)+c$ where $c$ is a constant. |
H: A power-exponential congruence equation
Let $n \in \mathbb{N}$ with $(n,\varphi(n))=1$ , where $\varphi$ is the Euler-totient function. Prove the equation $x^x \equiv c \pmod{n}$ has integer solution for all $c \in \mathbb{N}$
My thought: By Euler's theorem one has $\varphi(n)^{\varphi(n)} \equiv 1 \pmod{n}$, so I need to find an $x$ such that $x^x \equiv c \varphi(n)^{\varphi(n)} \equiv c \pmod{n}$. It seems one needs to consider the primitive roots $\bmod(n)$ somehow, but $c$ may not be coprime to $n$.
AI: First, note that the hypothesis $\gcd(n,\phi(n))=1$ implies that $n$ is a product of distinct primes (since $p^2\mid n$ implies $p\mid\phi(n)$).
Now we deduce that $${\rm if\ }\gcd(n,\phi(n))=1{\rm\ then\ }a^{1+k\phi(n)}\equiv a\pmod n{\rm\ for\ all\ }k.$$ Let $p$ be a prime dividing $n$. If $a\equiv0\pmod p$, then surely $a^{1+k\phi(n)}\equiv a\pmod p$, since both sides of the congruence are zero. If $\gcd(a,p)=1$, then $$a^{1+k\phi(n)}=a^{1+k'(p-1)}=a\bigl(a^{p-1}\bigr)^{k'}\equiv a\pmod p$$ for some integer $k'$. Thus we have $a^{1+k\phi(n)}\equiv a\pmod p$ for all primes $p$ dividing $n$. Since these primes are pairwise coprime, and since $n$ is their product, $a^{1+k\phi(n)}\equiv a\pmod n$ follows by the Chinese Remainder Theorem.
Now, given $c$, choose $k$ such that $1+k\phi(n)\equiv c\pmod n$. Again, the hypothesis $\gcd(n,\phi(n))=1$ guarantees $k$ exists. Now $$(1+k\phi(n))^{1+k\phi(n)}\equiv c^{1+k\phi(n)}\equiv c\pmod n$$ and we are done; we have shown we can take $x=1+k\phi(n)$. |
H: Area of Triangle inside another Triangle
I am stumped on the following question:
In the figure below $AD=4$ , $AB=3$ , and $CD=9$. What is the area of Triangle AEC?
I need to solve this using trigonometric ratios however if trig. ratios makes this problem a lot easier to solve I would be interested in looking how that would work.
Currently I am attempting to solve this in the following way
$\bigtriangleup ACD_{Area}=\bigtriangleup ECD_{Area} + \bigtriangleup ACE_{Area}$
The only problem with this approach is finding the length of AE or ED. How would I do that ? Any suggestion or is there a better method
AI: $AE+ED=AD=4$, and $ECD$ and $EAB$ are similar (right?), so $AE$ is to $AB$ as $ED$ is to $CD$. Can you take it from there? |
H: Subvariety of Product of Elliptic Curves
This is almost certainly known (and maybe written down somewhere?). Is there an example of two elliptic curves $C, E/k$ that are not isomorphic, yet there is an embedding $C\hookrightarrow E\times E$ as an abelian subvariety?
If this is known, is there a reference that talks about such things.
AI: Giving a morphism $C \to E\times E$ is the same as giving a pair of morphisms $p,q: C \to E$, by the definition of the product. Translating $p$ and $q$ appropriately by some points of $E$, it is no loss of generality to assume that $p$ and $q$ both take the origin of $C$ to the origin of $E$, and hence are not only morphisms of curves, but homomorphisms of elliptic curves.
The map $C \to E \times E$ will be injective if and only if $ker(p) \cap ker(q) = 0$.
Now if $q^{\vee}$ denotes the dual of $q$, then $q^{\vee}\circ p$ is an endomorphism of $C$. If $C$ (equivalently, $E$) is not CM, then this will be multiplication by some integer $n$, in which case we find (composing with $q$)
that $\deg(q) p = n q$, and a short argument shows that in fact there is an morphism $r: C \to E$ such that $p = m r$ and $q = n r$ for some integers $m$ and $n$. If $ker(p)\, $ and $ker(q)\, $ have trivial intersection, it follows that
$r$ must be an isomorphism, so that $C \cong E$.
On the other hand, if $E$ has CM, then the situation you ask about can occur.
The easiest case is to assume that $E$ and $C$ both have CM by the full ring of integers $\mathcal O$ in some imag. quad. field $K$, but are not isomorphic (which is possible iff $K$ has class number bigger than one).
Then $Hom(C,E)$ is an $\mathcal O$-module, invertible (i.e. locally free of rank one) but not free. Such a module can always be generated by two elements. If we take $p$ and $q$ to be a pair of generators, then they will embed $C$ into $E\times E$.
If $E$ has CM by a field of class number one, then the same kind of construction should be possible, I think, by choosing $C$ to have CM by some proper order of $K$ with non-trivial class number. |
H: Average of numbers in a specific range
The question is :
What is the arithmetic mean of all multiples of 10 from 10 to 190 inclusive.
Now I know how many nos there are by using $\frac{190-10}{10}+1 = 19$ but how do I get their sum ?
AI: 180+10 = 190;
170+20 = 190;
160+30 = 190;
...
how many numbers are there? 18.. we always add two numbers, so the sum is 190 * 9. now we add the last odd number - 190. so it's 190*10 -> 1900
gauss came up with that. pretty clever. thanks to him, you can do this in your head.
now the mean is just the sum divided by the number of parts, no?
we have 19 parts, and a sum of 1900.
1900/19..
holdon, lemme get muh calculatorz. |
H: Is $\mathbb{Z}[\sqrt{2},\sqrt{3}]$ flat over $\mathbb{Z}[\sqrt{2}]$?
Is $\mathbb{Z}[\sqrt{2},\sqrt{3}]$ flat over $\mathbb{Z}[\sqrt{2}]$? The definitions doesn't seem to help. An idea of how to look at such problems would be helpful.
AI: $\mathbb Z[\sqrt{2},\sqrt{3}]$ is freely generated as a $\mathbb Z[\sqrt{2}]$-module (exercise). Free modules are flat. QED |
H: Application of Cauchy's integral theorem
On page 97 of John B. Conway's Functions of one complex variable, the author states that:
"Suppose $G$ is a region (open connected subset) and let $f$ be analytic in $G$ with zeros at $a_1,a_2,...,a_m$. So we can write $f(z)=z(z-a_1)(z-a_2)...(z-a_m)g(z)$ where $g$ is analytic on $G$ and $g(z)\neq 0$ for any $z$ in $G$.."
The author use this result to show that:
$\frac{1}{2\pi i} \int_{\gamma} \frac{f'(z)}{f(z)} dz=\sum_{i=1}^{m} n(\gamma;a_k)$.
I just don't understand how to show that the number of roots of $f$ in $G$ is finite. Can someone explain this to me? Thank you.
AI: Conway assumes that the number of zeroes is finite. |
H: Continuous mappings pull back closed sets to closed sets
George F Simmons, Topology and Modern Analysis pg.79 Problem 4
Let $X$ and $Y$ be metric spaces. Show that an into mapping $f:X \rightarrow Y$ is continuous $\iff$ $f^{-1}\left(G\right)$ is closed in $X$ whenever $G$ is closed in $Y$.
I can prove the problem for open sets, and I have been trying hard for closed. However, seems like I am stuck somewhere missing something obvious. Please don't answer directly, just give a small hint if possible.
EDIT: I am using the definition that $f^{-1}\left(G\right)$ exists only when $f$ is onto and if it is not then $f^{−1}\left(G\right)$ is a loose term for $f^{-1}\left(H\right)$ where $H$ is the range of $f$ in $G$.
AI: So you can show that $f$ is continuous iff preimages of open sets are open.
Now you go from there. Closed sets are complements of open sets.
What is $f^{-1}[Y\setminus G]$? |
H: Given that $5n$ is a square and $75np$ is a cube, why is the smallest possible value of $n+p$ equal to $14$?
I can't solve this problem:
Suppose $n$ and $p$ are integers greater than $1$, $5n$ is the square of a number, and $75np$ is the cube of a number. What is the smallest value for $n+p$?
(Answer given is $14$)
I don't even understand if $5n$ is the square of the same number which has a cube of $75np$. Any suggestions? How would I solve this problem?
AI: $5n$ is a square implies that $5n = 5^2 \times a^2$. This gives us that
$n = 5a^2$, where $a \in \mathbb{Z}$.
Similarly, $75np$ is a cube implies that $75np = 3^3 5^3 b^3 \implies np = 3^2 5 b^3$, where $b \in \mathbb{Z}$.
Can you now conclude what $a$ and $b$ should be for $n+p$ to be a minimum? |
H: Question about the $\mathrm{Tor}$ functor
Assume we want to define $\mathrm{Tor}_n (M,N)$ where $M,N$ are $R$-modules and $R$ is a commutative unital ring.
We take a projective resolution of $M$:
$$ \dots \to P_1 \to P_0 \to M \to 0$$
Now does it matter whether we apply $-\otimes N$ or $N \otimes -$ to this? It shouldn't because we have $N \otimes P \cong P \otimes N$. Right? Thanks.
AI: I guess the point here is that $N \otimes_R -$ and $- \otimes_R N$ are naturally isomorphic functors. Therefore, you get an isomorphism of chain complexes $N \otimes_R P^{\bullet} \cong P^{\bullet} \otimes_R N$, which implies that the two complexes have isomorphic homology. So, it doesn't matter if you apply $N \otimes_R -$ or $- \otimes_R N$. |
H: Definition of 'closed relative to'
I know what's the definition of 'open relative'.
I googled 'close relative', but i couldn't find a definition of it.
How come every metric space $X$ is close relative to $X$?
If $p$ is a limit point of $X$, there exists a neighborhood $N_r(p)$ and $q\in X$ such that $q\in N_r(p)$ and $q≠p$.
What kind of definition makes $p$, a limit point of $X$ is a point of some set, thus $X$ is closed relative to $X$?
AI: Given a topological space $A$, we can endow a subset $B\subseteq A$ with the subspace topology.
We say that a subset $C\subseteq B$ is "open relative to $B\,$" when $C$ is open in the subspace topology on $B$; that is, by definition, when there is some open subset $U$ of $A$ such that $C=U\cap B$.
The definition is identical for "closed", i.e. a subset $C\subseteq B$ is "closed relative to $B\,$" when it is closed in the subspace topology on $B$, which (by definition) is when there is a closed subset $D$ of $A$ such that $C=D\cap B$.
Thus, it is a trivial result that every topological space $A$ is closed relative to $A$, because $A$ must be closed in its topology (that is part of the definition of a topology) and therefore $A=A\cap A$ is the intersection of a closed subset of $A$ with $A$.
Any metric space $X$, with distance function $d$, can be given a topology where the open subsets of $X$ are precisely those subsets $Y\subseteq X$ with the property that, for any $p\in Y$, there is an $\epsilon>0$ such that $B_\epsilon(p)\subseteq Y$, where
$$B_\epsilon(p)=\{q\in X\mid d(p,q)<\epsilon\}$$
is the open ball of radius $\epsilon$ centered at $p$. Thus, any metric space can be given the structure of a topological space.
Thus, we can simply apply the above result (which holds for all topological spaces) to a metric space, and obtain the statement that any metric space $X$ is closed relative to $X$. |
H: Going from binomial distribution to Poisson distribution
Why does the Poisson distribution
$$\!f(k; \lambda)= \Pr(X=k)= \frac{\lambda^k \exp{(-\lambda})}{k!}$$
contain the exponential function $\exp$, while its relation to the binomial distribution would suggest it's all about powers of $2$?
AI: The Binomial $(n,p)$ distribution puts weight $p_n(k)={n\choose k}p^k(1-p)^{n-k}$ on each integer $k$ such that $0\leqslant k\leqslant n$. Fix $k\geqslant0$ and let $n\to\infty$ and $p\to0$ in such a way that $pn\to a$.
Then, ${n\choose k}\sim\frac{n^k}{k!}$, $p^k\sim \left(\frac{a}n\right)^k$, $(1-p)^n\to\mathrm e^{-a}$ and $(1-p)^{-k}\to1$. Hence $p_n(k)\to\mathrm e^{-a}\frac{a^k}{k!}$ the weight at $k$ of the Poisson distribution of parameter $a$.
Finally, the appearance of $\mathrm e^{-a}$ is due to the fact that $(1-p)^n\to\mathrm e^{-a}$ when $n\to\infty$ and $p\to0$ in the regime $pn\to a$, that is, to the fact that $\left(1-\frac{a}n\right)^n\to\mathrm e^{-a}$ when $n\to\infty$. |
H: Aftermath of the incompletness theorem proof
This is somewhat of a minor point about the incompletness theorem, but I'm always a little unsure:
So one proves that there is a formula which is unprovable in the theory of consideration. Okay, at this point one is done.
Then, as that unproven sentence contains the claim that this (the unprovable-ness) would happen, one is in some sense justified to say "So the statement really is true, but still, it's unprovable within the theory". Now here is my problem: I'm very unsure in which sense this notion of truth which somewhat comes from outside the theory is sensible. There is a note on that point on the wikipedia page but I don't really comprehend it.
The word "true" is used disquotationally here: the Gödel sentence is true in this sense because it "asserts its own unprovability and it is indeed unprovable" (Smoryński 1977 p. 825; also see Franzén 2005 pp. 28–33). It is also possible to read "GT is true" in the formal sense that primitive recursive arithmetic proves the implication Con(T)→GT, where Con(T) is a canonical sentence asserting the consistency of T (Smoryński 1977 p. 840, Kikuchi and Tanaka 1994 p. 403)
So I can see that if one is technically aware that one is now talking in a meta language, that one has introduced a new "true". But then again (a) if one reflects on the fact that one draws such technicals conclusions outside of the initial freamework, then it seems to me one should really introduce another meta-meta-language. And (b) isn't it really just a little ambiguous to say "Gödels incompleteness theorems says that there are true statements, which can't be proven within a certain strong theory"?
I'd be thankful if someone could elaborate on that.
AI: There might be a bit of a chicken-and-egg happening here. For the common proofs of Gödel's Incompleteness Theorem, we put special priority on the structure $( \mathbb{N} , 0, S, + , \cdot , \mathrm{exp} )$ of elementary arithmetic. If something is true in this structure, then we often refer to that statement as true.
Peano Arithmetic is one attempt to axiomatize all truths of this structure. In more modern terms, it was perhaps hoped that the statement $$\mathbb{N} \models \phi \;\Leftrightarrow\; \text{PA} \vdash \phi$$ would have been true (or true). Gödel's Theorem tells us that this is impossible, and, more, that any attempt to "nicely" axiomatize truth in $\mathbb{N}$ is doomed to fail, by being either inconsistent, or incomplete.
As the structure $\mathbb{N}$ "exists" outside of the formal system PA (and "existed" well before this system was devised), we can argue about truth within the structure $\mathbb{N}$ using methods outside the formal system PA. In fact, there are several truths about $\mathbb{N}$ that we know cannot be proved within PA (e.g., Goodstein's Theorem and a certain strengthening of the finite Ramsey's Theorem). As long as there is no real controversy about these extra-PA methods used, the truth of these results is similarly uncontroversial. It is by using such methods outside of PA that we argue that if PA is consistent, then $G_{\text{PA}}$ is true (in $\mathbb{N}$).
While perhaps a bit ambiguous, this consequence of Gödel's Theorem could be restated (somewhat awkwardly) as "Given any "nice" consistent axiom system capable of expressing elementary arithmetic, there are number-theoretic statements which are true in $\mathbb{N}$ that cannot be proved within that system." |
H: Norm of compactly supported functions with disjoint support .
If $u_n \in C_c^\infty (\mathbb R)$ with $u_n=u(x+n)$ , $n\in \mathbb N$ , $u$ is not identically zero.
How do i prove that $||u_{n+k}-u_n||_{L^q}^q =2||u||_{L^q}^q$.
What my doubt is that even if we take $u_{n+k}$ and $u_n$ to have disjoint support, it doesn't seem that the equality holds .
$||u_{n+k}-u_n||_{L^q}^q=\int |u(x+n+k)-u(x+n)|^q$ , i don't see how i can relate the value of $u(x+n+k) $ and $u(x+n)$
looking forward for some hints and help.
Thanks
AI: If the supports of $v$ and $w$ are disjoint it holds that
$$
\int |v + w|^q = \int_{\mathrm{supp}(v)}|v+w|^q + \int_{\mathrm{supp}(w)}|v+w|^q.
$$
Since $w$ is zero on $\mathrm{supp}(v)$ it holds that
$$
\int_{\mathrm{supp}(v)}|v+w|^q = \int_{\mathrm{supp}(v)}|v|^q
$$
and vice versa. |
H: Applying the symmetric difference to multiple sets
I was working on set theory and I came across a rule:
$$(A\cup B) - (A\cap B) = (A\cap B)'$$
I have 4 sets $A$, $B$, $C$, $D$. How can I apply the rule above to all set at same time? is the below mentioned result mathematically correct?
$$(A\cup B\cup C\cup D) - (A\cap B\cap C\cap D) = (A\cap B\cap C\cap D)'$$
Regards,
AI: If by $(\cdots)'$ you denote the complement of $(\cdots)$,
$$
(A ∪ B) - (A ∩ B) = (A\cap B)'
$$
which is corresponds to the symmetric difference $A\,\triangle\,B\, $. As pointed by Martin your second equation is also just a way of writing the complement.
And, as already pointed out by Tapu, for multiple sets the $(\cap_k A_k )' \neq \triangle_k A_k$.
$\hskip2.7in$
Venn diagram of $~A \triangle B \triangle C$ |
H: Prove that $\mathop {\lim }\limits_{(x,y,z) \to (0,0,0)} \left( {\frac{{{x^2}y - x{z^2}}}{{yz - {z^2}}}} \right)=0$
Prove that:
For all $\epsilon>0$ exist $\delta>0$ which depends on $\epsilon$, such that:
$$\left| {\frac{{2{x^2}y - x{z^2}}}{{yz - {z^2}}}}-0 \right|<\epsilon$$ ever that
$$0 < \sqrt {{x^2} + {y^2} + {z^2}} < \delta $$
I find it very difficult to find $\delta$ in terms of $\epsilon$.
Any suggestions to prove this?
$$\mathop {\lim }\limits_{(x,y,z) \to (0,0,0)} \left( {\frac{{2{x^2}y - x{z^2}}}{{yz - {z^2}}}} \right)=0$$
thanks.
AI: Hint: Consider sequences
$$
\begin{align}
(x_n,y_n,z_n)&=(n^{-1/2},2n^{-1},n^{-1})\\
(x_n,y_n,z_n)&=(0,2n^{-1},n^{-1})
\end{align}
$$
then you get
$$
\begin{align}
\lim\limits_{n\to\infty}\frac{2x_n^2 y_n-x_nz_n^2}{y_n z_n-z_n^2}&=\lim\limits_{n\to\infty}(4-n^{-1/2})=4\\
\lim\limits_{n\to\infty}\frac{2x_n^2 y_n-x_nz_n^2}{y_n z_n-z_n^2}&=\lim\limits_{n\to\infty}0=0
\end{align}
$$
Thus we conclude that the limit
$$
\lim\limits_{(x,y,z)\to (0,0,0)}\frac{2x^2 y-x z^2}{y z-z^2}
$$
doesn't exist.
P.S.
I used approach from this answer. |
H: Counterexample of Compactness
Let $X$ be a metric space and $E\subset X$.
Let {$G_i$} be an open cover of $E$
For every open cover {$G_i$}, there exists a finite subcover {$G_{i_n}$} of $E$ such that $G_{i_n} \in${$G_i$}.
For every open cover {$G_i$}, there exists {$M_n$}, a finite family of open sets, such that $E\subset$$\bigcup M_n \subset \bigcup G_i$.
As you know, if 1 is true, then $E$ is compact.
I think 1 and 2 both have the same meaning, but can't prove the equivalence. (1→2 is trivial, but 2→1 is not)
If 1&2 are not equivalent, please give me some counterexamples..
AI: $2$ does not imply $1$. Take for example $E=M=\mathbb{R}$, with the open cover $\{ G_n=(-\infty,n)|n \in \mathbb{N}\}$. This cover has no finite subcover, but we can just take $M=\mathbb{R}$ and $\{ M \}$ is a set satisfying the conclusion of $2$. |
H: Good problem book on Abstract Algebra
I am currently self-studying abstract algebra from Artin. In that background, I am looking for a problem book in a spirit somewhat similar to Problems in Mathematical Analysis by AMS so that I have a lot of problems to solve.
AI: I don't know if I would call it a "problem book" but Herstein's fantastic "Topics in Algebra" has some amazing basic problems in abstract algebra. If you want more commutative algebra, Atiyah-Madonald have a lot too, but you should probably wait with that one.
For a problem book, I would recommend:
Exercises in Algebra: A Collection of Exercises, in Algebra, Linear Algebra and Geometry (Algebra, Logic and Applications , Vol 6) .
Look for old quals in Algebra too and old exams. I can not, if you're learning abstract algebra, recommend Aluffi's book "Algebra: Chapter 0" enough. I used it as a first introduction and I fell in love with the subject.
Good luck! |
H: How do I work out a new constrained scale height based on a new width?
This is super simple I guess but my mind has gone blank.
I have these dimensions below, and I am trying to scale this rectangle to 610 wide
Width 560
Height 315
From these figures, how do I work out the percentage to scale the height so the rectangle is still at the same aspect ratio?
Thanks
Think I got it...
560 / 100 = 5.6
610 / 5.6 = 108.92
Answer 108.92%
AI: Since $\frac{610}{560} \doteq 1.089$, you have to scale by approximately $8.9$% (from 100% to 108.9%). |
H: Existence statement and Axiom of Choice
Let $I$ be an infinite set.
Suppose that for every $i\in I$, there exists a set $S_i$ satisfies a statement $\psi(S_i)$.
Here, is constructing a family of such $S_i$ (i.e. $\{S_i\}$ for $i\in I$) using Axiom of Choice??
For example, $\{V_i\}$ is given, suppose for every $i\in I$, there exists $G_i$ such that $V_i$ = $Y\cup G_i$. Then constructing $\{G_i\}$ is a result of AC?
AI: It varies.
If you can prove that there exists a unique $S_i$ then there is no need to use the axiom of choice.
On the other hand, think of $\psi(x) = \exists y(y\in x)$, then every non-empty set has this property, so one can easily construct an example where the axiom of choice is essential.
In the example, since $Y$ is a fixed parameter it is simple to define $G_i=V_i\setminus Y$. You can also take $G_i=V_i$ for all $i$.
For example, if $\cal A=\{A_i\mid i\in I\}$ is a family of non-empty and pairwise disjoint sets then $\varnothing$ can be in at most one element of the family. Without loss of generality, it is in none of the members. If $\cal A$ does not have a choice function then the sentence $\psi(x)=\exists y(y\in x)$ has at least one element in every $A_i$ which satisfy it, in fact $\{a\in A_i\mid \psi(a)\}=A_i$, so we cannot use this to choose from this family. |
H: Quotient of a free $\mathbb{Z}$-module
I'm trying to find the quotient of a free $\mathbb{Z}$-module, but somehow I don't really find the right procedure on how to get the right quotient (nor have I found any sources). I've read What does it mean here to describe the structure of this quotient module? already, but I didn't manage to reproduce the second matrix from that specific post.
In my specific example I have $\mathbb{Z}^3$ as a left-$\mathbb{Z}$ module. Lets call the generators $e_1, e_2, e_3$. My submodule is given by
$$K:=\operatorname{span}_{\mathbb{Z}}\left<i_1:=\left(\matrix{0\\-2\\1}\right), i_2:=\left(\matrix{-2\\0\\1}\right)\right>$$
My idea to get the quotient module $\mathbb{Z}^3/K$ thus far was to use the equalities
$$i_1 = -2e_2+e_3\\
i_2 = -2e_1+e_3
$$
Which yields (for $z_1,z_2,z_3,l_1,l_2\in\mathbb{Z}$):
$$ z_1e_1+z_2e_2+z_3e_3 \equiv z_1e_1+z_2e_2+z_3e_3+l_1i_1+l_2i_2\\
= z_1e_1+z_2e_2+z_3e_3+l_1(-2e_2+e_3)+l_2(-2e_1+e_3)\\
= (z_1-2l_1)e_1 + (z_2-2l_2)e_2 + (z_3+l_1+l_2)e_3$$
So for $z_1$ even, the first summand should be identical to zero. For $z_1$ uneven it should be identical to one. Same goes for $z_2$ and the second summand, whereas the third summand should be identical to zero in any case (though i'm not really sure about how to deal with that one, since I've already fixed $l_1$ and $l_2$). Hence I'd say my quotient should be $\mathbb{Z}_2^2$. However, letting sage do the calculations, it claims that the result should be $\mathbb{Z}\oplus\mathbb{Z}_2$
So my question is: Is my result wrong? If so: Where's my mistake and is there a better way to do it?
AI: I would tackle the problem in two steps. In the first one we try to find the answer heuristically and in the second step we prove our claim rigorously. So take any class in $\mathbb Z^3/K$ represented by $(x,y,z)$, the first relation now tells you that
$$e_3=2e_2,$$
so we may write our element as $(x,y+2z,0)$. In fact $y+2z$ can be any integer so we may assume by substitution that our class is represented by $(x,y',0)$. Now a combination of the two relation tells us $$2e_1=2e_2,$$ so we may write our class as $(x+2\lfloor\frac{y'}{2}\rfloor,y' \mod 2,0)$, again the first entry can be any integer. Thus any class in $\mathbb Z^3/K$ may be represented uniquely by a pair $(x',y'')\in \mathbb Z\times \mathbb Z/2$.
Now for the rigorous proof, we have to find a surjection $\mathbb Z^3\to \mathbb Z\times \mathbb Z/2$ with kernel $K$. Then by a isomorphism theorem this map factors over the desired isomorphism. Such a map is given by
$$(x,y,z)\mapsto (x+2z+2\lfloor\frac{y}{2}\rfloor,\bar y).$$
Obviously you have to check that this is a modul hom and that the kernel is indeed $K$. |
H: For Maths Major, advice for perfect book to learn Algorithms and Date Structures
Purpose: Self-Learning, NOT for course or exam
Prerequisite: Done a course in basic data structures and algorithms, but too basic, not many things.
Major: Bachelor, Mathematics
My Opinion: Prefer more compact, mathematical, rigorous book on Algorithms and Data Structures. Since it's for long-term self-learning, not for exam or course, then related factors should not be considered (e.g. learning curve, time-usage), only based on the book to perfectly train algorithms and data structures.
After searching at Amazon, following three are somehow highly reputed.
(1)Knuth, The Art of Computer Programming, Volume 1-4A
This book-set contains most important things in Algorithms, and very mathematically rigorous. I prefer to use this to learn algorithms all in one step.
(2)Cormen, Introduction to Algorithms
It's more like in-between (1) and (3).
(3)Skiena, The Algorithm Design Manual
More introductory and practical compared with (1), is it ok for this to be warm-up then read (1), and skip (2) ?
Desirable answer: Advice or more recommended books
AI: Knuth is an interesting case, it's certainly fairly rigorous, but I would never recommend it as a text to learn from. It's one of those ones that is great once you already know the answer, but fairly terrible if you want to learn something. On top of this, the content is now rather archaically presented and the pseudocode is not the most transparent.
Of course this is my experience of it (background CS & maths), but I have encountered this rough assessment a few times. |
H: Finding Nash Equilibria with Calculus
The problem is summarized as:
There are two players. Player 1's strategy is h. Player 2's strategy is w. Both of their strategy sets are within the range [0,500].
Player 1's payoff function is:
$
P_h(h, w) = 50h + 2hw-\frac{1}{2}(h)^2
$
Player 2's payoff function is:
$
P_w(h, w) = 50w + 2hw - \frac{1}{2}(w)^2
$
Find a Nash Equilibrium.
I was taught to solve these problems in the following way. Find the first derivative of Player 1's payoff function with respect to h, equate it to 0, then solve for h, and then repeat for Player 2 but with respect to w and solving for w instead. However, I found the first derivatives to be:
$
P_h(h, w)^\prime = 50 + 2w - h
$
$
P_w(h, w)^\prime = 50 + 2h - w
$
Now after equating these first derivatives to 0 and solving for h and w, we get that h = -50 and w = -50. The issue now is that these strategies aren't within the strategy set [0,500] as mentioned in the problem question. Where am I going wrong?
AI: (I have not studied Nash equilibria before, but I'll take a stab at this anyway).
Ok, so say h is on the x axis, and w is on the y axis.
Both players are trying to maximize the profit, i.e.
they change their strategies according to the derivatives of the payout functions.
Plotting this stream plot, we get the graph below:
We see that in the entire region, at least one of the derivatives is always positive. Thus, at least one player gains on increasing w or h.
(We see this because each arrow points either right or up or both).
From this, we can see that the point (h,w)=(500,500) is an equilibrium,
and you can verify this by seeing that both the derivatives in this point are positive, and even more, $P_h(h,500)>0$ for all $h\in[0,500]$, and similarly $P_w(500,w)>0$ for all $w \in [0,500].$ Thus, no player would gain on changing the strategy if they are in (500,500). |
H: Check if below limits exist $\lim_{(x,y,z) \to (0,0,0)} \left( {\frac{{2{x^2}y - x{z^2}}}{{y^2 - xz}}} \right)$?
Check if below limits exist $$\lim_{(x,y,z) \to (0,0,0)} \left( {\frac{{2{x^2}y - x{z^2}}}{{y^2 - xz}}} \right).$$
Is there any succession to prove that this limit is not zero?
AI: Taking the sequence $x_n = z_n = \frac{1}{n}$, and $y_n = \frac{1}{n} + \frac{1}{n^2}$, you get
$$ \lim_{n\to \infty} \frac{\frac{1}{n^3}+\frac{2}{n^4}}{\frac{2}{n^3}+\frac{1}{n^4}} = \frac{1}{2}.$$
You can easily find a sequence for which the limit is $0$, and hence the limit does not exist. |
H: Open relative and choice
I'm using ZF as my axiom system.
Let $X$ be a metric space and $K\subset Y \subset X$. Suppose $K$ is compact relative to $X$.
Let $\{V_a\mid a\in I\}$ be a family of open sets relative to $Y$ such that $K\subset \bigcup V_a$. Then for every $a\in I$, there exists $G_a$, an open subset of $X$, such that $V_a = Y\cap G_a$.
"I want to choose such $G_a$ for every $a\in I$ without the Axiom of Choice."
I have tried to show whether closure of $X\setminus V_a$ is such $G_a$, but it didn't work well so maybe it's not the way of choosing.
Here's what I have proved.
If $A\subset B$ is open (or closed) relative to $B$, then $B\setminus A$ is closed (or open) relative to $B$.
AI: I suppose that you are trying to show that $K$ is compact relative to $Y$ as well.
You don't need to choose $G_a$ and it might be impossible too, instead consider the open cover of $K$ which consists of all such sets, namely:
$$\{U\subseteq X\mid U\text{ is open, and }\exists a\in I: U\cap Y=V_a\}$$
This is an open cover of $K$ in $X$ and therefore has a finite subcover $U_1,\ldots, U_n$. Now show that $V_i=U_i\cap Y$ is a finite subcover of $\{V_a\mid a\in I\}$.
Generally speaking, the whole point in compactness is that it allows us to avoid the axiom of choice by generating finite sets from which we can choose without the axiom of choice. The trick is often "if you can't choose, take anything that fits, and let compactness choose for you". |
H: How high will the water rise
I need to know where I am going wrong since I am getting the wrong answer
12 litres of water are poured into an aquarium of $50$ cm length , $30$ cm breadth and $40$ cm Height.How high in cm will the water rise. (Ans 8 cm)
Edit:
So I was making some very illogical assumptions however according to advice given which suggested that I keep the changing dimension value variable I got the answer which is
$1000cm^3$ = $50 \times 30 \times x$
so $x = \frac{1000}{1500}$. so $1$ litre has height $\frac{1000}{1500} = \frac {2}{3}$ so $12$ litres will give $12 \times \frac{2}{3} = 8 cm$
AI: Congratulation: your answer is correct! |
H: Doubling the Sides of the Cube
For the following question
A cubical block of metal weighs 6 pounds.How much will another cube of the same metal weigh if its sides are twice as long ? Ans=48
Here is how I am solving it but I am getting 12 as the answer
For the current cubical block $12$ edges of cube weigh = 6 pounds
$1$ edge weighs = $\frac{6}{12} = \frac{1}{2} pound$
Doubled edge would weigh $2 \times \frac {1}{2} = 1$ so 12 edges would weigh $12 \times 1 = 12$
AI: HINT:
Write down the volume of a cube with edge length $x$. Think of an aquarium with all sides equal.
As 1. with length $2x$.
What's the ratio of the volumes?
How heavy is the larger cube? |
H: Is it true that if $\alpha \in \operatorname{Frac}(A)$ and $s\alpha \in A$, then $\alpha \in S^{-1}A$?
In the proof of Proposition 1.9 in Chapter VII of Algebra by Serge Lang, it seems to me that the following property is used.
Let $A$ be a commutative entire ring, $S$ a multiplicative subset of $A$, $0 \not \in S$. Let $\alpha$ be an element of the quotient field of $A$. If $s \alpha \in A$ for some $s \in S$, then $\alpha \in S^{-1}A$.
(in a proof that $A$ integrally closed implies $S^{-1}A$ integrally closed).
Is this bold statement true ?
Since $\alpha$ is expressed as $\alpha=a/s'$, where $a, s' \in A, s' \neq 0$, $s \alpha \in A$ implies that $s a = s' b$ for some $b \in A$.
From this, how can I prove that $s' \in S$ ?
Any help would be appreciated.
AI: What is $\,S^{-1}A\,$ ? By definition,
$$S^{-1}A:=\{a/s\;:\;a\in A\,,\,s\in S\}\,$$with the "usual" operations (the definition is there).
Thus, we can in fact say that for $\,\alpha:=\frac{a}{b}\in F\,\,,\,a,b\in A\,,\,b\neq 0$ , as
$$\alpha\in S^{-1}A\Longleftrightarrow \frac{as}{b}=\alpha s\in A\,\,,\,\text{for some}\,\,s\in S$$
In fact, there's hardly anything to prove here...
Added: Let's see if the following clarifies a little:
$$\exists\,\,s\in S\,\,s.t.\,\,s\alpha=a\in A\Longrightarrow \alpha=\frac{a}{s}\in S^{-1}A\,\,,\,\text{per definition of fractions ring}$$
and we don't care what the "original" form of $\,\alpha\,$, as element of the fractions field of $\,A\,$ , is/was. |
H: Generating random numbers with the distribution of the primes
I would like to generate random numbers whose distribution mimics that of the primes.
So the number of generated random numbers less than $n$ should grow like $n / \log n$,
most intervals $[n,n+n^\epsilon]$ should contain approximately $n^\epsilon / \log n$
generated numbers (Selberg's short intervals), etc. Does anyone know of a computationally feasible method for generating such "look-a-like primes"?
Addendum. I implemented Henry and Xoff's suggestions.
Here are several instances of the first ten "pseudoprimes":
$$ 4, 5, 9, 10, 17, 23, 27, 28, 31, 44 $$
$$ 7, 8, 9, 10, 12, 15, 18, 19, 27, 34 $$
$$ 6, 11, 15, 16, 23, 26, 27, 29, 45, 49 $$
And here is the cumulative distribution pseudoprimes up to $10^6$ (red), together with
a plot of $n / \log n$ (purple), for one random run:
AI: You could generate random numbers $U_2,U_3,U_4,\ldots$ independently and uniformly in $[0,1]$ and include $n$ in your random set if $U_n \lt \dfrac{1}{\log_e n} - \dfrac{1}{2(\log_e n)^2}$ or some similar expression. |
H: Weyl's unitarity trick
Weyl's unitarity trick creates from an irreducible representation of a compact group a unitary representation by averaging with a Haar measure.
Does anyone know a reference to the paper (or book, with page number) where Weyl introduced his unitarity trick?
AI: The first instance I'm aware of is in Hermann Weyl, Theorie der Darstellung kontinuierlicher halb-einfacher Gruppen durch lineare Transformationen. I, Math. Z. 23 (1925), pages 271-309. (Freely available on GDZ)
See in particular §5, "Der Satz von der vollen Reduzibilität", starting on page 288. |
H: Solving linear congruences by hand: modular fractions and inverses
When I am faced with a simple linear congruence such as
$$9x \equiv 7 \pmod{13}$$
and I am working without any calculating aid handy, I tend to do something like the following:
"Notice" that adding $13$ on the right and subtracting $13x$ on the left gives:
$$-4x \equiv 20 \pmod{13}$$
so that $$x \equiv -5 \equiv 8 \pmod{13}.$$
Clearly this process works and is easy to justify (apart from not having an algorithm for "noticing"), but my question is this: I have a vague recollection of reading somewhere this sort of process was the preferred method of C. F. Gauss, but I cannot find any evidence for this now, so does anyone know anything about this, or could provide a reference? (Or have I just imagined it all?)
I would also be interested to hear if anyone else does anything similar.
AI: $bx\equiv a\pmod{\!m}$ has a unique solution $\!\iff\!b\,$ is coprime to the modulus $m$. If so, by Bezout $\,b\,$ is invertible $\!\bmod m,\,$ so scaling $\,bx\equiv a\,$ by $\,b^{-1}\,$ we obtain the unique solution $\,x\equiv b^{-1}a =: a/b.\,$ We can quickly compute $\,b^{-1}\pmod{\!m}\,$ by the extended Euclidean algorithm, but there are often more convenient ways for smaller numbers (e.g. here and here are a handful of methods applied).
We describe a few such methods below, viewing $\, x\equiv b^{-1}a \equiv a/b\,$ as a modular fraction.
[See here for the general method when the solution is not unique, i.e. when $\gcd(b,m)>1$].
The first, Gauss's algorithm, is based on Gauss's proof of Euclid's lemma via the descent $\,p\mid ab\,\Rightarrow\, p\mid a(p\bmod b).\,$ Generally it only works for prime moduli, but we can also execute the general extended Euclidean algorithm in fraction form too (using multi-valued "fractions").
It works by repeatedly scaling $\rm\:\color{#C00}{\frac{A}B}\overset{\times\ N} \to \frac{AN}{BN}\: $ by the least $\rm\,N\,$ with $\rm\, BN \ge 13,\, $ then reducing mod $13$
$$\rm\displaystyle \ mod\ 13\!:\,\ \color{#C00}{\frac{7}9} \,\overset{\times\ 2}\equiv\, \frac{14}{18}\, \equiv\, \color{#C00}{\frac{1}5}\,\overset{\times \ 3}\equiv\, \frac{3}{15}\,\equiv\, \color{#C00}{\frac{3}2} \,\overset{\times\ 7}\equiv\, \frac{21}{14} \,\equiv\, \color{#C00}{\frac{8}1}\qquad\!\! $$
Denominators of the $\color{#c00}{\rm reduced}$ fractions decrease $\,\color{#C00}{ 9 > 5 > 2> \ldots}\,$ so reach $\color{#C00}{1}\,$ (not $\,0\,$ else the denominator would be a proper factor of the prime modulus; it may fail for composite modulus)
Simpler: $ $ using $\rm\color{#0a0}{least}$ residues: $\displaystyle\ \ \frac{7}9\,\equiv\, \frac{7}{\!\color{#0a0}{-4}\!\ \,}\,\overset{\times\ 3}\equiv\,\frac{21}{\!\!-12\ \ \ \!\!}\,\equiv\, \color{#c00}{\frac{8}1}$
This optimization using $\rm\color{#0a0}{least\ magnitude}$ residues $\,0,\pm 1, \pm 2.\ldots$ often simplifies mod arithmetic. Here we can also optimize by (sometimes) cancelling obvious common factors, or by pulling out obvious factors of denominators, etc. For example
$$\frac{7}9\,\equiv\, \frac{\!-6\,}{\!-4\,}\,\equiv\frac{\!-3\,}{\!-2\,}\,\equiv\frac{10}{\!-2\,}\,\equiv\,-5$$
$$\frac{7}9\,\equiv\,\frac{\!-1\cdot 6}{\ \ 3\cdot 3}\,\equiv\,\frac{\!\,12\cdot 6\!}{\ \ \,3\cdot 3}\,\equiv\, 4\cdot 2$$
Or twiddle it as you did: $ $ check if quotient $\rm a/b\equiv (a\pm\!13\,i)/(b\pm\!13\,j)\,$ is exact for small $\rm\,i,j,\,$ e.g.
$$ \frac{1}7\,\equiv \frac{\!-12}{-6}\,\equiv\, 2;\ \ \ \frac{5}7\,\equiv\,\frac{18}{\!-6\!\,}\,\equiv -3$$
When working with smaller numbers there is a higher probability of such optimizations being applicable (the law of small numbers), so it's well-worth looking for such in manual calculations.
Generally we can choose a congruent numerator giving an exact quotient by Inverse Reciprocity.
$\bmod 13\!:\ \dfrac{a}{b}\equiv \dfrac{a-13\left[\color{#0a0}{\dfrac{a}{13}}\bmod b\right]}b\,\ $ e.g. $\,\ \dfrac{8}9\equiv \dfrac{8-13\overbrace{\left[\dfrac{8}{\color{#c00}{13}}\bmod 9\right]}^{\large\color{#c00}{ 13\ \,\equiv\,\ 4\ }}}9\equiv\dfrac{8-13[2]}9\equiv-2$
Note that the value $\,\color{#0a0}{x\equiv a/13}\,$ is exactly what we need to make the numerator divisible by $b,\,$ i.e.
$\qquad\quad\bmod b\!:\,\ a-13\,[\color{#0a0}x]\equiv 0\iff 13x\equiv a\iff \color{#0a0}{x\equiv a/13}$
This is essentially an optimization of the Extended Euclidean Algorithm (when it takes two steps).
Note $ $ Gauss' algorithm is my name for a special case of the Euclidean algorithm that's implicit in Gauss' Disquisitiones Arithmeticae, Art. 13, 1801. I don't know if Gauss explicitly used this algorithm elsewhere (apparently he chose to avoid use or mention of the Euclidean algorithm in Disq. Arith.). Gauss does briefly mention modular fractions in Art. 31 in Disq. Arith.
The reformulation above in terms of fractions does not occur in Gauss' work as far as I know. I devised it in my youth before I had perused Disq. Arith. It is likely very old but I don't recall seeing it in any literature. I'd be very grateful for any historical references.
See here for further discussion, including a detailed comparison with the descent employed by Gauss, and a formal proof of correctness of the algorithm.
Beware $ $ Modular fraction arithmetic is valid only for fractions with denominator coprime to the modulus. See here for further discussion. |
H: If n balls are thrown into k bins, what is the probability that every bin gets at least one ball?
If $n$ balls are thrown into $k$ bins (uniformly at random and independently), what is the probability that every bin gets at least one ball? i.e. If we write $X$ for the number of empty bins, what is $P(X=0)$?
I was able to calculate the $E(X)$ and thus bound with Markov's inequality $P(X \geq 1) \le E(X)$ but I don't how to work out an exact answer.
http://www.inference.phy.cam.ac.uk/mackay/itprnn/ps/588.596.pdf
AI: What is the chance that all $k$ bins are occupied?
For $1\leq i\leq k$, define $A_i$ to be the event that the $i$th bin stays empty.
These are exchangeable events with $P(A_1\cdots A_j)=(1-{j\over k})^n$ and so
by inclusion-exclusion, the probability that there are no empty bins is
$$P(X=0)=\sum_{j=0}^k (-1)^j {k\choose j}\left(1-{j\over k}\right)^n.$$
Stirling numbers of the second kind can be used to give
an alternative solution to the occupancy
problem. We can fill all $k$ bins as follows: partition the balls
$\{1,2,\dots, n\}$ into $k$ non-empty sets, then assign the bin values $1,2,\dots, k$ to
these sets. There are ${n\brace k}$ partitions, and for each partition $k!$ ways to
assign the bin values. Thus,
$$P(X=0)={{n\brace k}\,k!\over k^n}.$$ |
H: Decomposition of $l$ in a subfield of a cyclotomic number field of an odd prime order $l$
Let $l$ be an odd prime number and $\zeta$ be a primitive $l$-th root of unity in $\mathbb{C}$.
Let $K = \mathbb{Q}(\zeta)$.
Let $A$ be the ring of algebraic integers in $K$.
Let $G$ be the Galois group of $\mathbb{Q}(\zeta)/\mathbb{Q}$.
$G$ is isomorphic to $(\mathbb{Z}/l\mathbb{Z})^*$.
Hence $G$ is a cyclic group of order $l - 1$.
Let $f$ be a positive divisor of $l - 1$.
Let $e = (l - 1)/f$.
There exists a unique subgroup $G_f$ of $G$ whose order is $f$.
Let $K_f$ be the fixed subfield of $K$ by $G_f$.
Let $A_f$ be the ring of algebraic integers in $K_f$.
Let $\mathfrak{l} = (1 - \zeta)A$.
$\mathfrak{l}$ is a prime ideal lying over $l$.
Let $\mathfrak{l}_f = \mathfrak{l} \cap A_f$.
My question: Is the following proposition true? If yes, how would you prove this?
Proposition
(1) $lA = \mathfrak{l}^{l-1}$.
(2) $lA_f = \mathfrak{l}_f^e$.
(3) $\mathfrak{l}_fA = \mathfrak{l}^f$
AI: Note: I am going to use $p$ everywhere instead of $l$.
Yes, and it all follows from the fact that $p$ is totally ramified. To see this, show that $N_{K/\mathbb{Q}}(1-\zeta) = p = [A : \mathfrak{p}]$ and notice that $p = \prod_{k \in (\mathbb{Z}/p\mathbb{Z})^\times} (1 - \zeta^k) = \epsilon (1 - \zeta)^{p-1}$ for some cyclotomic unit $\epsilon$. Hence $pA = \mathfrak{p}^{p-1}$.
$(2)$ and $(3)$ are immediate consequences of $(1)$. |
H: Largest positive integer $k$ such that $\mu(n+r)=0$ for all $1\leq r\leq k$
Find the largest positive integer $k$, such that $\mu(n+r)=0$ for all $1\leq r\leq k$ where $r,n$ are positive integers.
As far as I could make out, we need to find out the maximum range(if nay) of numbers where each has a square divisor.
I have gone through the theory of square-free numbers here and there, but could not proceed much.
AI: There is no such largest positive integer. Given $k$ primes, by the Chinese remainder theorem we can find a number $m$ that has remainders $1$ through $k$ with respect to their squares. Then $m-k$ through $m-1$ all have square divisors. |
H: Does anyone recognize this function?
I am looking for a function $f(n)$ that satisfies the following two conditions at the same time $$ \frac{f(n-1)}{f(n)}=(-1)^n\quad ,\quad \frac{f(n+1)}{f(n)}=(+1)^n\equiv 1,\quad \forall n\in\mathbb{N}\ . $$
As I am not even sure if such a function exists, I'd appreciate any help or comment. Thank you very much in advance!
AI: No such function exists. For the requirements to be well-defined, we need $f(n)\neq 0$, so in particular $f(1)=k\neq 0$. Then $f(2)=f(1+1)=(+1)^1f(1)=k$, but this gives
$$k=f(1)=f(2-1)=(-1)^nf(2)=-k$$
hence $k=0$, a contradiction. |
H: non-residually finite group
Let $G$ be the subgroup of $\text{Bij}(\mathbb{Z})$ generated by $\sigma : n \mapsto n+1$ and $\tau$ which switches $0$ and $1$.
How can we prove that $G$ is not residually finite? Is it hopfian?
AI: $G$ is not residually finite. Specifically, any map $f:G\to H$ with $H$ finite sends $g=(123)$ to the identity. To see this, pick $n$ with $n!>\max(5,|H|)$ and note that $f$ restricts to a map $S_n\to H$, which must have a non-trivial kernel, but the only normal subgroups of $S_n$ are $1$, $A_n$ and $S_n$.
$G$ is hopfian. Consider an epimorphism $f:G\to G$. If $\ker(f)$ contains an element $g$ of infinite order, write $g=g'\tau^k$ where $g'$ is a finitary permutation. There exists $h$ with $f(h)=\tau$; write $h=h'\tau^\ell$ similarly. Then $h^kg^{-\ell}$ is a finitary permutation, so $f(h^kg^{-\ell})=\tau^k$ has finite order, but this is a contradiction. If $\ker(f)$ contains a non-trivial element $g$ of finite order then the restriction $f:S_X\to G$ has a non-trivial kernel for some finite set $X=\{-n,\dots,n\}$. But we can choose $|X|\geq 5$ and derive a contradiction as before unless the kernel is the group of even permutations. But in the last case, since $n$ can be chosen arbitrarily large, $\ker(f)$ is the group of even permutations, so $G/\ker(f)$ is isomorphic to $\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$ (I think), contradicting the surjectivity of $f$.
So $f$ must be an isomorphism. |
H: Anticommutative operation on a set with more than one element is not commutative and has no identity element?
An anticommutative operation on a set X to be a function $\cdot:X\times X\rightarrow X$ satisfying two properties:
(i) Existence of right identity: $\exists r\in X:x\cdot r=x$ for all $x\in X$
(ii) $x\cdot y=r\iff(x\cdot y)\cdot(y\cdot x)=r\iff x=y$ for all $x,y\in X$
Then, why is it true that an anticommutative operation on a set $X$ of more than one element is not commutative and has no identity?
An obvious counterexample that I found is the set $X=\{r,a\}$ where $r\cdot r=r, a\cdot r=a, r\cdot a=a, a\cdot a=a$. If I am right then this is probably a typo since other sources seem to define an anticommutative operation to be one where $x\cdot y=y\cdot x \iff x=y$ for all $x,y\in X$.
Is there anything obvious that I am missing?
AI: If we prove that there is no identity, then it will follow that the operation is not commutative, since we already have a right identity.
Note that by (ii), $xx=r$ for all $x$. Now suppose that $rx=x$. Then:
$$(xr)(rx) = xx = r$$
hence by (ii) we conclude that $x=r$.
Therefore, if $X$ has more than one element, $x\neq r$, then $rx\neq x$, hence $r$ is not a two-sided identity, and moreover, $rx\neq x=xr$, so the operation is not commutative. |
H: Convex function on Banach space
Let $(Y,\|\cdot\|)$ a Banach space and $b\colon Y\to \mathbb{R}$ a nonnegative convex function such that, for some $\mathcal{E}>0$, the set $\{y\in Y\,:\, b(y)<\mathcal{E}\}$ is nonempty and bounded.
I need prove that exist $z\in Y$, $c>0$ and $C>0$ such that $b(y)-b(z)\geq c\|y-z\|$ for $y\in Y\backslash B_C(z)$.
Note: $B_C(z):=\{y\in Y\,:\, \|y-z\|\leq C\}$.
Thanks in advance.
AI: Let $\epsilon_0$ be the fixed positive constant such that $F_0 := \{ y \mid b(y) < \epsilon_0\}$ is nonempty and bounded. Fix $z\in F_0$. Then we have that if $y \in Y \setminus F_0$, $b(y) - b(z) \geq \epsilon_0 - b(z) =: \delta_0 > 0$.
By assumption $F_0$ is bounded, hence there exists $C > 0$ such that $B_C(z) \supset F_0$. (Here I take $B_C(z)$ to be the open ball; a simple modification of the following argument also allows for closed balls.)
Let $c = \frac{\delta_0}{2C}$. Now we verify this choice of $C$ and $c$ is good.
Given $y \in Y\setminus B_C(z)$, let $\tilde{y} = \frac{C}{\|y-z\|} (y-z) + z$. By convexity of the function $b$ we have that
$$ \frac{C}{\|y-z\|} (b(y) - b(z)) + b(z) \geq b(\tilde{y}) \geq b(z) + \delta_0 $$
Hence we have that
$$ b(y) - b(z) \geq \frac{\delta_0}{C} \|y-z\| > \frac{\delta_0}{2C} \|y-z\| = c \| y - z\| $$
as claimed. |
H: Harmonic coordinates for Ricci flow
It is customary to use DeTurck's argument (or Hamilton's original one involving the Nash-Moser iteration) for proving local existence of the Ricci flow. I am wondering why one cannot use harmonic coordinates for this purpose, as can be done for the Einstein equations.
AI: There are at least two problems if you naively approach the problem from that point of view:
How do you propose to time-evolve the harmonic coordinate system? For the Einstein's equation the coordinate system is more properly called "wave coordinates" as the coordinate functions solve the wave equation (compare to the Riemannian case when the coordinate functions solve the Laplace equation).
Harmonic coordinate systems are not necessarily global. For hyperbolic equations like Einstein's equation, finite speed of propagation implies that you can cut up into small neighborhoods, and solve in the domain of dependence o that neighborhood in the (local) wave coordinate system. For parabolic equation one cannot use the same method.
That said, the DeTurck trick can be interpreted as the "harmonic coordinate" version of the proof. In the DeTurck trick you solve the modified Ricci flow with a vector field $X^a = g^{bc}\Gamma_{bc}^a$ where the $\Gamma$ are the analogues of the Christoffel symbol "relative to a fixed background metric". Now, the choice of harmonic coordinate system is one in which $0 = g^{bc}\Gamma_{bc}^a$ (the real Christoffel symbol relative to the coordinates now). So we can see DeTurck's trick as circumventing the difficulty in choosing a global, truly harmonic coordinate system, by compensating it with a time-dependent diffeomorphism that "gets rid of" that extra non-harmonicity.
The correct analogue for the DeTurck trick in Einstein's equations is not the simple o' wave coordinate system. Instead, it is what Choquet-Bruhat calls the "$\hat{e}$ wave gauge" and what is sometimes known as the "wave-map gauge"; for more details you should consult chapter VI.7 of her recent monograph General Relativity and the Einstein Equations. |
H: Is the axiom of choice needed to show that $a^2=a$?
A comment on this answer states that choice is needed for the statement that $a^2=a$ for all infinite cardinals $a$. In Thomas Jech's Set Theory (3rd edition), his theorem 3.5 proves this statement when $a = \aleph_b$ for some $b$.
It's not clear to me in the proof where the axiom of choice is used - can someone clarify that?
I would also like to know if this statement is true without choice when we are working with a set that is already well-ordered. Specifically: suppose we have a well-ordered set $A$, then is it true that for any infinite $B \subset A$, we have
$$ \text{card}(B)^2 = \text{card}(B) ? $$
I believe that Jech's proof of 3.5 would hold in that case without the need for AC, but I would like to check this.
AI: For [infinite] every well-ordered set $A$ it holds that $A^2\sim A$. Furthermore if $B\subseteq A$ and $A$ can be well-ordered then $B$ can be well-ordered as well, and therefore this property is true for $B$ as well (if it is infinite).
However it is possible that for a non well-ordered set it might not be true anymore, in fact we can easily generate such set from every non well-orderable set. If $A$ is a non well-ordered set and $\aleph(A)$ is an ordinal such that $\aleph(A)\nleq A$ (e.g. the Hartogs number of $A$) then $A+\aleph(A)$ has the property: $$\big(A+\aleph(A)\big)^2>A+\aleph(A)$$
One should remark that it is not true for all non well-ordered sets. Even if $\mathbb R$ cannot be well-ordered it still holds that $\mathbb R^2\sim\mathbb R$.
Some proofs:
To see the proof that $A^2\sim A$ implies choice, see my answer: For every infinite $S$, $|S|=|S\times S|$ implies the Axiom of choice.
To see that for well-ordered sets it holds, see my answer: About a paper of Zermelo.
Let me add a bit of history, and refer to the Jech citation. Zermelo formulated the axiom of choice and proved (without it) that every [infinite] well-orderable set has this property in $1904$. The proof appears in the second link above. When one proves that the axiom of choice is equivalent to the fact that every set can be well-ordered, one immediately sees that the axiom of choice implies that every infinite set is bijectible with its square.
On the other hand, Tarski proved in $1923$ that the opposite holds, in particular the proof relies on an interesting lemma which abuses the example I gave for a set without this squaring property. The details appear in the first answer I have linked.
Interestingly enough, in the early $1970$'s two proofs were announced that $A+A\sim A$ does not imply the axiom of choice. I only know that one was published, it was in the Ph.D. dissertation of Sageev, and my advisor told me that he (an undergrad at the time) remembers Sageev sitting on the bench and working on this proof, and that it took him a long time to finish it. (Every footnote referring to this model suggests that the proof is very hard.) |
H: let $f$ be any meromorphic function then $\operatorname{ord}_p(f)=0$
let $f$ be any meromorphic function at $p$ whose laurent series is $\sum_n c_n(z-z_0)^n$, define the $order$ of $f$ at $p$ $$\operatorname{ord}_p (f)= \min\{n:c_n\neq 0\}$$
let $f=\frac{p}{q}$ be a rational function, considered as meromorphic function on a riemann sphere, we may factor $p,q$ and write $$f(z)=c\prod_i (z - \lambda_i)^{e_i}$$ c is a nonzero constant, $\lambda_i$ are distinct complex number and $e_i$ are integers, Then $\operatorname{ord}_{z=\lambda_i}(f)=e^i$ but why $\operatorname{ord}_\infty = \deg(q) - \deg(p)=-\sum_i e_i$? and finally $\operatorname{ord}_x (f)=0$ unless $x=\infty$ or $x$ is one of $\lambda_i$? and why $\sum_{x\in X}\operatorname{ord}_x(f)=0$?
Thank you for your help.
AI: By definition, $\operatorname{ord}_\infty(f) = \operatorname{ord}_0(\hat f)$ where $\hat f(z) := f(\frac{1}{z})$. In particular, if $f(z) = a_n z^n + \ldots + a_1 z + a_0$ is a polynomial with $a_n \neq 0$ then $\hat f$ has a pole of order $n$ at $0$, hence $\operatorname{ord}_\infty(f) = \operatorname{ord}_0(\hat f) = -\deg f$. Since the order function satisfies $\operatorname{ord}(fg) = \operatorname{ord}(f) + \operatorname{ord}(g)$ it follows that
$$\operatorname{ord}_\infty(p/q) = \deg q - \deg p$$
for polynomials $p$ and $q$.
If $f(z) = c \prod_i (z-\lambda_i)^{e_i}$ and $x$ is different from all $\lambda_i$ and $\infty$ then
$$\operatorname{ord}_x(f) = \operatorname{ord}_x(c) + \sum_i e_i \, \operatorname{ord}_x(z-\lambda_i) = 0$$
since the functions $z \mapsto z-\lambda_i$ are holomorphic and don't vanish in $x$.
The relation $\sum_x \operatorname{ord}_x(f) = 0$ follows from
$$\operatorname{ord}_\infty(f) = -\sum_i e_i = - \sum_i \operatorname{ord}_{\lambda_i}(f).$$ |
H: 5 digit no combination problem
How many 5-digit numbers can be created based on the following conditions:
Starting from the left, the first digit is even and nonzero, the second is odd, the third is an odd prime and the fourth and fifth are two random digits not used before in the number?
Note: First digit cannot be a $0$ (Zero)
AI: The first digit is even, and since the number must be a 5-digit number we have 4 choices, namely 2,4,6,8. The second digit is odd, so there are 5 choices, 1,3,5,7,9. For the third digit, it must be 3, 5, or 7.
For the final two digits, the number of choices will depend on how many distinct digits we used for the first three. The only possible overlap is if the second and third digits are the same, so we'll treat that case separately.
If they are different, there are 4 choices for the second digit, since it can't be the same as the third digit. Then we have 7 remaining digits, from which we need to pick 2. There are 7*6=42 possible choices, so multiplying this with 4*4*3=48 gives 2016. I am assuming here that the fourth and fifth digits must be distinct, which seems reasonable given the wording.
In the case that the second and third digits are the same, we have only one choice for the second digit (it must equal whatever we pick for the third), and still 3 choices for the third. In this case there are 8 choices to pick 2 digits from, so there are 8*7=56 choices. Multiplying by the 4*1*3=12 choices for the first three digits gives 672 possibilities.
Adding these together, we have 2688 5-digit numbers with all of these properties. |
H: Show $A$ is "real-equivalent" to its transpose
The problem statement:
Let $A$ be a real $9\times 9$ matrix with transpose $B$. Prove that the matrices are real equivalent in the following sense: There exists a real invertible $9\times 9$ matrix $H$ such that $AH=HB$.
Unfortunately, I don't have much of an attempt at a solution. My first thought was to notice that $A$ and $B$ have the same Jordan normal form so they're similar. But the Jordan form may have complex entries as the eigenvalues may be complex. Also confusing is why the matrix is $9\times 9$. How could this matter? Nine is a perfect square and it's odd, but so are a lot of other numbers.
Any help is welcomed.
AI: The key is to prove the following:
Theorem. Let $A$ and $B$ be matrices with real coefficients. If $A$ and $B$ are similar over $\mathbb{C}$, then they are similar over $\mathbb{R}$.
This problem appears, among other places, in Berkeley Problems in Mathematics, by de Souza and Nuno Silva (Problem 7.7.10 in the second edition; the problem has been used many times in the Prelim exam at Berkeley). The book offers two answers, though I think that one of the solutions (using the rational canonical form) is a bit sloppy.
Here is the other approach. Assume $A = UBU^{-1}$, with $U$ a complex matrix. Write $U=K+iL$, where $K$ and $L$ are real matrices, and assume that $L\neq 0$. Then
$$A(K+iL) = AU = UB = (K+iL)B$$
and so the real parts and imaginary parts of these matrices are equal. That is, $AK = KB$ and $AL = LB$ (the problem is we don't know that $K$, by itself, is invertible).
Now, for any complex number $z$ we have $A(K+zL) = (K+zL)B$. Let $p(z) = \det(K+zL)$. Prove that $p$ is not identically zero, and take a real number that is not a root.
Once you have this theorem, your observation that $A$ and $B$ are similar over $\mathbb{C}$ implies the result you want.
(For what it is worth, the other argument given by the book is that the Rational canonical forms of $A$ and $B$ have to be the same, since they are the same over $\mathbb{C}$, and so $A$ and $B$ are similar over $\mathbb{R}$; the reason this is sloppy is that the rational canonical form is usually defined, for uniqueness purposes, in terms of the irreducible factors of the minimal/characteristic polynomials, and the rational blocks are companion matrices of powers of these irreducible factors. But the irreducible factors over $\mathbb{R}$ are not necessarily the same as the irreducible factors over $\mathbb{C}$, so that the "uniqueness" of the rational canonical form does not directly apply; an argument needs to be made that the equality over $\mathbb{C}$ implies an equality when the irreducible factors are taken over $\mathbb{R}$ instead, and this argument is missing in the book). |
H: Expressing ratio of functions as a constant number
Suppose that we have a function $f$ where $Q(\tau)$ is modified by multiplying $Q(\tau)$ by a real number. Let $f'$ be the modified function, and let $k \in \mathbb{R}$ or $k \in \mathbb{C}$. Take the ratio:
$\frac{f}{{f'}} = k$
$f = \exp \left( { - {{\int\limits_0^\tau {\frac{\omega }{{2Q(\tau ')}}\left( {\frac{\omega }{{{\omega _h}}}} \right)} }^{\frac{{ - 1}}{{\pi Q(\tau ')}}}}d\tau '} \right)\exp \left( {i\int\limits_0^\tau {\left( {{{\left( {\frac{\omega }{{{\omega _h}}}} \right)}^{\frac{1}{{\pi Q(\tau ')}}}} - 1} \right)\omega } d\tau '} \right)$
$f' = \exp \left( { - {{\int\limits_0^\tau {\frac{\omega }{{4Q(\tau ')}}\left( {\frac{\omega }{{{\omega _h}}}} \right)} }^{\frac{{ - 1}}{{2\pi Q(\tau ')}}}}d\tau '} \right)\exp \left( {i\int\limits_0^\tau {\left( {{{\left( {\frac{\omega }{{{\omega _h}}}} \right)}^{\frac{1}{{2\pi Q(\tau ')}}}} - 1} \right)\omega } d\tau '} \right)$
Note that the only thing changed between $f$ and $f'$ is that $Q(\tau)$ in $f$ has become $2Q(\tau)$ in $f'$.
Is it possible to find a constant value $k$ for this ratio? For example, $k = 1/2$. Why or why not?
AI: Take for instance, for $f:\mathbb{R}\to\mathbb{R}$ and $f':\mathbb{R}\to\mathbb{R}$, let us also let our scale factor be $2$:
$$f(x)\equiv\int_{0}^{x}{\sin{x}\:dx}=\left[-\cos{x}\right]_{0}^{x}=-(\cos{x}-1)$$
And:
$$f'(x)\equiv\int_{0}^{2x}{\sin{x}\:dx}=\left[-\cos{x}\right]_{0}^{2x}=-(\cos{2x}-1)$$
So we have:
$$\frac{f(x)}{f'(x)}=\frac{\cos{x}-1}{\cos{2x}-1}=\frac{1}{2(1+\cos{x})}$$
Which is clearly dependent on $x$. |
H: Elementary Geometry Nomenclature: why so bad?
A long-ish wall of text, and I apologize.
Some background: when I was a first-year university student, my chemistry professor was lecturing and was trying to find the word to describe a shape. A student piped up and said, "that's a rhombus." The professor stopped mid-stride, looked at him squarely, and said, "rhombus? That's a stupid word. What's a rhombus? I don't even think that's a word. The word I was thinking of was 'parallelogram'." This was shocking, because this was an American professor, at an American university, and in my American public education, I was taught what a rhombus was in the second or third grade.
Recently, however, I was thinking that maybe my professor wasn't wrong. Consider the naming system for quadrilaterals. The term "quadrilateral" makes some sense: "quad" from Latin for "four", and "lateral" meaning side. And then you get parallelogram, with "parallel" meaning "parallel" and "gram" from Greek meaning "drawn". But then a rectangle is a special case of a parallelogram where the angles are all right angles, which follows clearly enough, and a square is a special case of a rectangle, and important enough to merit its own term.
But then a quadrilateral with only two parallel sides is a trapezoid, which derives from Greek for "table shaped". And then a rhombus is the complement to the square in the special cases of parallelograms -- its angles are anything but right angles!
Confusing yet? We've got the following suffixes describing shapes: -lateral, -gram, -zoid.
We also have triangles, which makes sense because it's "three angles." Yet a "quadrangle" is a region in a university campus.
Increasing the number of sides in the shape, we go from "quadrilaterals" to "pentagons". Ok, now we've gone from the Latin prefix for "four" and a suffix meaning "side" to the Greek for "five" and a totally different suffix. Sometimes we describe the word using a root that means "drawn", and sometimes we describe it by the way that it looks.
And still "rhombus" fits in nowhere in this crazy, convoluted scheme!
To bring this all back to mathematics, and to ask my original question:
Individually, I can find the etymology of each of these terms. But why did the mathematics community adhere to these terms, particularly in elementary education? Did these terms get translated haphazardly from Elements? Is this one of those consequences of the somewhat insular nature of the mathematical community during the Renaissance era? The mathematics community has evolved to be fairly precise in its use of terminology. Why is the terminology surrounding elementary geometry so fragmented?
AI: We also have, for example, add/sum/negation vs. multiply/product/reciprocal. As with natural language (be/is/was, go/went, speak/spoke), the oldest terms tend to be the most irregular, because they became established before the currently used structure emerged. |
H: What's polynomial composition useful for?
I've made this question here. But I got no answer then I'll make a question with it:
I remember of studying:
Sum of two polynomials;
Difference of two polynomials;
Product of a constant and a polynomial;
Product of two polynomials;
And they kinda make sense for me, but I have a doubt on composition of two polynomials, what is composition of two polynomials useful for?
EDIT: This image made me understande it more easily, I've found it on wikipedia.
AI: Composition of polynomials is usually used when we have two polynomial functions, say $f(x)$ and $g(x)$, and we wish to perform something like:
$$f(g(x))=(f\circ g)(x),$$
If we needed to know the result of this for some value of $x$, we could of course simply compute $g(x)$ and then use this as our argument $x$ in $f(x)$.
However, say we are writing a computer program, and we need to calculate $f(g(x))$ several thousand times over the course of the program. Computing $g(x)$ for each value of $x$ first, and then computing $f(g(x))$ based on the output of $g(x)$ can be expensive. A much more efficient approach would be to sit down and actually perform the composition of the two polynomials (which will result in a polynomial of degree $\deg{f} + \deg{g}$), $(f\circ g)(x)$.
Of course, the difference in efficiency here would be fairly negligible, but it is just a simple example of where you could encounter polynomial composition.
EDIT: Incorrect. The output degree is $\deg{f} \times \deg{g}$. In general, the composition written out contains much more terms than the two separate, so it is not useful to expand them. In fact, there are algorithms that explicitly rely on iteration or composition because it is an efficient and numerically stable way to construct polynomials of very high degree. |
H: Are there larger than countable chains in the join-semilattice of Turing degrees?
Recall that Turing degrees are equivalence classes of subsets of $\mathbb{N}$ under Turing equivalence (mutual Turing reducibility). They are partially ordered by Turing reducibility and form a join-semilattice of cardinality $2^{\aleph_0}$, which is also known to contain antichains of cardinality $2^{\aleph_0}$. Countable chains can be easily constructed by iterated Turing jump.
Question: Does it contain chains of cardinality $>\aleph_0$?
AI: Yes, there are $\omega_1$-like chains, which can be obtained by iterating the Turing jump into the transfinite. These are uncountable linearly ordered chains of order type $\omega_1$, uncountable chains with all proper initial segments being countable.
Let $0$ be the computable degree; if $0^{(\alpha)}$ is defined, then let $0^{(\alpha+1)}$ be the jump of $0^{(\alpha)}$, and at limit ordinals simply find any degree above, such as $\oplus_{\beta<\alpha} 0^{(\beta)}$, using fixed representatives of $0^{(\beta)}$ and a fixed representation of $\alpha$.
Indeed, by filling out this chain into a dense order, one can embed a copy of the long rational line into the Turing degrees.
Every chain must have all countable initial segments, however, since the cone below any degree is countable, as there are only countable many Turing machine programs. |
H: The sum of powers of two and two's complement – is there a deeper meaning behind this?
Probably everyone has once come across the following "theorem" with corresponding "proof":
$$\sum_{n=0}^\infty 2^n = -1$$
Proof: $\sum_{n=0}^\infty q^n = 1/(1-q)$. Insert $q=2$ to get the result.
Of course the "proof" neglects the condition on $q$ for this formula, and the sum really diverges. However I now noticed an interesting fact:
If you use two's complement to represent negative numbers on computers, $-1$ is represented by all bits set. Also, sign extending to a larger number of bits (that is, getting the same number in two's complement representation on more bits) works by copying the left-most bit (also known as sign bit) into the additional bits on the left.
Now imagine that formally you sign-extend the number $-1$ to infinitely many bits. What you get is an infinite-to-the-left string of $1$s. Which, using the normal base-2 formula $n = \sum_k b_k 2^k$ (where $b_k$ is the bit k positions from the right, i.e. $b_0$ is the rightmost bit), that infinite string of $1$s translates into exactly the sum above! So in some sense we have an independent re-derivation of that equation.
Now my question is: Is there something deeper behind this? Somehow I cannot imagine it is just coincidential.
AI: Yes. What you are doing is known as working in the $2$-adic numbers.
The $2$-adic numbers are equipped with a curious notion of distance given by the $2$-adic metric. In this metric, two numbers are close together if their difference is divisible by a large power of $2$. In particular, large powers of $2$ are very small. So relative to the $2$-adic metric the geometric series you wrote down really does converge, and the value it converges to really is $-1$. |
H: How to solve this system of fractional equations?
I have to complete a summer packet of 90 Algebra 2 questions. I have completed 89 of them, the only one I could not get was this. I know the answer is $y = \frac {47}2$, $\frac 17$ according to WolframAlpha, but I have no idea how to reach that answer, my Algebra 2 Honors teacher couldn't figure it out.
Use substitution or linear combination to solve the following system:$$\begin{cases}\dfrac 3{x-1} + \dfrac 4{y+2} = 2\\
\dfrac 6{x-1} - \dfrac 7{y+2} = -3 \end{cases}$$
AI: How about this approach: It's not hard to see that $2 \cdot \left(\frac{3}{x-1}\right) = \frac{6}{x-1}$. Therefore, we can do a linear combination approach:
Add $-2$ times the first equation to the second equation. We get:
$$ \left(\frac{6}{x-1} - \frac{7}{y+2} \right) - 2 \left( \frac{3}{x-1} + \frac{4}{y+2} \right) = -3 - 2\cdot 2.$$
Simplifying gives $$\frac{-15}{y+2} =\frac{-7-8}{y+2} + \frac{6-6}{x-1} = -7.$$
Can you take it from here? |
H: If $\phi \in C^1_c(\mathbb R)$ then $ \lim_n \int_\mathbb R \frac{\sin(nx)}{x}\phi(x)\,dx = \pi\phi(0)$.
Let $\phi \in C^1_c(\mathbb R)$. Prove that $$
\lim_{n \to +\infty} \int_\mathbb R \frac{\sin(nx)}{x}\phi(x) \, dx = \pi\phi(0).
$$
Unfortunately, I didn't manage to give a complete proof. First of all, I fixed $\varepsilon>0$. Then there exists a $\delta >0$ s.t.
$$
\vert x \vert < \delta \Rightarrow \vert \phi(x)-\phi(0) \vert < \frac{\varepsilon}{\pi}.
$$
Now, I would use the well-known fact that
$$
\int_\mathbb R \frac{\sin x}{x} \, dx = \pi.
$$
On the other hand, by substitution rule, we have also
$$
\int_\mathbb R \frac{\sin(nx)}{x} \, dx = \int_\mathbb R \frac{\sin x}{x} \, dx = \pi.
$$
Indeed, I would like to estimate the quantity
$$
\begin{split}
& \left\vert \int_\mathbb R \frac{\sin(nx)}{x}\phi(x) \, dx - \pi \phi(0) \right\vert = \\
& = \left\vert \int_\mathbb R \frac{\sin(nx)}{x}\phi(x) \, dx - \phi(0)\int_\mathbb R \frac{\sin{(nx)}}{x}dx \right\vert \le \\
& \le \int_\mathbb R \left\vert \frac{\sin(nx)}{x}\right\vert \cdot \left\vert \phi(x)-\phi(0) \right\vert dx
\end{split}
$$
but the problem is that $x \mapsto \frac{\sin(nx)}{x}$ is not absolutely integrable over $\mathbb R$. Another big problem is that I don't see how to use the hypothesis $\phi$ has compact support.
I think that I should use dominated convergence theorem, but I've never done exercises about this theorem. Would you please help me? Thank you very much indeed.
AI: Note that
$$
\small
\int\limits_{\mathbb{R}}\frac{\sin(nx)}{x}\phi(x)dx-\pi\phi(0)=
\int\limits_{\mathbb{R}}\frac{\sin(nx)}{x}\phi(x)dx-\phi(0)\int\limits_{\mathbb{R}}\frac{\sin(nx)}{x}dx=
\int\limits_{\mathbb{R}}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx
$$
Denote
$$
\small
I_{m}(n):=\int\limits_{[-\pi m,\pi m]}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx\qquad
I(n):=\int\limits_{\mathbb{R}}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx
$$
We claim that $I_m(n)$ converges to $I(n)$ uniformly by $n\in\mathbb{N}$ when $m\to\infty$. Indeed, since $\phi$ is compactly supported
$$
\small
\lim\limits_{m\to\infty}\sup\limits_{n\in\mathbb{N}}\left|I_m(n)-I(n)\right|=
\lim\limits_{m\to\infty}\sup\limits_{n\in\mathbb{N}}\left|\;\int\limits_{\mathbb{R}\setminus[-\pi m,\pi m]}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx\right|=
$$
$$
\small
\lim\limits_{m\to\infty}\sup\limits_{n\in\mathbb{N}}\left|\;\int\limits_{\mathbb{R}\setminus[-\pi m,\pi m]}\frac{\sin(nx)}{x}(-\phi(0))dx\right|=
\lim\limits_{m\to\infty}\sup\limits_{n\in\mathbb{N}}\left|\phi(0)\int\limits_{\mathbb{R}\setminus[-\pi mn,\pi mn]}\frac{\sin(y)}{y}dy\right|=
$$
$$
\small
|\phi(0)|\lim\limits_{m\to\infty}\left|\;\int\limits_{\mathbb{R}\setminus[-\pi m,\pi m]}\frac{\sin(y)}{y}dy\right|=0
$$
Since convergence is uniform by $n\in\mathbb{N}$ we can say
$$
\small
\lim\limits_{n\to\infty} I(n)=
\lim\limits_{n\to\infty} \lim\limits_{m\to\infty} I_m(n)=
\lim\limits_{m\to\infty}\lim\limits_{n\to\infty} I_m(n)=
\lim\limits_{m\to\infty}\lim\limits_{n\to\infty}\int\limits_{[-\pi m,\pi m]}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx
$$
Since $\varphi\in C_c^1(\mathbb{R})$, then the function $x^{-1}(\varphi(x)-\varphi(0))$ is in $L^1([-\pi m,\pi m])$ for all $m\in\mathbb{N}$. Then by Riemann–Lebesgue lemma
$$
\small
\lim\limits_{n\to\infty}\int\limits_{[-\pi m,\pi m]}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx=0
$$
so $\small\lim\limits_{n\to\infty}I(n)=0$. This is exactly what we wanted to prove. |
H: Help with radical equation
Please, help me to solve this equation. No advanced math should be needed.
$$
3x^2 - 4x + \sqrt{3x^2 - 4x - 6} = 18
$$
I'm clueless. It should be simple.
AI: Hint : set $u:= \sqrt{3 x^2-4x-6}$ then your equation becomes $u^2+6+u=18$ or $$u^2+u-12=0$$ |
H: A characteristic of intersection with cartesian product
Fix some binary relation $f$.
Does there necessarily exist a set $C$ such that $(x\times x)\cap f\ne \varnothing \Leftrightarrow x\cap C\ne \varnothing$ for all sets $x$?
AI: Not necessarily.
Say $f=\{(a,b)\}$, with $a\neq b$, and assume such a set $C$ exists.
In particular, $\{a,b\}\cap C\neq\varnothing$, since $\{a,b\}\times\{a,b\}\cap f\neq\varnothing$. Hence either $a\in C$ or $b\in C$ (or both). If $a\in C$, then $\{a\}\cap C\neq\varnothing$, but $\{a\}\times\{a\}\cap f=\varnothing$. If $b\in C$, then $\{b\}\cap C\neq\varnothing$ but $\{b\}\times\{b\}\cap f=\varnothing$. So no such $C$ exists.
By the above argument, if $f$ contains a pair $(a,b)$ such that (i) $a\neq b$; and (ii) $(a,a)\notin f$; and (iii) $(b,b)\notin f$; then no such set $C$ can exist.
Conversely, suppose that for all $a,b$, if $(a,b)\in f$ then either $(a,a)$ or $(b,b)$ lies in $f$. I claim that then the set $C$ does exist, by letting $C$ consist precisely of those elements $r$ for which $(r,r)\in f$.
Indeed, suppose that $x$ is a set such that $x\times x\cap f\neq\varnothing$. Then there exist $a,b\in x$ such that $(a,b)\in f$; by assumption, either $(a,a)\in f$, and hence $a\in C$ so $x\cap C\neq\varnothing$; or else $(b,b)\in f$ and hence $b\in C$, so $x\cap C=\varnothing$.
Conversely, if $a\in x\cap C$, then by definition of $C$ we have $(a,a)\in f$, so $x\times x\cap f\neq \varnothing$.
In summary:
A set $C$ as given exists if and only if $f$ satisfies the condition
$$(a,b)\in f\implies (a,a)\in f\text{ or }(b,b)\in f.$$
If that is the case, then $C=\{a\mid (a,a)\in f\}$ is the desired set; one can justify that $C$ is a set since it is contained in $\mathrm{dom}(f)\cup\mathrm{range}(f)$. |
H: Does this qualify as a proof of $\sum _{k=0}^{l}\binom{n}{k}\binom{m}{l-k} = \binom{n+m}{l}$? (Spivak's 'Calculus')
I'm working through Spivak's 'Calculus' at the moment, and a question about series confused me a bit. I think I have the solution, but I'm not sure if my "proof" holds.
The question is:
Prove that $\sum\limits_{k=0}^{l}\binom{n}{k}\binom{m}{l-k} = \binom{n+m}{l}$
Hint: Apply the binomial theorem to $(1+x)^{n}(1+x)^{m}$
So I expanded $(1+x)^{n}(1+x)^{m}$, then expanded that again, and compared that to $(1+x)^{n+m}$, to get the equality:
$$\begin{align*}
&\binom{n}{0}\binom{m}{0} + \left [ \binom{n}{0}\binom{m}{1} + \binom{n}{1}\binom{m}{0} \right ]x \\
&+ \left [ \binom{n}{0}\binom{m}{2} + \binom{n}{1}\binom{m}{1} + \binom{n}{2}\binom{m}{0} \right ]x^{2} + \cdots\\
&\qquad = \binom{n+m}{0} + \binom{n+m}{1}x + \binom{n+m}{2}x^{2} + \cdots
\end{align*}$$
Now, my question is, is the original statement proven, just because the terms match up on both sides of the equality, or is this insufficient?
Thanks.
AI: Two polynomials are equal iff they have equal degree and equal coefficients.
This statement directly follows from the following fact: if
$$
f(x)=\sum_{k=1}^m a_k x^k
$$
then
$$
a_k=\frac{f^{(k)}(0)}{k!}
$$ |
H: A trigonometric identity
If one sees the simplification done in equation $5.3$ (bottom of page 29) of this paper it seems that a trigonometric identity has been invoked of the kind,
$$\ln(2) + \sum _ {n=1} ^{\infty} \frac{\cos(n\theta)}{n} = - \ln \left\vert \sin\left(\frac{\theta}{2}\right)\right\vert $$
Is the above true and if yes then can someone help me prove it?
AI: Hint 1: Use that
$$
\log(1-z)=-\sum\limits_{n=1}^\infty\frac{z^n}{n}
$$
Hint 2: Set
$$
z=r e^{i\theta}
$$
Hint 3: Take a real part
Hint 4: Take a limit $r\to 1-0$ and use Abel's summation formula |
H: Discontinuous functions that are continuous on every line in ${\bf R}^2$
This is exercise 7 from chapter 4 of Walter Rudin Principles of Mathematical Analysis, 3rd edition. (Page 99)
Define $f$ and $g$ on ${\bf R}^2$ by: $$f(x,y) = \cases {0,&if $(x,y)=(0,0)$\\
xy^2/(x^2+y^4) &otherwise}$$
$$g(x,y) = \cases {0,&if $(x,y)=(0,0)$\\
xy^2/(x^2+y^6) &otherwise}$$
Prove that:
$f$ is bounded on ${\bf R}^2$
$g$ is unbounded in every neighborhood of $(0,0)$
$f$ is not continuous at $(0,0)$
Nevertheless, the restrictions of both $f$ and $g$ to every straight line in ${\bf R}^2$ are continuous!
This is one of the few specific problems I remember from my university career, which ended some time ago. I remember it because I toiled over it for so long and when I finally found the answer it seemed so simple.
AI: Consider $0\le {(x\pm y^2)}^2 = x^2 \pm 2xy^2 + y^4$, so $\pm 2xy^2\le x^2+y^4$, and $(\pm2)f(x,y) = (\pm 2){xy^2\over x^2+y^4}\le 1$ since $x^2+y^4$ is non-negative. Taking the negative sign gives $f(x) \ge-\frac 12$ and taking the positive sign gives $f(x)\le\frac12$.
Let $B$ be given, and consider the value of $g$ at the point $((1/3B)^{-3}, (1/3B)^{-1})$. At this point $g$ takes the value $\frac{1\vphantom{(1/3B)^{-5}}}{2\vphantom{(1/3B)^{-6}}}\frac{(1/3B)^{-5}}{(1/3B)^{-6}} = {\vphantom{(1/3B)^{-5}}3B\over \vphantom{(1/3B)^{-5}}2}$, which is strictly larger in value than the given $B$.
Consider the restriction of $f$ to the semiparabola $P=\{(y^2, y), y>0\}$. $f$ is easily seen to have the constant value $\frac12$ everywhere on $P$, which includes points arbitrarily close to the origin. But $f(0,0) = 0$, so $f$ is not continuous at $(0,0)$.
$f$ and $g$, being rational functions with nonzero denominator, are continuous everywhere except at the origin, so their restrictions to lines that do not pass through the origin must be continuous. We therefore need only consider lines through the origin.
Every line through the origin is either the $y$-axis, with $x=0$, or has $y=mx$ for some $m$. We want to show that each of these restrictions takes values close to 0 in the vicinity of the origin, to agree with $f(0,0) = g(0,0) = 0$.
The restriction of $f$ to the $y$-axis is the constant function $f(0,y) = 0/y^6 = 0$ everywhere except at the origin, which agrees with $f(0,0)=0$, so $f$ is continuous on this line; $g$ is similarly zero everywhere on the $y$-axis.
For other lines $y=mx$, we get $f(x,mx) = \frac{m^2x^3}{x^2+m^4x^4} = \frac{m^2x}{1+m^4x^2}$, and it is easy to see that $\lim_{x\to0} f(x,mx) = 0$, since the numerator goes to zero and the denominator to 1. $g(x,mx)$ is similar: $\lim_{x\to0}g(x,mx) = \lim_{x\to0}\frac{m^2x^3}{x^2+m^6x^6}= \lim_{x\to0}\frac{m^2x}{1+m^6x^4} = 0$.
I remember I dealt with sections 1 and 4 quickly, but then suffered over section 3 for hours before I hit upon the idea of looking at the parabola $P$. After this, section 2 was easy, because the idea is basically the same: look at the restriction of $g$ to the set $\{(y^3,y), y\in{\bf R}\}$. |
H: Are all $p$-adic number systems the same?
After just having learned about $p$-adic numbers I've now got another question which I can't figure out from the Wikipedia page.
As far as I understand, the $p$-adic numbers are basically completing the rational numbers in the same way the real numbers do, except with a different notion of distance where differences in more-significant digits correspond to small distances, instead of differences in less-significant digits. So if I understand correctly, the $p$-adic numbers contain the rational numbers, but not the irrational numbers, while the non-rational $p$-adic numbers are not in $\mathbb{R}$ (someone please correct me if I'm wrong).
Now the real numbers do not depend on the base you write the numbers in. However the construction of the $p$-adic numbers seems to depend on the $p$ chosen. On the other hand I am sure that the construction of the real numbers can be written in a way that it apparently depends on the base, so the appearance might be misleading.
Therefore my question: Are the $p$-adic numbers the same for each $p$ (that is, are e.g. $2$-adic and $3$-adic numbers the same numbers, only written in different bases), or are they different (except for the rational numbers, of course). For example, take the $2$-adic number $x := ...1000001000010001001011$ (i.e. $\sum_{n=0}^\infty 2^{n(n+1)/2}$), which IIUC isn't rational (because it's not periodic). Can $x$ also be written as $3$-adic number, or is there no $3$-adic number corresponding to this series?
In case they are different, is there some larger field which contains all $p$-adic numbers for arbitrary $p$?
AI: No, the different $p$-adic number systems are not in any way compatible with one another.
A $p$-adic number is a not a number that is $p$-adic; it is a $p$-adic number. Similarly, a real number is not a number that is real, it is a real number. There is not some unified notion of "number" that these are all subsets of; they are entirely separate things, though there may be ways of identifying bits of them in some cases (e.g., all of them contain a copy of the rational numbers).
Now, someone here is bound to point out that if we take the algebraic closure of some $\mathbb{Q}_p$, the result will be algebraically isomorphic to $\mathbb{C}$. But when we talk about $p$-adic numbers we are not just talking about their algebra, but also their absolute value, or at least their topology; and once you account for this they are truly different. (And even if you just want algebraic isomorphism, this requires the axiom of choice; you can't actually identify a specific isomorphism, and there's certainly not any natural way to do so.)
How can we see that they are truly different? Well, first let's look at the algebra. The $5$-adics, for instance, contain a square root of $-1$, while the $3$-adics do not. So if you write down a $5$-adic number which squares to $-1$, there cannot be any corresponding $3$-adic number.
But above I claimed something stronger -- that once you account for the topology, there is no way to piece the various $p$-adic number systems together, which the above does not rule out. How can we see this? Well, let's look at the topology when we look at the rational numbers, the various $p$-adic topologies on $\mathbb{Q}$. These topologies are not only distinct -- any finite set of them is independent, meaning that if we let $\mathbb{Q}_i$ be $\mathbb{Q}$ with the $i$'th topology we're considering, then the diagonal is dense in $\mathbb{Q}_1 \times \ldots \times \mathbb{Q}_n$.
Put another way -- since these topologies all come from metrics -- this means that for any $c_1,\ldots,c_n\in\mathbb{Q}$, there exists a sequence of rational numbers $a_1,a_2,\ldots$ such that in topology number 1, this converges to $c_1$, but in topology number two, it converges to $c_2$, and so forth. (In fact, more generally, given any finite set of inequivalent absolute values on a field, the resulting topologies will be independent.)
So even on $\mathbb{Q}$, the different topologies utterly fail to match up, so there is no way they can be pieced together by passing to some larger setting. |
H: Find numbers in a set whose sum equals x
Given $S$, a set (unsorted) of $n$ real numbers;
and a real number $x$;
what is the fastest algorithm to determine if $S$ contains two numbers whose sum exactly equals $x$?
We could add each element of $S$ with each other element to determine if a pair whose sum is $x$ exists, but that would be $O(n^2)$ and I wonder if a faster approach exists.
AI: I don't know if it is optimal, but I sorted the numbers ($O(n \log n)$), then added the largest to the smallest. If the sum is too small, move the lower one up. If the sum is too large, move the larger one down. Continue until you find a pair or the ends meet. The second phase is $O(n)$ as you only make $n-2$ changes at most. |
H: Discreteness of Preimages of points, map between compact riemann surfaces
Let $f:X\rightarrow Y$ be a non-constant holomorphic map between compcat riemann surfaces, we need to show $f^{-1}(y),\forall y\in Y$ is finite and discrete subset of $X$.
What if $X$ and $Y$ are non compact?
well, $f$ is onto clearly, and I understand some how I need to use the fat that $Zeros$ of $f$ is a discrete set, but I am not able to write rigoriously the answer.
AI: a) If $X$ and $Y$ are non compact it certainly is not clear that $f$ is onto since it is false: think of the inclusion of a disk into $\mathbb C$.
b) It is not true either that the fibers $f^{-1}(y)$ are finite: think of the sinus function $\sin:\mathbb C\to \mathbb C$ with $\sin^{-1}(0)=\pi \mathbb Z$, an infinite set.
c) It is however true that the fibers $f^{-1}(y)\; (y\in Y)$ are discrete closed subsets of $X$: closedness follows from continuity of $f$ and discreteness can be checked locally at points of $x\in X$.
This means that in order to check it you may assume that $X$ and $Y$ are disks containing the origin and that $x=0, f(x)=0$.
You may then invoke the result that zeros of non-constant holomorphic functions are isolated. |
H: A (probably trivial) induction problem: $\sum_2^nk^{-2}\lt1$
So I'm a bit stuck on the following problem I'm attempting to solve. Essentially, I'm required to prove that $\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2} < 1$ for all $n$. I've been toiling with some algebraic gymnastics for a while now, but I can't seem to get the proof right. Proving it using calculus isn't a problem, but I'm struggling hither.
AI: As often happens with induction proofs, the easiest approach to proving this statement (which doesn't seem inducable at all - after all, how does knowing the sum for $n$ is less than $1$ tell you anything about the sum for $n+1$?) via induction is to transform it into a stronger one:
$$\mathrm{For\ all\ } n\geq2, \frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2} \lt 1-\frac{1}{n}.$$
Now, the answer becomes a matter of simple algebra:
$$\sum_{i=2}^{n+1} \frac{1}{i^2} = \sum_{i=2}^{n} \frac{1}{i^2} +\frac{1}{(n+1)^2}\lt 1-\frac{1}{n}+\frac{1}{(n+1)^2}\lt 1-\frac{1}{n}+\frac{1}{n(n+1)} = 1-\frac{1}{n+1}.$$ |
H: Intersection of 2 $p$-simplices is a finite union of some $p$-simplices
I'm looking for a non-painful proof of this assertion.
A p-simplex is defined as the set of all sums $\sum_{i=0}^p t_i x_i$ with $0\leq t_i\leq 1$, $\sum_{i=0}^p t_i=1$ for a geometrically independent set of points $x_i\in \mathbb{R}^{n}$ for $n\geq p-1$.
Additionally, is it possible to choose the finely many simplices so that each two of them share no interior points?
AI: Let $s_1,\dots,s_p$ and $t_1,\dots,t_p$ be two sets of affinely independent vectors in $\mathbb R^{p-1}$, and $\sigma$ and $\tau$ their convex envelopes:
$$\sigma=\lbrace \sum_1^px_i s_i|x_1,\dots,x_p\geq 0,~x_1+\dots+x_p=1\rbrace=conv(s_1,\dots,s_p)\\ \tau=\lbrace \sum_1^px_i t_i|x_1,\dots,x_p\geq 0,~x_1+\dots+x_p=1\rbrace=conv(t_1,\dots,t_p).$$ For any $1\leq i\leq p$, let $\sigma_i=conv(s_1,\dots,s_{i-1},s_{i+1},\dots,s_p)$ be the $i^{th}$ face of $\sigma$, and similarly for $\tau$. For a non-empty convex set $C$, define $\mathrm{Ext}(C)=$ the set of its extremal points i.e. the set of points in $C$ that aren't midpoints of proper segments contained in $C$.
A famous theorem asserts that for compact convex sets $K\subset\mathbb R^{p-1}$,
$$K=\overline{conv}(\mathrm{Ext}(K)),$$
that is, $K$ coincides with the closed convex envelope of its extremal points.
We'll show that
$$\mathrm{Ext}(\sigma\cap\tau)\subset\lbrace s_1,\dots,s_p,t_1,\dots,t_p\rbrace\cup\bigcup_{1\leq i,j\leq p}\mathrm{Ext}(\sigma_i\cap\tau_j)~~~~(*)$$
Since for $p=1$, the number of extremal points of a compact convex set is $1$ or $2$, by induction on the dimension $p$, the set of extremal points of an intersection of two $p$-simplices is finite.
So let's prove $(*)$. First of all, it is clear that if $x\in C\subset C'$ is an extremal point of $C'$, then it is an extremal point of $C$ aswell. So in order to prove the assertion, we only need to prove that any extremal point of $\sigma\cap\tau$ that is not a vertex of $\sigma$ or $\tau$ is contained in some $\sigma_i$ and some $\tau_j$. This is indeed true, for otherwise, such an extremal point would belong to (for instance) $\sigma\setminus\cup_1^p\sigma_i=\mathrm{Interior}~\sigma$ and to $\tau\setminus\lbrace t_1,\dots,t_p\rbrace=\tau\setminus\mathrm{Ext}(\tau)$, so it could be represented as a mid point of a proper segment $I$ included in $\tau$, and thus as a midpoint of a proper segment (actually a subsegment of $I$) included in $\sigma\cap\tau$ because $\sigma$ contains an open ball centered around that point, which contradicts the extremality of that point in $\sigma\cap\tau$. Thus every extremal point of $\sigma\cap\tau$ that is not a vertex of either $\sigma$ or $\tau$ belongs to an intersection of two faces $\sigma_i\cap\tau_j$, and by what precedes is an extremal point of that convex set.
So by induction on $p$, the set of extremal points of the intersection of two $p$-simplices is finite, and since the convex envelope of a compact set is compact (corollary to Caratheodory's theorem), we have
$$\sigma\cap\tau=\overline{conv}(\mathrm{Ext}(\sigma\cap\tau))=\overline{conv}(\mathrm{finite~set})=conv(\mathrm{finite~set})\\ =\bigcup p\mathrm{-simplices~built~from~that~finite~set}$$
The last equality uses Carathedory's theorem, according to which, for any nonempty subset $S\subset\mathbb R^{p-1}$,
$$conv(S)=\lbrace \sum_1^pt_is_i|s_1,,\dots,s_p\in S\rbrace.$$ This concludes the proof that the intersection of two $p$ simplices is a finite union of $p$-simplices. |
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