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H: What is $\mathfrak{gl}(\infty)$
As title says, I know what is $\mathfrak{gl}(n,\mathbb{C})$, but what is $\mathfrak{gl}(\infty)$? Where can I find good reference for this?
AI: This is the finitary Lie algebra
$$
\mathfrak{gl}(\infty)=\bigcup_{n\in \mathbb{N}}\mathfrak{gl}(n).
$$
The simple ones have been classified:
Theorem(Baranov): There are only three nonisomorphic finitary simple Lie algebras of countable dimension:
$\mathfrak{sl}(\infty)$, $\mathfrak{so}(\infty)$, and $\mathfrak{sp}(\infty)$.
The paper of Baranov here contains a number of good references: http://www2.le.ac.uk/departments/mathematics/extranet/staff-material/staff-profiles/ab155/research/sdiag.pdf. Finitary Lie algebras have been studied by many authors (e.g., Penkov, Bahturin, Strade, Zalesskii, and many others). |
H: Reducing ordinals in the representation of a limit ordinal as $\bigcup_{\beta < \alpha} \beta$
Let $\alpha$ be a limit ordinal such that $\kappa$ < $\alpha$ < $\kappa^{+}$
where $\kappa$ is initial ordinal (so $|\alpha| = \kappa$).
I want to know whether or not I can find a sequence {$\beta_{i}$} where $ i < \kappa$ such that $\beta_{i} < \beta_{j} $ iff $i<j$, and $\bigcup_{i < \kappa} \beta_{i} = \alpha$.
AI: Not necessarily: you cannot do it if $\kappa>\omega$ and $\alpha=\kappa+\omega$. For regular $\kappa$, like $\omega_1$, you can do it if and only if the cofinality of $\alpha$ is $\kappa$. |
H: Showing $\sum_{r=1}^\infty \frac{1}{(r-z)^2}$ is holomorphic on $\mathbb{C}\setminus\mathbb{N}$
Ok, so as per the question title I'm wanting to show that
$\displaystyle\sum_{r=1}^\infty \dfrac{1}{(r-z)^2}$ is holomorphic on $\mathbb{C}\setminus\mathbb{N}$.
I'm thinking that if I show that
$$\int_{\gamma} \sum_{r=1}^\infty \frac{1}{(r-z)^2} \ \text{d}z = 0 $$
Then by Morera's theorem, the sum is holomorphic. The issue is that the series' convergence isn't uniform on all of $\mathbb{C}\setminus\mathbb{N}$. Although the question does hint that for any $z \in \mathbb{C}\setminus\mathbb{N}$ the series is uniform on some neighbourhood of $z$
Any help greatly appreciated!
AI: The holomorphicity of the sum is a direct consequence of Weierstrass' theorem:
Suppose that $f_n(z)$ is holomorphic in the region $\Omega$, and that the sequence $\{f_n(z)\}$ converges to a limit function $f(z)$ in $\Omega$, uniformly on every compact subset of $\Omega$. Then $f(z)$ is holomorphic in $\Omega$. Moreover, $f_n'(z)$ converges uniformly to $f' (z)$ on every compact subset of $\Omega$. |
H: The Mean of a multivariate function
Jane and Jack each toss a fair coin twice.
Let X be the number of heads Jane obtains.
Let Y be ther number of heads Jack obtains.
Define U = X + Y.
Find the mean and variance of U.
I have tried to first find f(U) and am unable to.
Would it be:
$$ {4 \choose u}(0.5)^{u}(0.5)^{4-u} $$
Thank you!
AI: Your $P(U=u)$ is correct. Here is a hint, but I encourage you to do the problem your way too, it would be an instructive exercise.
Hint:
$E(X+Y)=E(X)+E(Y)$
If $X$ and $Y$ are independent then $Var(X+Y)=Var(X)+Var(Y)$ |
H: Are there p-adic manifolds?
Is there anything resembling a manifold on the field of p-adic or complex p-adic fields?
If so is there a connection to algebraic geometry as rich as in the reals?
AI: Yes, there are. One source to learn about them is the second half of Serre's book Lie Algebras and Lie Groups, and another is Peter Schneider's recent book $p$-adic Lie Groups, which is very nice.
They are called locally $p$-adic analytic manifolds, or more generally, if $k$ is any non-Archimedean field, locally $k$-analytic manifolds. The definition is formally the same as a real or complex analytic manifold, but one uses the word "local" because, for non-Archimedean fields, functions given locally by power series do not need to be given globally by a power series. This is in contrast to rigid analytic spaces, which are more akin to schemes, and have as their coordinate rings certain "affinoid" algebras of convergent power series. Much of the theory for complex manifolds carries over to locally analytic manifolds (as is made clear in Serre's book), although of course there are some differences. For example, paracompact locally analytic manifolds are "strictly paracompact," meaning that every open cover admits a refinement by pairwise disjoint opens. But one has versions of the inverse and implicit function theorems, tangent bundles, Lie algebras for locally $k$-analytic groups, etc.. The notes mentioned in the comments are on various theories of non-Archimedean analytic spaces, which are somewhat more analogous to complex analytic spaces in general.
If $k$ is a non-Archimedean field and $X$ is a smooth, separated $k$-scheme of finite type, then $X(k)$, the set of $k$-rational points, has a canonical locally $k$-analytic structure, and $X\rightsquigarrow X(k)$ is functorial in $X$. This is the same as for schemes over $\mathbf{C}$. I'm not sure what other connections with algebraic geometry you're referring to. |
H: vector problem in parallelopiped
I want to find out absolute volume the parallelopiped
I have not got that how they did with this vector notation, $h= \vec{A}\cdot \vec{n} $
The volume will be $\vec{A}\cdot (\vec{B} \times \vec{C}) $
AI: Note that $|\vec{n}|=1$ and $\vec{A}\cdot \vec{n}=|\vec{A}||\vec{n}|cos(\alpha)=|\vec{A}|cos(\alpha)=h$ |
H: $(X,d)$ m.e., with $Y \subset X$: $Y$ is open, $Y$ is connected it's equivalent to another property.
Let $(X,d)$ be a metric space and let $Y \subset X$: $Y$ is open. Prove that $Y$ is connected if and only if there aren't $A,B \subset X$ non-empty such that $Y=A \cup B$ and $A \cap \overline B=\emptyset$ and $\overline A \cap B=\emptyset$.
My attempt at a solution:
I could prove that the second statement implies $Y$ is connected (here I've used that $Y$ is connected if and only if the only clopen sets are $Y$ and $\emptyset$:
Suppose the second condition holds but $Y$ is disconnected. Let $S \subset Y$ such that $S$ is clopen and suppose $S\neq Y$ and $S \neq \emptyset$. Clearly, $Y=S \cup S^c$. $S=\overline S$ and $S^c=\overline S^c$, so $S \cap \overline S^c=S \cap S^c=\emptyset$, analogously, $\overline S \cap S^c=\emptyset$. But this is absurd by the hypothesis, this means that the only clopen sets are $Y$ and $\emptyset$, it follows that $Y$ is connected.
I couldn't do much with the other implication:
My hypothesis is that $Y$ is connected and I want to prove that this implies the second condition holds. I've tried to prove it by the absurd, i.e, I suppose there exist $A,B \subset X$ non-empty such that $Y=A \cup B$ and $A \cap \overline B=\emptyset$ and $\overline A \cap B=\emptyset$. I should conclude that $Y$ is disconnected. In some part of the proof I have to use the fact that $Y$ is open in $X$.
AI: Hint: Since $A\cap\overline B=\emptyset,$ then $X\setminus\overline B$ is an open superset of $A,$ so since $B\subseteq\overline B$ and $Y=A\cup B,$ then $A=Y\cap(X\setminus\overline B),$ so $A$ is open in $Y.$ Similarly, $B$ is open in $Y.$ What can we then conclude? (Can you justify these claims?) |
H: how the monotonicity of $\frac{1}{(logn)^n}$ ,with induction or an other way??
Could you tell me how to show the monotonicity of $\frac{1}{(logn)^n}$ ? With induction or with an other way?
AI: Let us define $a_{n}=\frac{1}{(log(n))^{n}}$
$log(n+1) > log(n)$ for all natural numbers $n$ as the function $f(x)=log(x)$ is strictly increasing.
Therefore we have $(log(n+1))^{n+1} > (log(n+1))^{n} > (log(n))^{n}$, as clearly $n+1>n$.
Thus $a_{n+1}=\frac{1}{(log(n+1))^{n+1}} < \frac{1}{(log(n))^{n}} = a_{n}$, and we have $a_{n}$ is monotonically decreasing. |
H: Show that $\lim\limits_{n\rightarrow\infty} e^{-n}\sum\limits_{k=0}^n \frac{n^k}{k!}=\frac{1}{2}$
Show that $\displaystyle\lim_{n\rightarrow\infty} e^{-n}\sum_{k=0}^n \frac{n^k}{k!}=\frac{1}{2}$ using the fact that if $X_j$ are independent and identically distributed as Poisson(1), and $S_n=\sum\limits_{j=1}^n X_j$ then $\displaystyle\frac{S_n-n}{\sqrt{n}}\rightarrow N(0,1)$ in distribution.
I know that $\displaystyle\frac{e^{-n} n^k}{k!}$ is the pdf of $K$ that is Poisson(n) and $k=\sum\limits_{j=1}^n X_j$ is distributed Poisson(n), but I don't know what I can do next.
AI: Hint: $e^{-n} \sum_{k=0}^n \dfrac{n^k}{k!}$ is the probability of what? |
H: Summation of a floored square root
I am working on a little something and have hit a roadblock of sorts. I have arrived at this equation:$$\sum_{n=1}^{r}\left\lfloor\sqrt{2nr-{n}^{2}}\right\rfloor$$
I am attempting to find some way of solving this summation as a function of r. If this is possible, could I get some help?
Also, please do not mark this as a duplicate; I searched the existing questions and had trouble finding an answer.
Thanks for any help. :)
AI: We can rewrite your sum as follows.
$$f(r)=\sum_{n=1}^{r} \lfloor \sqrt{2nr-n^2} \rfloor = \sum_{n=1}^r \lfloor \sqrt{r^2-(r-n)^2} = \sum_{n=0}^{r-1} \lfloor \sqrt{r^2-n^2} \rfloor$$
Thus, this counts the number of Gaussian integers $z=a+bi$ satisfying $|z|\le n$, $a>0$, $b\ge 0$.
This is http://oeis.org/A036698 .
Counting Gaussian integers with norm $|z|\le n$ is known as the Circle Problem. Only bounds are available, not exact formulas. We can prove that
$$f(r) = \frac{\pi}{4} r^2 + O(r).$$
See the links on the Wikipedia page for proofs. |
H: Square root in Banach algebra
Suppose we are given a unital Banach algebra $A$ and an element $a\in A$ such that the spectrum is a subset of the positive reals $\mathbb{R}_{>0}$. Then by a theorem (see for example W. Rudin 10.30) we know that there exists a "square root" of $a$, i.e. an element $b\in A$ such that $b^2 = a$.
Question: does there exist such a $b$ with a spectrum $\sigma(b)$ also contained in the positive reals $\mathbb{R}_{>0}$?
My thoughts:
From the spectral mapping theorem we know that since the square root is holomorphic in a neighborhood of the spectrum, $\sigma(\sqrt a)=\sqrt{\sigma(a)}$. However, is it trivial that $\sqrt a= \pm b$ as in the complex case?
Also, if the spectrum of $b$ had to be connected then it would be clear, but I don't think this is true in general.
AI: Question: does there exist such a $b$ with a spectrum $\sigma(b)$ also contained in the positive reals $\mathbb{R}_{>0}$?
Yes, that's an important point that you have a positive square root of a positive element.
You take the principal branch of the square root, defined in $\mathbb{C}\setminus (-\infty,0]$, with positive real part to construct it. |
H: Simple logarithm properties proof
I was reading a school algebra book about logarithm function (on $\mathbb{R}^+$). There were several properties without proof. So I decided to prove 2 of them myself.
The first property:
$log_a(x\cdot y) = log_ax + log_ay$
Proof
By definition of logarithm we know, that: $a^{log_ax} = x$ (axiom 1)
So $x\cdot y = a^{log_ax} \cdot a^{log_ay} = a^{log_ax\;+\;log_ay}$
Therefore
$log_a(x\cdot y) = log_a(a^{log_ax} \cdot a^{log_ay}) = log_a(a^{log_ax\;+\;log_ay}) = log_ax\;+\;log_ay$
question
I am not sure about this equality: $log_a(a^{log_ax\;+\;log_ay}) = log_ax\;+\;log_ay$, because we don't have an axiom, that $\forall x\;log_aa^x = x$, despite the fact, that we use it here. It must be a conclusion from the definition of logarithm:
$log_ax$ is a power to which we must take a to get x. If we want to know, to which power we must take a to get $a^x$ --- obviously, it's x.
This trivial axiom was missing in the book. Shouldn't we explicitly denote it?
The second property also aroused a question:
$log_ax^n = n \cdot log_ax$
Proof
$log_ax^n = log_a\prod_{i=1}^{n} x$
According to the previously proven property
$log_a(x\cdot y) = log_ax + log_ay$
Therefore $log_a\prod_{i=1}^{n} x = \sum_{i=1}^{n}log_ax = n \cdot log_ax$
question
Equality $x^n = \prod_{i=1}^{n} x$ is true only for $n \in \mathbb{Z}^+$, so formally speaking our proof is valid only for $n \in \mathbb{Z}^+$.
While throughout the book the equality $log_ax^n = n \cdot log_ax$ is considered to be true $\forall n \in \mathbb{R}$.
I don't understand, how to prove it.
AI: To the first question, as you stated, it is a consequence of the definition of the logarithm, so there's no need for an axiom for it.
For the second, realize that:
$$a^{n\cdot\log_a x}=(a^{\log_a x})^n=x^n=a^{\log_a x^n}\Longrightarrow n\cdot\log_a x=\log_a x^n, \forall n\in \mathbb{R},$$
since the property used is valid for real $n$ and the exponential function is injective. |
H: Limit at infinity involving $e$
I am to find the limit of $$\lim_{x \to \infty} \left(1+\frac{x}{5x^3+x^2+8}\right)^ {\dfrac{x^3+8}{x}}$$
I could not find the proper substitution here. I would be happy if someone could shed some light. Thanks.
AI: $$
\begin{align}
\lim_{x\to\infty}\left(1+\frac{x}{5x^3+x^2+8}\right)^{\frac{x^3+8}{x}}
&=\lim_{x\to\infty}\left(1+\frac{1}{5x^2+x+8/x}\right)^{x^2+8/x}\\
&=\lim_{x\to\infty}\left(1+\frac{1}{5x^2+x+8/x}\right)^{(5x^2+x+8/x)\frac{x^2+8/x}{5x^2+x+8/x}}\\
&=\left(\lim_{x\to\infty}\left(1+\frac{1}{5x^2+x+8/x}\right)^{(5x^2+x+8/x)}\right)^{\lim\limits_{x\to\infty}\frac{x^2+8/x}{5x^2+x+8/x}}\\[6pt]
&=e^{1/5}
\end{align}
$$ |
H: Is identity map one to one and onto?
Im reading a chapter of compactness in Real Analysis, Carothers, 1ed. Actually, identity map has been involved in and I've captured its definition:
Equivalent Metrics
As a last topic related to both compactness and uniform continuity, we discuss several notions of equivalence for metrics (and norms). Throughout, we will suppose that $d$ and $\rho$ are two metrics on the same set $M$. We will write $i:(M,d)\to(M,\rho)$ as the identity map and $i^{-1}:(M,\rho)\to(M,d)$ as its inverse (also the identity map, but in the other direction).
So is identity map one to one and onto?
AI: 1-1 means $f(a) = f(b)$ only if $a=b$. Clearly, for identity map this holds.
onto means for each $a \in M$ there is a $b \in M$ such that $f(b) = a$, which is trivially true for the identity map, let $b = a$. |
H: Prove that subsequence converges to limsup
Given a sequence of real numbers, $\{ x_n \}_{n=1}^{\infty}$, let $\alpha =$ limsup$x_n$ and $\beta = $ liminf$x_n$.
Prove that there exists a subsequence $\{ x_{n_k}\}$ that converges to $\alpha$ as $k \rightarrow \infty$.
Not sure how to start this without since I'm not given that the subsequence is bounded..
AI: Since $\alpha=\limsup x_n$, by the definition of $\limsup$, there is some $x_{n_1}$ with $|x_{n_1}-\alpha|<{1\over 2}$. (That's the crucial step, so be sure you understand why.)
Similarly, there is some $x_{n_2}$ with $|x_{n_2}-\alpha|<{1\over 2^2}$. Continuing, for each $k\in\mathbb{N}$, there is some $x_{n_k}$ with $|x_{n_k}-\alpha|<{1\over 2^k}$.
Then $\{x_{n_k}\}\subset \{x_n\}$ and $x_{n_k}\to \alpha$ as $k\to\infty$. |
H: For a sequence of non negative numbers, if the series converges, then the series of the sequence raised to p also converges if p>=1
Let $p \geqslant 1$ and let $(a_n)$ be a sequence of non-negative numbers. Then if $\sum\limits_{n=1}^\infty a_n$ converges, so does $\sum\limits_{n=1}^\infty a_n^p$. Prove this statement.
Sorry, we have just started learning series in class, but I have this homework problem and can't get anywhere. Any help/hints appreciated!
AI: Eventually, $a_n<1$, then $a_n^p<a_n$ so...? |
H: Euler numbers grow $2\left(\frac{2}{ \pi }\right)^{2 n+1}$-times slower than the factorial?
Stirling's approximation of the factorial for even numbers is given by
$$ (2n)! \sim \left(\frac{2n}{e}\right)^{2n}\sqrt{4 \pi n}. \tag{1} $$
Further, the Euler numbers grow quite rapidly for large indices as they have the following lower bound
$$ |E_{2 n}| > 8 \sqrt { \frac{n}{\pi} } \left(\frac{4 n}{ \pi e}\right)^{2 n}=
2\left(\frac{2}{ \pi }\right)^{2 n+1} \sqrt { 4n\pi} \left(\frac{2 n}{ e}\right)^{2 n}. \tag{2}$$
Why do these formulas look so similar? Or is $(2)$ just a way to say that Euler numbers grow $2\left(\frac{2}{ \pi }\right)^{2 n+1}$-times slower than the factorial?
AI: There's more to it, we have
$$\frac{1}{\cos z} = \sum_{n=0}^\infty (-1)^n \frac{E_{2n}}{(2n)!}z^{2n}.\tag{1}$$
Since $\dfrac{1}{\cos z}$ has poles in $(k+\frac12)\pi$ for $k\in\mathbb{Z}$ and is holomorphic everywhere else, the series $(1)$ has a radius of convergence of $\dfrac{\pi}{2}$. The Cauchy-Hadamard formula now says
$$\limsup_{n\to\infty} \sqrt[n]{\frac{\lvert E_{2n}\rvert}{(2n)!}} = \frac{2}{\pi}.$$
More precisely, from the partial fraction development of $\dfrac{\pi}{\cos \pi z}$ (see below) we obtain
$$\frac{\pi}{\cos \pi z} = \sum_{n=0}^\infty \left(\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^{2n+1}}\right)2^{2n+2}z^{2n}.\tag{2}$$
Replacing $z$ with $\pi z$ in $(1)$ and multiplying with $\pi$ yields
$$\frac{\pi}{\cos \pi z} = \sum_{n=0}^\infty (-1)^n\frac{\pi^{2n+1}E_{2n}}{(2n)!}z^{2n},\tag{3}$$
and comparing coefficients with $(2)$ yields
$$(-1)^n\frac{\pi^{2n+1}E_{2n}}{(2n)!} = 2^{2n+2}\underbrace{\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^{2n+1}}}_{s(2n+1)},$$
or
$$E_{2n} = (-1)^n 2 \left(\frac{2}{\pi}\right)^{2n+1}s(2n+1)\cdot(2n)!$$
Since $s(2n+1) \to 1$ for $n\to \infty$, we have
$$\lvert E_{2n}\rvert \sim 2\left(\frac{2}{\pi}\right)^{2n+1}\cdot(2n)!$$
$\cos \pi z$ has zeros in $a_k = k + \frac12,\: k \in \mathbb{Z}$ and nowhere else. All zeros are simple. The residue of
$$f(z) := \frac{\pi}{\cos \pi z}$$
in $a_k$ is therefore
$$\operatorname{Res}\left(f;a_k\right) = \frac{\pi}{\frac{d}{dz}(\cos \pi z)\bigl\lvert_{a_k}} = \frac{1}{-\sin \pi a_k} = (-1)^{k+1}.$$
The series
$$g(z) = \sum_{k\in \mathbb{Z}} (-1)^{k+1}\left(\frac{1}{z-a_k} + \frac{1}{a_k}\right)$$
is absolutely and locally uniformly convergent, and therefore $g$ is an entire meromorphic function with the same principal parts as $f$, thus $f-g$ is an entire holomorphic function.
$$g'(z) = \sum_{k\in\mathbb{Z}} \frac{(-1)^k}{(z-a_k)^2}$$
is easily seen to be an entire meromorphic function with period $2$ and the property that $\lim\limits_{\lvert \operatorname{Im} z\rvert\to \infty} \lvert g'(z)\rvert = 0$ uniformly in $\{ z : \lvert \operatorname{Re} z\rvert \leqslant 1\}$. The same holds for
$$f'(z) = \frac{\pi^2\sin \pi z}{\cos^2\pi z},$$
and furthermore $f'$ and $g'$ have the same poles and principal parts. Hence $f' - g'$ is an entire holomorphic function with period $2$ and $\lim\limits_{\lvert \operatorname{Im} z\rvert\to \infty} \lvert f'(z) - g'(z)\rvert = 0$ uniformly in $\{ z : \lvert \operatorname{Re} z\rvert \leqslant 1\}$, thus bounded, ergo constant, and the vanishing of $f'-g'$ for growing imaginary part determines that constant as $0$.
Therefore $f-g$ is constant, and since $g(0) = 0$, we have
$$\frac{\pi}{\cos \pi z} = \pi + \sum_{k\in\mathbb{Z}} (-1)^{k+1}\left(\frac{1}{z-a_k} + \frac{1}{a_k}\right).$$
Grouping, for $k \geqslant 0$, the terms for the two indices $k$ and $-(k+1)$, noting $a_{-(k+1)} = -a_k$, we obtain for $\lvert z\rvert < \frac12$
$$\begin{align}
\frac{\pi}{\cos\pi z} &= \pi + \sum_{k=0}^\infty (-1)^k \left(\frac{1}{a_k-z} + \frac{1}{a_k+z} - \frac{2}{a_k}\right)\\
&= \pi + \sum_{k=0}^\infty \frac{(-1)^k2}{a_k}\left(\frac{a_k^2}{a_k^2-z^2}-1\right)\\
&= \pi + \sum_{k=0}^\infty \frac{(-1)^k2}{a_k}\sum_{n=1}^\infty \left(\frac{z}{a_k}\right)^{2n}\\
&= \pi + \sum_{n=1}^\infty \sum_{k=0}^\infty \frac{(-1)^k2}{a_k^{2n+1}}z^{2n}\\
&= \pi + \sum_{n=1}^\infty \left(\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^{2n+1}}\right)2^{2n+2}z^{2n}.
\end{align}$$
Writing $\pi$ as the Leibniz series $\frac{\pi}{4} = 1 - \frac{1}{3} +\frac{1}{5} - \dotsb$ then yields $(2)$. |
H: Show that there is a number between 1 and 1000 such that there is a perfect square
Show that there exists an integer $n \in S = \{1,2, \ldots, 1000\}$ such that
$$\prod_{i\in S-\{n\}}i!$$
is a perfect square.
I was thinking in trying to prove it by contradiction using the Pigeonhole Principle.
AI: First observe that
$$\prod_{n\in S}n!=\prod_{n=1}^{1000}\prod_{k=1}^nk=\prod_{k=1}^{1000}\prod_{n=k}^{1000}k=\prod_{k=1}^{1000}k^{1001-k}\;.$$
For odd $k$ the exponent $1001-k$ is even, so
$$\prod_{k=1}^{500}(2k-1)^{1001-(2k-1)}$$
is a square, and we need only consider
$$\prod_{k=1}^{500}(2k)^{1001-2k}=\prod_{k=1}^{500}2^{1001-2k}\prod_{k=1}^{500}k^{1001-2k}\;.$$
Now $\sum\limits_{k=1}^{500}2k=500\cdot501$, so
$$\prod_{k=1}^{500}2^{1001-2k}=2^{500(1001-501)}=2^{500^2}$$
is a square, and we’ve reduced the problem to considering
$$\prod_{k=1}^{500}k^{1001-2k}=500!\cdot\prod_{k=1}^{500}k^{1000-2k}\;.$$
Each of the exponents $1000-2k$ is even, so $\prod\limits_{k=1}^{500}k^{1000-2k}$ is a square. Thus
$$\prod_{n\in S\setminus\{500\}}n!$$
is a square. |
H: Evaluating the integral of a delta function multiplied by an exponential
How does one evaluate the integral below? I don't think we can use integration by parts here. I am basically stuck.
$$ \int_{-1}^{6}{(2+e^{-t})\delta(t-2)} dt $$
$$ = \int_{-1}^{6}{2\delta(t-2)dt} + \int_{-1}^{6}{e^{-t}\delta(t-2)dt} $$
AI: As indicated by muffle, the answer is the value of $f(t)=2 + e^{-t}$ at point $t=t_0=2$, that is
$$
2 + e^{-2}
.
$$ |
H: Concrete Mathematics - Towers of Hanoi Recurrence Relation
I've decided to dive in Concrete Mathematics despite only doing a couple of years of undergraduate maths many years ago. I'm looking to work through all the material whilst plugging gaps in my knowledge no matter how large they are.
However, solving recurrence relation 1.1 - The Towers of Hanoi is causing me issues.
The recurrence relation is as follows:
$T_0 = 0$
$T_n=2T_{n-1}+1$ for $n>0$
To find the closed form the book adds 1 to both sides of the equations to get:
$T_0+1=1$
$T_n+1=2T_{n-1}+2$
Then we let $U_n=T_n+1$. Now I get lost. How do I get from the line above to the second line below after the substitution of $U_n$. It seems like I'm missing something simple.
$U_0=1$
$U_n=2U_{n-1}$
Then the book goes on to say:
It doesn't take genius to discover that the solution to this recurrence is just $U_n=2^n$; hence $T_n=2^n-1$.
Again, I'm lost, how do I get to $U_n=2^n$?
A bit disheartening considering this is meant to be so easy.
Any help is appreciated. Thanks.
AI: You’re just missing a little algebra. You have $U_n=T_n+1$ for all $n\ge 0$, so $U_{n-1}=T_{n-1}+1$, and therefore $2T_{n-1}+2=2(T_{n-1}+1)=2U_{n-1}$. Combine this with $T_n+1=2T_{n-1}+2$, and $U_n=T_n+1$, and you get $U_n=2U_{n-1}$, with $U_0=1$.
Now notice that $U_n$ is just doubling each time $n$ is increased by $1$:
$$\begin{align*}
U_1&=2U_0\\
U_2&=2U_1=2^2U_0\\
U_3&=2U_2=2^3U_0\\
U_4&=2U_3=2^4U_0
\end{align*}$$
The pattern is clearly going to persist, so we conjecture that $U_n=2^nU_0$ for each $n\ge 0$. This is certainly true for $n=0$. Suppose that it’s true for some $n\ge 0$; then $$U_{n+1}=2U_n=2\cdot2^nU_0=2^{n+1}U_0\;,$$ and the conjecture follows by mathematical induction.
Now we go back and use the fact that $U_0=1$ to say that $U_n=2^n$ for each $n\ge 0$, and hence $T_n=U_n-1=2^n-1$.
In my answer to this question I solved another problem using this technique; you might find the explanation there helpful. |
H: Using euclidean algo to find d (RSA encryption)
The questions says "let p = 5, q =11, n = 55 tocient(n) = 40. e=7.
Use the Euclidean algo to find the value of d.
This is driving me crazy. Here's what I did:
40=5(7) +5
7= 1(5) + 2
5= 2(2) +1
2 = 2(1) + 0
So the GCD is 1.
Now to get 1 = de +f*tocient(n) to find d.
1 = 5 - 2(2) => 1 = 5 - 2(7-5) => 1= -2(7) + 3(5) => 1 = -2(7) +3(40 - 5(7))
=> 1 = -17(7) +3(40)
Therefore d = -17. But how can that be when in order to decode a message I have to use c^d mod n.
I don't think d can be negative but I don't know what I'm doing wrong. Any help would be great!
Regards
OSFTW
AI: Mathematically, it's fine for $d$ to be negative, but because positive powers don't require computing the inverse, you may wish to make it positive. To do so, add and subtract $7 \cdot 40$ from the rightmost side:
$$
\begin{align}
1 &= -17 \cdot 7 + 3 \cdot 40 \\
&= -17 \cdot 7 + \color{red}{40 \cdot 7} + 3 \cdot 40 - \color{red}{7 \cdot 40 }\\
1 &= \ \qquad 23 \cdot 7 \qquad+ \quad -4 \cdot 40
\end{align}
$$ |
H: Finding the moment generating function of the product of two standard normal distributions
The following question is on my homework assignment that I cannot figure out:
Let U and V be independent random variables, each having a normal distribution with mean zero and variance one. Find the moment generating function of the random variable W = UV .
I have looked around online, and cannot find an answer to this question. In fact, the only answers I can find that even relate to the product of standard normal random variables are using techniques that we never covered in my class. We covered in class how to find the MGF for linear combinations of random variables, by W isn't linear, its a product of two normals. So that technique won't work.
What am I supposed to do? I am completely at a loss. I have tried multiplying the MGFs of U and V together, but that leaves me with something ugly that I can't reduce.
AI: The MGF is, by definition, $M(t) = E[e^{tUV}]$. Try integrating
$$ \dfrac{1}{2 \pi} \int_{-\infty}^\infty \int_{-\infty}^\infty e^{tuv} e^{-u^2/2} e^{-v^2/2}\ du\ dv$$
by completing the square. |
H: How to split the difference in game theory?
There are two parties in talks to settle a law suit.
The expected value of going to court for the plaintiff is $\$3,155$
The expected value of going to court for the defendant is $-\$10,244$
The defendant could give up $3,155 + \$1$ and settle the matter - and at first blush it seems the plaintiff ought to accept.
The plaintiff could demand $10,244 - \$1$ and settle the matter - and at first blush it seems the defendant ought to accept.
Assuming every dollar is equally valued by each party - what's the equilibrium point here? How is it calculated? Is it just a matter of splitting the difference between $-3,155$ and $-10,244$?
Thank you
AI: The game description is a bit incomplete: what happens if they cannot agree on a settlement? I assume that the defendant makes one settlement offer, and if this is rejected by the plaintiff, they go to court. I assume the expected values are common knowledge, and that both parties are risk neutral (expected utility maximizers).
There are many equilibria to this game. One is:
The plaintiff rejects any offer $x<10244$, the defendant offers $x=10244$. Outcome: plaintiff receives $10244 as a settlement.
Proof: This is a sequential game (defendant offers, then plaintiff accepts/rejects). We solve backwards. The plaintiff rejects any offer $x<10244$. Anticipating this, the defendant cannot offer $x<10244$, because that leads to trial, where he expects a greater loss. On the other hand, he will not offer $x>10244$, because he knows (in equilibrium) that the plaintiff accepts 10244.$\Box$
Indeed, there is a continuum of equilibria: a Nash equilibrium is characterized by an acceptance threshold $a$ of the plaintiff (he rejects any offer that is less than $a$), and an offer $x$ by the defendant.
Any $10244\ge a\ge 3155$ and $x=a$ is a Nash equilibrium.
Proof: Exact same reasoning as above.$\Box$
Because of this multiplicity of equilibria in sequential games, Reinhard Selten proposed the subgame perfect Nash equilibrium concept. It is a stronger concept, which basically rules out empty threats, thereby reducing the set of equilibria.
The only two subgame perfect equilibria are $a=3155$ or $a=3156$ and $x=a$.
Why? Because it is an empty threat of the plaintiff to reject an offer $x>3156$. After all, the plaintiff only expects $3155$ in trial, and the defendant knows this. Given that the offer, say, $x=4000$ is made, and given that both know after rejection there will be a trial where the plaintiff expects less, the defandant just doesn't believe empty threats like above. We have two equilibria, because the plaintiff is indifferent between accepting and rejecting the offer $x=3155$. |
H: epsilon delta approach to a problem
The following is what I would like to show.
$$\lim_{x \to5}\frac{1}{x-3}=1/2$$
Given any $\epsilon>0$, $|\frac{1}{x-3}-\frac{1}{2}|\le\frac{|x-5|}{|x-3|}\le2\epsilon$
How do I get rid of $|x-3|$ here to get to
$|x-5|<\delta$ ?
My friends told me something about $\delta=\min\{{1,2\epsilon\}}$ being a part of the answer, but I did not get it.
////////////////////
EDITED////////////////////
This is how I would do it.
We are given that $$|\frac{1}{x-3}-\frac{1}{2}|<\epsilon$$ we want to have
$$|5-x|<2\epsilon|x-3|\le \delta$$.
Since $|5-x|<1$ implies $|x-3|<3$, we can choose $\delta=\min\{1,6\epsilon\}$
because if $\delta=1$ then $1<6\epsilon$ so
$$|\frac{1}{x-3}-\frac{1}{2}|=\frac{|x-5|}{2|x-3|}<\frac{1}{6}<\epsilon$$
and if $\delta=6\epsilon$ then
$$|\frac{1}{x-3}-\frac{1}{2}|=\frac{|x-5|}{2|x-3|}<\frac{6\epsilon}{6}<\epsilon$$.
Therefore $$\lim_{x \to5}\frac{1}{x-3}=1/2$$.
Does this look okay ?
I am still a bit worried because I got $6\epsilon$ instead of $2\epsilon$
AI: We start by noting that $$\left|\frac1{x-3}-\frac12\right|=\left|\frac{2}{2(x-3)}-\frac{x-3}{2(x-3)}\right|=\left|\frac{5-x}{2(x-3)}\right|=\frac12\cdot\frac{|x-5|}{|x-3|}$$
We're going to be looking for an upper bound for $|x-5|$ already, and the constant won't give us any trouble, but we need to do something about that $|x-3|.$ In order to ensure an upper bound on the far-right expression above, we need to ensure a lower bound for $|x-3|.$ Fortunately, we're letting $x\to 5,$ so that won't be a problem. In particular, when $|x-5|\le 1,$ we have $x\ge 4,$ so that $|x-3|=x-3\ge 1,$ and so $\frac1{x-3}\le 1.$ Hence, if we put $\delta=\min\{1,2\epsilon\},$ then we have for all $x$ such that $|x-5|<\delta$ that $$\left|\frac1{x-3}-\frac12\right|=\frac12\cdot\frac{|x-5|}{|x-3|}<\frac\epsilon{|x-3|}$$ since $|x-5|<\delta\le2\epsilon,$ and so since $|x-5|<\delta\le1,$ then $$\left|\frac1{x-3}-\frac12\right|<\frac\epsilon{|x-3|}\le\epsilon.$$ |
H: Cardinality of the union of disjoint set, each of which have a cardinality of $\aleph_{0}$?
Consider three sets $A$,
$B$, and
$C$
such that $A\cap B=\emptyset$
, $A\cap C=\emptyset$
, $B\cap C=\emptyset$
($A,B,C$
are pairwise disjoint) and $\overline{\overline{A}}=\overline{\overline{B}}=\overline{\overline{C}}=\aleph_{0}$
(i.e. each set has a cardinality of $\aleph_{0}$). What is the cardinality of $A\cup B\cup C$? I feel like it should be $\aleph_{2}$, but every book I have read so far on transfinite arithmetic, perhaps necessarily, makes this question look overly complicated. Thanks.
AI: It’s simply $\aleph_0$: the union of three countable sets, whether pairwise disjoint or not, is countable, and if at least one of the sets is infinite, the union is countably infinite.
Suppose, for instance, that $f_A:\Bbb N\to A$, $f_B:\Bbb N\to B$, and $f_C:\Bbb N\to C$ are bijections. Define
$$f:\Bbb N\to A\cup B\cup C:n\mapsto\begin{cases}
f_A\left(\frac{n}3\right),&\text{if }n\equiv 0\pmod 3\\\\
f_B\left(\frac{n-1}3\right),&\text{if }n\equiv 1\pmod 3\\\\
f_C\left(\frac{n-2}3\right),&\text{if }n\equiv 2\pmod 3\;;
\end{cases}$$
it’s not hard to check that $f$ is a bijection. |
H: $\forall x,y,z \in \mathbb{R} (x\cdot 0 + y \cdot 0 + z \cdot 0=0) \to (x=y=z=0)$
$\forall x,y,z \in \mathbb{R} (x\cdot 0 + y \cdot 0 + z \cdot 0=0) \to (x=y=z=0)$?? If is true, why? Thanks in advance!
AI: If $+$ and $\cdot$ mean usual addition and multiplication of real numbers, and if $0$ means the particular zero element of the reals, then the predicate $P(x,y,z)=0$ holds for any triple $(x,y,z).$ In that sense, this predicate does not mean the same as $x=y=z=0$, as the latter predicate is only true at $(0,0,0)$ and yours is true at any triple $(x,y,z).$
Edit: I should have used predicate notation for both. That is, $P(x,y,z)$ is the predicate $x\cdot 0 +y \cdot 0 + z \cdot 0=0$, and it happens that $P(x,y,z)$ holds for any choice of the variables $x,y,z$. And we can define $Q(x,y,z)$ to be the predicate $x=y=z=0$, so that only $Q(0,0,0)$ is true, while $Q$ is false for other choices like $Q(1,3,2).$ The relation is that if $Q$ holds then $P$ also holds, but not conversely. Note that the same effect as $P$ could be obtained by defining $P$ to be $x=x.$ This is also true for any triple $x,y,z.$ |
H: When does equality hold in the Minkowski's inequality $\|f+g\|_p\leq\|f\|_p+\|g\|_p$?
I would like to see a proof of when equality holds in Minkowski's inequality.
Minkowski's inequality. If $1\le p<\infty$ and $f,g\in L^p$, then $$\|f+g\|_p \le \|f\|_p + \|g\|_p.$$
The proof is quite different for when $p=1$ and when $1<p<\infty$. Could someone provide a reference? Thanks!
AI: For $p = 1$, the proof uses the triangle inequality, $\lvert f(x) + g(x)\rvert \leqslant \lvert f(x)\rvert + \lvert g(x)\rvert$, and the monotonicity of the integral. You have equality $\lVert f+g\rVert_1 = \lVert f\rVert_1 + \lVert g\rVert_1$ if and only if you have equality $\lvert f(x) + g(x)\rvert = \lvert f(x)\rvert + \lvert g(x)\rvert$ almost everywhere. That means that almost everywhere at least one of the two functions attains the value $0$, or both values "point in the same direction", that is, have the same argument. You can simply formalise that as $f(x)\cdot\overline{g(x)} \geqslant 0$ almost everywhere.
For $1 < p < \infty$, in addition to the triangle inequality, the proof of Minkowski's inequality also uses Hölder's inequality,
$$\begin{align}
\int \lvert f+g\rvert^p\,d\mu &\leqslant \int \lvert f\rvert\cdot\lvert f+g\rvert^{p-1}\,d\mu + \int \lvert g\rvert\cdot\lvert f+g\rvert^{p-1}\,d\mu\tag{1}\\
&\leqslant \lVert f\rVert_p \lVert f+g\rVert_p^{p-1} + \lVert g\rVert_p \lVert f+g\rVert_p^{p-1}.\tag{2}
\end{align}$$
You then have equality $\lVert f+g\rVert_p = \lVert f\rVert_p + \lVert g\rVert_p$ if and only equality holds in $(1)$, and for both terms in $(2)$.
Equality in $(1)$ is nearly the same as for the $p=1$ case, that gives a restriction $f(x)\cdot\overline{g(x)} \geqslant 0$ almost everywhere, except maybe on the set where $f(x)+g(x) = 0$ (but equality in Hölder's inequality forces $f(x) = 0$ and $g(x) = 0$ almost everywhere on that set, so at the end we really must have $f(x)\cdot \overline{g(x)} \geqslant 0$ almost everywhere if $\lVert f+g\rVert_p = \lVert f\rVert_p + \lVert g\rVert_p$). For equality of the terms in $(2)$, if $f+g = 0$ almost everywhere, we must have $f=g=0$ almost everywhere, and if $\lVert f+g\rVert_p > 0$, we have equality if and only there are constants $\alpha,\beta \geqslant 0$ with $\lvert f\rvert^p = \alpha \lvert f+g\rvert^p$, and $\lvert g\rvert^p = \beta\lvert f+g\rvert^p$ almost everywhere. We can of course take $p$-th roots, and together with the condition imposed by the triangle inequality, we obtain that
$$\lVert f+g\rVert_p = \lVert f\rVert_p + \lVert g\rVert_p$$
holds if and only if there are non-negative real constants $\alpha,\beta$, not both zero, such that $\alpha f(x) = \beta g(x)$ almost everywhere. |
H: Problem concerning Numerical Solutions of Nonlinear Systems of Equations (Burden and Faires)
I need help with this particular question:
The nonlinear system
$3x_1 - \cos (x_2 x_3) - \frac{1}{2} = 0$
$x{_1}^2 - 625x{_2}^2 - \frac{1}{4}=0$
$\exp ^{-x_1x_2} + 20x_3 + \frac{10\Pi -3}{3}=0$
has a singular Jacobian matrix at the solution. Apply Newton's method with x$^{(0)} = (1,1,-1)^t$. Note that convergence may be slow or may not occur within a reasonable number of iterations.
My attempt on Maple:
What am I doing wrong here?
Solving it by-hand is a pain.
AI: In your NewtonMV procedure, try changing,
A := subs(s,f);
to,
A := eval(J, [s]);
or to,
A := subs(s, J);
Also, change the call LinearSolve(...) to :-LinearAlgebra:-LinearSolve(...) so that it gets the right export from the right top package, regardless of whether you've loaded any subpackage of Student.
Personally, I'd rather see evalf wrapped around each of the two arguments to LinearSolve than around its call, if only so that the very first iteration doesn't attempt an exact subcomputation.
So, something like,
dx := :-LinearAlgebra:-LinearSolve( evalf(A), evalf(-fx) );
Please note that you can inline code into your Questions and Answers here see the menubar above the editing region. It's not so helpful to paste images of code.
[edited, after all the typos have been resolved]
You might want to consider additional stopping criteria. Apart from a maximal number of iterations to allow, you could stop when either the increment dx got small enough, or when the evaluation of f got close enough to zero (ie. x is considered a root).
restart:
NewtonMV:=proc(f,v,x0,n,xtol,ftol)
local i,result,J,x,fx,A,dx,s;
J:=VectorCalculus:-Jacobian(f,v);
x:=x0;
dx:=Vector([infinity,infinity,infinity]);
fx:=Vector([infinity,infinity,infinity]);
result:=convert(x0,list);
for i from 1 to n
while :-LinearAlgebra:-Norm(dx)>xtol
and :-LinearAlgebra:-Norm(fx)>ftol do
s:=[seq(v[i]=x[i],i=1..nops(v))];
fx:=evalf(eval(f,s));
A:=evalf(eval(J,s));
dx:=:-LinearAlgebra:-LinearSolve(evalf(A),evalf(-fx));
if indets(dx,name)<>{} then
dx:=subs([seq(p=0,p=indets(dx,name))],dx);
end if;
x:=x+dx;
result:=result,convert(x,list);
end do;
printf("\nnumber of iterations: %a\n\n",i-1);
printf("final increment dx, and its norm:\n %a , %a\n\n",
convert(dx,list), :-LinearAlgebra:-Norm(dx));
printf("substitution of solution x into f, and its norm:\n %a , %a\n\n",
convert(eval(f,s),list), :-LinearAlgebra:-Norm(fx));
return [result];
end proc:
lhs1:=3*x1-cos(x2*x3)-1/2;
lhs2:=x1^2-625*x2^2-1/4;
lhs3:=exp(-x1*x2)+20*x3+(10*Pi-3)/3;
NewtonMV(Vector([lhs1,lhs2,lhs3]),[x1,x2,x3],
Vector([1,1,-1]), 400, 1e-5, 1e-7);
That is not intended to be the epitome of great coding, but it might give you some ideas. |
H: Question about automorphisms
Let $G$ be a finite abelian group and let n be a positive integer relatively prime to $|G|$.
a. Show that the mapping $ϕ(x)=x^n$ is an automorphism of $G$.
b. Show that every $x ∈ G$ has an $n^{th}$ root, i.e., for every $x$ there exists some $y∈G$ such that $y^n=x$.
a.To show that the mapping $ϕ(x)=x^n$ is an automorphism we need to show that $ϕ$ is a homomorphism, and $ϕ$ is one-to-one and onto. Let $ϕ:G->G$ be given by $ϕ(x)=x^n$ and let $x,y∈G$.
First we need show $ϕ$ is a homomorphism, that is $ϕ(xy)=ϕ(x)ϕ(y)$. So, $ϕ(xy)=(xy)^n = (xy)(xy)...(xy)=(xn)(yn)=ϕ(x)ϕ(y)$, since the group is abelian, so $ϕ$ is a homomorphism.
Next, suppose $ϕ(x)=ϕ(y)$, then $e=ϕ(x)ϕ(y^{-1})=ϕ(xy^{-1})=(xy^{-1})^n$, here we see that $ord(xy^{-1})$ divides $n$. Since $xy^{-1}∈G$, it’s order must divide $|G|$(Theorem 10.4), but $n$ and $|G|$ are relatively prime so, $ord(xy^{-1})=1$. Therefore $xy^{-1}=e$ and thus $x=y$. This shows that $ϕ$ is one-to-one. Since $ϕ$ is one-to-one and $G$ is finite, then this implies that $ϕ$ is also onto.
b. I don't know how to word the correlation between part a and part b.
AI: So, my attempt at a proof is: $\varphi(xy)=(xy)^n=x^n*y^n=\varphi(x)\varphi(y)$, so $\varphi$ is a homomorphism.
Good. (Note I've changed your formatting for readability. So these aren't quite direct quotes.)
If $\varphi(a)=\varphi(b)$ then $a^n=b^n \implies a=b$, so $\varphi$ is 1-1
How do you know $a^n = b^n \implies a=b$? This is false for an arbitrary finite abelian group $G$ and an arbitrary $n$. For example, $1^2 = (-1)^2$ is true, but $1 \neq -1$ in the group $\{\pm1\}$ (or in any group of $2N$-th roots of unity. You need to use the fact that $n$ and $|G|$ are relatively prime.
if $y\in G$ then $\varphi(y^-1)=(y^-1)^n=(y^n)^{-1} \in G$ [Check?]
This is unclear, and I think that's partially because of a typo. (Did you mean $\varphi^{-1}(y)$, instead of $\varphi(y^{-1})$?) Basically, you need to show that every element in $G$ is an $n$-th root, i.e., for each $y \in G$ you need to find $x \in G$ such that $x^n = y$. Why can you do this? Again, it depends on the fact that $n$ and $|G|$ are relatively prime. Try to figure out how.
and for (b), I'm not sure how to show that $x\in G$ has an $n$-th root? I think this is where the $n$ being relatively prime to $|G|$ comes into play. That means that $an\neq |G|$ for any integer $a\neq 1$.
Another way to think about this is that two integers $a$ and $b$ are relatively prime (a.k.a. coprime) if their GCD is $1$.
Now how do we see that for every $x$ there exists some $y\in G$ such that $y^n=x$.
As I explained above, this follows from what you proved in part (a).
Edit: Your new proof looks good. The reason you're not seeing part (b) from your proof of (a) is because you proved indirectly that $\varphi$ is a surjection. So for (b), think about what what "$\varphi$ is a surjection" means in terms of the elements of $G$. For all $y \in G$, surjectivity of $\varphi$ implies there exists $x \in G$ with $\varphi(x) = y$. In other words, for every $y \in G$, there exists $x \in G$ such that
$$
y = x^n,
$$
which (by definition) says that $y$ is an $n$-th root of $x$. |
H: How to show integral is related to complementary error function?
I have to somehow show that:
$$
\int_0^\infty{\frac{e^{-a t}}{t^{1/2}(t+x)}} \textrm{d}t=\frac{\pi}{\sqrt{x}} e^{a x} \textrm{erfc}(\sqrt{ax})
$$
I've tried substituting $u=\sqrt{t}$ to get
$$
2 \int_0^\infty{\frac{e^{-a u^2}}{u^2+x}} \textrm{d}u
$$
which at least has the Gaussian but I don't know where to go from here.
I've also tried looking at the problem backwards but I don't seem to be able to get the polynomial in the denominator.
What's the approach for these problems involving the error function?
Thank you!
AI: Consider the integral
$$\int_0^{\infty} du \frac{e^{-a u^2}}{u^2+x} = e^{a x} \int_0^{\infty} du \frac{e^{-a (u^2+x)}}{u^2+x} = e^{a x} J(a)$$
Then
$$J'(a) = \int_0^{\infty} du e^{-a (u^2+x)} = \sqrt{\pi} e^{-a x} a^{-1/2}$$
Integrating
$$J(a) = \sqrt{\pi} \int_0^a da' \, a'^{-1/2}\, e^{-a' x} = \sqrt{\frac{\pi}{x}}\int_0^{\sqrt{a x}} dt\,e^{-t^2} = \frac{\pi}{2 \sqrt{x}} \operatorname*{erf}{\left ( \sqrt{ax}\right)}$$ |
H: Proving a function is bounded given that the function is uniform continuous over a bounded, closed domain
Let's say I have a function $f : I \rightarrow \mathbb{R}$ that is uniformly continuous, and $I$ is a colsed and bounded interval. Then $f(I)$ is closed by the following:
Assume $f(I)$ is not closed, and that its supremum is not in $\{f(I)\}$. The proof is similar for the assumption that the infimum is not in $\{f(I)\}$.
Then $\exists\ x_n \subseteq I$ such that $x_n\rightarrow x_0 \in I$, but $f(x_n)\rightarrow f(x_0) = \sup\{ f(I) \}$ that is not in $\{f(I)\}$
However because $I$ is closed and bounded, $x_0 \in I$
Therefore $f(x_0) \in \{f(I)\}$
So $\sup\{f(x_n) \} \subseteq \{ f(I)\} $ which contradicts the assumption
So $\{ f(I) \}$ is closed.
I'm just wondering whether this is proof has enough rigor, or whether I've missed something, as I feel my proofs need a little work.
Thanks!
AI: This is an entirely different approach than you took, but you could do something like the following:
If $f:I \to \mathbb{R}$, where $I \subset \mathbb{R}$ is compact, is uniformly continuous on $I$, then the Extreme Value Theorem shows that $f$ attains its minimum and maximum values on $I$. It then directly follows that $f$ is bounded.
Note: If you want to get technical you could let $m_1,m_2$ be the minimum and maximum value respectively. Then let $M=max(|m_1|,|m_2|)+1$. Then conclude that $|f(x)|<M ~\forall x \in I$. |
H: Why are these relations not posets?
I was hoping you guys could help me clarify why these relations are or arent posets. I gave my thought process that resulted in the wrong answer.
a)(Z, =) poset
b)(Z, !=) not a poset
c) (R,==) poset
d) (R, <) not a poset
A) I thought it wouldn't be a poset, but it is. I had assumed it wasn't because no integer would relate to anything but itself. It would satisfy reflexive but thats it not asymmetric or transitive right? Im not sure how it is a poset if anyone can clarify
B) I thought it was a poset since any integer could relate to any integer, but the answer was that it was a poset.
C)Similar to A
D) I thought it was a poset since it could be drawn similarly to <= posets. On the sheet it was graded that it wasnt reflexive, if anyone can explain this a little bit more clearly.
AI: The relation of equality on any set is a partial order: it is reflexive, antisymmetric, and transitive. Transitivity, for instance, requires that if $x=y$ and $y=z$, then $x=z$, and that’s certainly true. In fact you should recognize that equality is an equivalence relation and is therefore reflexive and transitive; showing that it’s a partial order then requires only verifying that it’s antisymmetric. This is easy: is it true that if $x=y$ and $y=x$, then $x=y$? Yes, of course. It’s not asymmetric, and it’s not supposed to be asymmetric.
$\langle\Bbb Z\ne\rangle$ is clearly not reflexive: it’s never the case that $n\ne n$. Right away this tells you that it’s not a partially ordered set. In fact the relation $\ne$ fails on all three counts: it’s also not antisymmetric and not transitive. For the former, note that $1\ne 2$ and $2\ne 1$, so that antisymmetry would require that $1=2$, which is obviously not the case. For the latter, note that $1\ne 2$ and $2\ne 1$, so that transitivity would require that $1\ne 1$, which is also obviously not the case.
If the double equals sign is just equality again, as in (a) but on $\Bbb R$ instead of on $\Bbb Z$, then everything that I said about (a) applies equally well to (c).
In (d) just check the definition: if $x\in\Bbb R$, is it true that $x<x$? Definitely not, so $<$ is not a reflexive relation. (It’s vacuously antisymmetric, since it’s never the case that $x<y$ and $y<x$, and it’s transitive.) |
H: How to prove that if n and k are integers with 1 ≤ k ≤ n, then k*(n C k)=n(n−1 C k−1) combinatorally?
I am having with combinatorial proofs. My professor says to come up with a scenario so that we can connect both sides by double counting but I am clueless.
AI: There is a group of $n$ people. We want to give $k$ people in the group medals, one of them gold, and the others plastic. How many ways can we do this? Here are two ways of counting:
(i) We can choose the $k$ medal winners, and then from these choose the one who will get gold.
(ii) Or else we can choose the gold medal winner, and then the people who will get plastic. |
H: Prove that $\left | \cos x - \cos y \right | \leq \left | x - y \right |, \forall x,y \in \mathbb{R}.$
This is one of the problem in my text book where the section which the problem is stated talks about the mean value theorem and Rolle's theorem. By looking at this, I have no idea where to start. Can I have some hints??
AI: $\cos(x)$ is differentiable in $\mathbb{R}$.
By Mean value theorem $|\cos(x) - \cos(y)| = |sin(\xi)||x - y|$ where $\xi \in (x,y)$ or $(y,x)$.
$|\sin(x)| < 1$ $\forall$ $x \in \mathbb{R}$.
Now get your answer. |
H: Known probability for one month, what is the probability for 100 months?
This is a pretty famous probability problem:
The probability that a driver will have an accident in 1 month equals 0.02. Find the probability that in 100 months, he will have 3 accidents (Papoulis, 1984).
AI: the probabilities of outcomes in n Bernoulli trials with the probabilities $p$ (hit) and $q$ (miss)for a single trial are given by the appropriate term in the binomial expansion of:
$$ (p+q)^n $$
e.g. with 3 out of a hundred you want the term $\binom {100} {3} p^3 q^{97}$
the binomial coefficient represents the number of ways the 3 hits can be selected from the 100 trials |
H: Correctly representing a $2^n < n!$ statement
$$2^n < n!$$ After an inductive proof I determined that $2^n < n!$ is valid only for values greater than or equal to $4$.
So. How do I represent this conclusion?
Is this correct?
$$\forall n \in \mathbb{Z}^+, n \geq 4 \implies 2^n < n!$$
AI: Yes is correct! But better write in this way
$$\forall n\in\mathbb{N}\left(n\geq 4\Rightarrow 2^n<n!\right)$$ |
H: Application of Sylow's theorem to direct products
Question: Let $p $ be a prime number. Suppose that $|A|=N $ and $|B|=M $, and let $p^n $ be the largest power of $p $ that divides $N $ and $p^m$ is the largest power that divides $M $. Consider the group $A \times B$. Let $P$ be a p-Sylow subgroup of A and Q be a p-Sylow subgroup of B.
(a) What is the order of $A \times B$?
(b) What is the largest power of p that divides the order of $A \times B$?
(c) Why is $P \times Q$ a p-Sylow subgroup of $A \times B$?
(d) Show that every p-Sylow subgroup of $A\times B$ is a Cartesian products of a p-Sylow subgroup of $A$ and $B$.
Attempt at a solution:
(a) What is the order of $A \times B$?
If $p^n |N $ then $N=p^n l $. If $p^m |M $ then $M=p^m k $. So $|A \times B|=p^{m+n} kl $.
(b) What is the largest power of $p$ that divides the order of $A \times B$?
$p^{m+n} $
(c) Why is $P \times Q$ a p-Sylow subgroup of $A \times B$?
This is because $P$ is a subgroup of $A$ and $Q$ is a subgroup of $B$ such that $|P|=p^n$ and $|Q|=p^m$. Hence $P \times Q$ is a $p$-Sylow subgroup because $|P \times Q|=p^{n+m}$ and $P \times Q \leq A \times B$.
(d) This one I'm a bit stuck on.
AI: For (d), consider $\iota_A : A\hookrightarrow A\times B$, and $\pi_B : A\times B\to B$ to be the natural maps. If $H < A\times B$ is a $p$-Sylow subgroup, then $|H| = p^{m+n}$, so $H_1:= H\cap \iota_A(A) < \iota_A(A)$ is a p-group and $|H_1| \mid p^n$ since $|\iota_A(A)| = N$. Furthermore,
$$
H/H_1\cong \pi_B(H) \text{ and } |\pi_B(H)| \mid p^m
$$
Hence it must happen that $|H_1| = p^n$. Now $\iota_A : A\to \iota_A(A)$ is an isomorphism, so choose a subgroup $P < A$ such that $\iota_A(P) = H_1$.
Do the same for $B$ to obtain a subgroup $Q<B$ such that $\iota_B(Q) = H\cap\iota_B(B)$.
Now check that $H = P\times Q$ |
H: smooth lie group action
Let $\theta:G\times M\to M$ be a smooth left action of a Lie group $G$ on the manifold $M$.
Suppose $G$ is compact and $M$ is Hausdorff. Let $K$ be a compact set in $M$. Is it true that $G_K:=\{g\in G:(g\cdot K)\cap K\neq \emptyset \}$ is closed?
AI: If $y_n \in G_K$ and $y_n \to y \in G$. Let $k_n, p_n \in K$ such that
$$y_n p_n = k_n$$
That can be found as $y_n \in G_K$. As $K$ is compact, by passing to subsequences we can assume that $k_n \to k$ and $p_n \to p$, where $k, p\in K$. Let $n\to \infty$, then $yp = k$ and $y\in G_K$. |
H: Evaluation of $\int_\Gamma e^{-z^2}\ dz$
My question is simple. Prove the following equality
$$\int_\Gamma e^{-z^2}\ dz = \int_{-\infty}^\infty e^{-x^2}\ dx$$
where $\Gamma = \{ z\in {\bf \mathbb C}| \ {\rm Im}\ (z) = c \}$
AI: Consider
$$\oint_C dz \, e^{-z^2}$$
where $C$ is a rectangle having vertices $-R$, $R$, $R+i c$, $-R+i c$. By Cauchy's Theorem, this integral is zero. On the other hand, it is also equal to
$$\int_{-R}^R dx \, e^{-x^2} + i \int_0^c dy \, e^{-(R+i y)^2} -\int_{\Gamma} dz \, e^{-z^2} -i \int_0^c dy \, e^{-(-R + i y)^2} $$
As $R\to\infty$, the 2nd and 4th integrals vanish because each integral is bounded by the value
$$e^{-R^2} \int_0^c dy \, e^{y^2} \le c e^{-(R^2-c^2)}$$
Thus, we are left with the equality
$$\int_{-\infty}^{\infty} dx \, e^{-x^2} = \int_{\Gamma} dz \, e^{-z^2}$$
as was to be shown. |
H: Generalized distributive laws proof feedback
I'm currently learning proofs and elementary set theory. I would like to have feedback on my proof since I'm self-studying. Are some part superfluous or unclear? My proof goes as follows:
I will prove that $(\bigcap_i X_i)-A = \bigcap_i(X_i - A)$, where "$-$" is the set difference:
Suppose that $x \in (\bigcap X_i) - A$, then $x \in \bigcap X_i$ but $x \notin A$. Therefore $x \in X_i$ for all $i$, and $x \notin A$, which means that for all $i$, $x \in X_i$ and $x \notin A$. Thus for all $i$ then $x \in X_i - A$, and as such $x \in \bigcap_i (X_i - A)$.
I then proceed in the same way to show that $x \in \bigcap_i (X_i - A) \Longrightarrow x \in (\bigcap X_i) - A$.
Since all elements in $(\bigcap_i X_i)-A$ are also in $\bigcap_i(X_i - A)$ and vice versa, then $(\bigcap_i X_i)-A = \bigcap_i(X_i - A)$.
In (1.), I say that $x \notin A$ and $(\forall i) [x \in X_i]$ is equivalent to $(\forall i) [x \notin A$ and $x \in X_i]$. Is it ok to make such a step or should it be proved too? And if so, how could it be proved?
Thank you for your help!
AI: Your proof is correct and well-written, so good job.
As for your second question, it is certainly okay to use this fact without proof here. The fact that you can move the $x\in A$ inside the scope of the universal quantifier without changing the truth value is a special case of the fact from first-order logic that if $\phi$ and $\psi$ are formulas such that $x$ is not free in $\phi$, then $\forall x (\phi \wedge \psi)$ is equivalent to $\phi \wedge \forall x(\psi)$. Proving that this holds in general is not too hard, but requires a bit of formal-logic machinery to do precisely. Don't worry too much about proving facts like that (as long as they are true, of course!). |
H: Group of order 35, and normal subgroup of order 7
G a group of order 35, H a normal subgroup of order 7. Prove that if g in G has order 7, then g is in H.
What I'm doing is invoking Sylow's third theorem to show that there exist a single subgroup of order 5 and a single subgroup of order 7. And then the direct product of these has order 35, and then somehow... this means G has an element of order 35, which would mean G is cyclic, and if G is cyclic, each of its subgroups are cyclic. So H has an element of order 7 and if there's an element g in G of order 7, then it generates the subgroup H?
This is my train of logic? I could use some help on the somehow... part, or if that's even a valid way of proving that a group of order 35 is cyclic. Please help steer me back on track.
AI: Since $|G|=35=7\cdot 5$, a subgroup $H$ of order $7$ is a Sylow-$7$-subgroup. Moreover, given that $H$ is normal, it must be the only Sylow-$7$-subgroup of $G$. Now, if $g\in G$ has order $7$, then $\langle g\rangle $ is a $7$-subgroup of $G$ and hence is contained in a Sylow-$7$-subgroup of $G$. But $H$ is the only Sylow-$7$-subgroup, so $\langle g\rangle\subseteq H$. Therefore, $g\in H$. |
H: Can we determine which statements are incomplete due to Godel?
Due to Godel's incompleteness theorems we know that there are true statements in a system that cannot be proven with that system. My questions are 1) can we tell which statements in a system are the ones that cannot be proven, and 2) can we choose through the design of the system which statements will not be provable?
The motivation for my questions is simply the observation of how many results depend on the Riemann conjecture and how important to mathematics it is, along with the fact that it is almost universally believed to be true, yet it resists proof. Similar things can be said about the twin prime conjecture and many other unproven conjectures. Therefore, I ask the questions: is is possible that the Riemann conjecture (or twin prime conjecture) is one of those statements that is true but can never be proven? And can we know whether it is one of those questions?
AI: If first order Peano arithmetic is consistent, there is no algorithm that will determine, given input $\varphi$, whether or not $\varphi$ is neither provable nor refutable from first-order PA.
At our current state of knowledge, it is possible that the twin prime conjecture is true but not provable in first order PA. |
H: Proving a set is countably infinite $\left \{q \in \Bbb Q:q=\dfrac{a}{b}\ \text{where $a$ is even and $b$ is odd} \right \}$
Proving a set is countably infinite
$$\left \{q \in \Bbb Q:q=\dfrac{a}{b}\ \text{where $a$ is even and $b$ is odd} \right \}$$
I am not sure how to go about solving this problem. I know it has something to do with Cantor Diagonalization, but I am not sure how to exclude the items that do not meet the criteria of the set.
AI: First the set is countable because is a subset of $\mathbb{Q}$. Now is infinite because the set $A=\{\frac{2n}{2n+1}:n\in\mathbb{N}\}$ is infinite. This follows from the fact that there is a bijection between $\mathbb{N}$ and $A$ ($\mathbb{N}$ is infinite). And from the fact that $A$ is a subset of your set. |
H: When proving a statement by induction, how do we know which case is the valid 'base'?
For example proving 2^n < n!, 4 is the 'base' that works for this exercise, then starting from there we prove p + 1 considering p has to be at least 4 and we have our result. However, I believe determining the first valid value without a previous proof is kind of odd.
The question is, how?
I don't want to upload my homework with assumptions justified like "it's obvious just try using 0,1,2,3 and see that it doesn't work".
AI: The base of induction is the lowest element satisfying the criteria that you are proving, the least element in the set of naturals for which your property is assumed to hold. Perhaps it would be useful to give you a more beefy idea of induction aside from the usual "Base, Hypothesis, Inductive Step" method you were most likely taught.
For induction, we utilize a set of axioms for $\mathbb{N}$. Among these axioms are: $0 \in \mathbb{N}$, $\forall \text{ nonzero } n \in \mathbb{N} \text{ there exists a successor } s(n) \text{ of the form } n+1$, and that $0$ is no number's successor.
Now, when we prove something by induction, what we are doing is assuming that there is some set $S \subseteq \mathbb{N}$ of natural numbers satisfying some property, showing that the set is non-empty (showing for a base), and showing that if some arbitrary natural is in the set, its successor is also in the set. If we manage to do this, we effectively show that:
$S = \mathbb{N}- \text{the elements not in our problem domain}.$
You may comfortably intuit why this is true! If we have shown that for any element in $S$ that its successor is also in $S$, then we have shown that for our base $b \in S$: $s(b) \in S, s(s(b)) \in S, \text{ and so on.}$ Think of it like the domino effect knocking its way down the naturals.
Now we shall consider your example $2^n < n!, \forall n \ge 4$. Remember that our base is the least element of the problem domain $n \ge 4$, hence 4. So, let $S = \{ n \in \mathbb{N} | 2^n < n!\}$. To show that $S$ is not empty, we explicitly show it for our base: $2^4 = 16 < 4! = 24$ hence $4 \in S$. Now we will assume that there is some $k \in S$ meaning that $k$ satisfies our desired property $2^k < k!$. To show that $s(k) \in S$ we do the usual manipulations of induction, which I'm sure you've already done for this problem, and show $2^{k+1} < (k+1)!$ and hence $(k+1) \in S$.
SUMMARY
I really hope this helps with induction. If it confused things, then ditch the ramble and just know that the basis is the least element of the given problem domain, which is usually given to you (in this case $\forall p \ge 4$). |
H: Proving $n^n \ge n!$, is induction necesary?
I am trying to prove that:
$n^n \geq n!$ is valid for a $x$ set of numbers.
So, I am trying an inductive process.
However, case $P(0)$ doesn't seem to work because I have read somewhere that $0^0$ is undefined, but I've also read that $0^0$ is $1$.
What approach do you recommend?
AI: $0^0$ is most definitely not undefined and $0^0 = 1$ as you've read. If you want a more thorough explanation this seems like a nice article.
EDIT: As pointed out in the comments by T. Bongers, my first sentence is misleading. Outside of the world of analysis, you are fairly safe in holding with the definition. As highlighted in the article there are very good reasons by we define $0^0 = 1$ in our usual context (here we are actually considering $f(x)^{g(x)}$ as $\lim_{x \to 0^+}{f(x)}$ and $\lim_{x \to 0^+}{g(x)}$). It is perhaps, a simplification, but a thoroughly discussed and justified simplification for our everyday application.
Induction is a simple method of proof:
Basis: $P(0) = 0^0 = 1 \ge 0!$.
Assume $P(k)$ holds or that $k^k \ge k!$. So, for $k+1$:
We know $(k+1)^{(k+1)} = (k+1)^k(k+1)$ and that $(k+1)! = k!(k+1)$.
So, $(k+1)^k \ge k^k \ge k!$ and so we multiply through by $(k+1)$, so $(k+1)^k(k+1) \ge k^k(k+1) \ge k!(k+1)$ hence $(k+1)^k(k+1) \ge k!(k+1)$. Using our earlier equalities, however, this is the same as $(k+1)^{(k+1)} \ge (k+1)!$. $\Box$ |
H: Does this sequence converge or diverge? $a_n = (-1)^n\frac{n^2+n+2}{2n^2+3n+4}$
We had this question on an exam and I don't know if I got it right or not. We were asked to justify our answer. I said the sequence diverges since the limit bounces between $-\frac{1}{2}$ and $\frac{1}{2}$ but now I think I got it wrong.
Again, here is the sequence.
$a_n = (-1)^n\frac{n^2+n+2}{2n^2+3n+4}$
AI: Well, $$\lim_{n \to \infty} \frac{n^2 + n + 2}{2n^2 + 3n + 4} = \lim_{n \to \infty} \frac{1 + \frac 1 n + \frac 2 {n^2}}{2 + \frac 3 n + \frac 4 {n^2}} = \frac 1 2 \ne 0$$
So $\lim_{n \to \infty} a_n$ does not exist, and in particular is not zero. |
H: How do I identify repeated irreducible factors?
I thought my solution was correct - but it seems like that's not the case. Can anyone possibly explain to me why I'm wrong?
AI: It's clear that $x^2 + 1$ is an irreducible quadratic. On the other hand, using the fact that $x^2 - 1 = (x - 1)(x + 1)$, we can rewrite the expression as
$$(x - 1)^2 (x + 1)^2 (x^2 + 1)^2$$
How many linear terms are there now? |
H: Proving discontinuity at $x=0$
I would like to prove that the function below is discontinuous at $x=0$. $$f(x)=\begin{cases}\sin\left(\frac{1}{x}\right) & \text{for } x\neq 0\\
0 & \text{for } x=0\end{cases}$$
What I have so far is this: (not much)
$$\left|\sin\left(\frac{1}{x}\right)-0\right|=\left|\sin\left(\frac{1}{x}\right)\right|$$
I'm not sure how to proceed. I think I need to think of some sort of bound that is helpful. I know that $|\sin(\cdot)|\leq1$, but I'm not sure how to use this fact properly. A hint would be appreciated.
AI: Recall that if a function is continuous at the origin, it is also sequentially continuous.
Now consider the following sequence: $x_n = \dfrac1{n \pi + \dfrac{\pi}2}$ and conclude what you want. |
H: Is $L^\infty(X, \mu)$ dual of $L^1(X, \mu)$ if X is discrete
Is $L^\infty(X, \mu)$ dual of $L^1(X, \mu)$ if X is discrete and $\mu$ is not finite?
Like, if X = {x,y} and $\mu({x}) = 1$ and $\mu({y}) = \infty $, then $L^\infty(X, \mu)$ is not dual of $L^1(X, \mu)$ since, Let $f(x) = 1$ and $f(y) = 0$ And I can define a linear operator to $\mathbb R$ by $ x*f(x) + y*f(y)$ but there is no element in $L^\infty(X, \mu)$ which can represent this.
Am I right?
AI: For your $X$ and $\mu$, the problem is not that $L^\infty(X,\mu)$ is too small, but rather that it is too big.
Because $L^\infty(X,\mu)=\mathbb C\oplus\mathbb C$ regardless of what $\mu$ is. While $L^1(X,\mu)=\mathbb C\oplus 0$. As the dual of a finite-dimensional vector space has its same dimension, $L^\infty(X,\mu)$ is too big to be the dual of $L^1(X,\mu)$ (it does contain the dual, though). |
H: Dirac Delta identity
Reading through the proof on the Helmholtz decomposition of a vector field, I came across the following identity:
$$\delta(x-x')=-1/4\pi*\nabla^2*(|x-x'|)^{-1}$$
Does anyone have any insight on how to prove/derive this identity? Thanks in advance. Here's the article for reference:
http://faculty.uml.edu/cbaird/95.657(2013)/Helmholtz_Decomposition.pdf
AI: There are a couple of ways to derive this, Ill sketch out what I think is the most intuitive.
If you write out the algebra for $\nabla \cdot \nabla (1/\vert \vec{x}-\vec{x}'\vert) $ you will find formally that it is zero. However the algebraic manipulations you perform are only valid in a neighborhood where the functions is continuous and differentiable. There is one glaring place where that isn't true and that is when $\vec{x}=\vec{x}'$.
The question is how can we define the divergence of the gradient at a point where it technically doesn't exist. The key is by thinking of the divergence of a vector field at a point as the flux per unit volume of the vector field at the given point. This is motivated by Gauss' Theorem,
$$ \int_{\partial V} \vec{A} \cdot d\vec{a} = \int_V \nabla \cdot V dV$$
So to find the divergence at $\vec{x}=\vec{x}'$ we calculate the flux of the gradient over the surface of a sphere centered at $\vec{x}'$ and then evaluate the limit of the flux as the radius of the sphere goes to $0$.
$$ Flux = \int_S \nabla \frac{1}{\vert \vec{x} - \vec{x}'\vert } \cdot \hat{n} da = -\int_S \frac{\vec{x} - \vec{x}'}{\vert \vec{x} - \vec{x}' \vert^3} \cdot \frac{\vec{x} - \vec{x}'}{\vert \vec{x} - \vec{x}' \vert } da \int_S \frac{\vec{R}}{R^3} \cdot \frac{\vec{R}}{R} R^2 d\Omega = \int_S d\Omega = -4\pi$$
So the flux per unit volume is $\lim_{R\rightarrow 0 } -4\pi/V = \infty$. We have that the lapacian of our function is zero everywhere $ \vec{x} \neq \vec{x}'$ and is infinite at $\vec{x}=\vec{x'}$ but integrates to a finite value. This is clearly a delta function so we conclude that,
$$ \nabla^2 \frac{1}{\vert \vec{x} - \vec{x}' \vert} = -4\pi \delta^{(3)}(\vec{x}-\vec{x}') $$
Thought I would add an explanation about what a delta function is:
A Dirac delta function can be thought of as a normalized distribution which is nonzero only at one point.
So we write,
$$ \delta^{(3)}(\vec{x}) = \begin{cases} 0 \quad \text{when } \vec{x} \neq 0 \\ \infty \quad \text{when } \vec{x} = 0 \end{cases} $$
with the understanding that
$$ \int_{V} \delta^{(3)}(\vec{x}) dV = \begin{cases} 1 \quad \text{if } 0 \in V \\ 0 \quad \text{if } 0 \notin V \end{cases}$$
You can construct such a "function" by taking the limit of some appropriate distribution. In 1 dimension we could write,
$$ \int \delta(x) f(x) dx = \lim_{n \rightarrow \infty} \int \delta_n(x) f(x) dx; \quad \text{where} \quad \delta_n(x) = \sqrt{\frac{n}{\pi}} e^{-n x^2}$$
Notice the importance of keeping the limit outside of the integral otherwise we wouldn't have something we could integrate. |
H: Good Textbook on Topology
I have one year calculus and one year linear algebra background. In addition, I have had one semester study in metric space analysis. Can anyone suggest some good textbooks on topology, please? A reader-friendly text with plenty of examples would be ideal. Thank you!
AI: I strongly recommend Munkres' Topology. It covers a lot of general topology in a very clear manner with quite a few examples, as well as covering some algebraic topology later in the book. |
H: How many ways are there for $2$ teams to win a best of $7$ series?
Case $1$: $4$ games: Team A wins first $4$ games, team B wins none = $\binom{4}{4}\binom{4}{0}$
Case $2$: $5$ games: Team A wins $4$ games, team B wins one = $\binom{5}{4}\binom{5}{1}-1$...minus $1$ for the possibility of team A winning the first four.
Case $3$: $6$ games: Team A wins $4$ games, team B wins $2$ = $\binom{6}{4}\binom{6}{2}-2$...minus $2$ for the possibility of team A winning the first four games; and the middle four (games $2,3,4,5$), in which case there would be no game $6$.
Case $4$: $7$ games: Team A wins $4$ games, team B wins $3$ = $\binom{7}{4}\binom{7}{3}-3$...minus $3$ for the possibility of team A winning the first four games; games $2,3,4,5$; and games $3,4,5,6$.
Total = sum of the $4$ cases multiplied by $2$ since the question is asking for $2$ teams.
Is this correct?
AI: We count the ways in which Team A can win the series, and double the result. To count the ways A can win the series, we make a list like yours.
A wins in $4$: There is $1$ way this can happen.
A wins in $5$: A has to win $3$ of the first $4$, and then win. There are $\binom{4}{3}$ ways this can happen.
A wins in $6$: A has to win $3$ of the first $5$, then win. There are $\binom{5}{3}$ ways this can happen.
A wins in $7$: A has to win $3$ of the first $6$, then win. There are $\binom{6}{3}$ ways this can happen. |
H: Optimizing bounds?
Pretty much trying to optimize the bounds of a function (or window when using a graphing calculator).
So the function is $=e^{-x}\sin(x)$, with $x\geq0$.
When taking the derivative I get $-\sin(x)+\cos(x)$. Setting it equal to $0$, I get $x=\pi/4$ and $x=5\pi/4$.
Graphing it on my calculator, it appears that the $y$ doesn't go more than about $.3$ when $x > 0$.
This isn't close to either of my critical points.
Any ideas?
AI: If $x = \pi/4$, then $e^{-x} \sin(x) \approx 0.322396942$, which agrees with your graphing calculator. |
H: Why, while checking consistency in $3\times3$ matrix with unknowns, I check only last row?
I would like to know whether my thinking is right. So, having 3 linear equations,
$$
\begin{align}
x_1 + x_2 + 2x_3 & = b_1 \\
x_1 + x_3 & = b_2 \\
2x_1 + x_2 + 3x_3 & = b_3
\end{align}
$$
I build $3\times 3$ matrix
\begin{bmatrix}1&1&2&b_1\\1&0&1&b_{2}\\2&1&3&b_3\end{bmatrix}
That reduces to
\begin{bmatrix}1&1&2&b_1\\0&1&1&b_1-b_2\\0&0&0&b_3-b_2-b_1 \end{bmatrix}
Now the thing is that if I prove that $b_3 = b_2 + b_1$ (which is obvious) then the system is consistent.
My doubt was why I don't need to prove the same about second row? Is it because there is no way that system of two equations (first and second row) that has contains unknowns ($x_1$ and $x_2$) be inconsistent?
AI: The second row corresponds to the equation $1x_2+1x_3=b_1-b_2$. This can be solved for any values of $b_1, b_2$. For example, we can take $x_2=b_1-b_2$, and $x_3=0$.
The third row corresponds to the equation $0x_1+0x_2+0x_3=b_3-b_2-b_1$. This cannot be solved if $b_3-b_2-b_1\neq 0$. No matter what $x_1,x_2,x_3$ are, you can't make 0 equal to 1.
However if $b_3-b_2-b_1=0$ then any values of $x_1,x_2,x_3$ will work. |
H: Definition of Left Translation of a function on a topological group
In Folland's A Course in Abstract Harmonic Analysis, he defines for a function $f$ on a topological group $G$, and $y\in G$,
$$L_{y}f:x\to f(y^{-1}x)\text{ and }R_{y}f:x\to f(xy)$$
He then remarks that the $y$ is inverted so that the map $y\mapsto L_{y}$ is a group homomorphism.
But it seems to me that $L_{yx}f = f((yx)^{-1}\cdot) = f(x^{-1}y^{-1}\cdot) = \left[L_{x}\circ L_{y}\right]f$, making the map an anti-homomorphism.
Am I misunderstanding the remark?
AI: For any $z \in G$,
$$
\begin{align}
\left( L_{yx}f \right)(z) &= f((yx)^{-1}z) \\
&= f(x^{-1} y^{-1} z) \\
&= \left( L_x f \right)(y^{-1}z) \\
&= \left( L_y L_x f \right)(z).
\end{align}
$$
So,
$$
L_{yx} f = L_y L_x f
$$
for
all $f \in G^*$. |
H: Write a grammar that generates the strings over {a,b} starting with a
The answer is:
S -> aA, A -> aA, A -> bA, A -> a, A -> b, S -> a
Any idea how they got this?
AI: Let's write a few strings from the language, some are:$a,aa,abbbb,aabaabbb,...$.
There is only one condition, every string in this language should start with an $a$, we don't care how the following $a$s and $b$s should be. In general the strings in this language are either a single $a$, or an $a$ followed by any sequence of $a$s and $b$s.
Now the rule $S\to a$ makes a single $a$. If something comes after that starting $a$, i.e. some sequence of $a$s and $b$s, then we should use $S \to aA$, Now $A$ produces all non-empty strings with $a$ and $b$ using $A \to aA,A \to bA,A\to a$ or $A\to b$.
Let's see how $ababb$ is made with this grammar:
$S \xrightarrow{S\to aA}aA\xrightarrow{A\to bA}abA\xrightarrow{A\to aA}abaA\xrightarrow{A\to bA}ababA\xrightarrow{A\to b}ababb$ |
H: good reference for the Laplace method
I was wondering if someone could suggest good reference about the Laplace method.
One source I have now is just a short section (appendix) of Stein's Complex Analysis textbook.
I wish it starts with basic materials and provide rather precise error bounds, and also treats multidimensional case.
Thank you very much.
AI: Bender & Orszag, Ch. 6, esp. Sec. 6.4, has a terrific treatment, ripe with detailed examples, error analysis, but at a level in which anyone with a decent background in calculus can be an expert. |
H: Sum of 2 Standard Guassian variables
We know that when X and Y are independent and normally distributed, then X+Y is also Normally distributed. But now, let's remove the hypothesis of Independence. Can anyone please prove or provide a counter example for the following statement
When $X$ and $Y$ are Gaussian random variables (on the same probability space) then $X+Y$ is also Gaussian.
Thanks
AI: Here is a counterexample: $Z$ has standard normal distribution. If $|Z|<1$ then $X=Y=Z$ otherwise $X=Z,Y=-Z$. Both $X$ and $Y$ have standard normal distributions but $X+Y$ lies between -1 and 1. |
H: Drawing Graph on 5 vertices.
Draw a graph on 5 vertices that contains no clique of size three (that is, no triangle) and no anti-clique of size three (that is, three vertices none of which is connected to any other).
Here are a few graphs which I made. The problem I'm having is I don't quite understand what the question is asking me for. Like user96614 suggested, a ring could work too. So am I suppose to just come up with some graph with 5 vertices where no 3 vertices are together and no 3 vertices are alone?
AI: The problem is probably asking you to find a graph such that for an three points , those three points don't form the triangle $C_3$ or its complement, the edgeless graph on 3 vertices. So the cycle $C_5$ works.
You can think of cliques as a collection of people where everyone is friends with each other and of anti-cliques as a collection of people where no two people in this collection are friends. |
H: Lottery Ball problem
There are two sets of numbered balls. One set consits of white balls numbered $1-10$ the other is blue balls numbered $1-20$. To play you select two white balls and two blue balls. What is the probability that your ticket contains exactly one matching white number and two matching blue numbers?
My 1st thought, $\dfrac{C(10,1)\cdot C(20,2)}{8550}$ but thats not right.
The total number of combinations I can make is $C(20,2) \cdot C(10,2)=8550$ (I took a shortcut). Now I want to choose 1 white ball that matches and 2 blue balls that match which gives me $1900$ when multiplied out. But this does not give me the answer. Why?
AI: Wouldn't the numerator just be 2 * 8?
There's only one way to pick the blue balls, C(2,2), two ways to pick the right number, C(2,1) and 8 ways to pick the wrong number C(8,1).
$\frac{16}{8550}$ |
H: Area of a right angled triangle is an even integer
Actual Question is :
Let $ABC$ be a triangle in the plane such that $BC$ is perpendicular to $AC$. Let $a,b,c$ be the lengths of $BC$, $AC$, and $AB$ respectively. Suppose that $a,b,c$ are integers and have no common divisor other than $1$. Which of the following statements are necessarily true?
a. Either $a$ or $b$ is an even integer.
b. The area of the triangle $ABC$ is an even integer.
c. Either $a$ or $b$ is divisible by $3$.
With this information, I could see that right angle is at $C$ and $AB=c$, $BC=a$, $AB=c$.
First of all I should check that either $a$ or $b$ is even.
Suppose not, then we would have $(2n+1)^2+(2m+1)^2=c^2$
i.e., $4n^2+4m^2+4n+4m+2=c^2 \Rightarrow 2(2n^2+2m^2+2n+2m+1)=c^2$
i.e., $2$ divides $c^2$ i.e., $2$ divides $c$ i.e., $2^2$ divide $c^2$
i.e., $4$ divides $c^2$ but then, $4n^2+4m^2+4n+4m+2=c^2$ so, going modulo $4$ we see that $\bar{2}\equiv\bar{0} \text{ mod } 4$ which is false.
So, at least one of $a,b$ has to be even integer.
This does not immediately imply that area is even integer (sq.units) because
$\text {Area} =\frac{1}{2} \cdot a \cdot b$ Suppose $a$ or $b$ has only $1$ as power of $2$ in its prime decomposition, then $2$ in numerator and denominator gets cancelled out which may not give an even integer.
I would be thankful if some one can help me to clear this.
Thank you
AI: For the even area problem, you had proved that one leg is even. But we can do better. We have $a^2+b^2=c^2$. Since $c$ is odd, $c^2\equiv 1\pmod{8}$.
Let $b$ be odd. Then $b^2\equiv 1\pmod{8}$. So we cannot have $a\equiv 2\pmod{4}$. For if $a\equiv 2\pmod{4}$ then $a^2\equiv 4\pmod{8}$, and therefore $a^2+b^2\equiv 5\pmod{8}$.
Thus $a\equiv 0\pmod{4}$, and therefore the area is even.
For the divisibility by $3$, show that if neither $a$ nor $b$ is divisible by $3$, then $a^2\equiv 1\pmod{3}$ and $b^2\equiv 1\pmod{3}$. Argue that this is not possible. |
H: Generally true that $\frac{\mathrm d}{\mathrm da} \int_{-\infty}^{a-y} f(x)\, \mathrm dx = f(a-y)$?
Is it generally true that
$$\frac{\mathrm d}{\mathrm da} \int_{-\infty}^{a-y} f(x)\, \mathrm dx = f(a-y),$$
where $a$ and $y$ in the expression are constants?
To give context to the question, I am reading about the function convolution of two independent random variables $X, Y$.
AI: Assuming $f$ is (for example) continuous, you are correct. To prove this, use the fundamental theorem of calculus in conjunction with the chain rule.
Let $$F(z)=\int_{-\infty}^{z} f(x) \ dx.$$
By the fundamental theorem, $F'(z)=f(z)$. In your problem, you want to evaluate the derivative of $F(a-y)$ with respect to $a$.
To use the chain rule, let's write $F(a-y)$ as a composition of two functions. Define $g(a)=a-y$. Then $F(a-y)=F(g(a))$. The derivative with respect to $a$, using the chain rule, is $F'(g(a))g'(a)$. We know $F'$ (it's $f$ -- see above), and we also know that $g'(a)=1$.
So $F'(g(a))g'(a)=F'(g(a))=f(a-y)$. |
H: Suppose that a cube is inscribed in a sphere of radius one. What is the volume of the cube? my reasoning vs answer
Now my reasoning is that, s^2 + s^2 = 2^2, where s is the side of the cube, giving, s^3 = 2 sqrt 2. But the answer and explanation here is different: http://math.acadiau.ca/aumc/hints4.pdf
how is the distance from the center to the corner that? sqrt(3(s/2)^2) ?
AI: Apply Pythagoras twice: The face diagonal $d$ of a cube of side length $a$ can be found from $d^2=a^2+a^2$, the spacial diagonal $D$ can then be found from $D^2=d^2+a^2$. |
H: Why is this binomial coefficient bounded thus?
Source: Miklos Bona, A Walk Through Combinatorics.
$$ \forall k\geq 2,\binom{2k-2}{k-1}\leq4^{k-1}.$$
The RHS is the upper bound of the Ramsey number $R(k,k)$.
How can I prove the inequality without using mathematical induction? I've merely expanded the LHS to obtain $\frac{(2k-2)!}{(k-1)!(k-1)!}$.
AI: Consider a set $S$ with $2k-2$ elements. The set $S$ has $2^{2k-2}=4^{k-1}$ subsets. The number of subsets of S that have exactly $k-1$ elements is $\binom{2k-2}{k-1}$. Clearly the number of subsets of $S$ with $k-1$ elements is less than the total number of subsets of $S$. (@Did's answer is effectively the same as mine.) |
H: $3+33+\dots+33\ldots3={10^{n+1}-9^n-10\over 27}$
I need help to show by Induction
$3+33+\dots+33\ldots3={10^{n+1}-9^n-10\over 27}$
Thank you.
AI: I think the problem has a typo.
It should be $$f(n):3+33+\cdots+\underbrace{33\cdots 33}_{n \text{ digits}}=\frac{10^{n+1}-9n-10}{27}$$
Let $f(n)$ holds true for $n=m$
$$3+33+\cdots+\underbrace{33\cdots 33}_{m \text{ digits}}=\frac{10^{m+1}-9m-10}{27}$$
$$\implies 3+33+\cdots+\underbrace{33\cdots 33}_{m+1\text{ digits}}=\frac{10^{m+1}-9m-10}{27}+\underbrace{33\cdots 33}_{m+1\text{ digits}}$$
$$=\frac{10^{m+1}-9m-10}{27}+\frac{10^{m+1}-1}3$$
$$=\frac{10^{m+1}-9m-10+9(10^{m+1}-1)}{27}$$
$$=\frac{10^{m+1}(1+9)-9m-19}{27}$$
$$=\frac{10^{m+2}-9(m+1)-10}{27}$$
So, $f(n+1)$ holds true for $n=m+1$ if $f(n)$ holds true for $n=m$
Establish the base case $n=1$ |
H: Is statistical dependence transitive?
Take any three random variables $X_1$, $X_2$, and $X_3$.
Is it possible for $X_1$ and $X_2$ to be dependent, $X_2$ and $X_3$ to be dependent, but $X_1$ and $X_3$ to be independent?
Is it possible for $X_1$ and $X_2$ to be independent, $X_2$ and $X_3$ to be independent, but $X_1$ and $X_3$ to be dependent?
AI: First problem: Toss a fair coin twice. Let $X_1=1$ if the first toss is a head, and $0$ otherwise. Let $X_3=1$ if the second toss is a head, and $0$ otherwise. Let $X_2$ be the number of heads in the two tosses combined.
Then $X_1$ and $X_2$ are dependent, as are $X_2$ and $X_3$, but $X_1$ and $X_3$ are independent.
Second problem: Again, two tosses of a fair coin. Let $X_1$ and $X_3$ each be $1$ if we get head on the first toss, and $0$ otherwise. Let $X_2$ be $1$ if we got a head on the second toss, and $0$ otherwise. Then $X_1$ and $X_2$ are independent, as are $X_2$ and $X_3$, but $X_1$ and $X_3$ are very much not independent. |
H: Application of pigeonhole principle
Select $11$ different numbers from $f\{1,2,...,20\}$. Prove that two of your numbers, $a$ and $b$, will differ by two.
Clearly this is an application of the pigeonhole principle. However, I'm not sure how to write up a coherent proof.
AI: Consider the following sets: $$\{1, 3\}, \{2, 4\}, \{5, 7\}, \{6, 8\}, \{9, 11\}, \{10, 12\}, \{13, 15\}, \{14, 16\}, \{17, 19\}, \{18, 20\}$$
Together, these $10$ sets account for all of the integers $\{1, \ldots, 20\}$. When we pick 11 numbers, by the Pigeonhole Principle, we will pick both numbers from at least one of the sets. Hence, these two numbers (which we can denote $a$ and $b$) will differ by two.
Hope this helps. Cheers! |
H: Show combination of affine functions and logs has at most one zero
For $x>0$, let
$$
f(x)=(x+2){\sf log}(x)-(x+1){\sf log}(x+1)
$$
Can anybody show that the equation $f(x)=0(x > 0)$ has at most one
solution.
AI: We compute
$$
f'(x) = 1 + 2/x + \log x - 1 - \log (x+1) = 2/x - \log (1 + 1/x).
$$
We want to show that this is positive. Putting $y = 1/x$, we just need to show $2y > \log (1+ y)$ for all positive y.
But these two functions are equal when $y = 0$ and the result is then clear by the concavity of the logarithm. |
H: Is $\mathbb{Z}$ isomorphic to a direct subproduct of the family $\left\{ \mathbb{Z}_{n}\right\} _{n>1}$?
Is $\mathbb{Z}$ isomorphic to a direct subproduct of the family $\left\{ \mathbb{Z}_{n}\right\} _{n>1}$?
AI: I'm assuming that $\mathbb Z_n$ means $\mathbb Z/n\mathbb Z$. I'm also assuming that "subproduct" means "subgroup of the product". If this is the case then yes, the subgroup generated by $(1, 1, \ldots) \in \prod_{n > 1}\mathbb Z_n$ is isomorphic to $\mathbb Z$.
To see that this is the case note that we can always define a homomorphism out of $\mathbb Z$ by picking an element $x$ and declaring $m \mapsto m\cdot x$. So what we need is for this to be injective. When $x = (1, 1, \ldots)$ we have $m\cdot x = (m, m, \ldots)$ and if $m \neq 0$ then the looking at the $|m| + 1$ coordinate gives $m\cdot x \neq 0$. Another way to see this: if $p$ is prime then the coordinate of $(m, m, \ldots)$ corresponding to $\mathbb Z_p$ is zero if and only if $p$ divides $m$. So if $m\cdot x = (0, 0, \ldots)$ then every prime divides $m$, so $m$ must be $0$. |
H: Can conclude that $N = f\left(K\right)\oplus f\left(L\right)$?
Let $K,L$ be two submodules of an $R$-module $M$ with the property that $K+L=M$ and $f:M\longrightarrow N$ is an $R$-epimorphism. Assume that $K \cap L = \ker f$. Can conclude that $N = f\left(K\right)\oplus f\left(L\right)$ ?
AI: It's obvious (from surjectivity) that $f(K)+f(L)=N$.
Now take $x\in f(K)\cap f(L)$. Then $x=f(y)=f(z)$ with $y\in K$ and $z\in L$. Since $y-z\in \ker f$ we get $y\in L$, so $y\in K\cap L$ and therefore $y\in\ker f$. Thus we get $x=f(y)=0$. |
H: Let $\left\{ f_{i}:M_{i}\longrightarrow N\right\} _{i\in I}$, can $Im\left(\oplus_{i\in I}f_{i}\right)=\sum_{i\in I}Im\left(f_{i}\right)$
Let $\left\{ f_{i}:M_{i}\longrightarrow N\right\} _{i\in I}$ be a family of $R$-homorphism from $R$-module $M_{i}$ to $R$-module $N$. Do we have $Im\left(\oplus_{i\in I}f_{i}\right)=\sum_{i\in I}Im\left(f_{i}\right)$?
AI: Yes, $x \in \sum_iim(f_i)$ if and only if you can write $x = \sum_if_i(m_i)$ where $m_i \in M_i$ and only finitely many $m_i$ are nonzero. But this is equivalent to $m = (m_i)_i \in \bigoplus_iM_i$ and $x = \left(\bigoplus_if_i\right)(m)$. |
H: Evaluate $\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots$
Question is to Evaluate :
$$\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots$$
what all i could do is :
$$\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots=\sum_{n=1}^{\infty} \frac{1}{(n+2)n!}=\sum_{n=1}^{\infty} \frac{n+1}{(n+2)!}=\sum_{n=1}^{\infty} \frac{n}{(n+2)!}+\sum_{n=1}^{\infty} \frac{1}{(n+2)!}$$
I have $$\sum_{n=1}^{\infty} \frac{1}{(n+2)!}=\sum_{n=0}^{\infty} \frac{1}{n!}-1-\frac{1}{2}=e-\frac{3}{2}$$
Now, I am not able to see what $$\sum_{n=1}^{\infty} \frac{n}{(n+2)!}$$ would be.
I would be thankful if some one can help me to clear this.
Thank you.
AI: $$\text{From }\frac{n+1}{(n+2)!},$$
$$=\frac{n+2-1}{(n+2)!}=\frac1{(n+1)!}-\frac1{(n+2)!}$$
Can you identify the Telescoping series?
The survivor will be $$\frac1{2!}$$ |
H: Area of triangle $OAB$
Question is :
Consider a circle of unit radius centered at $O$ in the plane. let $AB$ be a chord which makes an angle $\theta$ with the tangent to the circle at $A$ .find the area of triangle $OAB$
What all i could do was is just draw the picture and even in that i am not sure if he mean angle $BAP=\theta$ or $BAQ=\theta$. I am assuming for some time that $BAQ=\theta$
Now, I would have some hope if angle $OAB$ can be calculated from given data in terms of theta then i would use the area formula as
$\text{Area of triangle $OAB= \frac{1}{2}.OA.OB.\sin (\angle AOB)$}$ and as $OA=OB=1$ we would then get
Area of triangle $OAB=\frac{1}{2}.\sin (\angle AOB)$
But i am not sure how to relate $\theta$ with $\angle AOB$.
I am not even sure if there is any way to relate this.
I would be thankful if some one can help me with this.
Thank you.
AI: The angle $(\angle BAQ)$ is a tangent (chord) angle.
Theorem An Angle formed by a chord and a tangent that intersect on a circle is half the measure of the intercepted arc.
So $$\theta=\frac{1}{2} m \overset{\frown}{(AB)}$$
from there $(\angle AOB)=2\theta$ |
H: Why is $ 2\binom nm^2
$\forall n\geq2 \forall m\geq2,$ $$ 2\binom nm^2<n^{2m}.$$
Why is the above inequality, which is equivalent to $ \binom nm<\frac{n^m}{\sqrt 2}$, true?
AI: $$\binom nm =\frac {n (n-1) ... (n-m+1)}{m!} < \frac{n^m}{m!}$$
Thus, if $m \geq 2$
$$\binom nm^2 < \frac{n^{2m}}{m!^2} < \frac{n^{2m}}{2}$$
But your bound is very crude! |
H: Is the number 100k+11 ever a square?
Is the number 100k+11 ever a square?
Obviously all such numbers end in 11. Can a square end with an 11?
AI: Let $a^2=100\cdot k+11$. Then easily $a$ must be odd.
So $(2n+1)^2=100\cdot k+11$. It follows $2(n^2+n)=50\cdot k+5$, which is impossible. |
H: How prove this $f(x)=g(x)=h(x)=0$?
let polynomial $f(x),g(x),h(x)\in C[x]$, and such
$$f^2(x)=xg^2(x)+xh^2(x),$$
prove or disprove
$$f(x)=g(x)=h(x)=0$$
I know solve this follow problem
let polynomial $f(x),g(x),h(x)\in R[x]$, and such
$$f^2(x)=xg^2(x)+xh^2(x),$$
prove or disprove
$$f(x)=g(x)=h(x)=0$$
then I guess when $f,g,h\in C[x]$ is true? Thank you
AI: Actually, this won't be true at all in $\Bbb C$. It might even be the case that each one of $f,g,h$ is non zero. There follows an example:
$$
f(x) = 2e^{i\pi/4}x,\qquad g(x) = ix+1, \qquad h(x) = x+i.
$$ |
H: Restricting the ordering of a given permutation
Let $p_1p_2p_3...p_n$ be a randomly-selected n-permutation.
Why is $P(p_1>p_2>p_3)=\frac 16$?
(P denotes probability.)
AI: There are $n!$ ways to arrange $n$ distinct numbers. The condition $p_1 > p_2 > \dotsc > p_n$ singles out one of these ways. If all permutations are equally probable, each then has a probability of $\frac{1}{n!}$. |
H: Difference between quaternions and rotation matrices
This is a really simple question, I guess. Do quaternions cover the same set of rotations as rotation matrices? I assume the answer is yes, they both can represent SO(3), but I'm unsure about the sources which prove that.
Extending the question a bit further, where is the difference between quaternions and rotation matrices? It seems that the rotations are the same, there are inversion and multiplication operators for both. So what is the actual difference?
We are using quaternions and rotation matrices in a nonlinear optimization framework, where we estimate positions of a bunch of things in SE(3). Quaternions are appealing since there are only four coefficients to be estimated, and only three of those need to be actually stored (we have normalized quaternions). So far we have two versions of the code, using quaternions and using rotation matrices, and both of those versions give the same results (well, almost, thanks to the floating point precision). Is there a point where "quaternions will break"?
AI: The unit quaternions are isomorphic to the group $SU(2)$ of unitary matrices of determinant $1$. Geometrically, they form $S^3$, the $3$-sphere.
On the other hand, the group of rotations in $\Bbb{R}^3$ is $SO(3)$, consisting of orthogonal (orthonormal, really) matrices of determinant $1$. Geometrically, they form $\Bbb{R}P^3$, $3$-dimensional projective space.
The group $SU(2)$ forms a double cover of $SO(3)$, which means that there are exactly $2$ elements of $SU(2)$ (differing by a sign) that correspond to each element of $SO(3)$.
This has a geometric analogue, too. Pairs of antipodal points in $S^3$ are identified to form a single point in $\Bbb{R}P^3$.
You may enjoy the brief discussion here, and the more lengthy one here as well. |
H: recipe for infinitely many irrational numbers - or is it?
What if we write 0. and then throw a coin and depending on the result continue the number with 1 or 0 and continue this process indefinitely. It seems like a recipe for producing irrational numbers.
Are these numbers really irrational? Are they transcendental? Are they normal?
Maybe I should also ask: is anything well-defined by this recipe? :-)
AI: since it may happen that the coin keeps showing tail and in this case the number (base 2) is an integer, I think that a definite answer cannot be done.
UPDATE if you are interested in a probabilistic / philosophical answer (i.e., a supernatural being which performs a supertask tossing the coin at times $t$, $t+\frac{1}{2}$, $t+\frac{3}{4}$, $t+\frac{7}{8} \ldots$, at time $t+1$ she will with near certainty ($p=1$) obtain a normal number. But this is not a "recipe", in the sense she cannot be sure it is. (Ok, if she is a supernatural being maybe she has some other way to know it :-) ) |
H: How find this linear equation $-x_{1}-x_{2}-x_{3}-\cdots+(2^n-1)x_{n}=2^na$?
let $a\in R$,then solve this follow equation:
$$\begin{cases}
x_{1}-x_{2}-x_{3}-\cdots-x_{n}=2a\\
-x_{1}+3x_{2}-x_{3}-\cdots-x_{n}=4a\\
-x_{1}-x_{2}+7x_{3}-\cdots-x_{n}=8a\\
\cdots\cdots\cdots\cdots\cdots\cdots\\
-x_{1}-x_{2}-x_{3}-\cdots+(2^n-1)x_{n}=2^na
\end{cases}$$
Maybe this problem have more methods,Thank you
AI: Write $s = x_1 + x_2 + \dotsc + x_n$. Then the system becomes
$$2^k x_k - s = 2^k a;\quad 1 \leqslant k \leqslant n.$$
Rearranging that, we obtain
$$x_k - a = 2^{-k}s.$$
Summing these equations yields
$$\sum_{k=1}^n (x_k-a) = s - na = \left(\sum_{k=1}^n 2^{-k}\right)s = \left(1 - 2^{-n}\right)s.$$
From that we obtain
$$s = n2^na$$
and
$$x_k = a(1 + n2^{n-k})$$
for $1\leqslant k \leqslant n$. |
H: Let $\{a_n\}$ be a positive monotonic decreasing sequence of real numbers. Show $\lim_{k \rightarrow \infty} \frac 1 k \sum_{n=1}^k a_n = \inf a_n$
Let $\{a_n\}_{n=1}^{\infty}$ be a monotonic decreasing sequence of positive real numbers. Show that $\lim_{k \rightarrow \infty} \frac 1 k \sum_{n=1}^k a_n = \inf a_n$
Suppose we take the sum of the first $k =m$ numbers then the average will be greater than the smallest number. However per definition of $\inf a_n$ we know that $\inf a_n \le a$ where $a \in A = \{a_n \mid n \in \mathbb N\}$. This holds for any $m$. Can someone give a hint or explanation how to proof this ?
AI: We will show that $\lim_{k \rightarrow \infty} \frac 1 k \sum_{n=1}^k a_n = \inf a_n=a$ and $a_n\to a$ because it is a monotonic decreasing sequence of positive real numbers.
Because $a_n\to a$ there is a $M>0:|a_n|\leq M$ for every $n\geq 1$. Just to make it more simple suppose that $a=0$(because if we set $b_n=a_n-a$ we have that $b_n\to 0$ and thus it's enough to show that $\frac {b_1+b_2+...+b_n}{n}\to 0 $ if $b_n\to 0$).
Let $ε>0$.Because $\frac {M}{\sqrt n}\to 0$ there is a $n_o\in \Bbb N:|a_n|<ε/2$ and $\frac {M}{\sqrt n}<ε/2$ for $n\geq n_0$. So for $n\geq n_0^2$ we have $|\frac {a_1+a_2+...+a_n}{n}|\leq \frac {a_1+a_2+...+a_{\sqrt n}}{n}|+|\frac {a_{\sqrt {n+1}}+...+a_n}{n}|\leq \frac {M[\sqrt n]}{n}+\frac {nε}{2}\frac {1}{n}\leq \frac {M}{\sqrt n}+\frac {ε}{2}\leq \frac {ε}{2}+\frac {ε}{2}=ε$ and thus $\frac {a_1+a_2+...+a_n}{n}\to a$ for $n\to \infty$ |
H: Complex numbers inequality
Let $z$ , $w$ $ \ \in \mathbb{C}$ with $|z|$ , $|w| < 1$. Then prove that $|\frac{z-w}{1-z\bar{w}}| < 1$.
What I have noticed : $|z-w| = |w||\frac{z\bar{w}}{|w|} -1| $ but I don't know how to proceed.
AI: $$|\frac{z-w}{1-z\bar{w}}| < 1<=>|z-w|<|1-z\overline w|<=>(z-w)(\overline z-\overline w)<(1-z\overline w)(1-\overline zw)<=>|z|^2+|w|^2<1+|z|^2|w|^2<=>|z|^2(1-|w|^2)<1-|w|^2<=>|z|^2<1$$ which is true. |
H: Differential forms as functionals on curves
Please give me a reference to a book or lecture notes where the following stuff is studied.
Let $M$ be a Riemann surface with boundary $\partial M$ (but not necessarily, any smooth $n$-dimensional manifold suffices if the following statements make sense in this case).
What does it mean for $1$-form $\omega$ on $M$ to have $\int\limits_\gamma \omega = 0$ $(\operatorname{mod} 2\pi)$ for any closed loop $\gamma$?
If $\omega_1$, $\ldots$, $\omega_N$ are closed forms which form a basis of $H^1(M)$ (de-Rham cohomology) then there are non-homotopically equivalent loops $\gamma_1$, $\ldots$, $\gamma_N$ in $M$ such that $\int_{\gamma_i} \omega_j = \delta_{ij}$. How to prove this? Is it possible to say more about these $\gamma_1$, $\ldots$, $\gamma_N$? Do they have some interpretation on terms of $H_1(M)$ (singular homology)?
As in 2., if $\omega_1$, $\ldots$, $\omega_N$ are closed 1-forms which form a basis of $H^1(M,\partial M)$ there exist non-homotopically equivalent loops $\gamma_1$, $\ldots$, $\gamma_N$, non-homotopically equivalent to any component of $\partial M$, such that $\int_{\gamma_i} \omega_j = \delta_{ij}$. What can we say about these curves in comparison with question 2?
AI: What you are struggling with is called Poincare Duality for manifolds with boundary. If $M$ is a compact oriented $n$-dimensional manifold (in your case, $n=2$, $k=1$), the Poincare duality reads:
$$
H^k(M)\cong H_{n-k}(M, \partial M), H_k(M)\cong H^{n-k}(M, \partial M).
$$
In the case $n=2$, you also have the Kronecker duality
$$
H^k(M)\cong (H_k(M))^*, H^k(M, \partial)\cong (H_k(M,\partial))^*
$$
(in general, you have to use real coefficients or use the Universal Coefficients Theorem, which is more subtle). The best place to read about this that I know is the book
R. Bott, L. Tu, "Differential forms in algebraic topology". |
H: Can $f(x)>g(x)$ be implied from $\frac{df(x)}{dx}\gt \frac{dg(x)}{dx}$?
I am new to functions. My question is Can $f(x)>g(x)$ be implied from $\frac{df(x)}{dx}\gt \frac{dg(x)}{dx}$?
AI: Given that $f(a)>g(a)$ at $a$ and $f'(x)\geq g'(x)$ for all $x>a$, you can deduce $f(x)>g(x)$ for all $x>a$ inductively. But the derivative in and of itself has nothing to do with the value of the function at the point at which you are taking the derivative. |
H: What the symbol $\subseteq$ represents generally?
My book says that $\subset$ is used to represente any subset, proper or improper, needing in this case to show the anti symmetric property of sets. ($A = B \iff A \subset B \, \, \wedge \,\, B \subset A)$
And $\subseteq$ is used specifically to represent improper subsets. (In other words, $A \subseteq B \iff A = B) $
But i saw so many articles using $\subseteq$ and not $\subset$ then i am kinda suspicious about this definition.
If this definition is correct, why $\subseteq$ it's so used? Just because it helps to prove the equality betwen the sets?
Thanks for the help.
AI: It is common practice to use $\subset$ to simply denote a subset, proper or not. $\subseteq$ is used to mean the same thing. This is mainly due to reasons of convenience. I'm a pedant and prefer to use $\subset$ for proper subset, whereas $\subsetneq$ is convention for proper subsets.
I disagree that $\subseteq$ is used to represent improper subsets since we already have a good symbol for that: $=$
In reading though, you should interpret $\subset$ as subset, proper or not. Context typically makes it clear though.
(Now imagine if primary and secondary educators decided to go on protest and use $<$ instead of $\leq$ for reasons of convenience). |
H: Numerical Integration of $g(x)=4\sqrt{1-x^2}$
The limits of integration are $[0,1]$ and thus the result should be $\pi$. My book suggests an elegant way to evaluate the integral by Monte Carlo Integration but I was wondering, can we reach the same result with an easier way? Thank you.
AI: Hints:
$$\sin u:= x\implies du\cos u=dx\implies$$
$$\int\sqrt{1-x^2}\,dx=\int \cos^2u\,du=\frac{u+\sin u\cos u}2\;\ldots$$ |
H: On derivatives...
I have a quick question here. I hope someone can help. I haven't done calculus for a long time so I seem to missed out on details.
If $x=g^{-1}(y)$ and $g$ is monotonic and is differentiable for all $x$, how did it happen that $$\frac{d}{dy}g^{-1}(y)=\frac{1}{dg(x)/dx}|_{x=g^{-1}(y)}$$
I tried to reason out that $$\frac{d}{dy}g^{-1}(y)=\frac{d}{dx}\frac{dx}{dy}x=\frac{dx}{dy}$$ But I do not know where to proceed. I hope someone helps. Thanks.
AI: Highlights:
$$y=g(x)\implies x=g^{-1}(y)$$
so let us denote $\;g(x_0):=y_0\;$ , and assuming what must be assumed (what? For example, $\;g'(x_0)\neq 0\;$ ...!) , we get:
$$\frac{d}{dy}g^{-1}(y_0):=\lim_{y\to y_0}\frac{g^{-1}(y)-g^{-1}(y_0)}{y-y_0}=\lim_{g(x)\to g(x_0)}\frac{x-x_0}{g(x)-g(x_0)}=\frac1{g'(x_0)}$$
since, by continuity of $\;g^{-1}\;$ , we have that $\;g(x)=y\to y_0=g(x_0)\implies x\to x_0\;$ |
H: Show two graphs are not isomorphic
I know this graphs are not isomorphic. However they have the same number of vertex and edges, and the same degree sequence, is not the most easy case.
If im correct, the graphs are isomorphic if evey posible bijection between vertex preserve adjacencies, then if i find just one bijection in wich some adjacencies are not preserved, that will be enough to show they are not isomorphic?.
AI: In the first graph, none of the vertices with degree $2$ are adjacent. In the second graph, there are two pairs of adjacent vertices with degree $2$. |
H: Do Symmetric problems have Symmetric solutions?
There is a general notion of Symmetry in mathematics,
that of an object being constant under some transformation.
If we think of our object as being a "mathematical problem" we can see certain mathemtical problems have symmetries in them, in terms of how the problem is stated.
This thought came to be when I was trying to solve a combinatorics problem on this website.
The problem statement was "Given 11 people and one safe, how many locks do we need on the safe to ensure a subset of 5 people cannot open the safe but a susbet of 6 people can open the safe. For each lock we can get as many copies of its key as we like.".
Now This is just an example, I am not asking about this problem in particular.
I merely want to note that in this problem the group of 11 people all have the same constraints put on them.
There is no mentioning of a property fulfilled by a certain "distinguished" subset of these 11 people, if we were to write the problem as:
"Given a set of 11 people $\{p_1,\dots ,p_{11}\}$, how many locks do we need ..." (the rest of the problem statement is the same as before),
then we would have a symmetry in the sense that if we permute the set $\{p_1,\dots ,p_{11}\}$ in the problem statement, and relabel the people accordingly, the problem statement does not change.
Intuitively, I expect a solution to such a problem to be symmetric in some way, for example I would expect each person do have the same number of keys.
Intuition is not a proof.
The question is this:
Is there a general principle in mathematics which relates symmetries in problem statemenets to symmetries in the solution?
AI: It is not in general true that a solution to a symmetric problem must itself be symmetric. For example:
Minimize $x^4-2x^2$.
This is an even function, so it is symmetric in reflection about the $y$ axis. However, the solutions are $1$ and $-1$, and neither of these solutions is itself symmetric. (Physicists speak about "symmetry breaking" in such cases, and it more complex variants it plays a significant role in modern particle physics).
What we can say is that if there's a symmetry of the problem, we can apply that same symmetry to the solution, and get a (possibly different) solution back. That's more or less the generic definition of what it means for a problem to have a symmetry. So the set of all solutions will be a symmetric as the problem is.
If we know somehow that the solution is unique, it will need to have all of the symmetries the problem has. This is sometimes a useful shortcut for finding it. (But be careful to check whether "unique" really means "unique modulo such-and-such symmetries which are present in the problem too").
In the vein of your example we could say
Given a box and two people, what is the minimum number of locks and keys we need to buy such that (a) the two people can open the box together, and (b) the box can't be opened if neither of the two people is present?
That's kind of a silly problem, but it's certainly symmetric under permutation of the two people. But the solution does not have the kind of symmetry you're intuitively expecting: One lock with one key, and hand the single key to one of the trusted people. |
H: quotient of two entire functions
Suppose we have a quotient of two entire functions , i.e. $g(z)=\frac{f(z)}{h(z)}$ where $f,h$ are entire functions in complex plane. If $f$ and $h$ have the same set of zeros, what can we tell from the order of these zeros? For example, if we have $f(z_0)=h(z_0)=0$, what can we say about order of $z_0$ as a zero of $h(z)$?
Also, in order for $g$ to be entire, we only need to examine how $g$ behaves at the neighbourhood of $g^{-1}(0)$. In this case, what conditions are needed so that $g$ is entire?
AI: Hint: Expand the functions $f$ and $h$ in Taylor series about $z_0$. |
H: Lebesgue measure of algebraic irrational numbers in $\mathbb{R}$
Find all Lebesgue measurable subsets $A \subset\mathbb{R}$ such that all $B\subset A$ is measurable.
I argued that if the measure is positive then $A$ is an interval so we can construct the Vitali set and thus it'd have non-measurable subsets. So $A$ must have measure $0$. Is it correct to say that such $A$ is the power set of $\mathbb{Q}$? Or we need to worry about algebraic irrationals as well( since they're countable)?
AI: You are on the right track but what about the irrationals? The have measure >0 but certainly do not form an interval. As for your question: Any singleton set is measurable (since it has outer measure zero). Now suppose this set were $\{\pi\}$. Does your conclusion still hold? |
H: $K$ and $L$ homeomorphic, then $C(K)$ is isomorphic to $C(L)$
Can someone sketch the proof (or give me some reference) of the following fact : If $K$ and $L$, compact and Hausdorff spaces, are homeomorphic then the lattices $C(K)$ and $C(L)$ are isomorphic. (I am aware this is half of Kaplansky theorem, but I'm curious to know why)
Thank you.
AI: Let $\xi :K\longrightarrow L$ be a homeomorphism. Then $T_{\xi }: C(L)\longrightarrow C(K),$ $T_{\xi } (u) =u\circ \xi $ is an isomorphism. |
H: Confusion about extension of functionals
Assume that $\Omega\subset \mathbb{R}^N$ is a bounded set and take a function $f:\Omega\to\mathbb{R}$ which belongs to $L^2(\Omega)$ but does not belongs to $L^q(\Omega)$ for $q>2$.
Define a bounded linear map $T:L^2(\Omega)\to\mathbb{R}$ by $$\langle T,g\rangle=\int_\Omega fg,\ \forall\ g\in L^2(\Omega)$$
If $1<p<2$ then, Hahn-Banach theorem implies the existence of $h\in L^{p'}(\Omega)$ ($p'$ is the conjugate of $p$) such that the functional $\overline{T}:L^p(\Omega)\to \mathbb{R}$ defined by $$\langle \overline{T},g\rangle=\int_\Omega hg,\ \forall\ g\in L^p(\Omega)$$
satisfies $\overline{T}_{|L^2(\Omega)}=T$ (note that $L^2(\Omega)\subset L^p(\Omega)$). Now my confusion is: we must have that $$\int_\Omega fg=\int_\Omega hg,\ \forall\ g\in L^2(\Omega)$$
which implies that $f=g$, but this seem to be a absurd, because in this way $\overline{T}$ would not be a bounded linear functional. Well, where is my flaw?
AI: The flaw is that for $p < 2$, the functional $T$ is not continuous in the $L^p(\Omega)$ norm. Thus it is not one of the sort the Hahn-Banach theorem says can be extended.
$T\colon L^2(\Omega) \to \mathbb{C}$ is continuous in the $L^2(\Omega)$-norm, and you have a continuous inclusion $j_p \colon L^2(\Omega) \hookrightarrow L^p(\Omega)$. But $j_p$ is not an embedding, thus $T\circ j_p^{-1}$ is not continuous.
Actually, since $L^2(\Omega)$ is dense in $L^p(\Omega)$ for $p < 2$, Hahn-Banach isn't needed for the extension here, if $T$ can be continuously extended to $L^p(\Omega)$, the extension is unique and given by uniform continuity. |
H: meaning of the symbol $Z_n^*$ in discrete mathematics
I was reading discrete Mathematics, and i found a symbol $$Z_n^*.$$ I don't know what it means.
The text says that the "image" with the multiplication operator is an abelian group.
can any one explain.
AI: maybe the author is talking about the group of unit elements of $\mathbb{Z}_n$ ? |
H: Relatively prime polynomials
If $f(x)$ is relatively prime to $p(x)$ in $F[x]$ prove that there is a polynomial $g(x) \in F[x]$ such that $f(x)g(x) ≡ 1_F \pmod{ p(x)}$.
Now it has just occured to me that this is a field we are working in and not a ring because we are using an $F$. This is just an assumption, so I don't know if I am correct still. (I don't think it matters in the end.)
My attempt:
Let $g(x)$ be a polynomial in $F(x)$. If $f(x)$ is relatively prime to $p(x)$ then there exists a polynomial $f(x)a(x) + b(x)p(x) = 1_F \mod p(x)$.
But this means that $f(x)a(x) = 1_F \mod p(x)$. Let $a(x) = g(x)$. Done.
AI: The existence of an inverse of $f(x)$ modulo $p(x)$ is a direct consequence of the fact that, with the euclidean algorithm, one can prove
for all $f(x)$ and $g(x)$ in $F[x]$, there exists $a(x)$ and $b(x)$ such that $f(x)a(x)+g(x)b(x)=h(x)$, where $h(x)$ is a greatest common divisor of $f(x)$ and $g(x)$.
I use “a” because the greatest common divisor is determined up to a multiplicative constant. Saying that $f(x)$ and $p(x)$ are coprime means that $1$ is a greatest common divisor.
So you're correct, if the quoted theorem can be used in your test. |
H: How do i get $b$ from ${b\over2-3ab}=2$
I have this expression:
$${b\over2-3ab}=2$$
So now i try to get $b$
I started with:
$${2-3a} = {1 \over 2}$$
But how you can see $b$ gets lost, what did i wrong?
Thanks!
My teacher said the right answer would be $b = {4\over 6a+1}$
AI: As I see it, you did two steps:
$$
\frac{b}{2-3ab}=2=\frac{1}{2}
$$
so you inverted both sides
$$
\frac{2-3ab}{b}=\frac{1}{2}
$$
and then you 'tried' to cancel out the factor $b$. But here you made a mistake, since
$$
\frac{2-3ab}{b}=\frac{2}{b}-\frac{3ab}{b}=\frac{2}{b}-3a\mbox{, which does not equal: } 2-3a
$$
You should instead have done what Lucian suggests. |
H: How prove this $\frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}=0$
let $a,b,c\in \mathbb{R}$, if such
$$\dfrac{1}{bc-a^2}+\dfrac{1}{ca-b^2}+\dfrac{1}{ab-c^2}=0$$
show that
$$\dfrac{a}{(bc-a^2)^2}+\dfrac{b}{(ca-b^2)^2}+\dfrac{c}{(ab-c^2)^2}=0$$
Does this problem has nice methods?
My idea:let
$$(ca-b^2)(ab-c^2)+(bc-a^2)(ab-c^2)+(bc-a^2)(ca-b^2)=0$$
then I can't. Thanks
AI: $$0 = \left( \dfrac{1}{bc-a^2}+\dfrac{1}{ca-b^2}+\dfrac{1}{ab-c^2} \right)\left( \dfrac{a}{bc-a^2}+\dfrac{b}{ca-b^2}+\dfrac{c}{ab-c^2} \right) =$$ $$ \dfrac{a}{(bc-a^2)^2}+\dfrac{b}{(ca-b^2)^2}+\dfrac{c}{(ab-c^2)^2} + \frac{a(ca-b^2) +a(ab-c^2) + b(bc-a^2) +b(ab-c^2)+c(bc-a^2)+c(ca-b^2)}{(bc-a^2)(ca-b^2)(ab-c^2)}$$ $$=\dfrac{a}{(bc-a^2)^2}+\dfrac{b}{(ca-b^2)^2}+\dfrac{c}{(ab-c^2)^2}$$
The last equation is true since the numerator of the big fraction is $0$. |
H: Closed subspace of Hilbertspace
Let $X$ be a norm closed subspace of a Hilbert space $\mathcal H$. Is it true that if $x_n \in X$ converges weakly to $x \in \mathcal H$, then also $x \in X$ ?
AI: Yes. Let $Y=X^\perp.$ Then for every $y\in Y$ we have $\langle x,y\rangle=\lim_{n\to\infty} \langle x_n,y\rangle=0.$ Hence $x\in Y^\perp=X.$ |
H: Work done by gravitational force
In my calculus class we learned about line integrals, and for homework we have exercise to find work done by gravitational force on material dot with mass $m$ which follows path of the elipse $\frac{x^2}{a^2}+\frac{z^2}{c^2}=1$ in second quadrant in positive direction (clockwise).
If I understood physics part of problem, gravitational force field is $f(X)=(0,0,-mg)$ and it's conservative which means work only depends on change in $z$ coordinate. And using formula with dot product I got:
$W=\int_C-mgdz$ and since parametrization of curve $C$ is $x=a \cos t, z=c \sin t, t\in [\frac{\pi}{2},\pi]$ I have $W=-mgc$. It makes sense to me, but I'm not sure weather my conclusions are right. Can you check please and correct my mistakes if I have them.
AI: Yes, your conclusion is correct. Since gravitational force is conservative in nature, work done by it is simply $-mgc.$ |
H: The meaning of $Z_a$, where $Z$ is a partitioning of $A$ and $a \in A$
I am unsure about the common usage of subscripting a set with something, more precisely something which might be a member of that set or some other set (as opposed to, say, subscripting a set with an integer for reasons of indexing).
The text where I encountered this usage, for context:
Let $Z$ be a partitioning. Define the binary relation $\equiv_Z \subseteq A \times A$ by, for all $a_1,a_2 \in A$, $a_1 \equiv_Z a_2$ iff $Z_{a_1} = Z_{a_2}$.
($Z$ is presumably a partitioning of $A$.)
$\equiv_Z$ is an equivalence relation, as can be shown (it is one of the exercises).
I interpret this to mean that $Z_{a_1}$ is a set, $Z_{a_1} \in Z$ and that $a_1 \in Z_{a_1}$, i.e. that $Z_{a_1}$ is the set in $Z$ that contains $a_1$ from $A$. Is this correct, or in line with most conventional notation?
AI: This, i.e. $Z_a$ being defined as the unique set with $Z_a \in Z: a \in Z_a$, seems to be the only sensible option.
More so since $\equiv_Z$ is a symbol that looks like it's supposed to be an equivalence relation (which it is, given this definition).
In cases like these, the subscript effectively acts as an argument to a function. That is, we can think of $Z_a$ as an alternative notation to $Z(a)$.
I'm not sure that this notation is conventional, but then, most sources I have seen go the other way around, deriving the partitioning $Z$ from a given equivalence relation $\equiv_Z$. |
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