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H: Definition of a real-valued random variable
I have trouble understanding the definition of a random variable:
Let $(\Omega, \cal B, P )$ be a probability space. Let $( \mathbb{R}, \cal R)$ be the usual
measurable space of reals and its Borel $\sigma$- algebra. A random variable
is a function $X : \Omega \rightarrow \mathbb{R}$ such that the preimage of any set $A \in \cal R$ is
measurable in $\cal B$: $X^{-1}(A) = \{ w: X(w) \in A \} \in \cal B$. This allows us to define
the following (the first P is the new definition, while the 2nd and
3rd Ps are the already-defined probability measure on $ \cal B$):
$ P(X \in A) = P(X^{-1}(A)) = P(\{ w : X(w) \in A \}) $
$ P(X = x ) = P(X^{-1}(x)) = P(\{ w : X(w) = x \}) $
Does it mean that the first P (the leftmost one) is a new probability measure defined on $( \mathbb{R}, \cal R)$? (in contrast with the 2nd and 3rd Ps defined on $(\Omega, \cal B)$)
What is $(\Omega, \cal B, P )$ for a binomial distribution for example?
In which cases do 2 random variables share the same probability space $(\Omega, \cal B, P )$?
AI: The probability of an event $X \in A$ is, by definition, the $P$-measure of the set of "outcomes" $\omega$ for which $X(\omega)$ is in $A$. Strictly speaking, all events are measurable subsets of the sample space, but it's usually simpler to speak of
events involving random variables without explicitly mentioning the sample space.
For example, if your random variable $X$ is the number of successes in
$n$ independent Bernoulli trials with probability of success $p$ in each one
(and thus has the binomial($n$,$p$) distribution), the sample space $\Omega$
could be $\{0,1\}^n$ (where $0$ corresponds to failure and $1$ to success). $\mathcal B$ would be all subsets of $\Omega$, and
$P$ gives each outcome with $k$ 1's and $n-k$ 0's probability $p^k (1-p)^{n-k}$.
But then if you want to relate this to some other random variable $Y$ that is
not determined by those same Bernoulli trials, you'll need a bigger sample space, where each outcome consists not just of the outcomes of those $n$ trials but also something else that determines $Y$.
EDIT: This is what distinguishes probabilists from real analysts. The
analyst is studying real-valued functions on a given space $\Omega$ with
a given $\sigma$-algebra $\mathcal B$ and probability measure $P$.
The probabilist will use the rigourous definition of the random variable $X$ in terms of $\Omega$ and $\mathcal B$ if necessary, but really thinks of $X$ in terms of a quantity involved in some (actual or imagined) experiment, and he/she is willing to change $\Omega$ and $\mathcal B$ in midsentence if that becomes convenient. |
H: Double set difference
Let $A,B \subset X$ and $A \cap B = \emptyset$. What is $A \setminus (A \setminus B)$? Since $A,B$ are disjoint, $A \setminus B = A$, whence $A \setminus (A \setminus B) = A \setminus A$. Now what is this set? It's the set of elements of $A$ that are not elements of $A$. Well, there are no such elements in $A$, right? Thus $A \setminus A = \emptyset$.
Is this correct? This is related to an exercise (my first ever) in measure theory, if you're wondering where this question could possibly come from.
AI: Yes, this is just fine. You can show more generally that $A\setminus(A\setminus B)=A\cap B$. Intuitively this should be clear: $B$ splits $A$ into the disjoint pieces $A\setminus B$ and $A\cap B$, which are relative complements of each other in $A$. |
H: Can all functions be restricted to a dense set from which they may be continuously extended?
Given a function $f:\mathbb{R}\rightarrow\mathbb{R}$ ($f$ is not necessarily continuous), will there always be a dense set $S\subseteq\mathbb{R}$ such that, if we restrict $f$ to $S$, we may then extend it to a continuous function of $\mathbb{R}$? (If the answer is "no," I would like some sort of counterexample.)
I have been trying to approach this with the $\mathbb{R}$-specific definitions of continuity and dense sets, but it has occurred to me that this might be a bad angle. I did manage to prove (or at least outline a proof) that, for any interval $I$ in $\mathbb{R}$, and for any $\epsilon>0$, there exist $x,y\in I$ such that $|f(x)-f(y)|<\epsilon$. I was able to further deduce that there must exist an interval $J$ in $\mathbb{R}$ with length $\epsilon$ such that $|\{x\in I:f(x)\in J\}|=\mathfrak{c}$. However, I am not sure where to go from here, or if these results are even useful to answering the question.
Thanks for any help!
AI: Here's a counterexample: let $$f(x)=\begin{cases}1;&x\geq0,\\0;&x<0.\end{cases}$$ If you restrict this to a dense subset of $\mathbb R$ and then extend it back in any way, the left and right limits at $0$ still won't agree, so the extension won't be continuous.
Here's a somewhat weaker statement that you might like: https://mathoverflow.net/questions/72800/every-real-function-has-a-dense-set-on-which-its-restriction-is-continuous. |
H: Regarding basis for a topology
We know that $\{[a,b):a,b\in \mathbb{Q}:a<b\}$ is a basis for a topology which is strictly finer than the usual topology on $\mathbb{R}$. I have also shown that this topology is contained in the lower limit topology on $\mathbb{R}$. But I couldn't find an example to show that this topology is strictly contained in the lower limit topology on $\mathbb{R}$.
Any hint will be appreciated.
AI: HINT: $[\sqrt2,2)$ is open in the lower limit topology but not in this one; why? |
H: how to prove $f^{-1}(B_1 \cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2)$
I am given this equation:
$f^{-1}(B_1 \cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2)$
I want to prove it: what i did is
I take any $a \in f^{-1}(B_1 \cap B_2)$, then there is $b \in (B_1 \cap B_2)$ so that $f(a)=b$. Because of $b \in (B_1 \cap B_2)$, it is true that $b \in B_1$ and $b \in B_2$, so $a \in f^{-1}(B_1)$ and $a \in f^{-1}(B_2)$.
this means $f^{-1}(B_1 \cap B_2) \subseteq f^{-1}(B_1) \cap f^{-1}(B_2)$.
is it ok?
AI: Yeah...this can be actually written in this way;
$a\in f^{-1}(B_1\cap B_2)$, means $f(a)\in B_1\cap B_2$ and so $f(a)\in B_1$ and $f(a)\in B_2$. Hence, $a\in f^{-1}(B_1)$ and $a\in f^{-1}(B_2)$ |
H: How to find the general solution of $\sin \left(\theta +\frac{\pi }{6}\right) =\cos 3\theta $
Find the general solution of the equation.
\begin{eqnarray} \sin \left(\theta +\frac{\pi }{6}\right)\space
=\space \cos 3\theta \\ \end{eqnarray}
The answers in my book are $\left(6n+1\right)\frac{\pi }{12},\space \left(n-\frac{1}{6}\right)\pi $.
Should I first convert all the things to cos(xxx)=??? or sin(xxx)=???.
So I try :
\begin{eqnarray}
\sin \theta \cos \frac{\pi }{6}+\cos \theta \sin \frac{\pi }{6}&=&\cos 2\theta \cos \theta -\sin 2\theta \sin \theta \\
\sin \theta \cos \frac{\pi }{6}+\sin 2\theta \sin \theta &=&\cos 2\theta \cos \theta -\cos \theta \sin \frac{\pi }{6}\\
\sin \theta \left(\sin 2\theta +\cos \frac{\pi }{6}\right)&=&\cos \theta \left(\cos 2\theta -\sin \frac{\pi }{6}\right)\\
\\?...&=&...?\\
\end{eqnarray}
What should I do?
I do not know whether my direction is correct.
Thank you for your attention.
Update 1 :
I can get the first ans by using method 1 but I cannot find the next one$\space \left(n-\frac{1}{6}\right)\pi $.
AI: Method $1:$
As $\cos(\frac\pi2-x)=\sin x$
$$\cos3\theta=\sin\left(\theta+\frac\pi6\right)=\cos\{\frac\pi2-\left(\theta+\frac\pi6\right)\}$$
$$\cos3\theta=\cos \left(\frac\pi3-\theta\right)$$
Now, we know $\cos x=\cos A\implies x=2n\pi\pm A$ where $n$ is any integer
Method $2:$
As $\sin(\frac\pi2-x)=\cos x$
$$\sin\left(\theta+\frac\pi6\right)=\cos3\theta=\sin\left(\frac\pi2-3\theta\right)$$
Now, we know $\sin x=\sin B\implies x=m\pi+(-1)^m B$ where $m$ is any integer |
H: $ \{ x : ( \inf_{n \geq k} f_n)(x) > a \} = \bigcap_{n=k}^{\infty} \{x: f_n(x) > a \} $
Pick $x \in \{ x : ( \inf_{n \geq k} f_n)(x) > a \} \iff ( \inf_{n \geq k} f_n)(x) > a \iff f_n(x) > a \forall n \geq k$. Therefore, $x \in \{y : f_n(y) > a \}$ for all $n \geq k$. And this holds iff $x \in \bigcap_{n=k}^{\infty} \{y : f_n(y) > a \}$. Hence, we have $$ \{ x : ( \inf_{n \geq k} f_n)(x) > a \} = \bigcap_{n=k}^{\infty} \{x : f_n(x) > a \} $$.
Is this correct?
AI: If $\inf_{n\geqslant k}f_n(x)\gt a$, then for all $n\geqslant k$, we have $f_n(x)\gt a$. But the converse doesn't need to hold: take $f_n(x):=\frac 1n$ and $a=0$. |
H: Am I correctly determining the convergence or divergence of these series: $\sum \frac{1}{n(\ln(n))^{5/4})}$?
I'm doing online homework, which immediately tells me if I got it right or wrong, and I got the following question correct, but I just want to make sure I'm getting it correct for the right reasons.
It asks me to determine the convergence of divergence of the given series
$$\sum_{n = 1}^\infty \left(\frac{1}{n(\ln(n))^{5/4})}\right)$$
Literally all I'm doing is the p-series test, so I'm looking at the exponent in the denominator, which is 5/4 > 1 and thus I'm saying that it converges. However, I get the feeling that there is more to it than that and that I actually need to do some sort of evaluating, but I'm not sure because the question is online and it just asks if it is converging or diverging, so all I know is that my answer is correct, I don't actually know if my reasoning is correct. Is that are all these is to the p-series test?
AI: Your reasoning is not correct. Hint: Note that $\dfrac{1}{x(\ln x)^{5/4}} = \left(-4(\ln x)^{-1/4}\right)'$ and use the Cauchy-Maclaurin integral test. |
H: Divisible abelian groups.
From Hungerford's Algebra (with slight changes)
1) Let $p$ be a prime and let $Z(p^{\infty})$ be the following subset of the group $\Bbb{Q}/\Bbb{Z}$ $$Z(p^{\infty}) = \{ \overline{a/b} \in \Bbb{Q}/\Bbb{Z} | a,b \in \Bbb{Z}, b=p^i \operatorname{for some} i \geq 0\}.$$ Show that for each prime $p$, $Z(p^{\infty})$ is a divisible group.
2) No nonzero free abelian group is divisible.
1) I find this a bit confusing. According to the textbook "An abelian group $D$ is said to be divisible if given any $y \in D$ and $0 \not= n \in \Bbb{Z}$, there exists $x \in D$ such that $nx=y$."
So, for example, let's look at $p=3$. Let us choose $2/9 \in Z(3^{\infty})$ and $4 \in \Bbb{Z}$. Now we need to find an $x \in Z(3^{\infty})$ so that $4x = 2/9$. But then $x = 2/36$, and $36$ is not a power of $3$. How is this group divisible then?
2) Let's assume, for contradiction, that there is a free abelian group is divisible.
We have a lemma and a proposition in the textbook that states
Lemma. An abelian group $D$ is divisible if and only if $D$ is an injective (unitary) $\Bbb{Z}$-module.
Proposition. A direct product of $R$-modules $\Pi_{i \in I} J_i$ is injective if and only if $J_i$ is injective for every $i \in I$.
Since a free abelian group is a free $\Bbb{Z}$-module, the group must be isomorphic to a direct sum of copies of $\Bbb{Z}$. By the lemma, it must be injective; and by the proposition, the group is injective if and only if $\Bbb{Z}$ is injective. But, since $\Bbb{Z}$ is not injective, we have a contradiction.
Is my answer correct?
Thanks in advance.
AI: When you want to divide $2/9$ by $4$, you need to find some $x \in Z(3^\infty)$ such that $4x = 2/9 \pmod 1$.
Since $2/9 = 2/9 + 2 = 20/9$, we can pick $x = 5/9$.
Let $Z(p^n) = \{a/b \in Z(p^\infty), b \mid p^n\} = p^{-n} (\Bbb Z / p^n \Bbb Z)$.
If $q$ is prime number distinct from $p$, then multiplication by $q$ is a bijection from $\Bbb Z/p^n\Bbb Z$ to itself, so it is also a bijection from $Z(p^n)$ to itself.
Since $Z(p^\infty)$ is the reunion of the $Z(p^n)$, multiplication by $q$ is a bijection from $Z(p^n)$ to itself.
Since $Z(p^\infty)$ is obviously $p$-divisible, it is $q$-divisible for all primes $q$, so ti its divisible.
Your answer for (2) is not completely correct, since a countably infinite direct sum is not a direct product. However you can still pick a basis for $G$, pick one basis element, and mention that this element it is not divisible by $2$. |
H: Find the absolute value of the difference between the area of these triangles.
Let $\triangle ABC$ and $\triangle ABC'$ be two non congruent triangles with side $AB=4$, $AC=AC'=2$$\sqrt{2}$ and $\displaystyle\angle B=30^\circ$. Find the absolute value of the difference between the area of these triangles.
Diagram:-
http://prntscr.com/1vl225
AI: HINT:
Apply $$\cos 30^\circ=\frac{a^2+4^2-(2\sqrt2)^2}{2\cdot a\cdot 4}$$ where the values of $a$ represent $BC,BC'$
The area of $\triangle ABC=\frac12BC\cdot AB\cdot\sin30^\circ$ and similarly for $\triangle ABC'$ |
H: Is Dedekind completion of ${}^{\ast}\Bbb R$ a Archimedean field?
Here's Theorem 1.2 on page 6, Martin Andreas Väth's Nonstandard Analysis(See here on googlebooks)
The Dedekind completion $\overline{X}$ of a totally ordered field $X$
is a complete Archimedean field with $\Bbb{Q}_{\overline{X}}$ as the canonical copy of $\Bbb{Q}_{X}$.
$X$ has the Archimedean property. For each $x \in X$ there is some $n \in \Bbb{N}_{X}$
such that $n > x$.
Each totally ordered field X contains a “canonical copy” of the set $\Bbb{N}_{X}$, namely $\{1_X, 1_X +1_X, 1_X +1_X +1_X, \ldots\}$.
${}^{\ast}\Bbb R$ is a totally ordered field without Archimedean property. Isn't it the case that its Dedekind completion doesn't have Archimedean property?
AI: The Dedekind-completion is an order completion, and if the field is non-Arcihmedean then its Dedekind-completion is not a field at all.
To see this simply note that in the completion, there is a point $t$ which is the realization of the cut $R=\{x\mid\exists n\in\Bbb N. x<n\}$. And $t-1$ cannot exist. |
H: Why, in the Rabin cryptosystem, during decryption, do we get four possibilities instead of two?
The encryption algorithm : c=m^2 modn, should mean that we have two(or one) possibilities for m. Why do we get four squareroots?
AI: For a prime $p$ the congruence $x^2 \equiv a \pmod p$ has either $0$ or $2$ solutions for $a \neq 0$. For example, $3^2 \equiv 2^2 \equiv 4 \pmod 5$, and $2^2 \equiv 5^2 \equiv 4 \pmod 7$. In a hand-wavy way, we can get a solution to both congruences at the same time by choosing one from each. Two solutions and two choices leads to $4$ overall possibilities. (In a less handwavy way, you follow the Chinese Remainder Theorem).
For example, the four square roots of $4$ mod $35$ are $2, 12, 23,$ and $33$. In these, you'll notice a pair congruent to $2 \pmod 5$, a pair congruent to $3 \pmod 5$, a pair congruent to $2 \pmod 7$, and a pair congruent to $5 \pmod 7$ (these are the same as the solutions I mentioned earlier). This is not a fluke, but a byproduct of the Chinese Remainder Theorem in action. |
H: $(123)$ and $(132)$ are not in the same conjugacy class in $A_4$
Could you tell me how to show that $(123)$ and $(132)$ are not in the same conjugacy class in $A_4$?
I know that all 3-cycles can't be in the same class, because the order of each class must divide $|A_4| = 12$ and there are eight 3-cycles in $A_4$.
If $\sigma$ is a 3-cycle fixing $n$ and $τ \in A_4$ then $\tau \circ \sigma \circ \tau^{-1}$
is a 3-cycle
fixing $τ(n)$. (And there are 3 such permutations). If we conjugate $(123)$ by $(132)$ we will get $(123)$ and vice versa. So it would appear that it only makes sense to conjugate $(123)$ by something with a $4$. If we conjugated $(123)$ by $(124)$, we would need $(124)(123)(124)^{-1}$ to send 4 to 4, but it doesn't. It turns out that such $\tau$ cannot have anything "common" with $(123)$ - what I mean by this is that there can't be $(12...), (23...), (31...)$ in $\tau$. So the only cycles that remain are $(134), (142), (243)$ - and they are indeed ion the same class as $(123)$.
As you see, this isn't the best way to solve this problem.
I also know that if $g \in A_n$ commutes with an odd permutation, then all permutations with the same cycle type as $g$ are in one conjugacy class, but I have no idea how to show that in $A_4$ no odd permutation commutes with $(123)$.
Could you make it more mature?
Thank you.
AI: One way to see this is to work in $S_4$ instead. There $(1\,2\,3)$ and $(1\,3\,2)$ are conjugates, of course, since $(2\,3)^{-1}(1\,2\,3)(2\,3)=(1\,3\,2)$.
If we have a $\tau$ such that $\tau^{-1}(1\,2\,3)\tau=(1\,3\,2)$, it must factor as $\tau=\rho(2\,3)$ for some $\rho$, and we have $$(2\,3)^{-1}\rho^{-1}(1\,2\,3)\rho(2\,3)=(1\,3\,2)=(2\,3)^{-1}(1\,2\,3)(2\,3)$$
Canceling the $(2\,3)$s we find $\rho^{-1}(1\,2\,3)\rho=(1\,2\,3)$, and it's then easy to see that $\rho(1)$ determines $\rho$ and there are only three possibilities, which are all even permutations. So $\tau$ must be odd and so not in $A_4$.
This may or may not be easier to follow than Tobias' suggestion to brute-force the possible $\tau$s directly. Personally I think it's easier to think about possible stabilizers of a cycle than to keep things straight while trying to conjugate one cycle into another. |
H: About continued fractions as best rational approximations
I'm reading this notes about continued fractions:
http://www.math.jacobs-university.de/timorin/PM/continued_fractions.pdf
I had no problems understanding everything there, except one thing that has me stuck.
At page 9, the author proves that the "convergents" are the best rational approximations.
In that proof, he says that $|x-\frac{h_n}{k_n}|<\frac{1}{2k_n^2}$ (*).
Previously, he has shown that $|x-\frac{h_n}{k_n}|<\frac{1}{k_{n+1}k_n}$. And as $k_{n+1}>k_n$ (increasing denominators), I see that you can get $|x-\frac{h_n}{k_n}|<\frac{1}{k_n^2}$.
But I really don't understand where that 2 in (*) (which is vital for the proof) comes from. The author seems to justify that step saying that denominators increase and $k_{n+1}\geq 2$...
Anybody can help me with this? Thanks a lot in advance.
AI: Well, it's wrong. We have the equality
$$\left\lvert x - \frac{h_n}{k_n}\right\rvert = \frac{1}{k_n(\alpha_{n+1}k_n + k_{n-1})},$$
where $\alpha_{n+1} = [a_{n+1},\, a_{n+2},\, \dotsc\,]$ is the $n+1^{\text{st}}$ complete quotient, and the $a_k$ are the partial quotients. We know that $a_{n+1} < \alpha_{n+1} < a_{n+1}+1$ (unless the continued fraction ends at $a_{n+1}$, in which case we have equality on the left, or it ends at $a_{n+2}$ and that is $1$, then we have equality on the right).
So we only have
$$\left\lvert x - \frac{h_n}{k_n}\right\rvert < \frac{1}{2k_n^2}$$
if $\alpha_{n+1} + \frac{k_{n-1}}{k_n} > 2$. If $a_{n+1} \geqslant 2$, that is satisfied, also if $a_{n+1} = 1$ and $k_{n-1}$ is close enough to $k_n$, but not in general.
As an example, consider $x = \frac{137}{127} = [1,\, 12,\,1,\,2,\,3]$ and its convergent $\frac{13}{12} = [1,\,12]$. We have
$$\frac{13}{12} - \frac{137}{127} = \frac{7}{12\cdot 127} > \frac{1}{2\cdot 12^2}.$$
The result, that the convergents (except possibly the $0^{\text{th}}$) are best approximations, is true, but the given proof is invalid.
It is not hard to fix, though. We suppose $n \geqslant 1$, since as we said, the $0^{\text{th}}$ convergent need not be a best approximation. There are two possibilities,
$\frac{p}{q}$ and $\frac{h_n}{k_n}$ lie on the same side of $x$, then $$\frac{1}{k_n^2} > \left\lvert x - \frac{h_n}{k_n}\right\rvert \geqslant \left\lvert \frac{p}{q} - \frac{h_n}{k_n}\right\rvert = \frac{\lvert pk_n-qh_n\rvert}{qk_n} \geqslant \frac{1}{qk_n} \Rightarrow q > k_n,$$
or they lie on different sides of $x$, then $$\frac{1}{k_nk_{n-1}} = \left\lvert\frac{h_{n-1}}{k_{n-1}} - \frac{h_n}{k_n}\right\rvert = \left\lvert\frac{h_{n-1}}{k_{n-1}} - \frac{p}{q} \right\rvert + \left\lvert \frac{p}{q} - \frac{h_n}{k_n}\right\rvert \geqslant \frac{1}{qk_{n-1}} + \frac{1}{qk_n} = \frac{k_n+k_{n+1}}{qk_nk_{n-1}},$$ whence $q \geqslant k_n + k_{n-1} > k_n$. (We used that $\left\lvert x -\frac{h_n}{k_n}\right\rvert < \left\lvert x - \frac{h_{n-1}}{k_{n-1}}\right\rvert$.) |
H: Removable singularity theorems for manifolds
All the theorems on removable singularities are for functions defined on open domains $\Omega \in \mathbb{C}$. But what are the corresponding theorems for functions defined on Riemann surfaces? How do they differ and are there any extra issues we need to take into account?
AI: As is pointed out in the comments, for Riemann Surfaces, the following version of Riemann's Removable Singularity Theorem holds:
Theorem:Let $X$ be a Riemann surface, and $f:X-\{p\}\to \mathbb{C}$ a holomorphic function. If $f$ is bounded in some neighborhood of $p$, then $f$ extends uniquely to a holomorphic function $f:X\to\mathbb{C}$.
This is proven, as the comment suggests, by working some some coordiante disc around $p$ and applying the usual Riemann's Theorem.
A generalization of this to higher-dimensional manifolds has several possible forms. The most general that I know of is Hartog's Extension Theorem. This is a much deeper result, and has many implications. It's algebraic analogue is every useful in algebraic geometry as well. |
H: Tips for simplifying
I'm working on my homework for Uni. I'm stuck on some problems. I'm not looking for a solution, just some hints on how to do it.
$ \frac{3R^2-6Rv}{2v-R} \Leftrightarrow -3R$
I think here I'm also missing the same idea as in the problem above:
$\frac{R}{v}-\frac{2t}{R} = \frac{R}{v+3t} \Leftrightarrow R =\pm\sqrt{\frac{2}{3}v^2+2vt}$
Also: Sorry for my bad latex, I'm still learning. Any tips, advice or comments are highly appreciated.
AI: $$\frac{3R^2-6Rv}{2v-R}=3R\frac{R-2v}{2v-R}=-3R$$
or
$$\frac Rv-\frac{2t}R=\frac R{v+3t}\stackrel{\text{Mult. through by}\;R\cdot v\cdot(v+3t)}\implies R^2(v+3t)-2vt(v+3t)=R^2v\implies$$
$$\left(3t\right)R^2=2vt(v+3t)\implies R=\pm\sqrt{\frac{2v(v+3t)}{3}}\;\;,\;\;t\neq 0$$ |
H: Conditional probability in continuous distribution
I'm struggling understanding how conditional probability for a continuously distributed random variable is to be calculated. The task is as follows:
$f(t) = 1/8 * (4-t)$ for $0 < t <= 4 $ and $0$ elsewhere
We know that in this interval exactly one event $E$ will happen, with probability at a specified time $t$ given by the distribution. Find the probability that $E$ will happen in the interval $1<t<=1.5$ given that it has not happened in the interval $0<t<=1$.
I tried setting this up using Bayes' theorem, but it did not seem to be a good solution. How to solve this one?
AI: Try to write it out to make it more clear:
$$
Pr(1<t \leq 1.5| 1<t\leq 4)=\frac{Pr(1<t \leq 1.5, 1<t\leq 4)}{Pr(1<t\leq 4)}=\frac{Pr(1<t \leq 1.5)}{Pr(1<t\leq 4)}
$$
where the last equality holds because $(1, 1.5]$ is a subset of $(1, 4]$. Then you can simply calculate the numerator and the denominator using standard ways. |
H: Tips on solving precalc equation
Need some hints on my uni homework.
I'm sure I made a mistake somewhere, but I don't see it (I'm solving for x):
$$\frac{x+\frac{1}{y}}{\frac{2}{z}-x}=\frac{2-\frac{2z}{y}}{\frac{y}{z}-1}
\\Solution: x=\frac{3}{y+2z} $$
$$\frac{xy+1}{y}*\frac{z}{2-xz} = \frac{2y-2z}{y}*\frac{z}{y-z}\tag{1}$$
$$\frac{xyz+z}{2y-xyz}=\frac{2yz-2z^2}{y^2-yz} | *(2y-xyz)\tag{2}$$
$$xyz+z=\frac{4yz-2xyz^2-4z^2+2xz^3}{y-z} | *(y-z)\tag{3}$$
$$xy^2z-xyz^2+yz-z^2=4yz-2xyz^2-4z^2+2xz^3\tag{4}$$
$$xy^2z+xyz^2-2xz^3=3yz-3z^2\tag{5}$$
$$x=\frac{3yz-3z^2}{y^2z+yz^2-z^3}\tag{6}$$
$$x=\frac{3(y-z)}{y^2+yz-z^2}\tag{7}$$
I know I can go on from there, but I plugged that into the calculator and got conflicting results with the given solution (using two random values for $y$ and $z$.
Any suggestions or comments on the latex markup are highly appreciated!
AI: The equation gives $(x+ \frac 1 y) (\frac yz - 1) = (\frac 2z - x) (2 - \frac {2z} y)$ by the cross-ratio thing.
$\frac{xy+1}y \frac {y-z}z = \frac {2+xz}z \frac{2(y-z)}y$ Reducing fractions.
$xy+1 = 2 (2-xz)$ eliminate common factors.
$xy + 1 = 4 - 2xz$
$x(y+2z) = 3$
$x = \frac 3{y+2z}$
So it certianly does give the right results.
You make life complicated for yourself, by not eliminating common factors after writing equation 1. There are three common factors $y-z$, $y$, and $z$, this reduces equation $2$ to $\frac{xy+1}{2-xz}=2$, which follows on from what i do at 'eliminate common factors' above.
The mistake, though is in equation 6, where the $2xz^3$ is carried from eqn 5 to $xz^3$ in eqn 6. When the 2 is restored, the top and bottom are multiples of $y-z$, which gives the indicated result. |
H: Prove that for each prime $p$ there exists a nonabelian group of order $p^3$
Can anyone helo me out with this question: "Prove that for each prime there exists a nonabelian group of order $p^3$".
This is my attempt:
I know how to show that $|Z(G)|= p$.
So, since $C(x)$ contains $Z(G)$ for every $x$ not in $G$ then $|C(x)|> |Z(G)|$ i.e. $|C(x)|> p$. But $C(x)$ is always a subgroup of $G$ so it’s order must divide the order of $G$ and so $|C(x)= p^2$. (If $|C(x)|= p^3$ then $|G:C(x)|= |G|/|C(x)| = p^3/p^3 = 1$, so the number of elements in every conjugacy class would be 1, which would imply that every element is in the center of $G$, which in turn would mean that $G$ is abelian, which is not what we're after.)
If $|C(x)|= p^2$ then $|G:C(x)|= |G|/|C(x)| = p^3/p^2= p$.
Here’s where I don’t know how to go forward. I know that if there would be $p^2 – 1$ conjugacy classes then $|G |= |Z(G)|+ (p^2-1) \cdot|G:C(x)|= p + (p^2-1)p = p + p^3 – p = p^3$.
But how do I know that there actually exists a group with exactly $p^2 – 1$ conjugacy classes?
Greatful for input on my reasoning and also other ways of proving this, including how to construct such a group.
AI: Your approach doesn't seem fruitful. All of the techniques you are using could at best show that it's not impossible to have an abelian group of order $p^3$. :)
Basically this comes down to the fact that $|\text{Aut}(\mathbb{F}_p^2)|=(p^2-1)(p^2-p)$, and so, in particular is divisible by $p$. Thus, there is a non-trivial semi-direct product $\mathbb{F}_p^2\rtimes (\mathbb{Z}/p\mathbb{Z})$.
Explicitly you can construct a non-abelian group of order $p^3$ by looking at the Heisenberg group:
$$H(p):=\left\{\begin{pmatrix}1 & a & b\\ 0 & 1 & c\\ 0 & 0 & 1\end{pmatrix}:a,b,c\in\mathbb{F}_p\right\}$$
It's probably worth mentioning that there are always exactly $5$ groups of order $p^3$, two non-abelian, for each prime $p$. See here.
I feel like it's worth mentioning the following interest fact:
Call a number $n$ nilpotent if $n=p_1^{e_1}\cdots p_m^{e_m}$ such that for all $i,j$ and for all $1\leqslant k\leqslant e_i$ we have that $p_i^k\not\equiv 1\mod p_j$.
Theorem:
All groups of order $n$ are nilpotent if and only if $n$ is a nilpotent number.
All groups of order $n$ are abelian if and only if $n$ is a cubefree nilpotent number.
All groups of order $n$ are cyclic if and only if $n$ is a squarefree nilpotent number (this is equivalent to $(n,\varphi(n))=1$). |
H: Strictly convex function: how often can its second derivative be zero?
It's a basic fact that a twice-differentiable function from $\mathbb{R}$ to $\mathbb{R}$ is strictly convex if its derivative is positive everywhere.
The converse is not true: consider, e.g., $f(x) = x^4$, which is strictly convex, with $f ''(0)=0$.
Is there a partial converse, however?
Is it true, e.g., that a strictly convex twice-differentiable function from $\mathbb{R}$ to $\mathbb{R}$ can have zero second derivative at at most one point?
Thanks for your help!
AI: We have a partial converse. It's not quite as strong as allowing only finitely many or only isolated zeros of the second derivative, but it's strong enough.
For a function $f \colon I \to \mathbb{R}$, where $I \subset \mathbb{R}$ is an open interval, possibly $I = \mathbb{R}$, convexity can be formulated as
$$\bigl(\forall u,v,w \in I\bigr)\left(u < v < w \Rightarrow \frac{f(v)-f(u)}{v-u} \leqslant \frac{f(w)-f(v)}{w-v}\right)$$
and strict convexity with the strict inequality.
For a differentiable function $f$, these conditions can be seen to be equivalent to
$f$ is convex if and only if the derivative $f'$ is non-decreasing,
$f$ is strictly convex if and only if $f'$ is strictly increasing.
For a twice continuously differentiable function $f$, the above can be easily seen to be equivalent to
$f$ is convex if and only if $f'' \geqslant 0$ everywhere,
$f$ is strictly convex if and only if $f'' \geqslant 0$ everywhere and $f''$ does not vanish on any non-empty open interval $J \subset I$. |
H: A = {x/x^2+1 : x belongs to reals } Show that A is a subset of [-1/2, 1/2]
I am really stuck with this assignment and have very little idea on how to proceed. I think using derivative is not allowed. How should I proceed, could anyone give any tips?
I've thought something like.
If $x>0$
$$
x^2+1 > x \geq 1
$$
But this is really all I got, I've tried changing the form of the expression but nothing really. I would be grateful for tips!
AI: We will prove that
$$
\left| \frac{x}{x^2+1}\right|\le\frac{1}{2}
$$
for every real number $x$. Square both sides of the inequality then we get
$$
\frac{x^2}{(x^2+1)^2}\le\frac{1}{4}.
$$
In fact, these two inequalities are equivalent, and last inequality is easy to prove: it is equivalent to $0\le (x^2-1)^2$. |
H: How to find the general solution of $\tan \left(x+\frac{\pi }{3}\right)+3\tan \left(x-\frac{\pi }{6}\right)=0$
Find the general solution of the equation.
\begin{eqnarray} \tan \left(x+\frac{\pi }{3}\right)+3\tan \left(x-\frac{\pi }{6}\right)=0\\ \end{eqnarray}
The answers in my textbook are $n\pi $ and $n\pi +\frac{\pi }{3}$.
Previously, I compute the similar questions by using the following operations :
\begin{eqnarray}
\\\tan 3x&=&\cot 5x\\
\\\tan 3x&=&-\tan \left(\frac{\pi }{2}+5x\right)\\
\\\tan 3x&=&\tan \left(-\frac{\pi }{2}-5x\right)\\
\\3x&=&-\frac{\pi }{2}-5x\\
\\8x&=&n\pi -\frac{\pi }{2}\\
\\x&=&\frac{n\pi }{8}-\frac{\pi }{16}\\
\end{eqnarray}
But now there is a 3 in front of the second tan.
What should I do?
I do not know whether my method is correct. If you have any other methods, would you mind telling me the methods?
Thank you for your attention.
Update 1 : Found one of the answers, are my operations correct?
AI: If we assume that your question is to solve the equation $tan(x+\pi/3)+3tan(x-\pi/6)=0$.
if we let $A=x+\pi/3, B=x-\pi/6$, then $A-B=\pi/2$, so we can transform the term $tanA$ to $tan(\pi/2+B)$.
and because $tan(\pi/2+B) = -cotB$, the equation will be $-cotB+3tanB=0$, $tan^2B=1/3$ so the general solution of $B=x-\pi/6=n\pi+\pi/6$ or $n\pi-\pi/6$
so $x=n\pi +\pi/3$ or $n\pi$ |
H: Conditional expectation $E(X\mid XY)$ when $(X,Y)$ is standard normal
I have to calculate
$\mathbb{E}(X|X*Y)$ with X,Y being independent and standard normal distributed. I got at tip in this post (Conditional expectation on components of gaussian vector), that I should use the definition and Bayes Thm to solve the problem. I played around a bit, but I just don't get it :(
May anyone give me another hint?
AI: As every conditional expectation, $E[X\mid XY]=w(XY)$ for some measurable function $w$. Recall that:
Conditional expectations depend only on the joint distribution of the random variables considered, in the sense that, if $E[X\mid XY]=w(XY)$, then $E[X'\mid X'Y']=w(X'Y')$ for every $(X',Y')$ distributed like $(X,Y)$.
Choosing $(X',Y')=(-X,-Y)$ above, one gets $X'Y'=XY$ hence $$w(XY)=E[-X\mid XY]=-E[X\mid XY]=-w(XY).$$ Thus, $$E[X\mid XY]=0.
$$
One sees that $E[X\mid XY]=0$ for every centered gaussian vector $(X,Y)$, neither necessarily independent nor standard.
Still more generally:
Let $\Xi$ denote any centered gaussian vector, $u$ an odd measurable function such that $u(\Xi)$ is integrable and $v$ an even measurable function. Then, $$E[u(\Xi)\mid v(\Xi)]=0.$$ |
H: if X not finite then O is not a $\sigma$ - algebra
let O (family of sets) consist of those sets which are either finite or have a finite complement. then O is an algebra. I did this part!
my question now is:
if X (space) is not finite, then O is not a $\sigma$ - algebra.
The definitions of algebra and $\sigma$ - algebra are clear to me. For the first part of the question it didn't matter whether X was finite, because if it wasn't then $X^{c}$ would, and therefore would be included in O. And since $\sigma$ - algebra differ from algebra in that we have closure under countable unions, I assume that I have to look there. But I am not sure how to go abut it.
Any sorts of tip would be great, thank you very much.
AI: Hint: $X$ contains a countable subset with an infinite complement. |
H: If $F$ has characteristic $p$, then $pa$ =0 for all $a \in F$
I have to prove the statement in the title, i.e
If $F$ has characteristic $p$, then $pa = 0$ for all $a \in F$, $p$ prime.
From the definition of a characteristic of a field, we have that
If F is a field of characteristic p then the prime field P of F is isomorphic to $\mathbb{Z}_p$.
i.e $\exists \phi :P ->\mathbb{Z}_p$, a bijective ring map.
Do I have to prove that (p) is an ideal in F so that $pa=0$ in $F/I$ ? ($I = (p)$)
Thanks
AI: Since $F$ has a prime field $\Bbb{F}_p$ so $p1=0$ (where $1$ is a multiplicative identity of $F$.) So by distributive law we get
$$
pa=\underbrace{ a+a+\cdots+a }_{p \text{ times}}=a(\underbrace{ 1+1+\cdots+1 }_{p \text{ times}})=a0=0.
$$ |
H: Does $\det(I+A(I+B)^{-1})=\det(I+A^*(I+B)^{-1})$ hold for $A,B$ positive semi-definite matrices?
Is
$$\det(I+A(I+B)^{-1})=\det(I+A^*(I+B)^{-1})$$
where $I$ is identity matrix, $A,B$ are positive semi-definite complex valued matrices and $A^*$ is the conjugate (Hermitian) transpose of $A$.
Thanks a lot in advance.
Question related to Possible matrix-determinant identity
AI: For complex matrices, positive-semi definite implies hermitian. So $A^*=A$.
Edit: see my answer here for a proof. |
H: Help solving this Linear First order ODE
I'm trying to solve this ODE :- $x (dy/dx) + y \log(x) = e^x x^{(1-1/2 \log(x))}$
I divided the equation throughout by $x$, obtaining $(dy/dx) + y\log(x)/x = e^x x^{(\log(x^{-1/2}))}$.
Then, I obtained the Integrating factor as $e^{((\log (x)^2)/2)}$.
Then, $y e^{((\log(x)^2)/2)} = \int e^{(x + ((\log(x)^2)/2))} x^{(\log(x^{-1/2}))} dx$.
I'm not sure how to proceed from here. It seems the integral on the right has to be evaluated by parts but it is quite tedious and tends to get messy. Is there a simpler way to evaluate the integral? Or is there any other approach to this differential equation?
AI: You're almost there. You just need to recognize that
$$x^{\log{x^{-1/2}}} = e^{-(1/2) \log^2{x}}$$
The nasty factors cancel and the integral becomes trivial. |
H: Why do mathematicians use only symmetric matrices when they want positive semi-definite matrices?
Why do mathematicians use only symmetric matrices when they want positive semi/definite matrices?
I mean I haven't seen using non-symmetric positive semi/definite matrices. If non-symmetric positive semi/definite matrices exist can those be always written by a symmetric PSD matrix?
AI: First, one can argue that non-symmetric positive definite matrices are pathological, in the sense that when you move to the complex case all positive definite matrices are hermitian.
For a non-symmetric positive definite matrix you can say little more than the fact that it has positive eigenvalues. You don't have many of the nice properties that symmetry adds. For instance, without symmetry you don't even have that the singular values agree with the eigenvalues, nor diagonalizability.
Edit: here is why in the complex case, positive semidefinite implies hermitian. Actually, the proof implies that in the complex case $A$ is hermitian if and only if $x^*Ax\in\mathbb R$ for all $x$.
Assume $x^*Ax\in\mathbb R$ for all $x$. then
$$
\mathbb R\ni(y+\alpha x)^*A(y+\alpha x)=y^*Ay+\overline\alpha\,x^*Ay+\alpha\,y^*Ax+|\alpha|^2\,x^*Ax.
$$
As this expression is real, it equals its complex conjugate
$$
y^*Ay+\alpha\,y^*A^*x+\overline\alpha\,x^*A^*y+|\alpha|^2\,x^*Ax.
$$
So
$$
\overline\alpha\,x^*Ay+\alpha\,y^*Ax=\alpha\,y^*A^*x+\overline\alpha\,x^*A^*y.
$$
Taking first $\alpha=1$ and then $\alpha=i$, we get
$$
x^*Ay+y^*Ax=y^*A^*x+x^*A^*y,
$$
$$
-i\,x^*Ay+i\,y^*Ax=i\,y^*A^*x-i\,x^*A^*y.
$$
Multiplying the first equation by $i$ and adding, we get
$$
2i\,y^*Ax=2i\,y^*A^*x.
$$
As this works for any $x,y$, we deduce that $A=A^*$. |
H: Bijection question
I need to show that $(m,n) \mapsto 2^{m-1}(2n-1)$ is a bijection of $\mathbb{N} \times \mathbb{N}$ on $\mathbb{N}$
I think I need to show that the expression is both injective and surjective, but I am not sure how to do that.
Maybe a kind person can help me in some direction?
Thanks in advance.
AI: For surjectivity, you'll want to show that every natural number is an odd number (possibly $1$) times some non-negative integer power of $2$.
For injectivity, suppose that $$2^{m_1-1}(2n_1-1)=2^{m_2-1}(2n_2-1),$$ where $m_1\ge m_2,$ so that $$2^{m_1-m_2}=\frac{2^{m_1-1}}{2^{m_2-1}}=\frac{2n_2-1}{2n_1-1}$$ is an integer. Since a ratio of two odd numbers can't be even, it follows that $2^{m_1-m_2}=1,$ so $m_1=m_2,$ and likewise $n_1=n_2$. |
H: Finding inverse of functions[methods of]
I am now trying to understand functions, inverses and composites. I must admit am not getting a thing. But following some leads, I managed to work one as below. Is this a good understanding on hows and whys?
Find the inverse of function $ y=\sqrt{6+x}$
$ y=\sqrt{6+x}$ can be expanded as $y^2=6+x$
solving for $x$ we get $x=y^2-6$
swapping the position of $x$ and $y$ we get $y=x^2-6$
$ \therefore$ the inverse of $y=\sqrt{6+x}$ is $$y=x^2-6$$
AI: Careful here, your work is actually correct (the swapping method is in fact standard), except that you must keep track of your domain. The inverse of $y=\sqrt{6+x}$ is $y=x^2-6$, and your work is absolutely correct. However the domain on $\sqrt{6+x}$ is $x \geq -6$, hence on the inverse, we have that $y \geq -6$.
Now for the inverse $y=x^2-6$, since $y \geq -6$, it follows that $x \geq 0$ as,
$$x^2 -6 \geq -6 \Rightarrow x^2 \geq 0 \Rightarrow x \geq 0.$$ Now if you plot the two solutions you will have something that looks like this.
If you do not keep track of your domains and ranges like this, then your work is wrong as the inverse must be 1-1, and the inverse of the inverse must also be 1-1. The inverse of the inverse must give the function you started with on the correct domain. |
H: Find expected value of $f(x)=\frac{1}{\pi(1+x^2)}$ (Integral does not converge)
I want to find expected value of random variable given by pdf:
$$f(x)=\frac{1}{\pi(1+x^2)}$$
So I have:
$$E(X)=\int_{-\infty}^{\infty}xf(x)dx=\int_{-\infty}^{\infty}\frac{x}{\pi(1+x^2)}dx$$
Integral is equal to: $$\frac{\log(1+x^2)}{2 \pi}$$
So it will not converge for plus or minus $\infty$. Does it mean that there exist no expected value?
AI: Yes, the expected value does not exist. However, note that the principal value of the integral is indeed zero, i.e., if
$$F(R) = \int_{-R}^R \dfrac{x}{\pi(1+x^2)}dx$$
then $F(R) = 0$ for all $R$ and hence
$$\lim_{R \to \infty} \int_{-R}^R \dfrac{x}{\pi(1+x^2)}dx = 0$$
However, the median and mode are well-defined and is zero. This distribution is called the Cauchy distribution. |
H: How does this differentiation come about ?
The question is that: If $f(z)$ is analytic, show that $\frac{\partial f}{\partial \bar z} = 0$
Now, assuming $f(z) = u + iv$
$\frac{\partial f}{\partial \bar z} = \frac{\partial}{\partial \bar z}(u + iv)$
What the book does is this:
$$
(\frac{\partial u}{\partial x}.\frac{\partial x}{\partial \bar z} + \frac{\partial u}{\partial y}.\frac{\partial y} {\partial \bar z})
$$
for the $u$ part. Similarly for the $v$ part
Can someone please explain how that is done ?
AI: You have a function $u = u(x,y)$ of two variables. In turn, each variable is a function of two more variables, namely $z$ and $\overline{z}$. While you cannot write down $u$, you can write down $x$ and $y$ as
$$
x = \frac{z+\overline{z}}{2}, \qquad\text{ and }\qquad y = \frac{z-\overline{z}}{2i}
$$
Now you are looking to find $\frac{\partial u}{\partial \overline{z}}$. By the Chain rule, you get
$$
\frac{\partial u}{\partial x}.\frac{\partial x}{\partial \bar z} + \frac{\partial u}{\partial y}.\frac{\partial y} {\partial \bar z}
$$
In other words, you are finding the rate of change of $u$ w.r.t $\overline{z}$ : This will depend on the rate of change of $u$ w.r.t. $x$ and $y$, which in turn change w.r.t. $\overline{z}$. The chain rule helps you compute how these rates compound.
(Note : Just think of $z$ and $\overline{z}$ as symbols here, and not as complex numbers, if that is what is confusing you!) |
H: Interpreting the sign of fourier coefficients
I am studying the Fourier series right now. Hopefully it's going okay. Now I have been playing a little bit with taking the product of a wave function (a sine or cosine with some phase) with a sine and integrating the result (using a convolution sum) over the period of the function ($2 \pi$ in this case).
All is good and I believe I can make a good sense of the result. Generally the technique tells how much the input signal (a sine in my case) "contains" of the sine function of frequency $f$ and phase $\phi$. And we test the input signal with the cosine function, it indicates the same thing but for the cosine function.
Now what I haven't found in the references that I have on Fourier series, is an interpretation of the sign of the Fourier coefficients $a_n$ and $b_n$. I read that one possible interpretation of the coefficients is in terms of amplitude and phase. By my understanding is that this is true when you express the coefficients in terms of complex numbers, where the modulus $|z|$ of the number relates to the amplitude and the argument $arg(z)$ relates to the phase. Is that correct?
But what I am after really is to know if there's a possible interpretation of the sign of the coefficients $a_n$ and $b_n$. Can you tell something about the signal when for instance $b_n$ is negative instead of positive (doesn't it indicate something about the phase of our input signal in relation to the cosine for some given frequency, etc.)?
Thank you.
AI: Yes, you're correct that taking $z_n=a_n+ib_n$ will allow you to express your signal in terms of amplitude and phase using $|z|$ and $\operatorname{arg}(z)$ respectively.
You can always switch from $a_n$ and $b_n$ to $z_n$ and see how changing the former parameter set affects the latter. Namely, if you change sign of $a_n$but not $b_n$, or vice-versa, you effectively conjugate $z_n$ (and also multiply by $-1$ if it's $a_n$which you change). Then it's easy to see how $\operatorname{arg}(z)$ will change.
Note that if you use complex parameter $z_n$ instead of real parameters $a_n$ and $b_n$, then should still have a pair of parameters for each $|f|$ where $f$ is frequency (i.e. you have to work with signed frequencies): $a_n+ib_n$ and $a_n-ib_n$, which after doing inverse Fourier transform would cancel out the imaginary part of each other and give you real signal.
And, answering your comment, yes, $\operatorname{arg}(z_n)$ is the phase shift of the harmonic $n$ from its purely cosine version. |
H: How to find the general solution of $\sin \left(x+\frac{2\pi }{3}\right)=3\sin \left(x+\frac{\pi }{3}\right)$
Find the general solution of the equation.
\begin{eqnarray} \sin \left(x+\frac{2\pi }{3}\right)=3\sin \left(x+\frac{\pi }{3}\right)\\ \end{eqnarray}
The answers in my textbook are $n\pi -\tan ^{-1}\left(\sqrt{\frac{3}{2}}\right)$.
\begin{eqnarray}
\\\sin \left(x+\frac{2\pi }{3}\right)=3\sin \left(x+\frac{\pi }{3}\right)\\
\\cos \left(x+\frac{\pi }{6}\right)=3\cos \left(x-\frac{\pi }{6}\right)\\
\end{eqnarray}
What should I do?
Would you mind telling me the method for solving this question?
Thank you for your attention.
AI: $$\frac32\sin x+\frac{3\sqrt3}2\cos x=3\sin\left(x+\frac\pi3\right)=\sin\left(x+\frac{2\pi}3\right)=-\frac12\sin x+\frac{\sqrt3}2\cos x\implies$$
$$3\tan x+3\sqrt3=-\tan x+\sqrt3\iff4\tan x=-2\sqrt3\iff\tan x=-\frac{\sqrt3}2\ldots$$ |
H: Does $ x^p+y^p=kz^p$ have any solutions when $x,y,z,k,p>2, gdc(x,y,z)=1$?
Does the Diophantine equation
$$\displaystyle x^p+y^p=k(z^p)$$
have any solutions when $x,y,z,k,p>2, $ and $ x,y,z$ are co-primes?
AI: Let $a,b$ be two positive integers, and $p$ odd prime.
Then
$$(a^{p}-b)^{p}+(a^{p}+b)^{p} \equiv 0 \pmod{a^{p}}$$
Thus $k=\frac{(a^{p}-b)^{p}+(a^{p}+b)^{p}}{a^{p}}$ is integer and
$x=(a^{p}-b)^{p}, y=(a^{p}+b)^{p}, z=a, k, p$ is a solution.
And it is easy to generate relatively prime solutions( for example $a$ is odd prime, $b=2$).
In the case $p=3, a=3, b=2$ for example we get
$$25^3+29^3 =1482 \cdot 3^3$$ |
H: In my calculator why does $\sqrt4 -2=-8.1648465955514287168521180122928e-39?$
I have tried this on my Windows 7 calculator with $\sqrt9 -3 $ it too gives some weird answer- ie$1.1546388020691628168216106791278e-37$. And so for any $n$(positive) $\sqrt n^2-n= wierd_ .answer$
Why does this happen?
AI: Presumable, the Windows 7 calculator uses floating point arithmetic to do these computations.
The IEEE 754 floating point standard requires that basic arithmetic operations (which include sqrt) are correctly rounded: the result should be as if it was computed with infinite precision and then correctly rounded to fit in a double.
Apparently, the Windows 7 calculator has a bug: instead of the correct answer 2 (which is of course exactly representable in a double) it seems to come up with the largest double smaller than 2. If you then subtract 2, you get something very close to but not exactly equal to 0. |
H: Restricted permutations revisited!
In how many ways can we arrange $n$ different things at $r$ places (each of $r$ places can have any of the $n$ things)repetition allowed,such that $2$ of the $n$ things are always included?
Foe example if n=3 lets say {2,3,4} and r=3 and(2,4) being compulsory,we have $12$ permutations:-
({2,4,4} {2,2,4} {2,4,2} {4,4,2} {4,2,2} {4,2,4} {2,4,3} {2,3,4} {3,2,4} {3,4,2} {4,2,3} {4,3,2})
AI: Hint How many ways can you arrange $n$ objects at $r$ places, repetitions being allowed?
Of those, how many are bad? i.e. don't include at least one of the two compulsory numbers?
So, as you figured out, there are $n^r$ possible combinations. From these $(n-1)^r$ don't contain the first number and $(n-1)^r$ don't contain the second number. But we counted the $(n-2)^r$ combinations which don't contain both compulsory numbers into both $(n-2)^r$ so we need to add them back in. So our answer is
$$n^r-2(n-1)^r+(n-2)^r$$ |
H: The transform of a Hermitian matrix by a unitary matrix is Hermitian
In the following document, p91 (or p4 of the PDF) , section 6.03, it is stated that "The transform of a hermitian matrix by a unitary matrix is hermitian." Apparently the proof is obvious, but not to me... could someone elaborate?
AI: A matrix is hermitian if $ A^\ast = A$ where $A^\ast$ is the conjugated and transposed of $A$.
Unitary matrices have the property that $U\cdot U^\ast= Id$ where Id is the identity. So in special we have $U^\ast =U^{-1}$.
Now we look at the transformed hermitian:
$$ U^{-1} A U=U^\ast A U$$
if conjugate and transpose this we have
$$ (U^\ast A U)^\ast = U^\ast A^\ast (U^\ast)^\ast= U^\ast A^\ast U=U^\ast A U$$
which says that the transformed still hermitian. |
H: Multivariable Calculus question, show limit of equation DNE
Show that
$$\lim_{(x,y)\to(0,0)} \frac {3xy} {x^3+2y^2}$$ Does not exist.
This was on an old test. I am not sure what to show since it's intuitively obvious that as $x$ and $y$ approaches $0$, the denominator will become $0$ and thus DNE.
Am I wrong to argue in words?
AI: The numerator goes to zero too, so it may exist. But try to look at the case where $x=y=t$ and let $t$ go to zero.
A basic rule is, that the multiplicities of the zeroes in the numerator need to be strictly greater than the multiplicities of the denominator for having a continuous function. |
H: Simple question that I can't solve
Here is a relatively simple question that I'm unable to solve :/
There are $10000$ closed lockers in a hallway. A man begins by opening all $10000$ lockers. Next, he closes every $2^{nd}$ locker. Then he goes to every $3^{rd}$third locker and closes it if it is open or opens it if it is closed. After his $10000^{th}$ pass in the hallway, in which he toggles only locker number $10000$, how many lockers are open?
Any insight anyone?
AI: Hint: Locker $n$ will be toggled on pass $k$ if and only if $k$ divides $n$. For every $k$ that divides $n$, you can try to pair $k$ with $n/k$ so the locker toggles generally come in pairs, canceling each other out. For given $n$, almost all $k$ will be paired with a $n/k$ distinct from $k$. When can it happen that a $k$ dividing $n$ does not produce a distinct pair? |
H: How to show that $\mu$ is normally distributed
Suppose we have:
$$p(\mu \mid \sigma, \boldsymbol{w}, \boldsymbol{y}) \propto \exp\left[-\frac{1}{2\sigma^2}\sum_{t=1}^T\left(\frac{(y_t-\mu)^2}{w_t^2}\right) \right]$$
where $\boldsymbol{w} = (w_1, w_2, \cdots, w_t)$ and $\boldsymbol{y} = (y_1, y_2, \cdots, y_t)$
How do I show that:
$$p(\mu \mid \sigma, \boldsymbol{w}, \boldsymbol{y}) \propto \exp\left[-\frac{\sum_{t=1}^T(1/w_t^2)}{2\sigma^2}\left(\mu-\frac{\sum_{t=1}^T(y_t/w_t^2)}{\sum_{t=1}^T(1/w_t^2)} \right)^2 \right]$$
so that $\displaystyle \mu \mid \sigma, \boldsymbol{w}, \boldsymbol{y} \sim N\left(\frac{\sum_{t=1}^T(y_t/w_t^2)}{\sum_{t=1}^T(1/w_t^2)}, \frac{\sigma^2}{\sum_{t=1}^T(1/w_t^2)} \right) $?
Just can't seem to get the algebra right...
AI: Hint: Like every quadratic polynomial in $\mu$, the sum in the exponential is
$$
\sum_{t}\frac{(y_t-\mu)^2}{w_t^2}=A_T\mu^2-2B_T\mu+C_T=A_T\left(\mu-\frac{B_T}{A_T}\right)^2+C_T-\frac{B_T^2}{A_T},
$$
hence it suffices to identify the coefficients $A_T$, $B_T$ and $C_T$. Note that $A_T$ is the leading coefficient of the polynomial in $\mu$ in the LHS hence
$$
A_T=\sum_{t}\frac{1}{w_t^2}.
$$
Likewise, $C_T$ corresponds to the value of the LHS when $\mu=0$ hence
$$
C_T=\sum_{t}\frac{y_t^2}{w_t^2}.
$$
Finally $(-2B_T)$ is the value of the derivative of the LHS with respect to $\mu$ at $\mu=0$ hence
$$
B=\sum_{t}\frac{y_t}{w_t^2}.
$$
Finally,
$$
p(\mu\mid\sigma,y,w)\propto\exp\left(-\frac{A_T}{2\sigma^2}\left(\mu-\frac{B_T}{A_T}\right)^2\right).
$$ |
H: Determine all the extrema of a function subject to a non-linear constraint.
QUESTION
Determine all extrema of the function
$$f(x,y) = x+ 2y $$ subject to $$x^2 + y^2 - 80 = 0$$
ATTEMPT
I don't think I understand what I'm supposed to do. This was in a test and I ended up trying to "graphically"or ïntuitively" find out how the $f(x,y)$ would behave in a circle of radius $\sqrt80$
Which left me with some pretty random numbers that turned out to also be wrong.
What could I have done differently? (If possible could I get a bit of a detailed explanation or some links to that info.)
Lagrange Multipliers: I tried that too on the paper and it also went horribly wrong (but I did get some points, though) but I couldn't quite figure it out.
AI: This looks like a usual problem for Lagrange multipliers. You define a Lagrange function via
$$L(x,y,\lambda)= x+2y + \lambda \cdot (x^2+y^2-80)$$
Now you look for critical points of the function $L$.
But here is the lagrange part more explicit. A critical point is a point where the gradient of $L$ is
zero, as
\begin{align*}
\frac{\partial L}{\partial x} &= 1+ 2x\lambda \\
\frac{\partial L}{\partial y} &= 2+ 2y\lambda\\
\frac{\partial L}{\partial \lambda} &= x^2+y^2-80
\end{align*}
we have to solve
\begin{align*}
0&= 1+2x\lambda \\
0&= 2+2y \lambda \\
0&= x^2+y^2 -80
\end{align*}
Taking the first line times 2 we see that
$$ 4 x \lambda = 2y \lambda, $$
and as $\lambda$ surely isn't $0$ we have
$$y=2x.$$
Using this information in the third equation we have
$$ 80 = x^2+ 4x^2 \iff 16=x^2$$
so you have to check $x=\pm 4$.
Via compactness you know one is a minimum and one is a maximum. So the one where $x$ (and hence $y$ ) is positive must be the maximum. |
H: Can we identify the time if we know every angle between three hands of a watch?
Let $M, H, S$ be the minute hand, the hour hand, the second hand of a watch respectively. Also, let $A_{MH}, A_{MS}, A_{HS}$ be the angle between $M$ and $H$, $M$ and $S$, $H$ and $S$ respectively.
Then, here is my question.
Question : Can we identify the time of the watch if we know all of $A_{MH}, A_{MS}, A_{HS}$ of a watch? If we can, how can we identify the time?
Motivation : I came up with this question while I was repairing my watch:) The answer seems yes, but I'm facing difficulty. Can anyone help?
AI: The answer is yes.
If all three hands move continuously, then the angle of the hour hand is enough to completely determine the time, and therefore the angle of the other two hands.
Let $X$ be the angle of the hour hand from the $12$ o'clock position. Let $Y$ be the angle between the hour and minute hand. Let $Z$ be the angle between the minute and second hand.
The hour hand moves at a rate of $30$ degrees per hour, while the minute hand moves at a rate of $360$ degrees per hour. It we use a different frame of reference, it is possible to see that the minute hand moves at a rate of $330$ degrees per hour relative to the hour hand. This means that the value of $Y$ is equivalent every $360/330=12/11$ hours.
The second hand moves at a rate of $21600$ degrees per hour. Relative to the $360$ degrees per hour of the minute hand, it is traveling $21240$ degrees per hour. The value of $Z$ repeats every $360/21240=1/59$ hours.
Now, we need to find a number that is the smallest multiple of $12/11$ and $1/59$.
$$11*59 = 649$$
$$12/11=708/649$$
$$6/59=66/649$$
$$\text{LCM}(708,66)=7788$$
$$7788/649 = 12$$
As we can see, the unique combination of angle $Y$ and $Z$ only repeats every twelve hours. |
H: Picking Numbers for a Digit
How many three digits numbers are out there with two digits the same and the other different?
If I imagine I have three slots. __ __ __. The first slot is the only one with a restriction. The range of digits can be $[1,9]$ while the 2nd and 3rd can take any digit value from $[0,9]$ Overall I have 900 ways.
Since I am really picking two numbers (one of which will be the pair and the other the single digit. I can do ${10 \choose 1}{9 \choose 1}$. But the problem I am having is don't I have to take into account which I pick first, the two digit pair or the single digit? And once I do pick the digit for the pair and the digit for the single number, how do I specify that "hey, this number is for the pair, this number is for the single".
AI: 900 is too high. I'll try to show you how to calculate it by looking at different orders of the calculation. But first let's try to understand why $900$ is off.
The problem with $9 \cdot 10 \cdot 10$ is that you are performing a calculation without considering the effect of the choice you make for each digit. If you had 9 choices for the first digit, 10 choices for the next, and 10 choices for the next, you could do that in $9 \cdot 10 \cdot 10 = 900$ ways. Consider the case in which you chose the single digit in the hundreds place, there are 9 choices for it. Say you select a 3 for the hundreds place. You would not then have 10 choices for the doubled digit, you would have $10-1 = 9$ because 3 cannot be the doubled-digit and the single digit. Ok, so now you choose the number in the tens place you have 9 options. You choose to use the $4$ from your 9 remaining options. So in choosing the first two numbers you had 9 options and then 9 options. But now that you have decided those two and which is repeating you have only 1 option for the final digit, it must be a 4. So your choices were really 9 options, 9 options, 1 option. The $9 \times 10 \times 10$ assumes that a choice for one digit does not diminish the range of choices for a later digit.
Below I'll show a few ways you could solve the problem.
Choosing the double-digit first
If the double digit includes the first digit then you have ${9 \choose 1}$ choices for the double digit and ${10 -1 \choose 1}$ choices for the remaining digit since one of the digits was chosen for the doubled digit and we don't want to count combinations in which all three digits are the same. So ${9 \choose 1} \cdot {9 \choose 1} = 81$ ways this can happen. However, the double digit can be either in the second or third digit, so we have 2 choices of where to place it so ${9 \choose 1} \cdot {9 \choose 1} \cdot 2 = 162$.
If the single digit is in the hundreds place there are ${9 \choose 1}$ ways it can be chosen. Once selected that leaves ${10-1 \choose 1}$ ways to select the double digit in the tens/ones place. There is only one way to place the double digits into the tens/ones so we don't have to double the result. We have ${9 \choose 1} \cdot {9 \choose 1} = 81$ ways this can be done.
Combining these results we have 243 numbers with two digits the same and one different.
Choosing the single-digit first with a twist
Let the single digit [1-9] fall in the hundreds place there are ${9 \choose 1}$ choices. With this single-digit position, the doubled digit may fall in the range [0-9], so we have ${10 -1 \choose 1}$ choices. Therefore with the single-digit in the hundreds there are ${9 \choose 1} \cdot {9 \choose 1} = 81$ numbers.
Now let the single digit be in the tens or ones place. Here's the twist, in these cases if we consider choosing the single digit first we have 10 choices, but depending on which choice we make we have to consider the choices for the hundred's place differently. So for simplicity, let's choose the double-digit first in these cases there are ${9 \choose 1}$ choices for the double digit, and ${10 - 1 \choose 1}$ choices for the single digit as well as ${2 \choose 1}$ locations for the single digit. So, ${9 \choose 1} \cdot {2 \choose 1} \cdot {10 -1 \choose 1} = 162$ numbers. Again there are $81 + 162 = 243$ such numbers.
Really actually choosing the single-digit first
I've already looked at what happens if you choose the single-digit in the hundred's place: ${9 \choose 1} \cdot {10 - 1 \choose 1} = 81$ such numbers.
Now, if we are truly set on calculating the outcomes and choosing the single digit first, it's easiest to consider two non-overlapping cases. Case 1: the single digit is in the range [1-9] and case 2: the single digit is a 0.
If the single-digit is in the range [1,9], then there are ${2 \choose 1}$ positions it can occupy combined with ${9 -1 \choose 1}$ choices for the double-digit we have ${9 \choose 1} \cdot {2 \choose 1} \cdot {9-1 \choose 1} = 144$ numbers. If the single digit is a zero, then there are ${9 \choose 1}$ choices for the double digit so we have ${1 \choose 1} \cdot {2 \choose 1} \cdot {9 \choose 1} = 18$ such numbers.
Again there are $18 + 144 + 81 = 243$ such numbers. |
H: What is wrong with this reasoning about free modules?
Let $A$ be any set. Let $F(A)$ be the free $R$-module of base $A$ and the map $i: A \to F(A)$ be such that $i(a) = a$. Then for any $R$-module $M$, and a map $g:A \to M$, we can find an unique map $f: F(A) \to M$ by defining it as $f(a) = 0$ for $a \notin A$, and $f(a) = g(a)$ for $a \in A$. Then, we have $f(i(a)) = f(a) = g(a)$.
I think I might have misunderstood something here because it seems too easy to define the unique map $f$. Also, by defining $i(a) = a$, we need to have $A \subseteq F(A)$, and I am guessing that it need not be so? However, I just starting reading the section on Free Modules in Dummit & Foot (3rd Ed) and they defined $i$ as the inclusion map, which I take it to mean $i(a) = a$ for all $a \in A$.
AI: Where the characterization of a free module says „map“ an $R$-linear map is meant, and the function you define is not linear. |
H: A persistent difference
Here's a fun math problem. I wasn't able to get it - am curious what you guys have to say.
Pick a four-digit number whose digits are not all the same. From its digits form the smallest four-digit number $m$ and the largest, $M$. Find $(M-m)$. Keep repeating the procedure. (Treat, say $234$ as $0234$.) Eventually you will reach $6,174$ - permamently, since:
$7,641-1,467=6,174.$
For example, start with $4,818$:
$8,841-1,488=7,353;$
$7,533-3,357=4,176;$
$7,641-1,467=6,174;$ and so on.
Can you prove that you always reach $6,174$?
AI: This is known as Kaprekar's constant. We can partition the set of possible four digit combinations in to those with the same difference. Note that because $M$ and $m$ share the same digits, they are congruent modulo $9$ and so their difference is a multiple of $9$, and so the number of partitions we need to consider is smaller than one might imagine.
In fact, the set of four-digit integers which can be written as $M-m$ is only on the order of $50$. So, these partitions can then be fairly easily enumerated and placed in to a diagram much like the one below found on Wikipedia. It's a rather 'brute force' approach, but it works. |
H: Factoring of a Pro-$\mathcal{C}$ morphism
Let $\mathcal{C}$ be a category, $X$ an object of $\mathcal{C}$, and $p:I^{\circ}\rightarrow \mathcal{C}$ a projective system in $\mathcal{C}$.
Let $\alpha\in \mathrm{Mor}_{Pro(\mathcal{C})}(p,X)$, where $X$ is identified with the constant projective system of $X$. Can $\alpha$ factor through $p \rightarrow p(i) \rightarrow X$ for some $i\in I^{\circ}$?
AI: This is more-or-less built into the definition of $\mathbf{Pro}(\mathcal{C})$. Recall that, given inverse systems $A$ and $B$ in $\mathcal{C}$, we have
$$\mathrm{Hom}(A, B) = \mathop{\varprojlim_j} \mathop{\varinjlim_i} \mathrm{Hom}(A i, B j)$$
and so if $B$ is an inverse system indexed by the trivial diagram, then a morphism $A \to B$ necessarily factors through one of the projections $A \to A i$. The formally dual statement is true for $\mathbf{Ind}(\mathcal{C})$. |
H: $\int_{1}^{2}\frac{1}{(3-5t)^2}dt$
Let $g(t)=(3-5t)^2$, $f(x)=\frac{1}{x^2}$, $g(1)=-2$, $g(2)=-7$.
$$
\begin{align}
& \phantom{={}}\int_1^2 \frac{1}{(3-5t)^2} \, dt \\[6pt]
& =-\frac{1}{5}\int_1^2 \frac{1}{f(g(t))}g'(t) \, dt \\[6pt]
& =-\frac{1}{5}\int_{-2}^{-7}\frac{1}{x^2} \, dx \\[6pt]
& =-\frac{1}{5}\left[-\frac{1}{x}\right]_{-2}^{-7}
\end{align}
$$
I can't follow the whole process. Why we have to multiply $g'(t)$ and why $\dfrac{1}{f(g(t))}$?
Also in the third line, why does $g'(t)$ disappear? $-\frac{1}{x}$ is the correct one?
AI: The procedure outlined makes sense only if $g(t) = 3 - 5t$, and not, as you've written, $g(t) = (3-5t)^2$.
So given $u = g(t) = 3 - 5t,\;$ we know that $\;g(1) = 3 - 5(1) = -2,\;$ and $\;g(2) = 3-5(2) = -7$.
Then given $f(x) = \dfrac 1{x^2}$, it follows that $\;f(g(t)) = \dfrac 1{(g(t)^2)} = \dfrac 1{(3 - 5t)^2}$.
So the substitution being made is $g(t) = \color{red}{\bf u = 3 - 5t}$.
Then $g'(t) = du = -5 dt \iff \color{blue}{\bf -\frac 15 du = dt.}$
$$\int_1^2 \dfrac{1}{(\color{red}{\bf 3 - 5t})^2} \color{blue}{\bf\,dt} = \int_{-2}^{-7} \dfrac{1}{\color{red}{\bf u}^2}\,\cdot\left(\color{blue}{\bf -\frac 15 du}\right)= -\frac 15\int_{-2}^{-7} \dfrac{1}{u^2}\,du $$ |
H: Solving a second order homogenous ODE with double roots
I'm working on solving the following homogenous equation:
$$ y'' - 8y' + 16y = 0 $$
Seems like a straight forward $y=e^{rx}$ substitution and then solve for r1 and r2:
$$ y=e^{rx}=0 $$
$$ y=re^{rx}=0 $$
$$ y=r^2e^{rx}=0 $$
$$ r^2e^{rx} - 8re^{rx} + 16e^{rx}=0 $$
$$ e^{rx}(r^2-8r+16)=0 $$
Since $e^{rx}$ can't equal zero:
$$ r^2-8r+16 = 0 $$
$$ (r-4)(r-4)=0 $$
$$ r_{1,2} = 4 $$
How do I express the generaal solution from here? I've tried the following but it is incorrect:
$$ y=c_1e^{4x}+c_2e^{4x} $$
AI: Because of the repeated root, we have to express it as:
$$ y(x) =c_1e^{4x}+c_2 x e^{4x} $$
See Paul's Online Notes for theory as to why. |
H: $\mathbb Q+X\mathbb R[X]$ is not Noetherian
Let $A=\{q+r_1X+ \cdots +r_nX^n: q \in \mathbb{Q}, r_i \in \mathbb{R}\}$ be the polynomial ring with rational costant terms. I have to prove that $A$ isn't a noetherian ring. How can I prove it?
AI: consider the chain of ideals:
$\pi xA \subset \pi xA + \pi^2xA \subset\pi xA +\pi^2 xA +\pi^3 xA \subset ...$ |
H: Given $f(x+1)=x^2-3x+2$, how can I find $f(x)$?
Given $f(x+1)=x^2-3x+2$, how can I find $f(x)$?
AI: Let $y=x+1$. We then have $x=y-1$. Hence,
\begin{align}
f(y) & = f(x+1) = x^2 -3x+2 = (y-1)^2 - 3(y-1)+2\\
& = y^2-2y+1-3y+3+2 = y^2-5y+6
\end{align} |
H: Let $\,\displaystyle f \colon [0,1] \to [0,1]$ be continuous and $\,f(0)=0,f(1)=1.$
I am stuck on the following problem that says:
Let $\,\displaystyle f \colon [0,1] \to [0,1]$ be continuous and $\,f(0)=0,f(1)=1.$ Then $f$ is necessarily
injective ,but not surjective
surjective,but not injective
bijective
surjective
I have to determine which of the following options is correct. Can someone help? Thanks and regards to all.
AI: you can kick out 1. and 2. by considering $f(x)=x$. furthermore A function with a graph like the letter $N$ will be not injective. But the function is surjective due to intermediate value theorem. |
H: How many n digit numbers can be formed using 1,2,3 which contain at least one 1 and one 3?
How many n digit numbers can be formed using $1,2,3$ which contain at least one $1$ AND one $3 $?
For $n= 5$,
$11133$ is valid.
$12312$ is valid.
$11122$ is not valid.
$33333$ is not valid
Please suggest a formula which works for all $n$, $3\leq n\leq100$.
AI: $3^n - (2^{n+1}-1)$
$3^n$ is the total possible numbers that can be formed. From this, we subtract the following cases:
1) no 1 and no 3 : 1 way
2) atleast one 1 and no 3 : $2^n-1$ ways
3) atleast one 3 and no 1 : $2^n-1$ ways |
H: NFA from grammar productions
Based on this grammar:
\begin{align} G = (\{S,A,B\}, \{a,b, c\}, S, P) \end{align}
\begin{matrix}
\\P:
\\S → abaS | cA
\\A → bA | cB | aa
\\B → bB | cA | bb
\end{matrix}
I created this NFA:
I'm not sure about $q1 \to q2$ and $q1 \to q3$, if maybe someone can clarify if this is wrong and why.
AI: It looks okay. The language consists of all words of the form $(aba)^ncwaa$ with $n\ge 0$, $w\in\{b,c\}^*$, and $|w|_a$ even and all words of the form $(aba)^ncwbb$ with $n\ge 0$, $w\in\{b,c\}^*$, and $|w|_a$ odd, where $|w|_a$ is the number of $a$’s in $w$. |
H: Does this integral make sense in some way?
I have a very simple, possibly silly question...
Can this integral make sense in some way? $$ \int \frac{dx}{dx}$$
And does it actually mean something to write things like $$ \int f(x)$$ without the differential?
I just got an expression of this type and I probably made some mistake along the way, but I'm still curious... my intuition from my limited knowledge of differential forms suggests that the last integral should simply evaluate to $f(x)$. How much sense does this make? I pretty much have no idea what to do here, so any help would be appreciated, especially if it includes intuitively pleasing explanations! :)
EDIT: To clarify my question some more, consider an ordinary integral $$\int f(x) dx = F(x) + c$$ where $\frac{dF}{dx} = f$(x).
We could write it as $$\int \frac{dF}{dx}dx = \int dF = F +c$$
Then, going back to my original integral, I could write it as $$\int \frac{1}{dx} dx = \int 1$$
Or for an arbitrary function: $$\int f = \int \frac{f}{dx} dx$$
But does that mean anything or is it just (incorrect) notation gymnastics? Does it make sense for "$1$" to be a differential of something? What about $f$?
AI: In most fields the integrand needs a $\mathrm{d(something)}$ for it to make sense. This is what makes it in a hand-wavy way "infinitesimal".
However, in some fields of maths you are allowed to say things like $\omega = \sum_a w^a{\rm d}x_a$ and then you can have $\int \omega$. But this only works in fields where people carefully explained just that particular kind of integration means. One such field is differential geometry, and one important use of the notation is Stokes' Theorem
$$\int_{\partial R}\omega = \int_R {\rm d}\omega $$ |
H: ii) Show that $u$ is not a real part of the function which analytic on $\mathbb{C} \backslash \lbrace 0 \rbrace$
Suppose $u(x,y)=\ln(x^2+y^2)$
i) Show that $u$ is harmonic on $\mathbb{C} \backslash \lbrace 0 \rbrace$
ii) Show that $u$ is not the real part of a function which analytic on $\mathbb{C} \backslash \lbrace 0 \rbrace$
I manage to show the first part. For the second part, note that $u(x,y)=\ln(x^2+y^2)=\ln(|z|^2)=2 \Re \log(z)$
But this only show that $u$ is not a real part of $\log(z)$. I don know how to show $u$ cannot be the real part of a function which analytic on $\mathbb{C} \backslash \lbrace 0 \rbrace$
Can anyone guide me?
AI: $z=0$ is a branch point of the complex logarithm, so it cannot be defined continuously (so not analytically) on $\Bbb C\setminus\{0\}.$ Since the real part of an analytic function determines the imaginary part up to imaginary constant, we're done. |
H: Get x and y in quadrat equations system
I need help in solving following system of quadratic equations :
$$
2x^2+y^2=4$$
$$2xy-2x=-5$$
I used every known me equations solving methods, but no was helpful for me... Can you help me by giving step-by-step solution (that is homework and I really need it...) ?
Thanks in advance !
AI: There are no real solutions. The resultant of $2 x^2 + y^2 - 4$ and $2 x y-2 x+5$ with respect to $x$ is $4\,{y}^{4}-8\,{y}^{3}-12\,{y}^{2}+32\,y+34$, which has no real roots. |
H: write $\lvert 10+4i\rvert$ in the form $z=re^{i\theta}$
write $\lvert 10+4i\rvert$ in the form $z=re^{i\theta}$
I am not sure what to do with the absolute value in this case.
AI: Since the absolute value is a real number, $\theta = 0$ and $r = \sqrt{ 10^2 + 4^2 } $ |
H: Is the kernel of this group action the centralizer?
In Dummit and Foote, they state
"... let the group $N_G(A)$ (normalizer) act on the set $A$ by conjugation. It is easy to check that the kernel of this action is the centralizer $C_G(A)$."
From what I understand, the kernel of this action is the set
$$ \{ g \in N_G(A) : g a g^{-1} = a \text{ for all } a \in A \}. $$
However, the centralizer of $A$ is
$$ \{ g \in G : g a g^{-1} = a \text{ for all } a \in A \}. $$
If this is true, I see that the kernel is a subset of the centralizer, but how are they equal?
AI: We always have $C_G(A) \subset N_G(A)$, since
$$N_G(A) = \{g\in G : gAg^{-1} = A\} = \left\{ g \in G : \bigl(\forall a \in A\bigr)\bigl(gag^{-1}\in A\bigr)\right\},$$
and of course a group element that fixes $A$ elementwise under conjugation fixes $A$ as a set.
So the kernel of the group action is
$$N_G(A) \cap C_G(A) = C_G(A).$$ |
H: What is the result of $x^{\top}A\dot{x}+\dot{x}^{\top}Ax$?
What is the result of
$x^{\top}A\dot{x}+\dot{x}^{\top}Ax$, provided that $A=A^{\top}$?
Actually, I wanted to expand $d(x^{\top}Ax)\over{dt}$.
AI: I'll use $y^T$ for the OP's $y^{\top}$.
Observe that $x^TA{\dot x}$, being a scalar quantity, is automatically symmetric: $(x^TA{\dot x})^T = x^TA{\dot x}$, and $({\dot x}^TAx)^T ={\dot x}^TAx$ as well. Now $(x^TA{\dot x})^T = {\dot x}^TA^Tx$, so we obtain $x^TA{\dot x} = {\dot x}^TA^Tx$. Thus
$x^TA{\dot x} + {\dot x}^TAx = {\dot x}^TA^Tx + {\dot x}^TAx = {\dot x}^T(A^T + A)x. \tag{1}$
Now using $A^T = A$, (1) becomes
$x^TA{\dot x} + {\dot x}^TAx = 2{\dot x} ^TAx; \tag{2}$
is that sort of thing we seek? It can't be taken much further, I'll warrant. |
H: How to find $p$ and $q$ if we have $\operatorname{lcm}(p,q)=b$ and $p+q=a$ where ($a,b \in \mathbb{N}$) and $p>q$.
What is the general formula to find $p$ and $q$ if we have $\def\lcm{\operatorname{lcm}}\lcm (p,q)=b$ or $\gcd(p,q)$ and $p+q=a$ where ($a,b \in \mathbb N$) and $p>q$?
Example: $\lcm(p,q)=84$ and $p+q=54$ and $p>q$.
AI: GCD problem:
If $gcd(p,q)=b$ and $p+q=a$ the problem is relatively easy, but the solution is not unique.
In this case, there is a solution if and only if $b |a$, and this case the problem reduces to finding $p',q'$ relatively prime such that $p'+q'=\frac{a}{b}$. Then $p=p'b, q=q'b$ are the solutions.
A solution in this case is $p'=1, q'=\frac{a}{b}-1$ which leads to the solution $p=b, q=a-b$.
The exact number of solutions is the cardinality of
$$\{ n | 1 \leq n \leq \frac{b}{a}, gcd (n, \frac{b}{a}-n)=1 \} \,.$$
LCM problem
Let $c=gcd(p,q)$. Then $p=c p', q=cq'$ and $gcd(p',q')=1$. We also have
$$b=cp'q'$$
and $a=cp'+cq'$.
So the problem reduces to solving
$$b=cp'q'$$
$$a=cp'+cq'$$
and
$$lcm(p',q')=1$$
Edit (instead of $c | gcd(a,b)$) As vadim's answer shows, we must have
$$c = gcd(a,b) \,.$$
So the following is a simplification of vadim's answer using the old approach of this problem.
Indeed, as $gcd(p',q')=1$ we get that no prime number can divide both $p'q'$ and $p'+q'$, thus $gcd(p'q', p'+q')=1$. This shows that
$$gcd(a,b)=gcd(cp'q', cp'+cq')=c \,.$$
Thus the problem simply reduces to the following question:
Does the equation
$$x^2-\frac{a}{gcd(a,b)}x+\frac{b}{gcd(a,b)}=0$$
have relatively prime integer solutions.
In your example, $gcd(54, 84)=6$, which means we need to look to the following equations:
Does
$$x^2-9x+14=0$$
have integral relatively prime solutions? |
H: Proof that $n^3 + 3n^2 + 2n$ is a multiple of $3$.
I'm struggling with this problem:
For any natural number $n$, prove that $n^3 + 3n^2 + 2n$ is a multiple of $3$.
That $n^3 + 3n^2 + 2n$ is a multiple of $3$ means that: $n^3 + 3n^2 + 2n = 3 \times k$ where $k \in \mathbb N$
So I tried to find a the number $k$.
The best result I found was: $n^3 + 3n^2 + 2n = n(n+1)(n+2)$
But I'm lagging at the last step, to prove that it is a multiple of $3$.
(However, I got the intuition, If you see the multiples of $3$: $\{0, 3, 6, 9,\dotsc\}$ there is a difference of $3$ between them.
So $n(n+1)(n+2)$ incorporates that difference. This means that if you take a number $n$ then $n$ or $n+1$ or $n+2$ could be a multiple of $3$ and so their multiplication is a multiple of $3$) But I couldn't extend that idea into a consistent mathematical proof.
Also this problem doesn't help either:
Proof that $n^3+2n$ is divisible by $3$
AI: Among three consecutive integers, one must be a multiple of three. Reason: if $n=3k$, we're done. If $n=3k+1$ then $n+2=3j$ is a multiple of three. If $n=3k+2$, then $n+1=3m$ is a multiple of three. In any case, $3\mid n(n+1)(n+2)$. |
H: What is the correct spelling of Paul Erdős's name?
I am not quite sure that this question belongs to the math.stackexchange.com. What I am sure is that definitely there are people here who know the answer to my question.
On the Internet and in the literature I saw two different types of spelling of the last name of Paul Erdös (together with the wrong one "Erdos"):
Erdős
Erdös
In the former case it is called "long Hungarian umlaut (double acute)" and in the latter "umlaut, trema or dieresis". Hence the question: which choice is the correct one, or maybe they both can be used interchangeably?
AI: The former is the correct spelling, but outside of $\mathrm\LaTeX$ and its \H command it's hard to enter, so a lot of people just use umlauts which are easier to type on a keyboard. |
H: $\mathbb{R}$ \ $\mathbb{Q}$ and $\mathbb{R}^2\setminus\mathbb{Q}^2$ disconnected?
If I want to prove that $\mathbb{R} \setminus \mathbb{Q}$ is disconnected, does it suffice to say that there are two open disjoint sets that cover $\mathbb{R}\setminus\mathbb{Q}$, namely:
$$(- \infty, 0), (0, \infty)\text{ ?}$$
Along the same lines, I want to prove or disprove that $S = \mathbb{R}^2 \setminus \mathbb{Q}^2$ (points $(x, y) \in S$ that have at least one irrationais connected. I feel that it is also disconnected; does it suffice to say that there are two open disjoint sets that cover $S = \mathbb{R}^2 \setminus \mathbb{Q}^2$, namely $((- \infty, - \infty), (0, 0))$ and $((0, 0), (\infty, \infty))$? (Very iffy on my assertion and my notation, sorry.)
Thanks!
AI: The answer to your first question is yes: $\{(\leftarrow,0),(0,\to)\}$ is a clopen partition of the irrationals, and its existence shows that they are not connected.
$\Bbb R^2\setminus\Bbb Q^2$, on the other hand, is connected, and even path connected: you can get from any point of it to any other point along a path lying entirely within $\Bbb R^2\setminus\Bbb Q^2$. In fact, you can do it along straight line segments, using at most three of them; just make sure that each horizontal segment lies on a line $y=a$ with irrational $a$, and each vertical segment lies on a line $x=a$ with irrational $a$. See if you can work out the details for yourself; I’ve written them up below and left them spoiler-protected.
Suppose that $\langle a_,b\rangle$ is a point with at least one irrational coordinate. Without loss of generality let $a$ be irrational. Let $\langle c,d\rangle$ be any other point with at least one coordinate irrational. If $d$ is irrational, you can travel along the line $x=a$ to the point $\langle a,d\rangle$, and then travel along the line $y=d$ to $\langle c,d\rangle$; this path lies entirely in $\Bbb R^2\setminus\Bbb Q^2$. If $c$ is irrational, travel along the line $x=a$ to any $\langle a,u\rangle$ with irrational $u$, then along $y=u$ to the point $\langle c,u\rangle$, and finally along the line $x=c$ to the point $\langle c,d\rangle$; again the path lies entirely in $\Bbb R^2\setminus\Bbb Q^2$. |
H: Con ZF implies Con ZFC using set sized models
Can we use forcing to construct models of ZFC and ZFC + GCH starting from c.t.m s of ZF? The usual way to obtain the associated relative consistency results (Con ZF implies Con ZFC and Con ZF implies Con ZF + GCH)is to look at the class sized model L (of ZF) and I was interested in seeing if there was an alternate way to do this. Is there any reference material I could look up?
(I do realize that a model of ZF + GCH will model AC so the the question was phrased this way in case it was possible to do the first construction but not the second)
AI: Not always.
There are models of $\sf ZF$ which cannot be extended to models of $\sf ZFC$ without adding ordinals. Since forcing is a technique which does not add ordinals, this means that this is impossible. Note that when I say that, I include class-forcing as well.
Moreover, if you only limit yourself to set forcing, then the answer is an even simpler no. One can easily construct (using class forcing) models of $\sf ZF$ such that no set forcing extension can satisfy the axiom of choice. Similarly one can produce models of $\sf ZFC$ that have no set forcing extension which satisfy $\sf GCH$.
Do note, though, that given a set model of $\sf ZF$ one can produce a set model of $\sf ZFC+GCH$ by simply taking the $L$ of that model, which will be a model of $\sf ZFC$ satisfying $V=L$ as well. |
H: How to solve these series?
Can anyone help me understand how to solve these two series? More than the solution I'm interested in understanding which process I should follow.
Series 1:
$$
\sum_{i = 3}^{\infty} i * a^{i-1}, 0 < a < 1.
$$
Series 2:
$$
\sum_{i = 3}^{\infty} i\sum_{k = 2}^{i-1} a^{i-k} * b^{k-2} , 0 < a < 1, 0 < b < 1.
$$
These two series come as part of a long mathematical proof which I omitted for brevity, if you think it is relevant I will post it.
AI: Hint: For the first one if $\sum_{i = 3}^{\infty} x^{i}=f(x)$ then $\sum_{i = 3}^{\infty} i \times x^{i-1}=f'(x)$.
For the second one consider that:
$$
\sum_{k = 2}^{i-1} a^{i-k} b^{k-2}=a^{i-2}\frac{1-\frac{b^{i-2}}{a^{i-2}}}{1-\frac{b}{a}}= \frac{a^{i-2}-{b^{i-2}}}{1-\frac{b}{a}}
$$
Then you can decompose the series into two one and use the previous step. |
H: The sum of three square roots bounded below by $\sqrt{82}$
Let $a,b,c >0$ and $a+b+c \le 1$. Prove: $$\sqrt{a^2+1/a^2}+\sqrt{b^2+1/b^2}+\sqrt{c^2+1/c^2} \ge \sqrt{82}$$
Progress: I tried to have 3 vectors $(a,1/a)$, $(b,1/b)$ and $(c,1/c)$, played around with the vector inequalities but no success.
AI: Assume first $a+b+c=1$.
Let $x=(a,1/a)$, $y=(b,1/b)$, $z=(1/c)$. Then the triangle inequality for the Euclidean norm tells us that
$$
\|x+y+z\|\leq\|x\|+\|y\|+\|z\|.
$$
This looks like
$$
\sqrt{1+\left(\frac1a+\frac1b+\frac1c\right)^2}\leq\sqrt{a^2+\frac1{a^2}}+\sqrt{b^2+\frac1{b^2}}+\sqrt{c^2+\frac1{c^2}}.
$$
So to prove the inequality in the question we need to show that
$$
\frac1a+\frac1b+\frac1c\geq9.
$$
Using the Cauchy-Schwarz inequality:
$$9=\left\|( \sqrt{a}, \sqrt{b}, \sqrt{c})\cdot(\frac{1}{\sqrt{a}}, \frac{1}{\sqrt{b}}, \frac{1}{\sqrt{c}})\right\|^2\leq (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} . $$
Finally, when $a+b+c <1$, let $d=1-(a+b) $. Then $c <d <1$; as $\sqrt {x^2+1/x^2} $ is decreasing on $(0,1) $, we have $\sqrt {c^2+1/c^2}>\sqrt {d^2+1/d^2} $. Also, $a+b+d=1$ and so
$$
\sqrt{a^2+\frac1{a^2}}+\sqrt{b^2+\frac1{b^2}}+\sqrt{c^2+\frac1{c^2}}\geq
\sqrt{a^2+\frac1{a^2}}+\sqrt{b^2+\frac1{b^2}}+\sqrt{d^2+\frac1{d^2}}\geq\sqrt {82}.
$$ |
H: Determine whether the following function is continuous at $(0,0)$
Determine whether the following function is continuous at $(0,0)$:
$$f(x,y)=\begin{cases} \frac{xy}{\sqrt {x^2+y^2}} & \text{if $(x,y)$ $\neq(0,0)$} \\
1 & \text{if $(x,y)=(0,0)$} \\
\end{cases}$$
Here's what I did. I know that $f(0,0)=1$ and I think that I have to show either $$\lim_{(x,y)\to(0,0)} f(x,y) = f(0,0) \ \text{or} \neq f(0,0)$$
How should I go about this?
AI: look at $x=0$ and $y\to 0$ which gives you $0$ as limit.
Or look at $x=y$ |
H: Proving that there are n Equivalence Classes Modulo n
For $a,b,n \in \mathbb{Z}$ and $n \geq 2$, I want to prove that there are $n$ equivalence classes mod $n$. I'm not sure how to do it - would I do it inductively? Any help would be appreciated.
AI: There are certainly at least $n$ equivalence classes: Namely, $[0], [1], [2], ..., [n - 1]$. To prove these are distinct, note that if $0 \le k, l < n$, then
$$[k] = [l] \implies n | l - k \implies l - k = 0$$
To see that there are at most $n$ equivalence classes, note that for any integer $m$, we may write $m = nq + r$ with $0 \le r < n$ by the division algorithm, so that $[m] = [r]$ and the class is already on our list. |
H: Proof of Rudin theorem 2.34 compact subsets of metric spaces are closed
Rudin writes at a certain point in the proof:
$\textbf {Since $K$ is compact}$, there are finitely many points $q_1,...,q_n$ in $K$ such that:
$K \subset W_{q1} \cup W_{q2} \cup ... \cup W_{qn}$ where the $\textbf{$W_{qi}$'s are neighborhoods}$ of a point $q\in K$
Well the implication in bold is my problem, that's definitely not what the definition of a compact set states.
The definition says that for every open cover we can get an open finite subcover but we can't assume that they're neighborhoods right?? How does he make this jump??
Thanks for the help!
AI: The definition of a compact set means that every open cover of $K$ has a finite cover. For each $q \in K$, choose a neighborhood $W_q$ of $K$; then $\{W_q : q \in K\}$ is an open cover of $K$, since it clearly contains every point in $K$.
Hence, we may select a finite subcover, and just let $q_1, ..., q_n$ be the points corresponding to the open sets in our finite cover. |
H: Proving linear independence of vectors which are functions of other independent vectors
If the $n$-component vectors $a,b,c$ are linearly independent, show that $a+b, b+c, a+c$ are also linearly independent, Is this true of $a-b,b+c,a+c$?
What I did was write the new vectors as sums and set it equal to zero
$$\begin{align} \sum(\lambda_i(a_i+b_i))+\sum(\lambda_i(b_i+c_i))+\sum(\lambda_i(a_i+c_i))=0 \end{align}$$
Then I assumed you could just add these terms up, but I don't actually know if that's allowed in this case. Anyways you obviously get
$$\sum\lambda_i(2a_i+2b_i+2c_i)=0$$
Seems to me that since the problem tells you $a,b$,and $c$ are linearly independent that these respectiev $\lambda$'s on these new terms must also be $0$.
Again, I doubt that works so I'm asking here. Wish this book had answers...
AI: Your setup needs to be modified some.
The key idea here is that any equation of linear dependence for $a+b$, $b+c$, and $a+c$ results in an equation of linear dependence for $a$, $b$, and $c$.
When you wrote
$$
\sum\lambda_i(a_i+b_i) + \sum\lambda_i(b_i+c_i) + \sum\lambda_i(a_i+c_i) = 0,
$$
you're not multiplying your vectors $a+b$, $b+c$, and $a+c$ by scalars; you're actually computing the dot product $(\lambda_1,\ldots,\lambda_n)\cdot ((a+b)+(b+c)+(a+c))$. What you really want is to show that if
$$
\alpha(a+b)+\beta(b+c)+\gamma(a+c) = 0
$$
for scalars $\alpha,\beta,\gamma$, then $\alpha=\beta=\gamma=0$. You rewrite this as a sum of (different) scalars time the vectors $a,b,c$, and then use the linear independence of the latter to complete the argument.
An alternative "slick" way is as follows. Notice that
\begin{align*}
a &= \frac{1}{2}\lbrack (a+b) - (b+c) + (a+c)\rbrack,\\
b &= \frac{1}{2}\lbrack (b+c) - (a+c) + (a+b)\rbrack,\\
c &= \frac{1}{2}\lbrack (c+a) - (a+b) + (b+c)\rbrack.
\end{align*}
So the "span" (set of all linear combinations of) $a+b$, $b+c$, and $c+a$ contains the lienarly independent vectors $a$, $b$, and $c$. You can show, in general, that
three vectors $u,v,w$ are linearly independent if you can find three linear combinations of $u$, $v$, and $w$ that are linearly independent. So this completes the proof (assuming you accept this last fact...you should prove it if you haven't already).
For the second problem, notice that the third vector $c+a = (a-b)+(b+c)$ is a linear combination of the first two vectors. So no, they're not independent. |
H: Show $5 \cdot 4^{\log_{2}{n}}$ is $\Theta(n^{2})$.
I'm having trouble working out the algebra for this problem. I know that we need to show $\exists c$ s.t. $5 \cdot 4^{\log_{2}{n}} \leq c \cdot n^{2} \forall n \geq n_{0}$, and also the other direction.
AI: $$
4^{\log_2 n} = 2^{2\log_2 n} = 2^{\log_2 n^2} = n^2.
$$ |
H: How do I use quantifiers to specify a pair of elements?
I'm just learning about quantifiers in class, and my professor has so far only given us examples where he uses quantifiers to specify one element/variable at a time from a set.
So he would write something like: $\forall$ a $\in$ $\mathbb{R}$ ...
Am I limited to only defining one variable at a time? Would it be "legal" for me to write $\forall$ (a, b) $\in$ $\mathbb{R}$?
AI: Formally, you would want to write $$\forall a\in\Bbb R\forall b\in\Bbb R(\ldots)$$
But it is perfectly readable to write $\forall a,b\in\Bbb R\ldots$ |
H: How to find $x$ and $y$ in $(x+c_1)(y+c_2)=c_3$ where $x, y \in \mathbb{N}$ and $c_1,c_2,c_3$ are non-zero constants?
Is there a general formula to find $x$ and $y$ in $(x+c_1)(y+c_2)=c_3$ where $x, y \in \mathbb{N}$ and $c_1,c_2,c_3$ are non-zero constants?
Because I was struggling trying to find solutions to this equation: (Picking two positive integer and trying)
$$(x+1)(y+3)=35$$
AI: $$ (x+c_1)(y + c_2) = c_3 \Rightarrow x + c_1= \frac{c_3}{y+c_2} \Rightarrow x = \frac{c_3 - c_1y-c_1c_2}{y + c_2 } $$
$$ y + c_2 \mid c_3 - c_1(y+c_2) \Rightarrow y + c_2 \mid c_3 $$
so find the factors of $c_3$ and you'll find $ x $ and $ y $ |
H: Question about the closed set
Is there a topological space on $\mathbb{R}$ in which the interval $(0,1)$ or intervals like $(a,b)$, $a,b\in\mathbb{R}$ is a closed set?
AI: Yes, the discrete topology, where every subset of $\mathbb R$ is open (hence every subset of $\mathbb R$ is closed as well). |
H: Proof that aleph null is the smallest transfinite number?
The wikipedia page on the cardinal numbers says that $\aleph_0$, the cardinality of the set of natural numbers, is the smallest transfinite number. It doesn't provide a proof. Similarly, this page makes the same assertion, again without a proof.
How does one prove there is no smaller transfinite number? Equivalently (I think), why is there no smaller infinite set than the natural numbers?
AI: This is a consequence of the following theorem:
Suppose that $A$ is a set of integers, then either $A$ is finite, or $|A|=|\Bbb N|$.
Since we define $\aleph_0$ to be the cardinality of $\Bbb N$, this means that every infinite subset of a set of size $\aleph_0$ is itself of size $\aleph_0$, and so there cannot be a smaller infinite cardinal.
Note that the above proves that $\aleph_0$ is a minimal element of the infinite cardinals. There is no smaller. To prove that it is in fact the smallest of the infinite cardinals we need to use some other set theoretical assumptions (e.g. every two cardinals are comparable) which are commonly assumed throughout mathematics nowadays.
The proof of the aforementioned theorem is simple, by the way. Suppose that $A$ is infinite, then the map $a\mapsto |\{a'\in A\mid a'<a\}|$ is a bijection between $A$ and $\Bbb N$. The proof of that is by induction. |
H: Dervation of $\sum_{k=0}^n(r+1)^k= \frac{(r+1)^{n+1}-1}{r}$
How can one derive the following identity?
$\sum_{k=0}^n(r+1)^k= \frac{(r+1)^{n+1}-1}{r}$
I have playing around with binomial coefficients and index shiftings but wasn't able to get anywhere.
AI: Let's call $q:=r+1$ and look for $x$, then
$$1+q+q^2+\dots+q^n = x \\
q+q^2+\dots+q^n+q^{n+1} = qx\,,
$$
by multiplying both sides by $q$. Now subtract them. |
H: Inequality with argument of quotient of complex number
Show on a complex plane:
$$\left\{ z \in \mathbb{C}: \operatorname{arg} \frac{i}{z} \leq \frac{3 \pi}{4}\right\}.$$
After some easy steps I came to:
$$-\operatorname{arg}(z) =\operatorname{arg} \overline{z} \leq \pi/4.$$
What now?
AI: If $z=r e^{i \theta}$, with $r >0$ and $\theta \in [0,2 \pi)$, then $\frac{i}{z} = \frac{1}{r} e^{i(\frac{\pi}{2}-\theta)}$.
Hence $\arg \frac{i}{z} = \begin{cases}\frac{\pi}{2}-\theta, & \theta \in [0,\frac{\pi}{2})\\ 2 \pi +\frac{\pi}{2}-\theta, & \theta \in [\frac{\pi}{2}, 2 \pi)\end{cases}$.
Note: I am using $\arg z \in [0,2 \pi)$.
Addendum: Using $\arg z \in [-\pi, \pi)$ instead gives:
$\arg \frac{i}{z} = \begin{cases}\frac{\pi}{2}-\theta, & \theta \in [-\frac{\pi}{2}, \pi)\\ \frac{\pi}{2}-\theta - 2 \pi, & \theta \in [-\pi,-\frac{\pi}{2})\end{cases}$.
Using this formula with $\arg z \le \frac{3}{4} \pi$ gives $\theta \in [-\pi,-\frac{\pi}{2}] \cup [-\frac{\pi}{4}, \pi)$. |
H: Constructing a closed, convex subset of $X^{\ast}$ that is not weakly-* closed
I'm asked to show that if $X$ is a non-reflexive Banach space, there exists (norm) closed and convex subsets of $X^\ast$ that are not $w^{\ast}$-closed. In other words, there's no analogue of Mazur's Lemma for the $w^{\ast}$-topology. As a hint, I'm given that a functional in a TVS is continuous iff it's kernel is closed.
I thought about two main approaches to tackle the problem:
(1st) $X$ non-reflexive implies at least one $\psi \in X^{\ast \ast}$ s.t. $\psi \notin i[X]$, $i[X]$ being the canonical embbeding of $X$ in it's second dual. Therefore, $\ker \psi$ is a closed, convex subset of $X^{\ast}$, and I'd have to show that it's not $w^\ast$-closed. The problem here is that I just can't find any direct connection between $w^\ast$-closedness and $\psi \notin i[X]$, and also, I don't see where I could use $X$ Banach.
(2nd) $X$ Banach non-reflexive implies that $i[X]$ as above is a proper, closed subspace of $X^{\ast \ast}$. Therefore, I could use a separation theorem to find $\Gamma$ in $X^{\ast \ast \ast}$ s.t. $\text{Re } \Gamma (\psi_{0}) > \sup_{x \in X} \Gamma(i(x))$ for some $\psi_{0} \notin i[X]$, and then try to get some useful result from embedding $X^{\ast}$ in $X^{\ast \ast \ast}$, maybe by using Goldstine's theorem about $w^{\ast}$-density. The problem here is that it seems too confusing, and I just couldn't get very far with so many embeddings. Specially considering that embedding is an $w-w^{\ast}$ homeomorphism, so if I get results about $w^{\ast}$ in the embedding of the first dual in the third dual, it translates as a result about $w$ in the first dual.
Anyone could give me hints at least about which direction follow?
AI: Your 1st approach is good. You need to notice that your $\psi$ is not $w^*$-continuous, and that a functional is continuous if and only if its kernel is closed. |
H: Can a subset have $0$ elements?
Let $S = \{1, 2, \dots, n \}$. Must any subset of $S$ have at least $1$ element?
For example, suppose the subset $\{ \emptyset \} \in \mathcal{P(S)}$ . Is the cardinality of $\{ \emptyset \} = 0$ or $1$?
AI: The empty set is a subset of any set, and it has zero elements. That is, we have
$$\varnothing\in\mathcal{P}(S)$$
and the cardinality of $\varnothing$ is $0$. However you are correct that the cardinality of $\{\varnothing\}$ is $1$.
(It is important to understand that the set $\varnothing$ and the set $\{\varnothing\}$ are different. Think of $\varnothing$ as an empty bag, and think of $\{\varnothing\}$ as a bag containing an empty bag; their contents are different.) |
H: Factoring $a^m + 1$, an odd prime
Why is it that if $a^m + 1$, an odd prime, with $m = kl$, and $l$ odd.
We get:
$$a^m + 1 = (a^k + 1)(a^{k(l-1)} - a^{k(l-2)} + \dots + a^k + 1)?$$
What is the name of this property?
AI: If $n$ is odd we have $$\frac{1+x^n}{1+x}=\frac{1-(-x)^n}{1-(-x)}=\frac{1-r^n}{1-r}=\cdots?$$
In your case if $m=kl$ take $x=a^k$. What this is saying is that if $m$ is odd and $a^m+1$ is an odd prime, $m$ must be prime as well. |
H: Integral of function having modulus
Evaluate
$$ \int c e^{-|x-U|} dx$$ Assume $c$ and $U$ to be constant.
I am not used to do integrals since long time. I suddenly have to use this in a computer program.
AI: Let $F$ be a primitive such that $F(0)=0$. Then
$$
F(X)=\int_{U}^X c \, e^{-|x-U|}dx=\left\{\begin{array}{ll}\int_{U}^X c \, e^{-x+U}dx &X>U \\ &\\ \int_{U}^X c \, e^{x-U}dx & X\le U\end{array}\right.=\left\{\begin{array}{ll}c \,(1-e^{U-X})& X>U \\ &\\ c\, (e^{X-U}-1)&X\le U\end{array}\right.
$$ |
H: Question on Power Sets: if $|A|=n$, what is $|P(P(A))|$?
(Let $P$ be the symbol for Power Sets)
My exercise states, if $|A|=n$, what is $|P(P(A))|$?
The given answer is:
If $|A| = n$, then $|P(A)| = 2^n$ and $|P(P(A))| = 2^{2^n}$.
For $n = 2$, we have $|P(P(A))| = 2^4 = 16.$
Where does the 2 come from in the if statement?
Thanks,
Tony
AI: Every subset of a set with cardinality $n$ can be expressed as a sequence of $0$ and $1$ with size $n$. $0$ corresponds to the absence of an element in the subset and $1$ to its presence. There are $2^n$ of these sequences because for each member of sequence we have 2 choices: $0$ and $1$ and because the sequence has size $n$, we have $2\times 2\dots\times 2=2^n$ possibilities. |
H: Help me this proof! Related to RSA public key cryptosystem
Basically it is similar to the RSA algorithm.
Let p and q be distince primes and let e and d be the integers
satisfying $de≡1$ (mod (p-1)(q-1)).
Suppose further that c is an integer with gcd(c,pq)>1.
Prove that $x≡c^d$ (mod pq) is a solution to the congruence $x^e≡c$
(mod pq).
I have done with when gcd(c,pq)=1 which I found pretty easy.
But for when gcd(c,pq)>1, c and pq has same factor. So, $p$ or $q$ or $pq$ is a divisor of c.
And then I am stuck with it.
What is the next step of the proof?
Thank you.
AI: So, you're trying to show $c^{de}\equiv c\pmod{pq}$. It suffices to show $c^{de}\equiv c\pmod p$ and $c^{de}\equiv c\pmod q$. Consider two cases for the first of these, depending on whether or not $c$ is a multiple of $p$. Can you take it from there? |
H: Using integrals to prove that the mean of the sampling distribution is the population mean
Let the random variables $X_1, X_2, \dots X_n$ denote a random sample from a population.
The sample mean of these random variables is:
$\overline{X}=\frac{1}{n}\sum\limits_{i=1}^{n}X_i$
I would like to show that the mean of the sampling distribution of the sample mean is $\mu$, the population mean.
Here's what I have done:
$$\begin{align}
E(\overline{X}) &= \int\limits_{\overline{X}} \bar{x} f(\bar{x})\,\, d\bar{x} \\
&=\int\limits_{\overline{X}} \left(\frac{1}{n} \sum\limits_{i=1}^n X_i \right) f(\bar{x}) \, d\bar{x}
\end{align}$$
From here, I am not sure what to do anymore but anyway I end up with:
$$\begin{array} {cc}
&=& \frac{1}{n} \left( \int\limits_{\overline{X}}X_1f(\bar{x}) \, d\bar{x} + \int\limits_{\overline{X}}X_2f(\bar{x}) \, d\bar{x}+ \dots + \int\limits_{\overline{X}}X_nf(\bar{x}) \, d\bar{x}\right)
\end{array}$$
Now, I don't know how to complete this as I am unsure how to interpret the last equation. Somehow, the $\int\limits_{\overline{X}}X_if(\bar{x}) \, d\bar{x}$ is suppose to equal to $\mu$ but I don't see how that can be true.
I know the answer will be $\mu$ because of here but I would like to arrive at the answer using integrals instead.
AI: No problem giving a complete answer. Let's see
As already stated, the sample mean is a function of many random variables, and so the symbol $E$ refers to the expected value with respect to their joint distribution. Denoting $\mathbf X$ the multivariate vector of the $n$ r.v.'s, their joint density can be written as $f_{\mathbf X}(\mathbf x)= f_{X_1,...,X_n}(x_1,...,x_n)$ and their joint support
$D = S_{X_1} \times ...\times S_{X_n}$
The sample mean is a function of this multivariate vector, $\bar X = \frac 1n \sum_{i=1}^{n}X_i = g(\mathbf X)$. Using the Law of Unconcscious Statistician We have
$$E[\frac 1n \sum_{i=1}^{n}X_i] = \int_D g(\mathbf x)f_{\mathbf X}(\mathbf x)d\mathbf x$$.
Under convergence regularity conditions we can decompose the multidimensional integral into an n-iterative integral:
$$E[\frac 1n \sum_{i=1}^{n}X_i] = \int_{S_{X_n}}...\int_{S_{X_1}}\left[\frac 1n \sum_{i=1}^{n}x_i\right]f_{X_1,...,X_n}(x_1,...,x_n)dx_1...dx_n $$
and using the linearity of integrals we can decompose into
$$ = \frac 1n\int_{S_{X_n}}...\int_{S_{X_1}}x_1f_{X_1,...,X_n}(x_1,...,x_n)dx_1...dx_n \; + ...\\ ...+\frac 1n\int_{S_{X_n}}...\int_{S_{X_1}}x_nf_{X_1,...,X_n}(x_1,...,x_n)dx_1...dx_n $$
For each n-iterative integral we can re-arrange the order of integration so that, in each, the outer integration is with respect to the variable that is outside the joint density. Namely,
$$\frac 1n\int_{S_{X_n}}...\int_{S_{X_1}}x_1f_{X_1,...,X_n}(x_1,...,x_n)dx_1...dx_n = \\\frac 1n\int_{S_{X_1}}x_1\int_{S_{X_n}}...\int_{S_{X_2}}f_{X_1,...,X_n}(x_1,...,x_n)dx_2...dx_ndx_1$$
and in general
$$\frac 1n\int_{S_{X_n}}...\int_{S_{X_j}}...\int_{S_{X_1}}x_jf_{X_1,...,X_n}(x_1,...,x_n)dx_1...dx_j...dx_n =$$
$$=\frac 1n\int_{S_{X_j}}x_j\int_{S_{X_n}}...\int_{S_{X_{j-1}}}\int_{S_{X_{j+1}}}...\int_{S_{X_1}}f_{X_1,...,X_n}(x_1,...,x_n)dx_1...dx_{j-1}dx_{j+1}......dx_ndx_j$$
As we calculate one-by-one the integral in each n-iterative integral (starting from the inside), we "integrate out" a variable and we obtain in each step the "joint-marginal" distribution of the other variables. Each n-iterative integral therefore will end up as $\frac 1n\int_{S_{X_j}}x_jf_{X_j}(x_j)dx_j$.
Bringing it all together we arrive at
$$E[\frac 1n\sum_{i=1}^{n} X_i ] = \frac 1n\int_{S_{X_1}}x_1f_{X_1}(x_1)dx_1 +...+\frac 1n\int_{S_{X_n}}x_nf_{X_n}(x_n)dx_n $$
But now each simple integral is the expected value of each random variable separately, so
$$= E[\frac 1n\sum_{i=1}^{n} X_i ] = \frac 1nE(X_1) + ...+\frac 1nE(X_n) = \frac 1nE(X) + ...+\frac 1nE(X)$$
$$= \frac 1n nE(X) = E(X)$$ |
H: Does $f'(0)$ exist?
This is a problem from my introductory calculus homework.
\begin{equation}
f(x) = \begin{cases}
x^2\sin\frac{1}{x}, & x \neq 0,\\
0, & x = 0.
\end{cases}
\end{equation}
Does $f'(0)$ exist?
If I approach the problem algebraically, I obtain the answer \begin{equation}f'(x)=\lim_{x \rightarrow 0}x\sin\frac{1}{x}\end{equation}
This should be equal to 0 from the Squeeze Theorem. So, $f'(0)=0$.
However, something about this answer worries me. At class, the professor said to informally think of the tangent line at a point as the line that the curve begins to "look like" if we zoom around the point (or so I remember, I might be misquoting here). At the time this made a whole lot of sense to me. However, if I look at the problem from this angle, my geometric intuition goes beep, since I imagine the curve will never begin to look like a line :) So I wanted to know your thoughts on this one. Thank you in advance.
AI: By definition:
$$f'(0):=\lim_{x\to 0}\frac{f(x)-f(0)}x=\lim_{x\to 0}\frac{x^2\sin\frac1x}x=\lim_{x\to 0}x\sin\frac1x=0$$
since $\;\sin\frac1x\;$ is bounded |
H: $A \cap(A \cup B) = A$ for all sets $A$ and $B$
For all sets $A$ and $B$, $$A \cap(A \cup B) = A.$$
I get that this is true informally, but what would be the notation to formally prove this?
AI: Hint: Frequently, the easiest method for showing some set $X$ and some set $Y$ are equal is this: first, show that any element $x\in X$ also satisfies $x\in Y$ -- which proves that $X\subseteq Y$. Then, show that any element $y\in Y$ also satisfies $y\in X$, proving that $Y\subseteq X$.
If $X\subseteq Y$, and $Y\subseteq X$, then necessarily $X=Y$.
So, start off with an element $x$ which is in $A\cap(A\cup B)$, and show that $x\in A$. Then, show that any element $y\in A$ also satisfies $y\in A\cap(A\cup B)$.
It might help you to remember the following two facts:
Every element of $A\cap X$ is both an element of $A$ and an element of $X$ -- and so in particular is definitely a member of $A$.
If $x\in X$ and $x\in Y$, then $x\in X\cap Y$.
Edit: Here's a start.
Suppose that $x\in A\cap (A\cup B)$. Then by definition of the intersection, we must have $x\in A$ and $x\in A\cup B$. In particular, $x\in A$.
Now, see if you can show that any element $x\in A$ must also be in $A\cap(A\cup B)$. |
H: Find equation of tangent at given point given x and y equations
The question asks to find an equation of the tangent line to the
curve at the given point
x = t^2 + t
y = t^2 - t^3
Point (0,2)
I've only done questions like this where there is a given t value and not where there is an actual point given, could anyone point me in the right direction? Should I merge the x and y equations somehow?
Thanks
AI: Start by finding the value of $t$: You have $t^2 + t = 0$ and $t^2 - t^3 = 2$.
The first equation has solutions $t = 0$ (which clearly doesn't work in the second equation), and $t = -1$, which does. Hence, you just need to find the slope when $t = -1$.
Now in order to find the slope, it will be useful to know that the slope is just
$$\frac{dy/dt}{dx/dt}$$ |
H: Cardinality of product of two sets is the product of the cardinalities of both sets
Suppose $|X| = n$ and $|Y| = m$. We want to show $|X \times Y | = mn$.
MY attempt: By hypothesis, we can find bijections $f: X \to \{ 1,...,n\}$ and $g : Y \to \{1,...,m\}$. We want to find a bijection from $X \times Y \to \{ 1,....,nm\}$. I was thinking maybe of $h(x) = 2^x 3^y$. But it doesnt seem to be bijective. How can we find a bijection? Thanks a lot.
AI: It is trivial to see that if $m=1$ then it holds, as we can map $x\in X$ to $(x,\ast)\in X\times Y$ where $Y=\{\ast\}$ and this is a bijection.
Suppose this is true for $m=k$, we wish to show it is also true for $m=k+1$. Let $g\colon X\times Y_k\rightarrow \{1,\ldots kn\}$ be a bijection. We also let $h\colon X\rightarrow \{1,\ldots, n\}$ be a bijection.
We wish to find a bijection $f\colon X\times Y_{k+1}\rightarrow\{1,\ldots,(k+1)n\}$ where $Y_{k+1}=Y_k\sqcup\{\ast \}$.
We can define the bijection $f$ by setting
$$f(x,y) = \begin{cases}
g(x,y) & \text{if }y\neq \ast,\\
kn+h(x) & \text{if }y=\ast.
\end{cases}$$
and it is readily seen that this is indeed a bijection. This completes the induction. |
H: how to calculate the normal vector for a bezier curve
Say we have a cubic Bezier curve (so 4 control points) named Q. I understand how to calculate the tangent at by taking the derivative of Q and substituting but i'm not sure how to calculate the normal vector.
Thank you for the help!
AI: If the curve lies in some plane, and you want a normal vector that also lies in this same plane, just rotate the tangent vector by 90 degrees clockwise or counter-clockwise. So, if the tangent vector is $(u,v,0)$, the normal vector will be either $(v,-u,0)$ or $(-v,u,0)$. If you use a unit-length tangent vector, this will give a unit-length normal vector, too.
If the curve is not planar, there are an infinite number of normal vectors at each point, and you have to decide which one you want. If $\mathbf{t}$ is the tangent vector, and $\mathbf{v}$ is any other vector not parallel to $\mathbf{t}$, then the vector cross product $\mathbf{t} \times \mathbf{v}$ will be normal to the curve. This cross product will not necessarily have unit length, though, so you may want to unitize it before using it in further calculations.
The calculation in the first paragraph is a special case of the one in the second paragraph. We just used $\mathbf{v} = (0,0,\pm 1)$.
Finding normal vectors for Bezier curves is no different from finding normal vectors on any curve. You can read about this in any book on calculus or differential geometry. |
H: The Three-Number Problem
I am currently working on a little extra credit for my 9th grade math class and I am stuck on the one of the problems.
The Three-Number Problem
I have chosen three numbers. The second is twice the first, and the third is three times the second. The sum of the first two when multiplied by the sum of the last two happens to be the same as the first number multiplied by the quare of the second number.
I decided to approach this by first starting with 3 variables.
f, s, and t.
f standing for the first number.
s standing for the second number.
t standing for the third number.
f = #
s = 2f
t = 3s
The equation I came up with was:
(f + s) * (s + t) = fs2
Which can be simplified to:
fs + ft + s2 = fs2
However, from there on, I am stuck. I have no idea how to solve for f. However, i'm sure that i'm in the right place and that somebody here can help me out. Thank you!
AI: Welcome to Math.SE.
Your approach is quite correct so far, though you made a mistake. You wrote that:
$(f + s) \cdot (s + t) = fs^{2}$
Which can be simplified to:
$fs + ft + s^{2} = fs^{2}$
Whereas it really simplifies to:
$fs + ft + s^{2} + st = fs^{2}$.
Let's use your definition of the variables to do some substitution.
$fs + ft + s^{2} + st = fs^{2}$
Substitute $t$ for $3s$.
$fs + f3s + s^{2} + s(3s) = fs^{2}$.
$fs + f3s + s^{2} + 3s^{2} = fs^{2}$.
$fs + f3s + 4s^{2} = fs^{2}$.
Substitute $s$ for $2f$.
$f(2f) + f(6f) + 4(2f)^{2} = f(2f)^{2}$
$2f^{2} + 6f^{2} + 4(4f^{2}) = f(4f^{2})$
$2f^{2} + 6f^{2} + 16f^{2} = f(4f^{2})$
$24f^{2} = 4f^{3}$.
$6f^{2} = f^{3}$
Divide both sides by $f^{2}$.
$6 = f$ |
H: Isometric identification of $c_0^*$ and $ \ell^1$
Let $\{x_n\}_{n=1}^{\infty}\subset \ell_1$ be a sequence in $\ell_1$ with $x_n = (x_n(1),x_n(2), x_n(3),\ldots )$
I want to show that $$\lim_{n\to\infty}\sum_{j=1}^{\infty} x_n(j)y(j) = 0 $$
for all $y\in c_0$ if and only if $\sup_n \left\|x_n\right\|_1<\infty$ and $\lim_{n\to\infty}x_n(j) = 0$ for $j=1,2,3,\ldots$.
Apparently we can use the fact that there's an isometric identification of $c_0^*$ and $\ell_1$ via the canonical pairing between $c_0$ and $\ell_1$.
So how does this identification help us? With this identification, do we interpret the $x_n$ as functionals, in the sense that $y\mapsto \sum_{j=1}^{\infty}x_n(j)y(j) $ ? . To me this seems like proving that $x_n$ converges to the $0$ - funtional iff those $2$ conditions hold. How can we show this?
Can someone shed some light over this?
AI: For one of the directions, the identification makes things simple. If the functionals converge in a weak* sense to the 0 functional, what does it tell you about the norms of the functionals? Can you choose a simple $c_0$ sequence that can get you the limit of $x_n(j)$ for fixed $j$?
For the other direction, seek to bound the sum for a fixed sequence $y$ by a given $\epsilon > 0$. Try separating the sum into two parts, one where $y$ is small (you know $y \in c_0$) and the rest is just a finite sum where you can take advantage of linearity of limits. |
H: Post-concatenation of the languages represented by the null set
I have a small question regarding concatenation of regular languages:
Is it true that the concatenation $L\varnothing$, where $L$ is any regular language, result in $\varnothing$?
Namely, does $L\varnothing$ = $\varnothing$?
Thanks a lot!
AI: Yes. If $L$ and $M$ are 2 languages, then their concatenation is defined by $$LM = \{xy : x \in L, y \in M\}.$$ If $M = \varnothing$, then there is no $y \in M$, so the set on the right hand side is empty. |
H: If a sequence of summable sequences converges to a sequence, then that sequence is summable.
Let $(a_i)^n$ be a sequence of complex sequences each of which are summable (they converge). Then if they have a limit, the limit sequence $(b_i)$ is also summable. All under the sup norm for sequences.
Let $(a_i)^n$ sum to $c_n$. I.e. for all $\epsilon \gt 0$, there's $N$ such that $m \gt N \implies |\sum_{i=1}^m a_i^n - c_n| \lt \epsilon$. I want to show that there's $b$ such that $|\sum_{i=1}^m b_i - b| \lt \epsilon$ similarly. Let $b = \lim c_n$. Where to?
AI: (edited) This doesn't work, the example is pointwise convergent: Consider the alternating sequence $x=(-1,1,-1,1,\ldots)$ and let $x_n$ take the first $n$ terms of $x$ and be zero afterwards. $x_n$ converges pointwise to $x$, and is summable, but the (pointwise) limit $x$ is not summable.
Here is a uniform convergence counterexample: Take the sequence $x(i) = 1/i$, which is not summable. Now take $x_n$ to match $x$ for the first $n$ terms and be zero after. $x_n$ converges to $x$ uniformly, but the limit is not summable. |
H: Looking for guidance on a Fourier integral
Working with a Fourier transform problem, I've encountered the following integral:
$$
\int_{-\infty}^{\infty}\frac{\exp\left(-a^2x^2+ibx\right)}{x^2+c^2}dx
$$
where $a$, $b$, and $c$ are real coefficients. Mathematica claims that this has no closed-form solution, but I suspect (hope?) that's not the case. Unfortunately, my background in complex analysis is limited to some panic-studying I did to make it through a field theory course a decade ago. I've spent the weekend dusting that off, but am still stumped...
I know there's a pole at $z=ci$. If I take (to me) the obvious extension of the integral to the complex plane by simply replacing $x$ with $z$, I can even calculate its residue as:
$$
\frac{\exp\left(a^2c^2-bc\right)}{2ci}
$$
My hope was to integrate over a semicircle in the positive half-plane, and use this residue to get the integral along the real axis. However, with this extension, the integral over the semicircle of radius R doesn't seem to be zero (or possibly even converge) with R going to infinity -- or at least, Jordan's lemma doesn't give me any reason to believe so, since the $-a^2z^2$ part becomes unfriendly on the imaginary axis.
I vaguely recall there being a strategy for dealing with this sort of problem. I specifically recall the strategy not being to use $zz^*$, since explicit dependence on the complex conjugate was a no-go. For the life of me, though, I can't recall what the strategy was.
If anybody can (a) refresh my memory on this; (b) give a solution method; or (c) explain why no closed form solution exists, that would be much appreciated. Thanks in advance for any help you can offer!
AI: The gaussian term makes complex analysis iffy. This integral may be done much more easily using the convolution theorem.
To begin, note that
$$\int_{-\infty}^{\infty} dx \, e^{-a x^2} e^{i k x} = \sqrt{\frac{\pi}{a}} e^{-k^2/(4 a)}$$
$$\int_{-\infty}^{\infty} dx \frac{e^{i k x}}{x^2+c^2} = \frac{\pi}{c} e^{-c |k|}$$
The integral sought is the convolution of these transforms. That is,
$$\int_{-\infty}^{\infty} dx \frac{e^{-a x^2}}{x^2+c^2} e^{i b x} = \frac{1}{2 \pi} \sqrt{\frac{\pi}{a}} \frac{\pi}{c} \int_{-\infty}^{\infty} dk \,e^{-k^2/(4 a)} \, e^{-c |b-k|} $$
This integral is very doable, despite what Mathematica says. For the time being, I will spare you the details, which involve splitting the integral about $k=b$ and rescaling to get a nice form. The result is that the sought-after integral is
$$\frac{\pi}{c} e^{a c^2} \left [e^{-b c} \left (1+\text{erf}\left (\frac{b-2 a c}{2 \sqrt{a}} \right ) \right ) + e^{b c} \left (1-\text{erf}\left (\frac{b+2 a c}{2 \sqrt{a}} \right ) \right )\right ] $$
where erf is the error function.
NOTE
My $a$ is the OP's $a^2$ - this was not intentional, but to get an answer the OP wants, sub $a^2$ for $a$ in the above result. |
H: Graph Theory Connectivity and Diameter of Graphs
Let G be a graph of order n with κ(G) ≥ 1. Prove that n ≥ κ(G)[diam(G) - 1] + 2.
κ(G) meaning connectivity of G and diam(G) meaning diameter of G.
I know that κ(G) and diam(G) are inversely related, but I still need help approaching this problem.
AI: Have you had a theorem which says that, for any two vertices $x,y$ in a $k$-connected graph, there are $k$ internally disjoint paths connecting $x$ and $y$? All right, then. Choose vertices $x,y\in V(G)$ so that $d(x,y)=\text{diam}(G)$, and then construct a set of $\kappa(G)$ internally disjoint paths connecting $x$ to $y$. Each of those paths has length at least $\text{diam}(G)$, and so has $\text{diam}(G)-1$ internal vertices. Since they are internally disjoint, those $\kappa(G)$ paths have at least $\kappa(G)[\text{diam}(G)-1]$ internal vertices among them. Add to that the endpoints $x$ and $y$, that makes $\kappa(G)[\text{diam}(G)-1]+2$ vertices. |
H: Formula for the number of digits in the number $2^x$
I'm wondering if there is a formula for the number of digits in $2^x$.
For example if $x = 3$ then the number of digits is equal to $1$ because $2^3 = 8$
or for example if $x = 4$ then the number of digits is equal to $2$ because $2^4 = 16$.
In an attempt to solve this problem on my own I made the following table:
2^1 = 1
2^2 = 1
2^3 = 1
2^4 = 2
2^5 = 2
2^6 = 2
2^7 = 3
2^8 = 3
2^9 = 3
2^10 = 4
2^11 = 4
2^12 = 4
2^13 = 4
2^14 = 5
At first it looks like there is a pattern of threes but then $10$-$13$ ruins it. It's late and I can't think of any other way to approach this problem, any help would be awesome.
The question is $2^x \ge 100\,000\,000$ digits.
AI: Number of digits in $n$ is given by $1+\lfloor\log_{10}(n) \rfloor$. Hence, in your case, the number of digits in $2^x$ is given by
$$1+\lfloor x \log_{10}2 \rfloor$$
where $\lfloor y \rfloor$ is the greatest integer $\leq y$. For instance, $\lfloor 3.4 \rfloor =3$; $\lfloor 4.1244 \rfloor =4$; $\lfloor -6.0926 \rfloor =-7$; |
H: Why is the degree of the node pictured 5, not 4?
On the Wikipedia page for degree (graph theory), the bottom-right vertex pictured is said to have a degree of 5.
My understanding is that the degree of a vertex is defined as the number of edges incident to the vertex. In this case, it appears as if the bottom-right vertex has 4 edges incident to it: {2,5}, {3,5}, {2,5}, and {5,5}. So isn't the degree of the lower-right vertex 4, not 5?
AI: But it enters the vertex twice. I'll try to present a few additional reasons to count it twice.
What can you say about the total sum of degrees in a graph? It's an even number, equal to $2 \cdot \#E$, where $\#E$ is the number of edges. This would fall apart if you counted such edges only once.
Also, imagine that it was a directed graph. You'd need the info that the edge both enters and exists vertex $5$. |
H: Proving an inequality:
I have this inequality but I am unsure how to prove it:
0$<\alpha \leq$1
a$^\alpha$+b$^\alpha$ $\geq$ (a+b)$^\alpha$
$\forall a,b \geq 0$
I was given a hint: we can assume b$\geq$0
$(\frac{a}{b})^\alpha$+1 $\geq$ $(\frac{a}{c}+1)^\alpha$
so it suffices to prove f(x)$\geq$ 0:
f(x)=x$^{\alpha}$+1 - (x+1)$^\alpha$
AI: We have equality if $\alpha=1$. So we can assume that $\alpha\lt 1$. We have equality when $a=b=0$. So without loss of generality we can assume that $b\gt 0$. Then our inequality is equivalent to
$$\left(\frac{a}{b}\right)^\alpha+1\ge \left(\frac{a}{b}+1\right)^\alpha.$$
We will show that
$$x^\alpha +1\ge (x+1)^\alpha\quad\text{when}\quad x\ge 0.\tag{1}$$
Setting $x=\frac{a}{b}$ will yield the desired result.
Let $f(x)=x^\alpha +1-(1+x)^\alpha$. To prove Inequality (1), we will show that $f(x)\ge 0$ for all $x\ge 0$.
We have $f(x)=0$ when $x=0$. We will show that for $x\gt 0$, $f(x)$ is an increasing function. That implies that $f(x)\gt 0$ for all $x\gt 0$.
Note that
$$f'(x)=\alpha\left(\frac{1}{x^{1-\alpha}}-\frac{1}{(1+x)^{1-\alpha}}\right).$$
Thus $f'(x)$ is positive if $x\gt 0$, since $x^{1-\alpha}\lt (1+x)^{1-\alpha}$. It follows that for $x\gt 0$, $f(x)$ is an increasing function. Since $f(0)=0$, this completes the proof. |
H: For $a,n \in \mathbb{Z}$, where $n>a>0$, prove that $\nexists x \in \mathbb{Z}$ s.t. $nx=a$
Suppose $a,n \in \mathbb{Z}$, and $n>a>0$. How do I prove that $\nexists x \in \mathbb{Z}$ s.t. $nx = a$ ? I'm really not sure where to start on this one. I'd be happy if someone could give me a hint.
Edit: I've solved this by contradiction, but I will not be 'accepting' an answer from below because I did not use any one of them in a significant way to solve the problem.
AI: Assume there is such an $x$. Since $nx = a$, then $0 < nx$ and $nx < n$. Can you now prove that $0 < n$ and $n < 1$? And can you prove that that is a contradiction?
Edit: changed $p < q < r$ statements to $p < q$ and $q < r$ statements. Because hypothetically, I forgot what $<$ means.
Edit 2: Electric Boogaloo
So one way of defining $\mathbb{Z}$ is that it is a commutative ring with a subset $\mathbb{N}$ such that:
Non-trivialty: $\mathbb{N} \ne \emptyset$.
Closure: for all $a,b \in \mathbb{N}$, $ab \in \mathbb{N}$.
Trichotomy: For all $x \in \mathbb{Z}$, exactly one of the following is true: $x \in \mathbb{N}$, $x = 0$, $-x \in \mathbb{N}$.
Well-ordering principle: If $S \subseteq \mathbb{N}$ and is non-empty, there is an $x \in S$ such that for all $y \in S$, $x \le y$.
Now we can define $<$. We say $a < b$ if $b - a \in \mathbb{N}$.
So if you want to prove your statement from the ground up, you should prove:
$a < b$ and $b < c \implies a < c$ (after this point you can write $a < b < c$)
$ab > 0$ and $b > 0 \implies a > 0$ (useful for part 3)
$ac < bc$ and $c > 0 \implies a < b$ (division not allowed)
$0 < 1$, or equivalently $1 \in \mathbb{N}$ (master troll)
$\not\exists x \in \mathbb{Z} \ 0 < x < 1$ (this one uses well-ordering) |
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