In this page: mathoverflow.net/questions/23478/… , there is the phrase "The closure of open ball of radius $r$ in a metric space, is the closed ball of radius $r$ in that metric space." given as a false belief. And there is the comment giving a simple counter-example: the closure of a ball of radius $r$ is not the closed ball of radius $r$. I would say that this is the trivial example, right? Is there any others?
How can I solve the functional equation to find all involutions whose reciprocals are non trivial involutions?
I think this functional equation can be expressed as:
$$ f(x)=\frac{1}{f(x)}, $$ with the conditions that $f(f(x))=x$ and $\frac{1}{f(f(x))}=\frac{1}{x}.$
$f(x)=\frac{1}{x}$ does not ...
Hello guys, I have another quick question about mathematical statements. Is the following statement false? "There exists a x in the reals, for all y in the reals, such that x*y=0"?
Oh, I'm sorry. I wasn't aware of this. Thanks for the pointer. I think it's false because if, for instance, we pick x = 3, then it wouldn't work if y = 1 (since the statement is claiming that it should work for all y in the reals).
But isn't this statement "lets" us pick x that would make this statement true for all possible numbers, not only 1? Or is just picking one is sufficient to make this statement true? I am quite confused about this concept.
Oh, so this statement structure (more precisely, the 'there is' quantifier) allows us to be given a number for which we could make this statement true?
since the other part states that y can be any real number?
Let $V$ be an $n$-dimensional inner product space, where $n>0$, and let $F$ be the linear transformation on $V$ defined by $F(\textbf{u})=\langle \textbf{u},\textbf{c} \rangle \textbf{b}-\langle \textbf{b},\textbf{c} \rangle \textbf{u} $, where $\textbf{b},\textbf{c} \in V$ and $ \langle \textbf{b},\textbf{c} \rangle \neq 0 $. Show that $V$ has a basis consisting of eigenvectors of $F$ and find the matrix of $F$ with respect to some such basis.
Oh, I see. So just to clarify - basically because the statement is stating that there exists a number (which we're not allowed to choose since it's given), then by finding at least one number that would make this statement work (which is 0 in this case), then we could make this statement work?
The right x. It has to work for all y. Everything depends on the form of your particular statement. You have to pay a lot of attention to "there exists" and "such that" and "for all."
One would probably want to eliminate $\textbf{b}$ in $F(\textbf{u})=\langle \textbf{u},\textbf{c} \rangle \textbf{b}-\langle \textbf{b},\textbf{c} \rangle \textbf{u}$, no?
Or rather $\langle \textbf{u},\textbf{c} \rangle \textbf{b}$.
How to solve $ (2D^2 -3D + 2) =x e^{-\dfrac{x}{2}}$ by operator method, where D is differential operator. It is not the case that I am solving by this method for the first time, but in this problem I get stuck.
@TedShifrin Since the inner product is not defined (it could be the dot product or not), it is unclear what would make $\langle \textbf{u},\textbf{c} \rangle =0$, isn't it?
I guess you are confused with the method of variation of parameters with the operator method, I usually have to use the solution of auxiliary equation(hence characteristics equation) in the method of variation of parameters. But in this case I have to $y_p = (F(D))^{-1} (x e^{-\tfrac{x}{2}} )$
@TedShifrin Yes, any vector in $\mathbf c^\perp$ is an eigenvector, since $v \in \mathbf c^\perp$ would give $F(\textbf{v})=-\langle \textbf{b},\textbf{c} \rangle \textbf{v}$
For the simplest case, say the on the right side of the differential equation, if there is a function of the form $e^{ \alpha x}$, and let the equation is $F(D) y = e^{ \alpha x}$, and if $F( \alpha ) \ne 0$, then for $ (F(D))^{-1} = (F(\alpha))^{-1} $
@TedShifrin You mean there's a way to manipulate $F(\textbf{u})=\langle \textbf{u},\textbf{c} \rangle \textbf{b}-\langle \textbf{b},\textbf{c} \rangle \textbf{u}$ to find the last eigenvector?
The "inverse" can't depend on what's on the right-hand side like that. What if $\alpha$ turns into $\beta$?
You. :)
@schn: Your first strategy was to get rid of the $\mathbf b$ term. What is another strategy to get an eigenvector?
@Ajay: I think you need to find something more general. You're being confused by having pure exponentials, since any power of $D$ turns them back into themselves. You need to read about the general method.
Assuming, you have pointed out the last formula, $(F(D))^{-1} = (F(\alpha))^{-1}$, My prof. convinced us that for the case when $\alpha$ is constant, $F(D) e ^ {\alpha x} = F( \alpha ) e^ { \alpha x } $.
I tried that, but at the end, I am getting something like $ (D - 2.5)^{-1} x$, which normally doesn't happens witht he problem that I have solved yet. The problem is -2.5 part.
Hey guys, to prove the following statement, "The number $a^2$ is even if an only if $a$ is even", can I use a direct proof strategy to prove one way, and then the contrapositive of this statement to prove the other way?
The contrapositive of “if P, then Q” is “if not Q, then not P”. It is not equivalent to “P only if Q” @Abwatts
“If a is even then a^2 is even” is equivalent to “If a^2 is not even then a is not even.” “If a^2 is even then a is even” is equivalent to “if a is not even then a^2 is not even.” But the first two statements are not equivalent to the latter two. For the iff , all four need to be true.
How many ways are there to pick an ordered tuple of 5 elements from
the set of positive integers between 1 and 50 inclusive, provided that
there must be at least one consecutive sequence of 4 consecutive
elements?
So what I think is somehow we need to use inclusion exclusion principle ; So I pro...
(a) Prove that $\operatorname{gcd}(a, b) | \operatorname{gcd}\left(3 a+b, a^{3}\right)$
Since say $d = gcd(a,b)$ then $d|a$ and $d|b$ this will imply $d|(3a+b)$ and $d|a^3$ and therfore $d$ is divisor of those two.But how to show it is greatest common divisor.
Morning guys, when defining "maximal" Hilbert space my textbook says, "the idea behind maximality is that we can obtain $\mathfrak H$ as a limit of a finite-dimensional projections, i.e. $$\mathfrak H= \bigoplus_{k\in\mathbb N} \{e_k\}$$"
Suppose that $a |(4 b+5 c)$ and $a |(2 b+2 c) .$ Prove that if $a$ is odd, then $a | b$ and $a | c$
So since $a |(2 b+2 c)$ this imply $a |(4 b+4 c)$ so $a |(4 b+5 c)$ after subtraction of two dividend we directly get $a | c$ but I am unable to get why it should divide $b$ if $a$ is odd.
Hey! So I have a question but I'm not sure if it's more computer sciencey or mathematical... it has to do with a moving circle on a grid, and the question is about how many points are in that circle
I don't see the point of bothering with a non-satisfying elementary proof of quadratic reciprocity when there are very satisfying conceptual proofs with ANT
@ÍgjøgnumMeg actually we saw that $\theta(\frac{1}{x})=\sqrt{x}\theta(x)$ in the L-functions lecture. That can actually be used as a basis for a proof of QR, too
anyway the point of the proof Vogel just went through was as follows; use Gauß Lemma that says that $\left(\frac{a}{p} \right)= \varepsilon_1 \cdots \varepsilon_{\frac{p-1}{2}}$ where the $\varepsilon_i \in \{\pm 1\}$
ang on
test $\lfloor \rfloor$
okay
so one shows that $\left( \frac{a}{p} \right) = (-1)^{\sum_{i=1}^{p-1/2}\lfloor \frac{2ai}{p}\rfloor}$
this leads you to a proof of the 2nd Ergänzungssatz
then because $\left(\frac{2a}{p}\right) = \left( \frac{2}{p} \right)\left(\frac{a}{p}\right) = \left(\frac{2}{p}\right)(-1)^{\sum_{i=1}^{(p-1)/2}\lfloor \frac{ai}{p}\rfloor}$
and then cancel the 2/p and get some junk for a/p idk
the nice thing about the proof by the theta inversion formula is that it generalizes to arbitrary number fields!, I haven't looked into this, but I saw it mentioned somewhere
and define $M:=\lbrace 1, \cdots, \frac{p-1}{2 }\rbrace \times \lbrace 1, \cdots, \frac{q-1}{2}\rbrace$ and then $\ell_1 := \lvert\lbrace (i,j) \in M : qi > pj\rbrace\rvert$ and $\ell_2 := \lvert\lbrace (i,j) \in M : qi < pj\rbrace\rvert$. Then $qi > pj \iff j < qi/p \iff j \leq \lfloor \frac{qi}{p}\rfloor$
ergh
do it like that lol
and then also something analog for i
and then you get $\ell_1$ is the sum over these floors and $\ell_2$ is the sum over the other floors and then you get that $\ell_1 + \ell_2$ is exactly the number of elements in $M$, which is $\frac{p-1}{2}\frac{q-1}{2}$ so that $\left(\frac{p}{q}\right)\left(\frac{q}{p}\right) = (-1)^{\frac{p-1}{2}\frac{q-1}{2}}$
and then move the q/p to the other side...
hahahaha
I was just sitting there thinking "why are we bothering with this"
der hot an Vogel!
hehe
@Mathein I'll try to fill my notes from yesterday with some actual information about the various kinds of L-functions Rösner talked about, I didn't get down all the relevant information about all the different kinds because he speaks so fast hahahaha
I like the proof using the Albert–Brauer–Hasse–Noether theorem
The Albert-Brauer-Hasse-Noether theorem says that for any global field $K$ with sets of places $\Sigma$, there's a short exact sequence $0 \to \mathrm{Br}(K) \to \bigoplus_{v \in \Sigma} \mathrm{Br}(K_v) \to \Bbb Q/\Bbb Z \to 0$. That splits non-canonically. By LCFT we have a canonical isomorphism, called invariant map $\mathrm{Br}(K_v)=\Bbb Q/\Bbb Z$ if $v$ is non-archimedean and $\mathrm{Br}(K_v)=(\frac{1}{2}\Bbb Z)/\Bbb Z$ if $v$ is real and obviously $\mathrm{Br}(K_v)=0$ if $v$ is complex. The map $\bigoplus_{v \in \Sigma} K_v \to \Bbb Q/\Bbb Z$ is given by summing up those local invari…
@Alessandro hm, for Kan extensions, you could look at Riehl's "Category in Context" http://www.math.jhu.edu/~eriehl/context.pdf for derived functors, any text on homological algebra will talk about those
@Ultradark $\Bbb RP^n$ is the compactification of $\Bbb R^n$. You can take a colimit over inclusions of $\Bbb RP^n$ into $\Bbb RP^{n+1}$ and obtain $\Bbb RP^\infty$, which is the classifying space of $O(1)$-principal bundles
@Ultradark Stereographic projection is used to define charts on the Riemann sphere for example
@Ultradark It's just used for giving the topological spaces you get from compactification nice charts, and you can verify they chart transitions are smooth
let $V$ be a real vector space, $||v|| = d(v,v)$ be a norm with the parallelogram rule, i.e., $||u-v||^2 + ||u+v||^2 = ||u||^2 + ||v||^2$. prove that $<u,v> := \frac{1}{4}( ||u+v||^2 - ||u-v||^2 )$ is an inner product over V.
(hermitian) symmetry and positivity are obvious. linearity on the first component is not.
If you took the stereographic projection of $S^2$, but only the portion that maps to the unit disk, and looked at its chart, and then identified the points on the boundary of the disk, would the resulting shape, $S^2$ be an isometry of the sphere you started with
What sort of answer would one possibly expect to "Let $D$ be a divisor with $deg(D)=-2$. Describe all meromorphic 1-forms $\omega$ such that $(\omega)=D$."
Is there a nice description given by the one relation we have (coming from the change of charts) for arbitrary rational functions?
Oh, I forgot to say, that was explicitly for CP^1
Perhaps I can deal with a local description where $\omega$ is $f$ on $U_0$ and $g$ on $U_1$, and deduce that $$a\ne 0, ord_{[a:b]}\omega = ord_{b/a}f,\qquad a=0, ord_{[0:1]}\omega = ord_{\infty}f - 2$$
although I think I am being unrigourous in some sense, since $f:\Bbb C\to \Bbb CP^1$ rather than $f:\Bbb CP^1\to\Bbb CP^1$, meaning the order at $\infty$ may make no sense
My task is to write volume integral in 3 coordinate systems : Cartesian, cylindrical and spherical. This integral shows volume of intersection of 2 spheres, first with center at $(0, 0,-3)$ and radius $5$ and second with center at $(0,0, 3)$ and radius $\sqrt{13}$.I am able to do it for Cartesian...
Well, apparently $\Omega(D)=\{\omega\in M\Omega^1(\Bbb CP^1)| (\omega)\geq D, \text{ or }\omega = 0\}$ is finite dimensional, and perhaps this is equal to my space, since any differential $1$-form must have degree $-2$, so this perhaps this already equals $\Omega(D)=\{\omega\in M\Omega^1(\Bbb CP^1)| (\omega)= D, \text{ or }\omega = 0\}$
Go back and review how you set up double/triple integrals. You fix the outer variable at some random value and ask how the inner variable varies, depending on the value of the outer variable.
You should watch some of my YouTube lectures on multiple integrals.
Sanity check: I have $\int_0^1 u^{s/2}\omega(u)\frac{du}{u}$ and I substitute $u := \frac{1}{x}$. Then I get $-\int_{1}^\infty x^{-s/2}\omega(\frac{1}{x})\frac{dx}{x}$... correct? (the $\omega(\frac{1}{x})$ factor doesn't generate a $-1$ so the sign should be correct)
@TedShifrin So the most well-known meromorphic $1$-form is probably the one corresponding to $dz = -1/w^2 dw$, i.e. with $1$ on $\phi_0(U_0)$ and $-1/w^2$ on $\phi_1(U_1)$
And I take an arbitrary divisor $D$ with $deg(D)=-2$, and we've deduced that it has $dim(\Omega(D))=1$. So maybe I pick a concrete form for $D$ like: $$D= \sum_{i=1}^n m_i\cdot p_i, \sum m_i = -2$$
Since you guys are discussing differential-things: I’ve learned that the Riemann integral over a surface D. The formal definition uses a limit quantifying over arbitrary points of of all partitions of D. I can’t really see how the choice is irrelevant, but I can see this probably relates to analysis (in $\mathh{R}^n), so I skipped it. Should I be worried?
No, actually, surface integrals are very subtle. If you don't approximate correctly, you can get very wrong answers. There's a very famous example. It appears in Spivak's Calculus on Manifolds and (stolen) also in my multivariable math book.
The example is originally due to Schwarz.
It's giving a (finite) right circular cylinder with apparently infinite surface area.
It seems that everything about derivatives is easier than those about integrals... the whole idea of approximating things in calculus makes me anxious, lol. Studying calculus without analysis is kinda harrowing.
@Galois: Your meromorphic $1$-form looks like $f(z)\,dz$ (in the main chart), so I'm asking you to give an explicit formula for $f(z)$ given that $(\omega) = D$. You already picked a specific representative for $D$. Let's just suppose $D=-a-b$ ($a,b\in\Bbb C$).
Where unless I'm wrong, $ord_{[z_0:z_1]}\omega = ord_{z_1/z_0}f$ for $z_0\ne 0$ and $ord_{[0:1]}\omega = ord_\infty f -2$ (which may not make sense, since $f$ is not defined at $\infty$ hmm)
Let $n$ be a positive integer.
Conjecture
There are infinitely many prime twins of the form
$$ ( 3^n - 2, 3^n - 4) $$
Examples include
$$(3^2 - 2,3^2 - 4) = ( 7,5 ) $$
$$ ( 3^{37} - 2 , 3^{37} - 4 ) $$
$$ ( 3^{41} - 2 , 3^{41} - 4 ) $$
Notice $2,37,41$ are primes themselves.
Coincidence ?
...
How can I solve the functional equation to find all involutions whose reciprocals are non trivial involutions?
I think this functional equation can be expressed as:
$$ f(x)=\frac{1}{f(x)}, $$ with the conditions that $f(f(x))=x$ and $\frac{1}{f(f(x))}=\frac{1}{x}.$
$f(x)=\frac{1}{x}$ does not ...
