it's not a serious algorithm, it's quantum bogosort
1. Quantumly randomise the list, such that there is no way of knowing what order the list is in until it is observed. This will divide the universe into O(n!) universes; however, the division has no cost, as it happens constantly anyway. 2. If the list is not sorted, destroy the universe. (This operation is left as an exercise to the reader.) 3. All remaining universes contain lists which are sorted.
given $f \in \operatorname{Hom}(X,Y)$ and $x \in X$ and $f(x) \in V \subseteq Y$, now $f$ is open so $f^{-1}(V) \subseteq X$ is open. Pick $x \in U \subseteq C \subseteq f^{-1}(V)$ with $U$ open and $C$ compact, then I think $C$ and $V$ should be what I want...
Does someone know if this this pair of differential equations have a name, I mean that if they are somehow important in mathematics $(1-x^2)y''-xy'+n^2y=0$ and $(1-x^2)y''-3xy'+n(n+2)y=0$ ?
So, random thought/question: Are there magmas out there which are strange enough that you couldn't possibly have a real world system which said magma represents the behavior of?
Hmmm, fair enough. I suppose it's not necessarily defined in all cases. Though, I was thinking about dihedral groups, and how they represent the symmetries of a polygon.
Or perhaps how the quaternions have been used to represent rotations in 3D computer simulations.
I'm no philosopher, but I think if you choose a really big cardinal $\kappa$, then I don't think a group like $\Bbb Z^{\kappa}$ or $\Bbb R^{\kappa}$ represents any real world behavior unless you believe in some platonic realm of ideas
physicists always seem to model the universe as manifolds, so if we accept that, there are some severe cardinality restrictions on what groups can represent "real world behavior"
Okay, there is a distinction I would make between "cardinality restrictions" and "structure restrictions," in that case.
Though, it's really hard to describe what I mean when I say "structure"
Like, I would want say "it's the way the magma elements link together under the magma operation" in more clear and rigorous terms, but I don't know even how to start describing that.
I think if I am going to figure out the pattern that governs scientific breakthroughs, I might need to research the creativity theory literature
because the nature of creativity and serendipity are key to the emergence of scientific breakthroughs, the most well known process where an unknown unknown becomes a known via scientific investigation
(meanwhile the spiritual and religious counterpart is known as mystical experiences and divination)
I wonder... is there a mathematics to creativity, given that creativity is ultimately an inductive process which needs input ideas in order to churn out a new idea?
Let $(R,\mathscr T)$ is finite compliment topology. How do I prove it lindelof? My attempt: Let $\{U_{\alpha}:\alpha \in \Lambda\}$ be an open cover. We know that any $\alpha \in \Lambda$. Let $\mathscr \setminus U_{\alpha}=\{x_1,x_2,...,x_n \}$. Then we can choose opens sets from $\{U_{\alpha}:\alpha \in \Lambda\}$ containing $x_1,x_2,...,x_n$. Let it be $U_{\alpha_1}, U_{\alpha_2},..., U_{\alpha_n}$.
Hence required countable subcover is $\{U_\alpha, U_{\alpha_1}, U_{\alpha_2},..., U_{\alpha_n}\}$
Rudin doesn't require that a metric space be non-empty. I agree that there is no good reason for a convention which says otherwise. For example, for those of you who are convinced by such reasons, we want subspaces of metrizable spaces to be metrizable.
Hey, $x^\dagger x=\langle x,x\rangle$ is always real, yeah?
Right, yes
If $A^\dagger=A$, then is $x^\dagger Ax$ real?
Ah yeah I see it
I just needed to write this down somewhere rather than holding it in my head
Ignore me
I need to get used to the $\langle x\rvert A\lvert x\rangle$ notation, I'm watching a video series on quantum mechanics and the different notation is confusing me
But when I write it as $x^\dagger Ax$ I can see immediately that it's real
Also the eigenvalues of $A$ are real because if $x$ is an eigenvector of $A$ then the eigenvalue is $\dfrac{\langle x|A|x\rangle}{\langle x|x\rangle}$
(aka $\dfrac{x^\dagger Ax}{x^\dagger x}$)
Also if $x$ and $y$ are different eigenvectors then $\langle x|A|y\rangle$ equals both $\lambda_x\langle x|y\rangle$ and $\lambda_y\langle x|y\rangle$, so either $\lambda_x=\lambda_y$ or $\langle x|y\rangle=0$
Arright I'm getting the hang of this bra/ket thing
but also $\langle x|U|x\rangle=\langle U^\dagger x|x\rangle=\langle U^{-1}x|x\rangle= \langle\lambda^{-1}x|x\rangle=\overline{\lambda^{-1}}\langle x|x\rangle$
@LeakyNun yeah for locally profinite groups you basically always reduce integrals to actual sums, we did this a lot in the proofs for local Langlands for GL(2)
Gotcha. The blowdown map is an isomorphism of sheaves away from the exceptional divisor, of course. Forms near the exceptional divisors are forms on $\mathcal{O}_{\Bbb P^1}(-1)$ compactly supported on the fibers of $\mathcal{O}_{\Bbb P^1}(-1) \to \Bbb P^1$.
I guess I don't know how explicitly you want that.
We haven't really talked about divisors, but loch already convinced me that the sheaf of differentials should only be supported on the preimage of the origin, since the map is an isomorphism away from it
Apparently the relative viewpoint of working in the category of schemes over a base scheme is a prominent one. I don't really know why though, it's just something I've been told
In the general case if you have a morphism $f:X\to Y$ you can look at $\Delta:X\to X\times_Y X$ and the sheaf $\mathscr I=\ker(\mathcal O_{X\times_Y X}\to\Delta_\ast\mathcal O_X)$ and then define $\Omega_{X/Y}$ as $\Delta^\ast(\mathscr I/\mathscr I^2)$
Then you can check that if $V$ is an open affine in $Y$ and $U\subseteq f^{-1}(V)$ is open affine in $X$ then $\Omega_{X/Y}$ restricts to the module of Kähler differentials on $U$
the way i think of it is that working over a base scheme is essentially working with "families" - so for example a morphism $X\rightarrow Y$ is smooth is basically saying that all your fibers are smooth varieties (+ some more conditions)
so in particular even if $X$ and $Y$ can be singular, $X\rightarrow Y$ can be smooth
I guess I can ask this smoothly as well. If I have a fiber bundle with compact fiber and base, look at the subcomplex of differential forms tangent to the fibers.
it's kind of ironic that Chevalley introduced adeles to get rid of analysis in the proofs for CFT and then people took that concept to do ... analysis with it
I'm looking at $X=\mathrm{Spec} A$ with $A=\Bbb C[x,y]/(xy^2-x^2-y^2+x)$ and the projection $X\to\mathrm{Spec}\Bbb C[x]$. If I didn't mess up with basic algebra $X$ has two singular points, corresponding to $(x,y)=(1,\pm 1)$. I'm now asked to compute the fibers over $(x)$ and $(x-1)$
the idea to do analysis on adeles (or more precisely on the adelic points of an algebraic group over a global field, but that's a nitpick) is very fundamental for automorphic forms etc.
computing the fiber over $(x)$ seems simple unless I'm missing something: it's just $\Bbb{C}[x]/(x) \otimes_{\Bbb C[x]} \Bbb C [x,y]/(xy^2-x^2-y^2+x)= \Bbb C[y]/(y^2)$
Suppose $1 \le p < \infty$ and $f_n ,f \in L^p[0,1]$. If $f_n \to f$ almost everywhere, then prove that $||f_n-f||_p \to 0$ iff $||f_n||_p \to ||f||_p$.
The above problem does not come the real analysis text I am using as a reference (Real Analysis by Royden & Fitzpatrick). However, in that ...
The first thing to observe is that the initial condition means that you are integrating the right half of the function $f'$ (this can be show easily by plotting a graph of $f'$, and noting how it is an odd function)
Since anything $x > 0$ is positive, and your integration range is from 1/2 to 1, it means the definite integral hence the area under the curve must be positive
This means the smallest value $m$ can be must be zero
$f'$ is positive on the interval, so $f$ is monotonic, hence the integral is bounded above by $f(1)/2$. Using the MVT, this is equal to $f'(\xi)/4$ for some $\xi\in(1/2,1)$. Then you just need to estimate $f'$ with the most trivial bounds and this will yield one $M$ as in the options. I'm not sure about the corresponding $m$ though.
Could you please give me some reference(any book, article) where the author discuss this particular class of problems [ I mean, "this" class of approximation} ?
Oh, sorry for misunderstanding. This is the standard approximation $\int_a^bf\le(b-a)\lVert f\rVert_{\infty}$, which is a direct consequence of the monotony of the integral. In this case $\lVert f\Vert_{\infty}=f(1)$, because the function is monotonic.
It's called the supremum norm, but I'll try to simplify by assuming $f$ is continuous. I assume you know that the integral is monotone; i.e. that if $f(x)\le g(x)$ for all $x\in[a,b]$, then we have $\int_a^bf\le\int_a^bg$. This should be intuitive if you think of the integral as (signed) area under a graph.
Now let $F$ be the maximum value of $f$ on $[a,b]$ (that this value exists is not trivial, but I will take it for granted). Then we have $f(x)\le F$ for all $x\in[a,b]$, so monotony gives us $\int_a^bf\le\int_a^bF=(b-a)F$ since $F$ is a constant.
In this particular case, $f$ is monotone, so its maximum value on the interval $[1/2,1]$ will simply be its value at the right-most point, that is $f(1)$.
That is all the reasoning behind the inequality $\int_{1/2}^1f\le\frac{f(1)}{2}$. Intuitively, youre just approximating the graph of the function by a big enough rectangle. If you want me to elaborate on a part, just ask.
Not terribly hard. You can reason on degree---for any linear combination of the two polynomials, every monomial will either have y-degree 2 (because of the second polynomial generating the ideal), or will have both y and z present (because of the first polynomial), so y by itself can't be a linear combination
is there a fast way to find real roots of sparse polynomials? In my case the polys have exactly 4 non-zero terms and exactly 2 real roots. I am currently using bpaste.net/show/b8d27ccc46d0 numpy but it is very slow
Suppose $U \subseteq X$ is an open, path-connected subset such that the inclusion induced map $\pi_1(U) \to pi_1(X)$ is trivial. Suppose I have paths $\eta_1$ and $\eta_2$ are paths in $U$ with the same starting and ending point such that $\iota \eta_1 \simeq \iota \eta_2$. Hatcher claims that $\eta_1$ and $\eta_2$ are path homotopic in $U$. Why is that?
Here is a quote from Hatcher's Algebraic Topology:
Given a set $U \in \mathcal{U}$ and a path $\gamma$ in $X$ from $x_0$ to a point in $U$, let $$U_{[\gamma]} = \{[ \gamma \cdot \eta ] \mid \eta \text{ is a path in } U \text{ with } \eta (0) = \gamma (1) \}$$ As the notation indicates, $U_{[\...
Rudin doesn't require that a metric space be non-empty. I agree that there is no good reason for a convention which says otherwise. For example, for those of you who are convinced by such reasons, we want subspaces of metrizable spaces to be metrizable.
Suppose $1 \le p < \infty$ and $f_n ,f \in L^p[0,1]$. If $f_n \to f$ almost everywhere, then prove that $||f_n-f||_p \to 0$ iff $||f_n||_p \to ||f||_p$.
The above problem does not come the real analysis text I am using as a reference (Real Analysis by Royden & Fitzpatrick). However, in that ...
