@geocalc33 seems likely to me that any explanation will be a matter of physics as much as math. (For instance, the index of refraction of the plastic and its thickness, as well as the curvature of the container)
I'm trying to calculate $\int_{S} \mathbf{F} \cdot d\mathbf{S}$, with $\mathbf{F} = x\mathbf{i} + y\mathbf{j}$ and $S$ being the half-sphere $x^2 + y^2 + z^2 = 1$, $z \geq 0$ with the normal vector pointing up. I tried $d\mathbf{S} = \left(\frac{x}{\sqrt{1-x^2-y^2}}\mathbf{i} + \frac{y}{\sqrt{1-x^2-y^2}}\mathbf{j} + \mathbf{k}\right)dxdy$ but that didn't seem to work very well. Is there an easier way?
If you were on the whole sphere, you could exchange variables. It's clear that the integral of $x^2$ is the same as the integral of $y^2$ is the same as the integral of $z^2$.
But, even so, in spherical coordinates $dS = \sin\phi\,d\phi\,d\theta$ (using the American convention), so I'm integrating $\sin^2\phi dS$ and this integral is easy peasy.
The bit that I find interesting: suppose we’d started from the (rotation-invariant) sphere integral. Then there are two ways to try to proceed. One is to symmetrize the integrand over the sphere, so that $x^2+y^2\to \frac{2}{3}(x^2+y^2+z^2)$. The other is to use reflection symmetry across $z=0$. The former approach gives an easy answer, while the latter only makes things harder to see.
So to use spherical symmetry requires you to first ‘give up’ the reflection symmetry. (I don’t like how I’m saying it but I can’t think of anything better)
Let $F_{k+1}\subseteq F_k$ be a sequence of closed ,non empty sets in a metric space $X$ Show that if the diameter of $F_n$ tends to 0 then the itersection is non-empty
@TedShifrin I can think of two, actually. First one is to find the current passing through a loop by integrating the volume current density (which is a vector field)
@TedShifrin so idea is to choose $x_n\in F_n$ then hit that sequence of the condition that $diam(F_n)\right arrow 0$ and that will give me a point in the intersection?
If I were to build up an example, I’d probably make use of the continuity equation for the volume charge distribution: $\nabla\cdot\vec{J}=\partial \rho/\partial t$ where $\rho$ is the volume charge density
@LukasHeger If $(X,d)$ is a complete metric space and $F_{k+1}\subseteq F_k$ is closed and non-empty for each k, then the intersection is non-empty, is false?
In mathematics, a field K with an absolute value is called spherically complete if the intersection of every decreasing sequence of balls (in the sense of the metric induced by the absolute value) is nonempty:
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Let $\epsilon>0$ Since $diam(F_n)\rightarrow 0$, there exists $N$ such that $n\geq N$, $diam(F_n)<\epsilon$. Hence whenever $n>m\geq N$, we have $F_n \subseteq F_m$ and so $d(x_n,x_m)\leq diam(F_m)<\epsilon$.
@TedShifrin otherwise, if $x\not F_n$ for some n, then since $x_n$ $\in F_n$, we may find $x_{n+1}, x_{n+2}....$ in $F_n$. This is also a subsequence which must converge to $x$
I will prove that as well, and it doesn't seem hard. Thank you once again. I should stop worrying that the problems are too difficult, it gets to me lol and I start to kinda get over anxious
Let f be a smooth map X->Y, and D a distribution acting on smooth functions on Y. Further let g be a smooth function on Y. Is it true that (f* D)(f* g) = D(g) , where the first f* denotes the pullback of distributions, and the second is the pullback on functions (i.e. precomposition)? Looking at the obvious commutative diagram this seems to make sense, but I can't see how this can be justified precisely.
@geocalc33 sounds like a number which is strictly less than 1 and at the same time strictly greater than 1 IRCC the definitions of contraction and expansive mappings
i'm sure this is simple maths, but i'm just not very good at it. I have a worker that does jobs... each job takes somewhere between 5 and 10 seconds to do... let's say 7 seconds is the median. i have 900,000 of these jobs (for now, let's assume the jobs never grow) meaning the worker would take 72~ days to do all jobs.
how many workers would i need to clear them in 8 hours?
@TedShifrin Any idea why the same doesn't work for AsymptoticDSolveValue? I computed the first few coefficients for the Maclaurin series by hand and wanted to use Mathematica to see if there is some pattern
This the only way we know to go. Squad up never roll alone. And we gon' ride on forever. We, ride out together. Pull up right in your zone. Take over the street. That's how we roll. And we gon' ride on forever. We, ride out together.
If you love perpetual loxodromic transformations give me one clap! If you love isochronous monodromic isoclines give me two claps! If you enjoy torsive non-commuting expansive/anticontractive maps shake your hands in the air and say “totally complete isogenous metric space!”
Suppose r 00 < Ez. A central problem in applied numerical probability is the description of sub-naturally Noetherian subrings. We show that there exists a stochastic u-totally bounded, partially semi-algebraic ring. In [17], the main result was the characterization of domains. Recent interest in functions has centered on describing homeomorphisms.
"The goal of the present article is to describe quasi-covariant hulls. In [10], it is shown that the Riemann hypothesis holds. In future work, we plan to address questions of minimality as well as positivit"
@courge9 I challenge you to show this is false... Definition 3.1. Let $l$ be a holomorphic monoid. A Gauss topos equipped with a Hamilton–Landau monodromy is a functor if it is discretely infinite
How do you find the inverse for a complex number without using the notation with a + ib .. if you only have the definition of multiplication of (a,b), (a°,b°) where a° and b° should be inverses such that (a,b).(a°,b°)=(1,0). I tried to do this with only the definition of the multiplication and i do not get a correct result
the definition is (a,b).(a°,b°) = (a.a°-b.b°, a.b°- b.a°)
@MadSpaceMemer well, how do you translate between (a,b) and a+bi? if you know what the inverse of a+bi looks like, just translate this back into ordered pairs and you're done
@courge9 well take the 2-sphere for example. an infintesimal rotation leaves the sphere unchanged. but what if you start with a small number of points as subsets of the sphere, and have an iterative rotation process that increases that subset of points until the subset converges to satisfy the equation of a 2-sphere
well that is a terrible advice, Copying is by no means a good meassurment of what a good scientist is. But i guess he was trying to make apoint, so okay
@MadSpaceMemer Of course he meant copying from your own notes or from the text of an exercise
@MadSpaceMemer NOT copying in the sense of plagiarism
Because if you start with a wrong assumption which stems from the fact that you made a mistake when writing off some text, exercise, theorem - whatever - then everything gets wrong
And of course, if you constantly fail at reading and writing things properly, you cannot be a good scientist.
That was the point he wanted to make - half serious, half tongue-in-cheek
If you have a vector field $X$ partitioned into integral curves
A flow on a set $X$ is a group action of the additive group of real numbers on $X$. More explicitly, a flow is a mapping $\phi:X\times \Bbb R \to X$ s.t.
$\phi(x,0)=x$ and
$\phi(\phi(x,y),s)=\phi(x,s+t)$
Given x in X, the set ${\displaystyle \{\varphi (x,t):t\in \mathbb {R} \}}$ is called the orbit of x under φ. Informally, it may be regarded as the trajectory of a particle that was initially positioned at x. "If the flow is generated by a vector field, then its orbits are the images of its integral curves."
I don't understand that part in quotes
I thought the orbits would be exactly the integral curves
The point is that a flow is just an action of the kind you described above. If the flow is smooth then it is generated by a vector field and follows its integral curves, but in principle that's not necessarily the case
Suppose we have a finite dimensional vector space $V$ and we're proving some statement about linear functionals, $\{f_1, \ldots, f_k \}$. We want to induct on the number of functionals, $k$. Now consider a subspace $W \subset V$ and suppose $\{g_1, \ldots, g_{k-1} \}$ satisfy whatever conditions we need, can we apply the induction hypothesis to the $g_{i}$? The main difference is that we assumed the hypothesis for the space V.
the crux here is that the statement you are inducting on is "the property holds for all vector spaces $V$", so your hypothesis is that the statement with $r-1$ functionals is true for all vector spaces $V$, so in particular for the functionals when restricted to $N_k$ and the induction works
the order is important in cases like these, as this would not necessarily work if you had fixed $V$ beforehand
Let f be a smooth map X->Y, and D a distribution acting on smooth functions on Y. Further let g be a smooth function on Y. Is it true that (f* D)(f* g) = D(g) , where the first f* denotes the pullback of distributions, and the second is the pullback on functions (i.e. precomposition)? Looking at the obvious commutative diagram this seems to make sense, but I can't see how this can be justified precisely.
an old asymptotic question which always gives me a headache whenever I've need to compute it: $\binom{2n}{n+x}\Big/ \binom{2n}{n}$ for large $n$ and small $x$
should be a Gaussian distribution but I've always been bad at showing it
the frustration on my part is that it's always seemed to me like there should be a better way to do that than just "hit it hard enough with Stirling's approximation for $n!$"
You could just find the kindest person working in combinatorics (in my experience, they are all kind) in your department, then just ask them if they had any idea.
One of the nice things about math is that there is enough people doing it that you don't have to do it alone.
In a Hilbert space, finite dimensional subspaces are closed: I feel like it's not enough to use an orthonormal basis $(e_n)$ and then to express each member $x_i$ of a convergent sequence as $x_i = \sum_n \langle x_i,e_n\rangle e_n$...what am I missing?
Since convergent implies weakly convergent, the right hand side there converges to $\sum_n\langle x,e_n\rangle e_n$ (where $x_i\to x$).
@JackOhara Oh I forgot that moving a chat message doesn't put it at the end of the target room. But anyway, your question is about mathematica or at least certain primality test algorithms and so off-topic for the room you originally posted in.
The reason why $(V\otimes V)^*$ is not always $V^* \otimes V^*$ is because we want forms to be continous which is not always the case in infinite dimension
I've always seen the definition of linear applications in infinite dimensional spaces to actually be bounded linear applications, because otherwise weird things happen
