V is a set that serves as an parameter when constructing the operator. It can be anything. The set in the brackets is the set of solutions to the equation Tf where T is the operator being defined.
what are the minimum requirements on V and E so that it is well defined, the initial integral but the variance may not be, i'm thinking really not nice functions
Could some one help explain how the solution set is worked out from the graph of the inequality x^2 + x - 2 < 0. I got (-inf,-2] U [1, inf). But it seems it is actually (-2,1) i don't understand why
@shaihorowitz well, I think V can be anything, because it's just like the direction point form of a line. i.e: L = p + [v]. There's nothing really restricting what can be added. As long as a function is addable and multiplicative it fits in just fine.
and as far as I know, there is no function R to R that cannot be multiplied by a constant to make a new function. That would imply the function is undefined.
As for E, I suspect that large swaths of sets will just result in all possible functions falling into E.
because if x^2 is in E
then the abnormal integral will treat sqrt(x^2) as a constant
and depending on your algebra
we now have x itself treated as constant
and that will just result in the operator being the composition Tf = x*f.
@shaihorowitz oh yeah, the reason why this question is so important is because it gives me a way to formalize redefining the integral, which is the basis of implied calculus which has many ways of simplifying and solving various differential equations.
for instance, the piecewise constant one I linked not only makes solving linear homogenous differential equations with piecewise constant functions possible in the context of an implied differential equation (using the operators)
but it is also possible to determine the variance such that the equation is continuous
picking that particular solution yields the solution to the actual differential equation
needless to say, there are probably other implied integral operators that have the same affect
anyway
just a thought i had real quick that motivated me to hop on
Anybody here do math animations? I want to animated a 3D polytope. It's relatively simple, but I'd like to have it be somewhat nice - decent shading, etc.
hey @MikeMiller I would like to check with you something really quick.
Are you here ?
Let $V \rightarrow X$ be vector bundle of rank r with sections $\{\sigma_1,...,\sigma_r\}$ that generate each fiber. Prove V is isomorphic to trivial bundle of rank r.
so since each $\sigma_1,...,\sigma_r$ generate each fiber we have for each $v \in V$ it is of the form $\Sigma_i \alpha_i \sigma_i(p)$ where $p \in X$ and $\alpha_i \in \mathbb{R}$
consider the $f : X \times \mathbb{R}^r \rightarrow V$ given by $(p,\Sigma_i \alpha_i e_i) \mapsto \Sigma_i \alpha_i \sigma_i(p)$ then this works right ?
$$|x-z|\leq |x|+|-z|$$ $$|x-z|\leq x+z$$ $$(x-z)^2\leq(x+z)^2$$ Is here the fail? $$x^2-2xz+z^2\leq x^2+2xz+z^2$$ $$x^2\leq x^2+4xz$$ Which only is true if 4xz is positive(or 0)
ah well, nevermind, squaring on both sides brings cases with it
@ThomasAndrews I cannot direct you to a tool as the only tool I know of is a project for my computer graphics class, but what do you mean by a polytope?
is it a multivariate function in some coordinate system?
@WDUK an obvious sanity check on your result is to plug in a particular value. simplest one is x=0, and this gives x^2+x-2=-2 which satisfies the inequality. so the interval you suggested isn't right.
more algebraically, you can rewrite the inequality to $(x-1)(x+2)<0$. for that to be true, exactly one of the terms has to be negative.
so you either need $x+2<0<x-1\implies 1<x<-2$---which is not consistent, so we reject it---or $x-1<0<x+2\implies -2<x<1$, which is the solution.
ive read the first chapter. the concepts in the core material don't require any thing more than what i already have, but to solve the exercises, i need other tools which i havent learned
can someone check this? if $x<y<z$ it follows that $x-y<0<z-y$ To show: $$0\leq |x-y|+|y-z|-|x-z|$$ $$0\leq -x+z-y-|x-z|$$ $$0\leq z-x-|x-z|$$ $$|x-z|\leq z-x$$ $$z-x\leq z-x$$
hi what does this sentence mean "with no less than ten multiple choice questions and no less than five true-or-false questions)".... the "No less than" is annoying.
I am 99% certain that my results are correct and that I have been able to answer my own question'
however, is there any way that I can improve upon the wording of this question or make it easier to understand for those who aren't aware of the concepts I am describing: math.stackexchange.com/questions/2008976/…
usually if words aren't growing into several paragraphs or involving undefined concepts, you're fine.
now if you were writing an advanced set theory textbook involving von neumann ordinals and how to define the notion of equation... I'd expect you to write an essay.
but that would be several layers of abstraction below where you are, so that would be an issue of explaining all the layers and how they interact to make math work not the actual reasoning behind the solution.
if you do it for the sake of practice and it is mathematically worthwhile or interesting, you might actually find something you wish to explore down the road.
I'll be frank
I was trying to solve floor function integrals about 2 years ago as what could be classified as recreational math
and at some point i had a few ideas
and now i am still writing a 10,000+ word paper
most of it is fluff junk
but there are some interesting bits that I am slowly expanding
@prodprod well, in my class we never learned them. We just learned what a matrix and vector are and all the algebraic identities and terms. perhaps other places go deeper into things. there was a chapter 4 supposedly defining vector spaces very deeply that we blatantly skipped because "it is too difficult"
@saturatedexpo where did you read that? Heck no.
the determinant is just not used by itself through brute force
If you want to get interesting results about well ordered sets you usually need to go through ordinals
Like for every pair of well ordered sets $A$ and $B$ either they are isomorphic, or $A$ is isomorphic to an initial segment of $B$, or $B$ is isomorphic to an initial segment of $A$
They are also generally useful constructions, for example (assuming AC) you can define the cardinality of $X$ as the least ordinal $\lambda$ in bijection with $X$ and it immediately follows that given $X$ and $Y$ sets either $|X|=|Y|$, $|X|<|Y|$ or $|X|>|Y|$
Ah, I almost forgot, transfinite induction is prettt useful
@TheGreatDuck from what I've learnt in my numerical analysis course determinants are computationally a disaster to calculate so they're avoided if possible (but we didn't talk about algorithms to compute them so I don't know how bad the situation actually is)
Find $\sum _{n=1}^{\infty }\left(\cfrac{\left(2^n\left(\log\left(2\right)\right)^n\right)}{n!}\right)\:$
Answer is $3$, by the ratio test the series converges. I googled it, but stuck for procedure.
Can you explain it, please?
@Alessandro They are not actually that bad as long as you have exact values. Unfortunately, the algorithms that do it fast tend to propagate errors quite badly as far as I understand
@Tobias interesting, I think I'll look into it. I know that if you go through the $LU$ factorization you can compute the determinant in $O(n^3)$ (but that kind of pointless if you need it to decide whether a linear system has solutions), but I don't know other approaches
By map, I am of course assuming in a homotopically nontrivial way. Otherwise you can just send the whole 3-sphere to a point in the 2-sphere, which is boring.
The thing is that you decompose $S^3$ into torii, as follows: decompose $\Bbb R^3$ into torii of increasing radius, with the unit circle on the $xy$-plane as the center circle of all the torii. These will "limit" to the "circle at infinity", aka the $z$-axis.
The map $S^3 \to S^2$ sends each of these torii to a latitude in $S^2$.
@DHMO how do you prove that a^2+b^2+c^2=3abc when a+b+c=0? I know 1 method of proving it through matrices but I want to know if there are other methods too?
Describing (or rather, counting) maps $S^m \to S^n$ which are not homotopic to the constant map (which can indeed only exist if $m \geq n$) is a big open problem in algebraic topology. Hopf map is remarkable because it gives the simplest such example.
Find $\sum _{n=1}^{\infty }\left(\cfrac{\left(2^n\left(\log\left(2\right)\right)^n\right)}{n!}\right)\:$
Answer is $3$, by the ratio test the series converges. I googled it, but stuck for procedure.
Can you explain it, please?
