There was a tutor instruction in doing some maths. He is well known to tell his students that in order to study, copy the thing 999 times. During the copying, I realise it actually primed your attention towards the subject.
He then test the understanding for e.g. asking me how to find the perpendicular line passing through the centroid. Initially I think I should just calculate it, but later on I realised that by placing a weight on the object, it will tilt in the direction thus revealing where the centroid is and hence while there is no analytical expression for it, it is clear where it is and so is the line.
I then find this way of thinking very interesting and I jot it down in my notes with other 9 drawings of the unique ways of thinking I found using this method
Reality check:
Hmm... I should figure out how to abstract this thinking style somehow: The idea of exploiting the inherent properties of mathematical objects that are in a sense, given for free, in order to bypass the need to compute them precisely
Given a mathematical object M and some target property x, what is the equivalent to placing a string of weight on M in order to make x to be determined automatically...?
Does a function with two independent variables, i.e., $f(x(s, t), y(s, t))$ have a total derivative? Wikipedia seems to suggest this, but I don't think I've ever seen such a thing in examples.
"The term "total derivative" is primarily used when f is a function of several variables, because when f is a function of a single variable, the total derivative is the same as the derivative of the function." en.wikipedia.org/wiki/Total_derivative
$F[x]/(p(x))$ is an $n$ dimensional vector space over $F$ if $p(x)\in F[x]$ is an irreducible polynomial. Is it true that $F[x]/(p(x))$ has dimension strictly less than $n$ as an $F$-vector space if $p(x)\in F[x]$ is a reducible polynomial? (Here $F$ is a field.)
I mean, how does $F[x]/(p(x))$ look if $p(x)\in F[x]$ reducible?
Nope. I'm looking at Rudin right now; he writes, "...a convex set $A \subseteq X$ which is absorbing, in the sense that every $x \in X$ lies in $tA$ for some $t = t(x) > 0$."
Let $w=\{w(ij)\}_{1\le i,j\le m}$ be an $m\times m$ symmetric matrix with non-negative real entries such that $w(i,j)=0$ iff $i=j$. Show that $$d(i,j)=\min\left\{\sum_{j=0}^{k-1}w(i_j,i_{j+1}):k\ge1,i_0=i,i_k=j,i_j\in\{1,\ldots,m\}\right\}$$ is metric on $\{1,\ldots,m\}$.
Please help me with...
the tricky thing there is that it's hardly obvious to me what the axes would be
I suppose one approach there would be to find $a,b,c$ such that $aX^2+b XY+c Y^2$ is time-independent (where $X,Y$ denote the components of $\varphi_t$)
I guess one could equally well write that as $4X^2-8XY+5Y^2 = 4x^2-8 x y+5 y^2$
tediously, tbh
I wrote down $aX^2+b XY+c Y^2$ as I indicated
and demanded that its first and second derivatives at $t=0$ both vanish. (all derivatives should in fact vanish, but this was all I needed for a solution)
That gave me linear equations in $a,b,c$ which I had mathematica solve for
and $(a,b,c)=(4C,-8C,5C)$ was the general solution, with free variable $C$
That said, this seems like far from the best approach
hmm i don't have one good one liner to answer this question -- so maybe just some remarks: - calculus is built on inequalities - which are really open conditions! - open sets look at 'all directions' (manifold locally looks like \R^n) - closed subsets of manifolds can be arbitrarily bad behaved (closed subsets of \R^n can be crazy) - sheaves make sense on Grothendieck sites - so you really only need a notion of 'covering'. I guess this begs the question of why people dont consider 'closed coverings' - but that doesn't seem like a good idea -- if, for example, you define sheaves on all close…
what I'm getting is that either $\text{tr}M = 0$ or $\det M=0$
second one is probably not so interesting, since if $\det M=0$ then there exists $\vec{v}$ such that $M\vec{v}=0$
which tells you that any multiple of this $\vec{v}$ will be unchanged by the flow
The first case is less clear to me
I guess the upshot of the first case is this. If M has zero trace and positive determinant, your trajectories are ellipses. If it has negative determinant, they're hyperbolas instead
(zero determinant and zero trace means there's two zero eigenvalues and therefore things are annoying)
Hmmm, is that right? That a ring is always generated by a single element? I'm not familiar with that result. (Also, wouldn't a Noetherian PID be "every PID?")
Is it related to the fact that the additive group of a ring is abelian?
What I'm looking for is essentially just a little bit more understanding on ring theory. I wasn't sure if such a ring would be possible at all. Though, now I get the impression that it might be possible if the ring lacks a unit?
Well, perhaps I should clarify what sort of ring I'm thinking about.
A ring $R$ which is not uniquely generated (I assume this necessitates that it lack a unit?) where if $I\subsetneq R$ is a proper ideal, then there exists an $r\in R$ such that $I=(r)$
Suppose that $g \in F_2 = \langle a,b \rangle (the free group on two generators) is such that $\langle g \rangle$ is a normal subgroup. Can I say anything about $g$ in general?
@randomgirl: I think you should probably provide a bit more explanation. Also, the question confused me because the OP talk about area rather than volume. But I think it's clear you interpreted it correctly.
Computer science related question. I am trying to make a clock for an assignment. The clock numbers are not positioned properly. Theres something messed up on the math side of things.
Code:
for (int i = 0; i < 12; i++) {
shapes.add(new Text(
(int)(xcenter-radius*Math.cos(Math.PI/180 * (120 + i * 30))),
(int)(ycenter-radius*Math.sin(Math.PI/180 * (120 + i * 30))),
(i+1) + "")
);
}
Alright. You can avoid using contradiction here, btw. If h is in H, then it’s also in hH and so the intersection isn’t empty. Hence they must in fact be equal.
Yeah, so I said since $a^{-1}b$ $\in$ H. Then $a^{-1}b$ $=$ $m$ for some m $\in H$. Therefore $b=am$ and so $b$ $\in$ $aH$, by the above lemma, $b \in bH$ as well
hence the intersection is non-empty therefore they are equal
I was dealing with some exceptions to that earlier, though. Tensor product stuff. In that case I very much wanted an actual argument rather than just intuition.
But that’s as much a matter of convincing my collaborator as myself
i'm trying to prove that if K is a subgroup of H and H is a subgroup of G the [G:H]=[G:H][H:K] this seems to be straightforward, because a group must have atleast one element. So does it suffice to show that [G:A] = o(G)/o(A)
Problem: Assume $m(E) < \infty$. For $f \in L^\infty (E)$, show that $\lim_{p \to \infty} |f||_p = ||f||_\infty$. Proof: $||f||_p^p = \int_{E} |f|^p \le \int_{E} ||f||_\infty^p = ||f||_\infty^p m(E)$, so $||f||_p \le ||f||_\infty [m(E)]^{1/p}$ which implies $\lim_{p \to \infty} ||f||_p \le ||f||_{\infty}$...Not sure if this is legal, since, technically, I don't know that the limit exists, only that the sequence is bounded...
Even if I can work out that kink, how do I show that $||f||_{\infty} \le \lim_{p \to \infty} ||f||_p$?
I'm trying to prove that given $f(x) = x$ for rational $x$ and $f(x) = 0$ for irrational $x$ that $L(f,P) = 0$ for all partitions $P$ in $[0,1]$. Any hints?
It seems pretty clear that it's a result of density, but I want to explain it without being too hand wavey.
Let $w=\{w(ij)\}_{1\le i,j\le m}$ be an $m\times m$ symmetric matrix with non-negative real entries such that $w(i,j)=0$ iff $i=j$. Show that $$d(i,j)=\min\left\{\sum_{j=0}^{k-1}w(i_j,i_{j+1}):k\ge1,i_0=i,i_k=j,i_j\in\{1,\ldots,m\}\right\}$$ is metric on $\{1,\ldots,m\}$.
Please help me with...
