yeah this seems the case because the way I thought about it at first wouldn't made any sense at all because they would be elements of the power set not of X
Hey, can anyone help me clear something up with the statement of this problem? Suppose n in N. Given a pair of polynomials f_n, g_n s.t. deg(g_n) < deg(f_n) and deg(g_n) < n , the Euclidian Algorithm for computing gcd(f_n,g_n) takes at least n steps.
I'm making an example with f=x^3, g=x^2, so deg(g)=2 < deg(f)=3, but how does this part look: deg(g) < n ?
@theDoctor there are multiple ways to interpret your question, but they all have the same response: whether or not pi is a "normal number" (keyword to search) is an open question
People talk about that an infinite, non-repeating sequence (i.e. an irrational number) will have every number possible of finite length. I'm just wondering if you can interpret my question about whether pi can be found in the irrational number sqrt(2).
if pi is a normal number, in any base it contains any finite string of digits in that base. which would mean any finite segment in sqrt(2)'s representation in that base can be found somewhere in pi's representation in that base.
if we take the poset of all topologies on a set, ordered by inclusion, we're essentially saying the upward-closed open intervals all contain a minimum, which seems too orderly.
@theDoctor if you read what I wrote, you will understand
This is the world we live in, for many it is very convenient that you are on the left side of that page, to convince you to stay there, to be controlled a whole life, to stay in their shadows, and never dare to excel and try to be far beyond all of them in terms of performance.
From my experience, those that like to control the others (only) are also very poor in terms of performance, they excel by controlling, not by worthy deeds. A performant system never keeps them there up, but the environments, systems are not the way I wish them be.
Okay, suppose you've got a car that needs to cover a distance of 1km as quickly as possible. It starts with a velocity of 0 and must stop at the end of the track.
It can accelerate at a rate of 0.1ms^2 and decelerate at a rate of -0.5ms^2
How quickly can the car cover the track? N.B. it has no top speed.
I've been trying to plug this in to the equations of motion and to use my normal bag of tricks from calculus to minimize the time, but to no avail
I can't figure out how you call an operator that has this property: $T: X \to X$, $T$ is linear and $(X, \leq)$ is a partially ordered space. Now $T$ has this property $T(Y) \leq Y$ for all $Y$. Should I call this a contraction?
can your Mathematica find the closed form of this one? $$\int_0^1 \int_0^1 \int_0^1 \frac{1}{(1 + x) (1 + y) (1 + z) (1 - x - y - z + x y + x z + y z + 7 x y z)} \ dx \ dy \ dz$$
And somehow, this should be different between $\mathbb{C}$ and $\mathbb{R}$ but I don't see why
Ah I just noticed the difference lies in the holomorphicness of the maps
Okay, let $x \in M$ and $U$ be a closed subset containing $x$, then the coordinate function $\pi_k$ from $U$ to $\mathbb{C}$ is holomorphic and continuous and therefore $\pi_k(U)$ is compact and therefore $\pi_k: U \to \mathbb{C}$ is bounded thus constant. So $U = \{ x \}$.
So we're taking a connected subset $U$ of $M$ and looking at its coordinate function, and we know that $\pi_k(U)$ is compact. Then we are trying to prove that it is constant, so we assume that it isnt, then be the open mapping theorem, $\pi_k$ is an open map, i.e. sends open sets to open sets.
@Krijn Well, but $M$ is a manifold, those tend to have some nice properties. What properties of a topological space guarantee that the connected components are open?
Sorry, thats written really confusing, that I mean is:
Let $x \in M_i$ where $M_i$ is the connected component. Then we have a continuous chart to $\mathbb{C}$ and in $\mathbb{C}^n$ we can just take a small enough open ball around $x$, then its original in $M_i$ is open.
So, the connected components of $M$ are open (in $M$). Thus you can apply the open mapping theorem to the component $M_i$. They are also closed (connected components are always closed), hence compact.
@Krijn Well, since it's compact (and nonempty), it's not open.
Let $\textbf{$\gamma$}$ and $\textbf{$\tilde{\gamma}$}$ be two plane curves.
Show that, if $\textbf{$\tilde{\gamma}$}$ is obtained from $\gamma$ by applying an isometry $M$ of $\mathbb{R}^2$, the signed curvatures $κ_s$ and $\tilde{κ}_s$ of $\textbf{$\gamma$}$ and $\textbf{$\tilde{\gamma}$}$ ar...
I often think that the problems of the real life are so hard, even in the case with the puppies, I wasn't able to find a solution that makes me happy, they cannot ever be compared with the hardest math problems I ever met.
@AlecTeal turns out that when people ping you as @Ale they ping me instead! (but you were in the chatroom when it happened yesterday so you didn't miss any message that was addressed to you)
Let $F$ be a field and let $G = F \times F$. Define multiplication and addition on $G$ by setting $(a, b)+(c, d) = (a+c, b+d)$ and $(a, b) \cdot (c, d) = (ac, bd)$. I want to check whether these operation define a field structure on G. But how do I check associativity for example?
@Chris'ssistheartist There was a stray cat outside, one that we'd seen before around our house. It was really being friendly towards me and my wife, but we have three cats already, so we really can't bring in another. It is not around every day, so it probably lives somewhere else. It may have been locked out this morning and that is why it was so friendly to us today.
It was hard not to bring it in, but it did not seem to want to come near our dog anyway.
I put a small dish of water out in case it needed some.
@robjohn My dogs don't like the cats at all either. :-)
@robjohn The thing that hurts me more is somewhat a metaphysical one and it's related to the fact that we are limited to do the good (or I'm limited to see the ways to do the good in all circumstances - I failed with the puppies).
@robjohn not all people here are careful with the dogs, so this is an important detail. When you give them up you never know how they are going to be treated.
I used the Taylor expansion three times, once for $f(x+h)$, once for $f(x+2h)$ and once for $f(x+3h)$ and the $f(x_1), f(x_2), f(x_3)$ of the above expression are the remainders of each expansion. Do we know if they have all the same sign? @ro…
@evinda The courses for this semester are real analysis, linear algebra, intro to informatics and physics (I don't really like physics, but I can't avoid it).
@MaryStar not even if they have the same sign. Take for instance $f^{(4)}(x_1)=f^{(4)}(x_3)=1$ and $f^{(4)}(x_2)=0$ and $h=1$. Then the left side is $\frac{86}{24}$ and the right side is $\frac{22}{24}$.
