I found the integrating factor easy. After seeing a few examples I thought, isn't there a general method for this? And well, it already existed in that form
i dunno. maybe more context is needed. is this a logic or set theory class where the answer is technical, or is "finite arithmetic" just informal code for something like "induction on the positive integers" as distinguished from induction on some general totally ordered set?
Is it generally true that if $f(x)^n < f(x)^{n+1}$, then $1 < f(x)$?
(for $f(x) \neq 0$)
I feel it should be obvious, but I am open to do the idea that there exists a function $f:\mathbb{R}\rightarrow\mathbb{R}$ where that is generally not true
I'm studying out of Weibel. I bought the paper back. Any electronic form of Weibel is really badly typeset
it was a genius move with forsite to see that at this time one is forced to order a proper copy
@Koro, @shintuku I'm at page around 10 of Weibel, if you guys would like to switch from basic AA over to applications
Since we were at the same level in AA class
I figure you would do just fine in Weibel
Homology is quite simple in its basic definition. You're just quotienting abelian groups
So it attracts me because though I don't know much of it, at least I can get the basic definitions. Unlike in Galois theory where everything is confusing
I think the most amount of points that can fit inside a unit cube is 8 (each corner) if no points can be less than 1 unit away from each other. To prove this, can I imagine dividing the cube into eight 1/2 by 1/2 by 1/2 mini cubes and saying that if I had 9 points, I require 2 points to be in the same mini cube which can't be farther than $\sqrt{3}/2 \lt 1$ from each other?
Hii I've a doubt. What is $2024! - 1 \mod 2024$? It's 2023 but how? The actual question which I was solving is $1\cdot1! + 2\cdot2! + 3\cdot3! + ... + 2023\cdot 2023! \mod 2024$.
If my back of envelope computations are right, those two spheres intersect, as I said, in a circle of radius $1/2$. On that circle there are two points whose distance from $(1,0,0)$ is $1$. Of those, one is farther than $1$ from $(0,1,0)$ but less than $1$ from $(0,0,1)$; the other is vice versa. So, it would seem you are correct.
I thought it was just as simple as splitting the cube into 8, 0.5x0.5x0.5 cubes and realizing that 9 points is impossible since 2 will end up in the same mini cube putting them within 1 unit of each other
It is intuitively clear to me that in a triangle T if we take any two points (on T or inside T), then the distance between them cannot exceed the perimeter of T.
But how does one prove this fact?
Is there a quick way to see this?
I require this to prove Goursat's theorem.
ohh I managed to prove it.
Take a triangle ABC. Take two points P and Q in the interior of the triangle. Join P and Q and extend the line segment to touch the sides of the triangle at say D and E. (D is on AB, and E is on AC). DE<AD+AE, whence PQ<AD+AE<= perimeter of ABC.
Similar argument works in case both the points don't lie in the interior of the triangle.
@Koro How about extending the line segment between the points until it hits two sides of the triangle. Consider the triangle formed by the extended line and the two sides of the triangle. Now apply the triangle inequality.
@geocalc33 what else would one do in a LINE? where was that sign?
I'd start by mentioning how negative that sign is ;-p
Suppose that we have a function f from X (an open subset of C) to C, that is complex differentiable at every point of X, can we say that the image of f can't be a line?
Here X is not given to be connected.
Suppose that f=u+iv. If the image is a line, then there exist constants m and c, v=mu+c.
Differentiating and using CR equations, we get: $u_x=u_y=v_x=v_y=0$.
If X is connected, then we have f=0, hence a contradiction.
So it seems to me that the image could be a line in lack of connectedness. But I don't know any counterexample to this.
the correction in the second last line is: If X is connected, then we have f=c, where c is a constant map.
@Koro if X is not connected you should get by the same reasoning that f is constant on every connected components of X, but there can be only countably many
Alright, I've worked out so far that based on the number of events and the simplest behavior I could have, we might have something like $\sum_i^n a_i = \frac{n-i}{n^2}$ where $n$ is the total number of events.
Is there a term to refer to a subgroup with "trivial" normalizer? I.e., if $H \le G$ with $N_{G}(H) = H$, does such a subgroup have a name? Perhaps self-normalizing? Does it go by any other names?
if we consider the factor group $\mathbb R/\mathbb Z$, then what is meant by 'if we give the identification topology to $\mathbb R/\mathbb Z$ (the corresponding partition of $\mathbb R$ is that given by the cosets of $\mathbb Z$)'?
I know that. What I don't understand is: we want to give $R/Z$ the identification topology. So we should be talking about partitions of R/Z and not of R, right?
Or is it to be interpreted as follows:
We consider the map p: R-->R/Z: r--> r+Z. It is onto. And we declare U (subset of R/Z) open iff $p^{-1}(U)$ is open.
The partition of R is given by the fibers of the projection to R/Z, points in the same equivalence class are squished together in the quotient. The equivalence classes are exactly the cosets of Z in R
yeah, R/Z has a different meaning too (Z identified to a point).
which should look like loops connected at one point.
Suppose that G is a compact hausdorff space, that also satisfies group axioms. It is given that the group operation is continuous, then how do I show that the map $g\mapsto g^{-1}$ is continuous?
Let's call this map $F$. I should show that for any open set U in G, $F^{-1}(U)$ is open. But not sure how to proceed with this (and also it doesn't seem to be using the compactness of G).
So instead, I should consider closed sets, but still the same issue.
Suppose that G is a compact hausdorff space, that also satisfies group axioms. It is given that the group operation is continuous, then how do I show that the map $g\mapsto g^{-1}$ is continuous?
Yeah, like the standard analysis exercise of proving (with compactness and Hausdorffness for one direction) that a function is continuous iff its graph is closed.
Note to Koro: This is of course the graph of the function you want to show is continuous.
Certainly an open ball in $\Bbb R^n$ is an $n$-dimensional submanifold of $\Bbb R^n$; a closed ball is likewise a submanifold with boundary. Most often submanifolds we encounter have dimension less than that of the ambient space. — Ted Shifrin12 mins ago
Moreover (modulo Whitney), a manifold need not even be embedded in a larger space. There needn't be an "ambient space".
But I got the impression (from the question) that the asker didn't even really understand that manifolds don't have to be submanifolds of $\mathbb{R}^n$.
Like I said, I am confused enough and outside my depth enough that I didn't want to comment, but there seems to be a huge gap between what this person is studying and the mathematical foundation they have.
Not that this is terrible---one shouldn't have to study graduate level differential topology to do big data (or whatever), but the gap is sufficiently large for me to be useless.
Until I can get from LA to Mammoth without spending a long time finding a charging station along the 395 and an even longer time charging, I don't think an EV is for me
I was very disappointed to find out that my Civic hybrid barely gets better mileage than my 2002 5-speed Civic. On long trips, I get about 45, but around town with lots of highway, it's just 36-38. Very disappointing.
In any event, I have always kind of figured that the "right" way to road-trip with an EV is to plan for a longish break every 200-250 miles---you can get an almost full charge in 30 minutes (if the right equipment is installed), which gives you enough time for a meal.
@robjohn Ah, okay. I see. I was answering a different question. I was thinking about where the infrastructure should be in order to make the trip possible. I wasn't thinking about where the infrastructure currently exists.
I bought my car in 2015, the last year they produced the Civic hybrid. If I needed to replace it, it would have to be with an Insight. But I don't intend to buy another car.
@Ted There is a Tesla supercharger in Lone Pine. Not in the area we usually stop, but it is there. I'll have to investigate on how an EV does toting a heavy load up a fairly steep incline for a good distance. There are 3 charging stations in Mammoth, which is good.
