Let $f \in C[0,1]$ and $f(0)=f(1)$.
How do we prove $\exists a \in [0,1/2]$ such that $f(a)=f(a+1/2)$?
In fact, for every positive integer $n$, there is some $a$, such that $f(a) = f(a+\frac{1}{n})$.
For any other non-zero real $r$ (i.e not of the form $\frac{1}{n}$), there is a continuous fu...
What do you call it when two functions seem to assume the same values as each other within some range, but they don't actually converge to a number/limit.
@robjohn I’m not really sure. I like to think that it’s because I write clearly and usually have a pretty good idea of where people are likely to have trouble.
@IssacM Naah man. Very very long ago, some 8-9 years I guess, I heard my friend telling me something similar, but I did not take him seriously then and I do not remember now. I have a hunch you are correct, no guarantees though.
@IssacM I have not taken any econ classes, so I'm not privy to such arcana. Interestingly, there was a grad student Robert Johnson in econ at the same time I was at Princeton for math. He left a lot of overdue charges at the Fine Hall Library and at the dorms as well. I had to deal with all of that.
@robjohn You did the following using binomial theorem I hope? \begin{equation} \int_0^x(1-t^2)^{-1/2}\mathrm{d}t=\sum_{n=0}^\infty\tfrac{1\cdot2\cdot3\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\frac{x^{2n+1}}{2n+1} \end{equation}
@robjohn Not really, just the important ones. Like this one on the Universal Chord Theorem, of those dealing with density arguments and stuff. The relevant and non trivial ones.
Feynman used to choose a random time at the start of the day and would offset his clock by that much time, and then whenever people asked him time, he would give answers like "in 3 hours and 48 minutes, time would be this" and so on.
Australian Mathematics Axiom: For the root of $N=1$ we have a fap. For the root of $N = 2$, we have a couple. For the root of $N = 3$ we have a threesome. For the root of every $N > 3$ we have an orgy. $\square$ — Gustavo Bandeira1 min ago
2
=D
I guess this dude post this thing about root on every root question.
@GustavoBandeira :P By the way the real root of $x^3 - (1+c)x^2 + (t+c)x - tc$ is c and other roots are complex if $t>\frac{1}{4}$ Chose a random value of $t$ everyday and spook people.
This is a topic that is still no available at my level but, it seems that Abel and Galois proved that polynomials of degree > 5 couldn't be solved by ordinary algebraic operations.
So, what operation is used then?
I don't remember what degree is exactly, if it's 5 or 6.
Galois came up with a precise criterion which allows us to decide if a polynomial (with, say, integer coefs) can be solvable with roots
it is an algorithm, so given such a polynomial we can in principle decide if it is solvable by roots —but the algorithms become impracticable as soon as the degree is larger that 15 or so
it is rather difficult to explain what the K-L polynomials are without entering into technical details, because they are very technical gadgets
in any case, a Lie algebra is usually studied through its «representations», and the K-L polynomials encode (in a rather complicated way) very important information about those representations
Very few people realize the enormous bulk of contemporay mathematics. Probably it would be easier to learn all the languanges of the world than to master all mathematics at present known.
The languanges could, I imagine, be learnt at a lifetime; mathematics certainly could not. Nor is the subject static. Every year new discoveries are published. In 1951 merely to print brief summaries of a year's mathematical publications required nearly 900 pages of print. In january alone, the summaries had to deal with 451 new books and articles.
The publications here mentioned dealt with new topics; they were no restatements of existing knowledge, or very few of the were. To keep pace with the growth of mathematics, one would have to read about fifteen papers a day, most of them packed with technical details and of considerable length. No one dreams of attempting this task.
@MarianoSuárez-Alvarez btw, thanks for the advice.
The Virtuoso Pianist (Le Piano virtuose) by Charles-Louis Hanon, is a compilation of sixty exercises meant to train the pianist in speed, precision, agility, and strength of all of the fingers and flexibility in the wrists. First published in Boulogne, in 1873, The Virtuoso Pianist is Hanon's most well-known work, and is still widely used by piano instructors and pupils. However, the applicability of these nineteenth-century exercises has been questioned by some piano instructors today.
Overview
The exercises address common problems which could hamper the performance abilities of a stud...
@KannappanSampath Three different shades of $\color{#503000}{\Huge\text{ Wood Brown}}\color{#784800}{\Huge\text{ Wood Brown}}$$\color{#A06000}{\Huge\text{ Wood Brown}}$
@MaoYiyi I guess Corei5 does that. Corei7 goes to 80 GFlops.
I am not serious by the way. I mean the numbers are real, about the processors at least. I am joking about the mathematica stuff. But I think you should be okay with Core i5.
@MaoYiyi Hmm, then it is possible that mathematica might be slow. I have a dell which turned 4 three days ago. However, I have a huge L2 cache a good FSB and a good graphics processor among other things, due to which it still performs really well.
I don't know. Mathematica claims to use OpenCL, so I would not be surprised if it does though, since OpenCL is a hybrid structure for use of both GPU and multiple CPU cores as computing resources.
It advertises only GPU computing though. However, if it uses openCL then extending the capabilities to multiple cores is not very difficult.
Apart from that, I guess, it can still use multiple cores to handle different threads, so may be it is possible.
@JonasTeuwen They are in a lot of CPU's. Core i7 does has an improved method of prediction which uses two target buffers instead of a single one thus reducing the incorrect prediction, but the technology has been there for a long time.
I am trying to show that the function $$f(a) = \lim_{h \to 0} \frac{a^h - 1}{h} $$ equals $\log a$. Is it enough to show that the function satisfies $f(ab)=f(a)+f(b)$ and $f(1)=0$ ?
So I need to find a number "a", that satisfies that equation. And I do not want to plug in some formula for e and show that it fits. I want to derive either a closed familiar sum or limit.