If you're into a treatment of complex analysis which does some algebraic/differential topology to start with and frames certain things in that context, try Berenstein and Gay @Adeek
I mean, it might not say anything about the real world, sure, until you decide from physics that a mathematical construction is superimposed, but to say that theorems of math say nothing... thems fightin words
oh. as a small data point in the math/physics war, one the other condensed matter grad students I know is headed off to Chicago for a postdoc. @daminark
Demonark: Another powerful technique which you should see is to use Morera to prove something defined by an integral of a continuous function with holomorphic $z$ dependence is a holomorphic function of $z$ by switching order of integration in a double integral.
well, there's a parameter with respect to which you're integrating ($t$) ... So how do we show $\displaystyle\int_\gamma \dfrac{f(t)}{t-\z}\,dt$ is a holomorphic function of $z$? Usual way is to differentiate under the integral.
I think it acts as a way to reconstruct an analytic function from its asymptotic series? Something like that, but if I knew what I was doing it wouldn't be called research.
yes, R is isomorphic as a Q-vector space to a direct sum of c=|R|-many copies of Q, hence R and RxR are both isomorphic as Q-vector spaces (any Q-vector space map is in particular an additive group homomorphism)
I am trying to characterize all the group homomorphisms from $R$ to $R$.
I have characterized all the "continuous" group homomorphisms from $R$ to $R$. They are of the form $f(x) = f(1) x$. Now I claim that all the group homomorphisms from $R$ to $R$ are necessarily continuous. Is this claim co...
Well, in NASCAR at least there's both stuff inside the cars (roll cages, HANS devices, etc) and outside (e.g. safer barriers) which help to minimize the impact on the driver
Cars don't run well without oxygen ... although it's not like it was in 1974 when I tried to drive over Independence Pass in a far-pre-fuel injection car.
My main memory of it is seeing a line across a mountain range across a valley, realizing that it was a road, and asking..."we're not really going there, are we?"
Well, this is a different clientele and a very different course. I last taught it over 10 years ago. I think AoPS has a lot of clever problem-solving in it ... unlike college precalculus.
hey ive been researching PNT prime number theorem and ive come to the conclusion that x/(ln(x)-ln(ln(x))-ln(ln(ln(x)))-ln(ln(ln(ln(x))))-ln(ln(ln(ln(ln(x)))))) is a better bound then x/ln(x)
@PVAL-inactive of course they are ~ to each other, ln(ln(x)) etc. are all o(ln(x))
the iterated ln's are hardly relevant though, as they're superseded in approximation by a simple affine function in the denominator: $\pi(x)\sim x/(\ln x-1)$ (see Legendre's constant)
@KasmirKhaan Say I have $(x^3+x^4+x^5+\dotsb)(x^2+x^3+x^4)$. Think about what the coefficient of $x^n$ in the product would be
@LucasHenrique I guess they're not really necessary… but what it means is, if $x$ is an integer in $\Bbb Z$, then $[x]$ is the corresponding element in $\Bbb Z_n$
It's like $x$ mod $n$
So, we don't have $1=6$ in the integers, but in $\Bbb Z_5$, we have $[1]=[6]$
@AkivaWeinberger It's just a simple consequence of CRT. Defined the function, you can show all of that stuff does exist and is unique; therefore bijective.
That's why "$\mathbb{Z}_{ab} \to \mathbb{Z}_{a} \times \mathbb{Z}_{b}$ is a bijection" was so weird; I was wondering what function they were talking about
@KasmirKhaan You know how, if we have $(a+b)(c+d)$, the product ($ac+ad+bc+bd$) is the sum of every possible combination of something from $a$ and $b$ with something from $c$ and $d$?
With $(x^3+x^4+x^5+\dotsb)(x^2+x^3+x^4)$, we'd get the sum of every combination of a term from the first factor and a term from the second. For concreteness, let's look for the coefficient of $x^7$ in the product
How many ways can you get a seventh-degree thing as a product of some term from the first thing and some term from the second?
Well, you have $x^3\cdot x^4$
and you have $x^4\cdot x^3$
and you have $x^5\cdot x^2$
Those are the only ways to get an $x^7$ term as a product of a term from the first factor and a term from the second
Do you see why the two questions ("what is the $x^7$ term in $(x^3+x^4+\dotsb)(x^2+x^3+x^4)$"$~$ and $~$"how many ways can you write $7=x_1+x_2$ with $x_1\ge3$ and $2\le x_2\le4$") are the same?
@AkivaWeinberger well you kinda gave me the problem backwards >< if we start from x_1+x_2 =7 and those conditions $x_1\ge3$ and $2\le x_2\le4$, how to get those polynomials?
@AkivaWeinberger Yes I see it they answer same thign but in different magical way
I am trying to characterize all the group homomorphisms from $R$ to $R$.
I have characterized all the "continuous" group homomorphisms from $R$ to $R$. They are of the form $f(x) = f(1) x$. Now I claim that all the group homomorphisms from $R$ to $R$ are necessarily continuous. Is this claim co...
