a lot of those integrals are fairly close to the definition of various functions, where maybe it's not particularly informative to learn that one is equal to the other
that's really the thing. outside of a classroom context with very clearly delineated rules about what works and what doesn't, you don't "have" to make substitutions
@SineoftheTime OK... Let's try with the removal of one degree of freedom. Since $E$ is analytic, setting $u=v$ we will have $E(2v) + E(0) = 2 E( \sqrt{2}v)$. Now, $f$ is differentiable, so we could compute the $n$-th derivative for $n \geq 1$. How to do that?
I'm not understanding what you are doing, what is f and why do you want to compute the derivative? I'm really not that interested interestend is solving the functional equation, because without further assumptions I don't even know if it's solvable
plus, as I said before, $x\mapsto x^2$ is a solution
Wait, what is the domain and codomain of the map you are talking about Jakobian? Is it $M\to M$, $x\mapsto \det(U)x$, or $A\to A$, $x\mapsto\det(U)x$? I suppose it's the former, but then I'm not sure how it follows that $u$ is injective.
@Joe it shouldn't matter since we have a free module. From definition of regular element is hould be that the map from $A\to A$ is injective. But this applies to show that the map $M\to M$ is injective
@Pizza $\settsinh(x)$ and $\arcsinh(x)$ are two different names to indicate the same function. From what I remember, the most used notation is $\arsinh(x)$, while $\settsinh(x)$ is used in Italian notation. Are you italian?
Where is the 'c' in the spelling of the inverse hyperbolic function of sine? Isn't it just a hyperbolic version of $\operatorname{arcsin}(x)$? That is, why is it written $\operatorname{arsinh}(x)$ and not $\operatorname{arcsinh}(x)$?
The same question applies to the other functions as well.
if someone told me to replace arcsinh with arsinh i would tell them to go right to hell, but as a threshold matter, i deny that anyone is in the business of writing inverse hyperbolic trig functions often enough for this to arise
its just some weird autodidact basement internet people thing
@Gian'sPizzeria My suggestion is to multiply and divide by $x$. You will thus have $\int \frac{1}{x (1+ \ln x)} \cdot (x \ln x)$. You will see very soon that by the method of integration by parts this integral can be reduced to a very simple algebraic expression.
Consider a surface of revolution $S$ and an embedding $e:S \hookrightarrow X^3$ for $X^3=[0,1]^3$ with cone points $p,q$ elements of $\partial X^3$ where $\partial X^3=X^3-(0,1)^3$ for $\mathrm {sup}~ \mathrm{dist}(p,q)=\sqrt{3}$. It can be shown that there exists a unique $\rho_{\mathrm{max}}=\m...
I think I finally understand how the integral of a simple function is independent of representation of a simple function and how this follows from linearity of the integral of simple functions. Subtle, quite subtle.
@Jakobian we did kind of :D but it was not in regards to how Folland defines the integral of a simple function I think
he defines it in terms of the standard representation, and then never mentions anything about independence of representation
some authors have different definitions of standard representation
and some authors actually explicitly make this into a proposition that the integral is independent of representation
but you can use linearity of the integral of simple functions to show they are independent of representation. On the other hand, you can also go the opposite direction and first show independence of representation and then linearity.
maybe this is changing the subject, but part of defining something in terms of a unique representative is avoiding any question about dependence upon representation
like that's exactly why authors choose to go that route and not some kind of fault in their system
i still haven't checked to see how my favorite books do this
I feel that people who do complex geometry still study classical varieties a lot, but I get the feeling that this is not as "hot" a topic as it used to be. This is just an outsider's perspective, though.
i do remember stuff people were doing in grad school which was basically 'well someone way more talented than me proved X a few years ago and i know those definitions too, so who knows'
do irreducible algebraic sets have an intuitive interpretation without invoking their equivalence to prime ideals? or are they defined and useful because of their equivalence to prime ideals
hmm such a comment makes me think to not use hartshorne as a main reference 0.0
(also I ignored the domain business at the end of my abstract algebra course...)
Irreducibility can be though of as a very strong connectivity requirement. Indeed, sometimes the word "hyperconnected" is used for irreducible. An irreducible algebraic set is just an algebraic set which, when viewed with its Zariski topology, is irreducible.
hm I guess by intuitive I mean related to solving for zeroes of polynomials. e.g. algebraic set can be understood as the abstract object associated with the common zeroes of a set of polynomials
i started with the rising sea, but it was too much category theory and no algebraic geometry at the beginning, and i kind of just want a little taste of some of the subject
I think The Rising Sea is very good, but if you want a taste for algebraic geometry without going into the technicalities first, then it seems like you should study classical algebraic geometry first. The Rising Sea is about scheme-theoretic algebraic geometry
The problem arose in the context of kinetic energy, where it can be proven from symmetry principles that $E(2v)=4E(v)$ without assuming $E=mv^2/2$ (see e.g. physics stackexchange).
While one may do further physics from this point to prove the desired result (that $E$ is quadratic in $v$) -- cons...
Where is the 'c' in the spelling of the inverse hyperbolic function of sine? Isn't it just a hyperbolic version of $\operatorname{arcsin}(x)$? That is, why is it written $\operatorname{arsinh}(x)$ and not $\operatorname{arcsinh}(x)$?
The same question applies to the other functions as well.
