Now I have to heavily emphasize the fact that I have never studied differential algebra or the concept of other types of differentiation (which is what I believe is the concept behind a differential algebra). So, if I am abusing the terminology a little bit, please forgive me.
Let us define a di...
plus, except for a few advanced courses, I wanted a final exam to check that students knew the definitions and the basics, not super-tricky stuff like on problem sets.
I think I could only take litterature courses even though I never got good grades in French in my life and that I come from the scientific door into the ENS
It doesn't have to be a proof or anything. A short justification "this question differs from question blah blah in that here we limit ourselves blah blah" whatever.
at my institution all the math students have to take a year of algebra + a year of analysis + 2 upper division math electives (topology, algebraic topology, PDE, functional analysis, complex analysis) and so on
Regarding the duplicate. Yes, I know the other one has a lot of shared text, but those were just definitions/setup and I was being lazy. The core questions are still different unless you believe derivatives are weak derivatives in which case you might need to read up on them. I don't know the exa...
Yes,this is very similar to a previous question I asked. That was about normal solutions and not weak solutions.
We define the operator known as the implied derivative denoted as $I(f)(x)(g)$ to be:
$$I(f)(x)(g) := g(x) \left(\lim_{h\to 0^+} \frac{f(x+h)-f(x)}{h} \right) + (1-g(x)) \left(\lim_{...
@SirCumference I'm guessing your problem is about, like, why $\lim_{x \to 0} \frac{x}{x} = 1$ by cancelling out $x$, even though pluggin in $x = 0$ gives $0/0$ ?
@EricSilva They are very selective (note that I'm only talking about the ENS Ulm, there are 3 others) which partially explains their internationnal success
The intuitive idea is pretty much that a limit of a function in a point isn't (just) determined by that point itself, but largely by the surrounding points.
My second conjecture here does have some logic issues. The problem is that I am not sure how to express the idea of looking at integral equations under equivalency classes based on the solution sets being equal. Please see the comments of the current answer. Such logic is beyond my set theory abi...
im trying to use set theory to convey the idea that out of all sets corresponding to integral equation solution sets, there is only one that is a superset of the solution to a differential equation while having the smallest set magnitude.
and I know that has to have at least one loophole. XD
but if someone could please help rewrite the logic so that works correctly and makes sense, please edit my post
we have no exams but we submit college applications and they get reviewed by like teams of people holistically and honestly it seems like it's mostly just luck if you've got the stats to get in
my school is one of the most selective ones in the states but honestly it seems like admissions is like a black box that you dont really undesrtand unless youve seen directly how it works
If $f : \Bbb R^2\to \Bbb R^2$ is $1$-lipschitz for the eculidean norm, prove ${1\over n}f^{\circ n}(x)$ converges to a limit that does not depend on $x$
My second conjecture here does have some logic issues. The problem is that I am not sure how to express the idea of looking at integral equations under equivalency classes based on the solution sets being equal. Please see the comments of the current answer. Such logic is beyond my set theory abi...
It does happen to be the case that all elements in the derived subalgebra of a semisimple Lie algebra are commutators though, so it might follow from that
Argument: $1 - \frac{1}{n^2 - 1} = 1$ and $1 - \frac{n}{n^2 - 1} = 1$. But the sequence of the factors is strictly decreasing; therefore the limits of the factors must all be equal. Thus the limit equals to $1^n = 1$
@Semiclassical Some Googling shows that every determinant 1 matrix is in the commutator subgroup (unless your field is $\Bbb Z_2$, in which case it fails for two dimensions)
Let $n$ be the number of functions in the product $\lim_{x \to a} f_1(x) \cdot f_2(x)\dotsb f_n(x)$. Suppose by induction that it works for $n$. Then for $n+1$, it follows from case $n=2$ that $\lim_{x \to a} f_1(x) \cdot f_2(x)\dotsb f_{n+1}(x) =(\lim_{x \to a} f_1(x) \cdot f_2(x)\dotsb f_n(x)) \cdot \lim_{x \to a} f_{n+1}(x)$. I think you can see my point
There are definitely a good number of them, and if nothing else it'll help with pacing perhaps (some people before did not account for a lot of questions/comments, so were in a bit of a rush or didn't get through everything)