I have linear maps $f\colon A\rightarrow V$, $g\colon B\rightarrow V$ and $h\colon C\rightarrow V$ such that $(f,g)\colon A\oplus B\rightarrow V$ and $(g,h)\colon B\oplus C\rightarrow V$ are isomorphisms, does the image of $A\oplus B\oplus C\rightarrow V\oplus V,\,(a,b,c)\mapsto(f(a)+g(b),g(b)+h(c))$ and the diagonal together generate $V\oplus V$?
yeah (this is actually a transversality computation for the trisecant map, which I thought I did a couple days ago, but now I'm not seeing it anymore)
ah, I get it now
my assumptions are not strong enough
I need to also assume that $(f,h)\colon A\oplus C\rightarrow V$ is an isomorphism
if I identify $V\oplus V/\Delta_V\cong V$ via $[v,w]\mapsto v-w$, all I need surjectivity of the map $A\oplus B\oplus C\rightarrow V,\,(a,b,c)\mapsto f(a)-h(c)$
so instead of just defining a trisecant map on the complement of $\Delta_M\times M$ and $M\times\Delta_M$ in $M\times M\times M$, I also have to exclude the points where the first and third coordinate are equal!
on my phone but how can I use the pigeon hole principle to prove : For a given natural number a, there exists k and l such that a^k - a^l is divisible by 10. Im having trouble following the logic
How can we assume f:NxN to N, just because N^2 is "larger" than N or the same size then you have f a surjection?
I guess what he means is that whatever a is and for whatever the last digit a has, by taking multiple power you will eventually obtain a number that has the same last digit and in which case their differences mod 10.
using pigeon principle you then not need to worry about the specific last digit anymore, all you care is that there are only 10 possible last digits whenever you are taking the power. And in the 11th time, you will have at least two powers have the same last digit which implies ....
like base case is a = 1, then u say this works for any a, then a+1. Do cases of a even or odd, perhaps show that the last digits mod 10 must be in some set
and then show a^k and a^l uses the same set for last digits
a+1**
@oscarmetalbreak Is this then the extreme case where N^2 is larger than 10 so it must be a surjection
Is there a relation between the existence of roots of the minimal polynomial to the matrix of a specific order that you can find? For example, Let $F$ be a field with prime number $4k+3$ of elements, then a matrix of order 4 in $GL_2(F)$ is unique up to conjugation, but the polynomial $x^4=1$ has no roots other than 1, -1. On the other hand when $F$ a field with prime number $4k+1$ of elements you have 7 such conjugate classes.
Maybe this is a simple linear algebra question but I really forgot...
@TedShifrin I think I got it but can you confirm my arguement? Since we are looking for a matrix $A$ satisfy $x^4-1=0$, which means the minimal polynomial of $A$ must divide that. In the first case, $p=4k+3$, one can write $x^4-1=(x^2+1)(x^2-1)$, and (x^2+1) is the only option of the minimal polynomial since otherwise A will be order of 2. However, when $p=4k+1$, one can write $x^4-1=(x+ p)(x-q)(x-1)(x+1)$ or $(x- p)(x+q)(x-1)(x+1)$ because of Wilson's theorem.
In the latter case, we can find the possible minimal polynomial of degree 2 by combining the linear factors and excluding the repeated ones and $x^2-1$, we will get 7.
don't start me! still in the shop. apparently waiting for some part being shipped from another country. over 3 months since he was rear ended!
it is a bit of an issue. i will need a car soon, i'm getting my hip replaced in december, so using my bike for transport won't be an option for a while.
and the supply of (non geometry) convex problems has dwindled...
i tried to get my daughter to get her own insurance. but all the agencies she contacted are not taking new clients until the new year. except one, and that was close to $2k for a 6 month period.
on the drive to school this morning munchkin said "you said someone died? henry?" because she overheard me telling my wife about kissinger the previous day and i guess had been stewing about the unknown ever since. i was like, oh wow, i definitely can't explain this.
i said he was some guy who lived to 100, which then changed the subject of how you live to 100. i reminded her of her mom's uncle, whom she'd met on thanksgiving, and pointed out that he was 90. so if you want to know who henry kissinger is, just take uncle ralph and add 10 years.
ok so I suppose the image is extending to infinity at the right end in which case we get that the second fundamental form at a point q is either positive or negative.
otoh what I say here is for varying p: I was thinking that at the left end, the bend is away from the normal there so the fundamental form there is <0 and in the middle it is positive.
ok so I suppose the image is extending to infinity at the right end in which case we get that the second fundamental form at a point q is either positive or negative.
I have one confusion now: I take a cylinder defined by level curve of $f(x_1, x_2, x_3)= x_1^2+x_2^2$. Let's take a point on its equator, say p. I have calculated that the second fundamental form $II_p(v)= -v_1^2-v_2^$, where $v= (v_1,v_2,v_3)$.
That is, the form is negative definite.
But if we look at the vertical line through p, then clearly along this direction the cylinder doesn't bend with respect to the normal at p. So as per this observation, shouldn't the form be 0 here?
Thus concluding that the form is semi-definite.
and not definite.
i.e., from the diagram it is observed that the form is negative along e_1 but 0 along e_2. But this doesn't tally with the formula I got above.
If $f$ is a continuous function such that $|f(x)-f(y)|>=ln(1+|x-y|)$ then which of the following options are true. A) f is injective. B) f is surjective C) f is strictly increasing or decreasing.
Let
$$f(x) = \sum_{n=1}^{\infty} (-1)^{n+1} a_n x^n$$
where $a_n > 0$ and
For all real $x>0$ the only zero's of $f(x)$ are at the positive integers.
How fast can $|f(x)|$ grow for $x>0$ ?
Let $I(\lambda)=\int_1^2e^{x^2}e^{-i\lambda x^2}dx$ then is $I(\lambda)$ differentiable on R? And what is the value of $I(\lambda)$ when $\lambda$ tends to infinity?
I am on my mobile so there may be some latex issues
@TedShifrin I thought 5 at the beginning, but then I found out $(x-p)$ $(x+p)$ are the two other minimal polynomial and in the case of $GL_2(F)$, they are unique representatives to the conjugacy classes.
@mick The reasons that someone might have downvoted have been explained to you innumerable times, both in chat and in response to your questions. The questions that you ask tend to either be completely contextless (which is the case with the one you linked), or long, discursive, ramble-y messes. I would very much encourage you to actually listen to and implement the advice that many people have given you in the past.
As I have said before, you need to sell your questions. Explain to other people why they are interesting and why they should care about the answer.
For a lot of people, the motivation is something like "I am taking an introduction to calculus class, and was assigned this problem. It is from [text], in [section x], where we have just learned [tool]. I suspect that [tool] should be used, but it is not clear to me how."
Most of your questions seem to have something to do with your own particular interests, which means that you actually need to do a better job of selling the work. We can all understand why a student might be motivated to finish a homework problem. I am not particularly compelled by such motivation, but I understand it, at least. Why should anyone care about your questions?
@SoumikMukherjee yes and $\int_1^2 e^{x^2}\mathrm{d}x$
not $\int_1^2 e^{x^2}\mathrm{d}x$, sorry
Let $u = -x^2$ then $I(\lambda) = \int_{-4}^{-1} \frac{1}{2\sqrt{-u}}e^{-u} (\cos(\lambda u)+i\sin(\lambda u))\mathrm{d}u$ and now using Riemann-Lebesgue lemma this converges to $0$
@TedShifrin Yes and when $p=4k+1$, one can write $x^4-1=(x^2-1)(x^2+1)=(x-p)(x+p)(x-1)(x+1)$ where $p, -p$ are the two roots I found for $x^2+1$ in this case.
Don’t use $p$ here. It’s already the prime. I still don’t understand how the minimal polynomial can be $x-a$. Then the characteristic polynomial would have to be $(x-a)^2$.
Is there an "intrinsic" definition of real analytic spaces and analytification of finite type schemes over R? The definition I found is: take complexification, then complex analytification, then take Galois invariants
Let $f(x)=\sum_{n=1}^{\infty} \frac{nx-[nx]}{n^2}$ is $f$ continuous at the rationals or the irrationals and is it integrable on a bounded closed subset of R?
I have done the continuity part but stuck at the integrable part
Funny that there were 10 questions from analysis and 10 from linear algebra and I remember most of the analysis questions but barely any linear algebra questions
I have a question. For a finite field $F$ of characteristic $p$ a prime. Will someone end up to be the same field after finitely many times of taking frobineus endomorphism to $F$. For example,let $\phi$ be the endomorphism, for every $n>N$, then $\phi^n(F)$ will always be the same. If so what could it be? I am guessing if it could be the prime ring.
Let $f(x)=\sum_{n=1}^{\infty} \frac{nx-[nx]}{n^2}$ is $f$ continuous at the rationals or the irrationals and is it integrable on a bounded closed subset of R?
I suppose it would be more interesting to ask whether the descending chain of subfields obtained by repeated application of the Frobenius ever stabilizes for non-perfect fields
the Frobenius endomorphism? perfect fields? most likely, yes the specific question of whether the chain of subfields by repeated applications of the Frobenius stabilizies? no, but you can answer that question as an exercise
@Shaun I have no idea whether I am balding or not. I've been shaving my head since my early 20s. I went from hair down to my ass to shaved bald in an evening.
(I started shaving my head when I was fencing competitively---my hair kept getting caught in the velcro on the mask, and it seemed that no hair would solve that problem and be much easier to maintain).
@Shaun I have no idea whether I am balding or not. I've been shaving my head since my early 20s. I went from hair down to my ass to shaved bald in an evening.
It's easier to find dirt on Santos than on Trump. No joke.
That's saying something.
> Among other things, Santos told the New York Post that he had not worked “directly” for Goldman Sachs and Citigroup, saying that a company he did work for did business with both of them.
He also said he had not graduated from Baruch College, nor “from any institution of higher learning.”
Let
$$f(x) = \sum_{n=1}^{\infty} (-1)^{n+1} a_n x^n$$
where $a_n > 0$ and
For all real $x>0$ the only zero's of $f(x)$ are at the positive integers.
How fast can $|f(x)|$ grow for $x>0$ ?
