$$\lim_{x\to0}\left(\lim_{N\to\infty}\sum^N_{n=0}\sum^n_{k=1}\frac{\log^{n-k}x}{x^{n-1}(n-k)!(n-1)^{k-1}}\right)$$ Can anything be done to that limit?
I have tried beating it with a stick but it just moans.
Some elegant basic results relate total curvature to things like knottedness for curves and something called Euler characteristic for surfaces. There's also Gauss's result that you cannot map the earth faithfully onto a flat piece of paper.
Hi guys, this should be a quick proof but I want to convince myself that the parallelogram law ( $ |x + y|^2 + |x - y|^2 = 2 (|x|^2 + |y|^2)$ ) fails for $L^1$ where $L^1$ is the space of all absolutely integrable functions.
@TedShifrin exactly, and $L^1$ is not a Hilbert space. so my professor, in his notes asked us to show that $L^1 $ is not a Hilbert space by showing the p.law fails. Its not a big deal but I make it a point to do all the little exercises from his lecture
@TedShifrin It is called "Taller de Cálculo Avanzando" (Workshop of Advanced Calculus?) and it is meant as a previous course to "Advanced Calc" which is a nice course.
@PedroTamaroff, @TedShifrin i think this works: consider $f = 1_{[0, 1/2]}$ and $g = 1_{[1/2,1]}$ then we have $|f + g|^2 = |f -g|^2 = 1 $ but $|f|^2 = |g|^2 = (1/2)^2$
Exactly, I'm using MV to compute $\tilde{H}_n(X\cup CA)$ and I'm left with a long exact sequence that looks very similar to the one that I get for the relative homology of $A$ over $X$
I'd like to use something like the five lemma, but first I need a morphism from $H_n(X,A)$ to $\tilde{H}_n(X\cup CA)$ that commutes with the boundary morphisms from the two chains
I can write down the two long exact sequences if that's of any help
Is it true that $H^n(\mathbb{P}^n, \mathcal{O}(d-s) \oplus \mathcal{O}(d-t)$ is isomorphic to $H^n(\mathbb{P}^n, \mathcal{O}(d-s)) \oplus H^n(\mathbb{P}^n, \mathcal{O}(d-t)) $ ??
@robjohn the cdf of exponential distribution is $1 - e^{-\lambda x}$ if $x \ge 0$, 0 otherwise in my case $\lambda = 0.001$ so $P(X > 800) = e^{-0.001*800}$
@rob
@robjohn which means after 800 hours an item has that chance of failure
my guess is that if i have a sample of 6 items, then i have to use binomial distribution knowing p = 0.449, so q = 1 - 0.449, and n = 6
so i end up with ${0.449}^6$
@robjohn remember, 6 objects are tested and the time where failures occur is written. What is the probability that any objects fail before 800 hours? -> so i think the previous result makes sense
@DavidRobertJones The chance of failure is $0.449$, the chance of not failing is $.551$, the chance of 6 items not failing is $0.551^6$ the chance of at least one failing is $1-0.551^6$
@robjohn are you sure? i think you have it the other way around. $P(X > 800) = 1 - P(X \leq 800) = 1 - (1 - e^{-0.001*800})$, which is the chance of an item last longer than 800 hours
i'm confused now
i need to find the probability of 6 items not fail within 800 hours
@robjohn $X$, which has exponential distribution with $\lambda = 0.001$, is the length of one item. so, $P(X < 800)$ means the probability of one item last less than 800 hours, correct?
@DavidRobertJones if the chance of one item not failing in 800 hours is $0.449$, then the chances of 6 items not failing in 800 hours would be $0.449^6$.
@Chris'ssis you had more than 79 characters without whitespace. Chat adds a weird space to make sure things wrap. If you add your own whitespace, you can be sure the added whitespace doesn't mess things up
@robjohn I really appreciate Ron Gordon, but I saw in a comment that a guy suggested that one needs to be very experienced to do that. I think it is a trivial telescoping sum. It's natural to think of a telescoping sum.
@robjohn Indeed, the question is pretty enjoyable.
Why corollary of Area theorem that under the same hypothesis, $|a_1| \leq 1$ is obvious :) Area theorem: If $F$ is holomorphic in $\mathbb{D}\setminus \{0\}$, $F$ is one-to-one in $\mathbb{D}$, and $$F(z)=\frac{1}{z}+\sum_{k=0}^{\infty}a_k z^k, \quad z \in \mathbb{D}$$ then $$\sum_{k=1}^{\infty}k |a_k|^2 \leq 1.$$ And why we can say that in $F(z) \quad z \in \mathbb{D}$? What with $0$? Here $\mathbb{D}$ is unit disc.
Ah, yes, it is trivial :)
We have that $|a_1|^2+2|a_2|^2+ \cdots \leq 1$, all is positive, so $|a_1|^2 \leq 1 \implies |a_1| \leq 1$.
But what with definition of $F(z)$ at $0$?
I suppose that this is trivial for @DanielFischer ? :)
Oh yes, I see, there is no problem with $F(0)$. It is $\infty$, but who cares, or...
Someone told me that (x) in Z]x is not a maximal ideal. Eh...I don't see how this is true, what other ideal of other than Z[x] itself could contain (x)?
Can someone take a look at my proof here and see is it ok - math.stackexchange.com/questions/588254/… - I thought I needed to show a bit more but it seems that from the comment a guy made that I might have shown enough. Have I shown enough?
if$ f_1 \in \Re (\alpha) $ and $f_2 \in \Re (\alpha)$ on $[a,b]$ , then $f_1 + f_2 \in \Re (\alpha)$ . proof: if $f = f_1 + f_2$ and $P$ is any partion of $[a,b]$ we have: $L(P,f_1,\alpha) + L(P,f_2,\alpha) \leq L(P,f,\alpha) \leq U(P,f,\alpha) \leq U(P,f_1,\alpha) + U(P,f_2,\alpha)$ can someone pls explain how to arrive to the first and last inequality
@anon So we have that $r$ must equal $0$ and hence we have that $f = qg + r$ with $r = 0$ and hence satisfying the required property making this a Euclidean ring?
Ok. As F[x] is a field, let f, g be such that fg = 1. Then fgg' = 1g'. I.e. f = g'. Now we also have that r must be equal to 0. So consider f = 1.g' + 0. We have q = 1 and r = 0 so we have a Euclidean ring.
Sorry, I'm badly wound up as I have my exam in the morning so I am rushing through a lot of material!
I don't mean last-minute rush as I haven't done anything for the exam...I've been studying non-stop for the last week but I have some loose ends to clear up
@anon My head is spinning - 2 or 3 of the following theorems are coming up, I want to be ready for anything - % Field is integral domain Finite integral domain is field F[x] has no zero divisors and thus is integral domain F[x] units are constant polynomials Injective homomorphism <=> ker(*) = 0 - THIS Properties of ker(*) Z -> Q unique homomorphism Z is a PID - 1 F[x] is a PID - 1 % I prime <=> R/I is integral domain % I maximal <=> R/I a field Maximal ideal is prime - 3 - learn backup theorems %
Along with lots of other stuff I have to know. Would someone mind writing an answer to my Field is a Euclidean ring post as I don't have anymore time for figuring out, every minute counts now.
Asymptotically, the proportion of semiprimes $\leqslant N$ where the two factors have roughly the same size is negligible, if we say that two primes $p \leqslant q$ have roughly the same size if there is a constant $\alpha > 1$ such that $q \leqslant p^\alpha$.
The number of semiprimes $\leqslan...
