One thing we can do is say, there's no reason why we consider the set $\{1\mathbin{..}17\,000\}$ rather than the multiset $\{1\mathbin{..}17\}\cdot1\,000$
eg we can think of it as just a thousand copies of each residue class mod 17
my cat has spent the last 3 hours nesting in a pile of clean laundry i was planning on filing away this evening. i think my house has developed something of a culture of entitlement.
@robjohn Thanks. Yes, it makes sense. My question: was bringing $\log (\epsilon)^\frac 1{a+1}$ in $\lambda=1+\log (\epsilon)^\frac 1{a+1}$ obvious?
I understand that choosing $\lambda$ this way works but was there any way of choosing it (the term $\log \epsilon^{\frac 1{a+1}})$?
@OliverDíaz: Thanks for the answer and especially for the comment at the bottom of the post. I also tried with Bernoulli's inequality yesterday but everything for the lower bound seemed to converge to $0$ or to something negative.
I think I got how that term was selected: $\lambda $ should be $\sim \epsilon^{\frac 1{a+1}}$ (I have no problem with this). In the limiting situation $\lim_{a\to \infty}, \epsilon^{\frac 1{a+1}}$ becomes $1$. $\epsilon^{\frac 1{a+1}}=e^{\frac {\log \epsilon}{1+a}}\ge 1+\frac {\log \epsilon}{1+a}$
@Koro My first attempt was to use Bernoulli's inequality within the interval $[0,1/a]$. Passing to the limit gave $1/2$. I did not tried the interval $[0,1/\log a]$ *which would include a negative part. That may also do the trick.
@leslietownes I went to a coffee shop and a staff there asked me if I would like that to be topped up with hazelnut something. I asked if he knew how hazelnut looked like, he didn’t know.
@robjohn Context was trying to figure out why the number of subsets of $\{1\mathbin{..}17\,000\}$ whose sum is divisible by $17$ is $\frac1{17}(2^{17\,000}+16\cdot2^{1\,000})$
well, i mean something bijective. modular arithmetic doesn't tell me anything that it counts. anything simple that it counts would be great, even if it's not the main thing.
yes. my question was imprecise. it was meant as a lead to: is there something simpler than the task before us, that the sum can be said to represent. where like, yes it's an integer because it's a number of ways of doing something, anything combinatorial, not necessarily the thing we want.
it's just a weird kind of sum.
like i can write any integer as a sum of binomial coefficients in all sorts of ways due to the strong law of small numbers, but what is that formula trying to tell us in a way we can grasp.
combinatorics is all smoke and mirrors to me. witchcraft more than math.
i sometimes get calls because i donated to the campaign of a republican i personally know. i tell them, "i only donate to candidates i personally know."
one time in 2004 i was stopped on the street to donate to john kerry. i said, what kind of donations are you talking about and the guy opened with something like $300. i said, i think his wife has that kind of money but i don't. look at my shoes.
a gazillion years ago when folks thought i might amount to something the irish consular mission used to include me in their circle. one of the things that kept coming up was some fancy dinner at $500 a head. as a student i had to decline
A friend from undergrad asked me a basic question about conditional probability and I had to google each and every basic thing. I don't think I've ever so clearly seen how much I've forgotten
Probably an excuse to go through my old notes. Anyone got tips for remembering stuff from a past life?
@robjohn It's not ick! I agree it would be bad if Numpy claimed that it's mathematically valid to add matrices of different sizes, but it's not doing that. Numpy is a Python library for operating on arrays, and many of those operations run much faster than plain Python code. Yes, it supports various matrix & tensor operations, but it's not like you're manipulating matrix or tensor objects, you're just manipulating arrays.
The thing that micsthepick alluded to is called broadcasting. It makes array code easier to write & read, and more efficient in its use of RAM.
hi. is it true no matter what set of things (maybe simple set of numbers, maybe some exotic computer science data structures) I have and a binary operation on those things and it follows (a) closure, (b) associativity, (c) commutativity, (d) distributive, (e) it has some element such that the operation doesn't change the thing and (f) every element has it's pair which when binary operated with it's pair, it becomes the previous element. Then I can do all the normal algebra on it.
@Koro in case you do not know, I thought I would share: it is possible for a sequence of differentiable functions to have a pointwise limit which is differentiable, but the sequence of derivatives does not converge to the derivative of the limit. E.g. sin(nx)/n
:) cool. I wanted to say, the sequence of derivatives of sin(nx)/n does converge to zero, but only in a suitably weak sense (like the sense of distributions)
Is there a difference between symmetric transitive closure and transitive symmetric closure? I searched about it but couldn't find a verified source on it.
@CalvinKhor I don’t think that the sequence of derivatives converges if x is not an integral multiple of $2 \pi$. For, the derivative is cos nx. Or, am I missing something?
@Koro uniform convergence is stronger than pointwise convergence, right? There's something weaker than pointwise convergence. And it does converge in the weaker sense
trying to find a proof or counterexamplel to the following: Let $f$ be a $2\pi$-periodic square-integrable function. If $f(0)=f(\pi)=0$, must $f(x)=-f(-x)=-f(x+\pi)$? The converse is true but the statement itself seems dubious.
($f$ is the solution of a second-order ODE with periodic coefficients in the example of interest, which seems ilke it'd be a strong enough condition to make the statement true. but i'm curious if that's actually needed)
the relevant ode is $f''(x)+(a-2q\cos (2x))f(x)=0$ with $2\pi$-periodic boundary conditions. i'm not seeing it asserted but presumably they're assumed to be smooth
Can we prove that for a>1 it is a^n->infinity as n->infinity using the binomial theorem the way it follows? "Since a>1, it is a^n=(1+h)^n for some h>0 and so (1+h)^n=bin(n,0)h^n+bin(n,1)h^(n-1)+...+bin(n,n-1)h+bin(n,n)h^0>nh->infinity as n->infinity."
