i presumed you were defining it as 'sensory experience', which would blur the lines between hallucination/'reality'/experience
I suppose if you can't figure out your senses are not being consistent, whatever you're experiencing is non-illusion (or so I presumed you were sorta saying)
> Summary: Albert Einstein once quipped, "Reality is merely an illusion, albeit a very persistent one." The famous scientist might have added that the illusion of reality shifts over time. According to a new study in the journal Psychological Science, age influences how we perceive the future
i'm reminded of a mitch hedberg joke about reality. when i was on acid, i would see things like beams of light. and i would hear things that sounded an awful lot like car horns.
@user85795 Whoever said that doesn't understand exponential decay. Radium-226 has a half-life of 1600 years. It decays to radon-222 gas, which has a half-life of ~3.8 days.
the mitch hedberg clips on youtube are usually fairly stupid. there's definitely something about the experience of being bombarded with jokes, each stupider than the last, for over 45 minutes.
it wears you down and then eventually you're in pieces. my wife and i were talking about this the other night. i love what i refer to as "anti-humor" where the joke is just being as unfunny as possible for as long as possible. she doesn't.
@user85795 Ah, Business Insider. Not necessarily a reliable source of info on nuclear physics. They can't even spell "atomic". See the caption of the image at the top of the page: "Marie and Pierre Curie. Credit: Atmoic Heritage Fund"
In the early days, people just didn't get that you need to be careful with radioactive stuff. The Curies first isolated radium by processing around a ton of uranium oxide. In their kitchen.
we're also involved in a furious trademark dispute with the Atomic Heritage Fund. we think they switched a letter to deceive the public and trade on our good reputation.
the idea of something completely silent, invisible, and yet lethal is still counterintuitive. somebody had to take those risks for everyone to figure it out.
i'd have given them the nobel prize just for that.
I always thought that Extremes would be a good name for a nightclub. People could say "I go to Extremes". OTOH, maybe a name like that would just attract trouble.
used to be one near by called The Office. sorry honey, long day at the office.
there was a place in my hometown called the brass rail which should have been called the third rail, for what happened to most people who came into contact with it.
@TedShifrin I don't know this result yet Ted. I am allowed to use cosets. I will also tell exactly what is the problem that got me stuck so that my problem is more expressed.
@leslietownes Leslie, I'll take a look at that as well after I discuss my proof (not yet complete)
I managed to show that a group of order 15 (henceforth called G) must have atleast one element of order 5 (say b) and at least one element of order 3 (say a). It follows that $G=<a>\times <b>$
@TedShifrin Because external direct product of <a> and <b> is G as they have no element in common because $o(<a>\cap <b>)$ must divide 3 and 5 that is gcd(3,5)=1
Then I want to claim that <a> is normal subgroup of G. By doing so, I'll be able to prove at least that G is abelian.
Then I can try to work my way out to prove cycle ...
@leslietownes the first answer by prof. Eigen answers my question I think. Than you!
Ted, that's what's been confusing me a lot lately. You are right if we talk about internal direct product which by definition considers normal subgroups. But what confuses me is this: It's clear that $<a>\cap <b>$ is identity. Now we know that $|<a><b>|=\frac{|<a>||<b>|}{|<a>\cap <b>|}=|<a>||<b>|=15=|G|$, from here it seemes intuitive to me to conclude $G=<a>\times <b>$ (external direct product), which may be wrong. May be Leslie can enlighten me on this.
:58329475, group action has been used in the answer. I don't know that :'(
You haven't shown that the external direct product is isomorphic to $G$. You need to get an isomorphism. Otherwise you prove this in great generality when it’s false.
@Astyx Yea, this is indeed the argument (I expected it to be obscure, so I didn't even try to think about it ;v)
(My book defines an isogeny to be a rational map (and hence a morphism) between elliptic curves. This induces a map between function fields, and if the corresponding field extension is separable, the isogeny is said to be separable)
I was looking for primes of the form $3^n + n^3$ or equivalently of the form $9^m + 8 m^3$.
I was not able to find any ??
How many exist ?
What are the first few ?
It seems like a trivial thing I missed.
Or maybe not since it somewhat resembles questions about fermat primes or mersenne primes.
My...
Finite group theory is like a sandbox. There's tons of ways of playing around with what is there to reach the desired conclusion. I do believe a modestly non-trivial input is necessary, but Cauchy's theorem does that job. In short, you can certainly avoid the class equation.
My feeling is that there should be a way of writing $n$ as something funny (like $n = (n-1) + 1$, then factoring a $z$ out in a clever way to compare to a "nice" series.
But, I mean... what tools are you expected to use?
Suppose we rewrite the sequence as you suggested and as i have done. We do realiaze that the given hint sequence and our sequence both have the same limit point. How do we connect those two arguments together?
I dont know if they are per se equal, they do however have teh same limit point. We know that the right sum must have the same limit (using diff calculus ) as the left. The left one is trivial (geometric sequence squared)
Actually, they are not equal. i just calculated the first few terms and they are not the same.
the product of two complex sequences $\sum_n a_n$ and $\sum_n b_n $ is defined as the sequence $ \sum_n c_n $ with $ c_n = \sum_{k=0}^n a_k b_{n-k} $ is called the cauchy product
Do you suppose that out of the given hint (since it says consider teh cauchy product of the series) or do you like suppose that because it is squared which implies its the same series multiplied with itself twice
( i have been expanding the first terms and trying to figure it, which two other sequences i need to multiply in order to get the same results, but it didnt work out)
when i taught i worked very carefully at putting legible letters on the board. it had a side effect of slowing me down to a point where approximately half of the class complained about it on student evaluations.
as long as it was not more than half, i figured, job well done.
too fast/too slow is something where it's fine to split the difference. if half of the class complained about legibility i would see that as a problem to be fixed.
nobody complained about that.
angle counting thing of arabic numerals? that sounds made up. say more about that.
It is this idea that the number of angles in a numeral keeps track of the value of that numeral. If you squint your eyes real hard, it almost makes sense.
i'm not a historian but it rubs me the wrong way, where i have a gut feeling that there's no way that that's true. although gut feelings can be both wrong and right.
Definitely a stupid question but I cannot find a reference for it and I can't figure it out myself: suppose I have a n-dimensional Gaussian distribution $\eta := p(x;\Sigma,0)\,dx$ with covariance matrix $\Sigma$ and zero mean. Then supposedly $\int_{\mathbb R^n}x_ix_j\,\eta = \Sigma_{ij}$. I don't understand why the covariance matrix has this property. Also, does this property change if we have a more general Gaussian with non-zero mean $p(x;\Sigma,\mu), \mu\neq 0$?
Actually it might be as simple as defining covariance as $Cov(f(x),g(x)) = \int f(x)g(x)\eta$...
Maybe I did figure it out....
ya it's just expected value, nevermind :(
the benefit of asking questions here seems to be that i figure it out when i ask for help about it
i have a question. I posted a question here but then I realized that it is more suitable dsp stack exchange. Is there a way to transfer the question there?
@eet It is possible for moderators to migrate questions. However, we typically don't do this, unless we know that the question is likely to be well-received at the target site. On the other hand, you can delete your question here, and post it again on a different SE site.
I don't know what question you are asking about, and it is unlikely that I would migrate your question, as, again, I am not going to migrate a question unless I know that the question will be well received on the target site.
huge variation in what sites like and do not like in terms of questions. which is totally sensible to me but also hugely important in whether X might migrate to Y in its identical form.
There are injective functions where not all elements in the codomain are mapped to. Is it possible for not all elements of the domain to be mapped (kind of opposite of above case)? What would that be called?
Anyone who can link to some good lecture notes summarizing the properties of the little oh (and probably big oh as well) notation, in particular the algebra associated with it? I'd be grateful.
Hey y'all, flags should be used for offensive and disruptive messages - not for messages asking for help. @BobSmith This isn't the right place to ask about people willing to tutor you - that's not what this site, nor this network is for.
Specifically, please only use flags for messages that are spam or violate the CoC that should be removed immediately (and warrant the 30 minute suspension that comes with it) - if a message requires moderator action, use a mod flag, and if a message is off-scope for the room or a user is being disruptive in a room, it is better to let a room owner or local mod know instead.
@XanderHenderson Then your ROs or mods should move/delete them. Flags raise the attention of all users with 10k+ reputation on chat, and shouldn't be used for messages that are "borderline"
But this kind of "Please tutor me" is a common form of spam---I would get a couple of these every month while I was in grad school. The distinction between "spam" and "legitimate cry for help" is hard to spot.
@XanderHenderson I definitely agree. When I saw it I almost flag nuked it myself but I decided it was on the side of the borderline that I didn't consider spam. I was gonna write up a response telling them this is an inappropriate place but I forgot and then only remembered when the flag appeared.