@QED I don't know how to write this kind of stuff in "math" (yes, it's a foreign language for me). I can argument that M is already all linear combinations, and that linear combination of all linear combinations is still the same. Would this be ok? I don't know to express this in mathematical terms
yes you need to learn to express it all in math terms
just saying stuff isn't good enough, sometimes you can give very plausible arguments in english which turn out to be totally wrong when you look at them carefully.
Can you give me a tip maybe? Because I don't know if it's possible to express in math terms what [M] is, it's not at my book at least, just the definition
@QED that's not on my book though. it just says M is a subset from V, and V is a K-Vectorspace. The linear span from M are all linear combinations of vectors from M. but sure, all finite linear combinations
@AsafKaragila Do you know how Cantor defended his statement: In mathematics the art of proposing a question MUST be held of higher value than solving it?
$[M]\subset[[M]]$ pretty simply, and $[[M]]\subset[M]$ because a linear combination of linear combinations is again a linear combination. Am I missing something?
@robjohn [[M]]⊂[M], can I just write that what you wrote, is that enough? You would actually be getting all linear combinations of all linear combinations, which was my initial argument, when people said that it wasn't enough
@AsafKaragila In his actual doctoral dissertation do you know how Cantor defended the thesis: In mathematics the art of proposing a question MUST be held of higher value than solving it?
@HenningMakholm Is this what your talking about sir: Never ever ever ever ever behave like someone who is so anxious to get an answer so he asks every single person showing activity and pinging anyone. There is nothing worse than such behavior on a chat. If you don't believe me, just try to think about what it would look like in real life.
@Clash Suppose $v=a_1u_1+b_2u_2$ where $u_k\in[M]$ and $u_1=b_1w_1+b_2w_2$ and $u_2=b_3w_3+b_4w_4$ where $w_k\in M$, then $$\begin{align}v&=a_1(b_1w_1+b_2w_2)+a_2(b_3w_3+b_4w_4)\\&=a_1b_1w_1+a_1b_2w_2+a_2b_3w_3+a_2b_4w_4\\&\in[M]\end{align}$$
@AsafKaragila I just thought that you knew something about set theorists' and in particular Georg Cantor's doctoral dissertation and how he actually defended the thesis: In mathematics the art of proposing a question MUST be held of higher value than solving it?
@Clash let me try again: For any $v\in[[M]]$, we can write $v=a_1u_1+b_2u_2$ where $u_k\in[M]$ and $u_1=b_1w_1+b_2w_2$ and $u_2=b_3w_3+b_4w_4$ where $w_k\in M$, then $$\begin{align}v&=a_1(b_1w_1+b_2w_2)+a_2(b_3w_3+b_4w_4)\\ &=a_1b_1w_1+a_1b_2w_2+a_2b_3w_3+a_2b_4w_4\\&\in[M]\end{align}$$
I chose sums of two, but you see how it can generalize to any finite sum.
I see... well that's a lot to write just to show something that seems obvious in english, oh well... someone has mentioned though that expressing arguments in words tend to be flawed and not objective... thanks a lot robjohn!
@robjohn imo it's more conceptual to show that [M] is a vector space, and then prove that $f: [M] \to [[M]]$ defined by $f(x) = x$ (as a linear combination) is an isomorphism
@Skullpatrol You cannot complain that I am being impolite by hoping that you will leave and never come back, but keep asking me and badgering me to "help" you.
@JonasTeuwen The other day I wrote something wrong. My Spanish is so rusty. "el cava" can really only mean one thing, i.e. fizzy wine. "Cave" in Spanish would be "la caverna" or "la cueva" : )
@AsafKaragila I will "will leave and never come back" ... this is my last question: Do you know where I can find an English translation of Cantor's paper?
I'd tell you that I don't mind you staying here, but I have my doubts about how long before you start bothering people by repeating some question over and over.
You are bothering me by asking (or stating - I don't really see the difference) this question over and over. And I'm only casually glancing at this chat!
There should be a formula for showing the correlation between time spent studying vs time spent procastinating. Over a scale of how close the test is. Alas I think I would procastinate in creating such a table
Then a moderator should be brought in and deal with the situation, which is what the flag for moderator function is for. Like I said, spam/offensive flags only remove one message.
@AsafKaragila So ... let's see you come back to a chat room where there has been no activity for 28 minutes and chat about a chicken recipe ... yet you complain that you don't want to have to get sucked any further into this site that I already check about once every waking hour (at least!)
Yeah, you're pissed at me because I'm calling you an asinine moron. The result, clearly, is that you're seeing my action contradicting without the slight hint of humor.
Furthermore, it is obvious that you have read the entire transcript of the chat from the day it began. You clearly know that I talk a lot about cooking in this chatroom.
For the love of god, and all that holy. What is wrong with you? Are you being obtuse just for the antagonism? Are you so thick that you can't see that the chat keeps history? People can come in later and see the activity before. People like J.M. actually read the transcript when they rejoin.
I am chatting for future generations as well as current chatters.
How can you realize that you consistently (and frequently) ask questions that many view as poor? When you view the art of posing a question is not important?
@TheChaz don't remind me of garbage. Had a clash recently with someone introducing that tag. I suspect V. would be further up the list of Jeroen's link if his worst questions weren't deleted.
I see. Also, sometimes it seems like I don't lose rep when downvoting a question (which I do sparingly, considering my strong opinions). Does it kick-in after an hour, or is there a rep threshold after which downvoting doesn't affect your rep? Third option?
@TheChaz No, to encourage voting on questions, the rep-penalty for downvoting questions was removed a few months ago. Downvoting an answer still costs you -1 and it kicks in immediately.
Acording to this graph I made, if this trend continues I will die. Problem is that I eat so much my stomach hurts. So if I eat more I will die. Tough draw
@robjohn that looks even more complicated than what I feared... One minor point: Do you really need l'Hôpital in order to get (5)? I would use the chain rule which is simpler to establish.
@robjohn - How does $ \int_{\mathbf{R}^2} e^{-(x^2+y^2)}\,dA &= \int_0^{2\pi} \int_0^{\infin} e^{-r^2}r\,dr\,d\theta\$ eventhrough $\int_{\mathbf{R}^2} e^{-(x^2+y^2)}\,dA$ is a square?
@robjohn That's exactly my objection (hence "less of a cheat" -- sorry about the unfortunate wording because you don't cheat at all :) ). Thanks for your answer, that's about what I thought the answer should look like. I think I have a similar argument in some lecture notes somewhere, I'll check a bit later.
@Victor you read my answer to your question, I assume. So the only thing left to do is to convert $e^{-x^2-y^2}=e^{-r^2}$ and that is pretty simple, no?
@Victor No, it is a plane. You could call it an infinite square, or circle if you want to try to justify the change from polar to rectangular to yourself that way, but it is neither a square nor a circle truly.
Well if nobody has anything to add skullpatrol will sign out for the last time ... I checked my account and it has been deleted ... It has been a pleasure chatting with all of you, thank you and good bye.
