However, I don't necessarily support the closure purely on the basis that it is an employment question. Academic integrity is different in my eyes than employment integrity.
Quick question: are imaginary numbers considered irrational? They certainly cannot be expressed as the ratio of two integers, but something tells me that $\text{irrationals}\subseteq\mathbb{R}$.
@TedShifrin Somebody asked a hilarious question in class today. We were studying rectangular hyperbolas and their parametric forms. Now it is $y=c^2/x$ and somebody asked how $y=-1/x$ could exist.
Turns out rectangular hyperbolas need c to be positive by definition
I could learn a lot of mathematics by myself however it would be more difficult without guidance. I think having the opportunity to be guided is very important.
Even if it is the smallest of chat messages telling me that I am close to the answer.
Otherwise now I might be reading about fuzzy set theory because nobody has shown me a common path.
@Alizter Hmm, well, the AMS usually have readable introductions to even pretty advanced topics. Unless you already tried it, give this a shot: www.mat.univie.ac.at/~michor/kmsbookh.pdf
@EspenNielsen Well I have not studied differential equations yet (or analysis in detail even). I am learning Lie theory at the moment and I can barely get my head around galois theory of polynomial field extensions.
@user2692669 I mean such as (spoiler tag, somewhat graphic) >!http://assets.sbnation.com/assets/2610587/closecut_medium.gif
spoiler tag didn't work, but at least I picked a version of it where it doesn't show all the blood. The guy got an iceskate to the neck and had to be rushed to the hospital
they already do, but yea., some that protects the neck and face better than they already have.
Anyways... math.
I had fun with math.stackexchange.com/questions/1103430/… and the extra curiosity question in the comments. How many different paths starting from $(0,0)$ ending at $(n,n)$ exist if the movements are right,up,down, or left 1 unit each and the total path length is $3n$.
it makes me wonder if the sequence had been studied before, but I'm too lazy to compute for the first several values of n to check oeis.org
I calculated it to be $a_n = \begin{cases}\sum\limits_{x=0}^{\frac{n}{2}}\dfrac{(3n)!}{(n+x)!(n+(\frac{n-2x}{2}))!x!(\frac{n-2x}{2})!} ~~~ \text{if }n\text{ is even}\\ 0 \text{ otherwise}\end{cases}$
Meh, its hardly useful information though. Granted, even the brady numbers oeis.org/A247698 have an entry on oeis (standard Fibonacci recurrence relation with b(1)=2308 and b(2) = 4261, given a name as a joke on a numberphile video)
Atiyah-Macdonald question: Page 77, proof of Proposition 6.7: In the third to last line of the proof he says if the length of a chain is $l(M)$, then it must be a composition series by ii). I don't see how it follows from ii). Can someone explain?
I checked the errata and it appears there, but then in the comments, someone says it does in fact work. Can someone tell me which is correct? I can't seem to figure it out the way it's written in the book.
Can someone double check my reasoning: Say the chain has length $l(M)$, if it's not a composition series, it means extra submodules can be inserted, which creates a chain of length $> l(M)$, contradiction ii).
Can I ask this in math site? Given a series of 30 random numbers in the ascending order - 120, 140, 220, 250, 400, 422, 451, 800, 864 etc., How can I resize each number of another one so the sum of final list is exactly 1000. The resized number should be a whole number. The size of each resized number should be relative to its original value. For eg. If 120 is resized to 100 then 140 should be resized to 117.
Off topic: I am trying to determine what an open subset $O$ of $L^(p)(\mathbb{R})$ would look like. Does this simply mean for any $f \in L^p$ there is an $\varepsilon>0$ such that the collection of every function $g\in L^p$ in between $f-\varepsilon, f+\varepsilon$ is cotnained in $O$?
@MikeMiller @BalarkaSen i have a stupid question for u: i read somewhere that $\mathbb{Z}$ has no non-trivial finite subgroups. What is meant by this? aren't $\mathbb{Z}_2$, $\mathbb{Z}_3$, $\mathbb{Z}_5$,... non-trivial finite subgroups?
@iwriteonbananas in the original context of $(\mathbb{Z},+)$ and how addition is defined you have that $5+1=6\neq 0$ and so the "addition" in $(\mathbb{Z},+)$ is different from the "addition" in $(\mathbb{Z}_5,+)$
Indeed, there are subgroups, such as $(5\mathbb{Z},+)$, but there are an infinite number of elements in $5\mathbb{Z}$. Technically $\mathbb{Z}_5$ is not a subgroup, but rather a "factor group"
Which is isomorphic to $\mathbb{Z}/5\mathbb{Z}$
compare this to $(\mathbb{Q}(i),\times)$. The trivial subgroup of course, $(\{1\},\times)$ always exists and is finite, but there exists another: namely $(\{1,-1,i,-i\},\times)$ and $(\{1,-1\},\times)$
As I understand, $\mathbb{R}^n$ is retractable, so there is a deformation map, h, from $\mathbb{R}\to \{0\}$. Then $h\circ g$ should be what you are looking for,
of course, taking the time variable into account and such., I'm only just starting algebraic topology, so I still need to get into the habits of writing these parametrizations correctly
anyways., I need to get onto the road. My class is having a quiz this morning, so I can't be late. Happy studying
Hey @DanielFischer @ArthurFischer Could you take a look at this: http://stackoverflow.com/questions/27942364/solve-recurrence-relation-without-using-the-master-theorem ?
@ArthurFischer Did you take a look at it? Because it was out of topic and I deleted it..
@robjohn @DanielFischer I want to find the exact solution of $T(n)=2T(\sqrt{n})+1$ By setting $2^m=n$ we get $S(m)=2S(m/2)+1$ I found that $S(m)=2^i S(\frac{m}{2^i})+ \sum_{j=0}^{i-1} 2^i$
If we would have for example $S(1)=c$, we would say that the recursion ends when $\frac{m}{2^i}=1$.
Hi @cirpis Could you take a look at this? http://math.stackexchange.com/questions/1104010/recurrence-relation-there-is-no-initial-condition?noredirect=1#comment2250980_1104010
@evinda Let $x_n=T\left(a^{2^n}\right)$. Then $x_{n+1}=2x_n+1$ which has the solution $$\begin{align}x_n &=\frac{2^n-1}{2-1}\cdot1+2^nx_0\\ &=2^n(T(a)+1)-1\end{align}$$ Therefore, $$T(n)=\log_a(n)(T(a)+1)-1$$
if i take three measurements of someones height(this is a real life applicatiion question), and i know my tool is accurate to $\pm .5$ exactly. i get $199.8,199.7,200$ i now know it is within $[199.5,200.2]$ correct?
@Mike is that so?
if i get 199.5 and 200.5 i know his height is exactly 200 as well?
