@robjohn I think the ideal way is to flaf this and a moderator threatening to suspend the user when he continues to violate the norms and rules of MSE. If you look at the lowest voted questions of the OP, many of his questions were closed for not entirely fitting reasons by many people.
@Soham: I don't know what you mean by contention but the guy downvoted Joriki (because he didn't give the general solution) and myself (it's not the first time : the guy has an easy downvote pass... ;-))
Well looks like that. But apart from that, there are a few question:
1. Can you explain me his solution? How did he get it? 2. From the conversation between you and Joricki I understand there had been some thing about explicit and implicit function going on. Can you explain it to me
@Soham: not really! Joriki was not used to general solution like the one you proposed (and that I corrected). I think that, at the end, we were both ok.
Dear friends: I have a question. I am learning to code. I learn some useful things. How do I stock pile them on my computer (I could choose to write but I am too lazy to write after trying things out on my computer. Looking forward to your suggestions....
hmm... one more last request... I think I understood what joricki explained to me about the characteristics curve and the dependent variable but can you explain it again to me, I would like to cross check my understanding
@Soham: you may write the general solution ($\phi(a,b)=C$) or something explicit like $z=f(a)b$ (he preferred the explicit version) but in lectures the general solution is often preferred (like in your formulation)
umm... no, not really, at least not right now. I was looking for an explaination of jorickis comment to me 22hrs ago, where he explained z=f(x,y)... can you check it three big paragraph type comments addressed to me
@Soham: It's easy to get confused about partial derivatives because our notation for them is so bad. Things would be much easier if we always marked which quantities are being considered as functions of which quantities, which quantities are being varied and which are being held fixed. In the present case, you're mixing up two different things.
@KannappanSampath You are welcome. There are many many solutions, you might want to try them at least some and see what floats your boat. Anyway, ask me this question again after your linux is setup, then I can tell you all the options.
@JayeshBadwaik Well, you can pick up the rudiments by just knowing finite groups, but very strong foundation in linear algebra is desired. But, to do some nice things with characters, it is nice to have modules over group rings and algebras around, I guess. But, also, @Mariano must be in a better postion than I am to tell you what's required to be motivated enough to learn rep. theory.
well, that applies to rep thepry of groups; if you want to do representations of algebras, you need a little more, but starting with groups is an excellent idea
@MarianoSuárez-Alvarez Okay. Are there any good introductions to it which start out in a similarly low on prerequisites way? I am already learning algebra and have finished groups and rings now and would like to have one more line of study to pursue.
my question is if given a pde say $F(x,y,x,p,q)$ and a curve $C$, we try to represent $C$ parametrically and then try to find out the initial condn yes?|
I want to show that if $f$ is holomorphic on $\mathbb{C}$, but meromorphic on $\infty$, then $f$ is a polynomial. Given the fact that meromorphic functions on the Riemann sphere are rational, i.e. given by $g(x)/h(x)$, I want to show that $h$ is constant. Is it enough to say that if it were not constant then the roots of $h$ would be poles and therefore $f$ is not entire as supposed?
there would be a finite number of poles, and by considering the sum of the principle parts of the pole, you can show using Liouville that subtracting this sum from your function results in the constant function
Yeah, there's a website based on stackexchange software which is dedicated to research mathematicians. I want to give it to my adviser, who might like it.
Q4. Is it possible that degenerate Basis -solution has a basis that is unique. True or False? (about simplex -method)
Q1. Basis -matrix is not invertible if the matrix corresponds to a degerate basis -solution. T or F?
- MY ANSWERS Q1. False. Degecrazy means that there are more than $m-n$ negative basis -solutions where $m$ is the amount of constraints and $n$ is the amount of variables.
Q4. True. Degecrazy is not the same as the weak duality where the difference between primal and dual is not 0.
<--- I messed up here, could someone check them? (they are not fully correct)
do any of you guys ever get to the stage when you get so confused by all the abstract stuff that you feel like going back to an elementary text for a while just get back to feeling that you understand it all?
@shoda What does the line `"If $i\not = B(1),...,B(m)$, then $x_i = 0$" on p.53 (Bertsimas) mean?
$i$ is scalar? And $B(1),...,B(m)$ is some vector? -No? It corresponds to the indices of the basis but what does it actually mean? What does the implication mean in practise? Cannot understand this not-equal -sign here.
i.e. I interpret this so that statements $(A)$ and $(B)$ are equivalent but probably not, they are two conditions to have a basic solution.
Does $i\not = B(1),...,B(m)$ mean $i\not = B(1)$, $i\not = B(2)$ ... and $i\not = B(m)$?
Let $S$ be the "open unit upper half disc", that is the area defined by the joint conditions $|z|<1$ and Re$(z)>0$. I want to define a bijection of this onto the open unit disc. My idea is to map $re^{i\theta}\mapsto re^{2i\theta}$, but of course you end up missing the positive real part of the unit disc with this map. Can I fix my idea?
Say $F$ is our field and $P$ is our "cone" then we should have that $P$ is closed under addition and multiplications. Additionally we should have $x^2 \geqslant 0$ and $-1$ not in $P$.
So now one would say that $x \leqslant y$ if $y - x \in P$.
Heya folks. Gotta quick question, I think i got the answer just need a lil' confirmation. Let $h:\mathbb{C}\to\mathbb{R}$ be a non-constant harmonic function. I want to show that it has a zero. My idea is to take $f=h+iv$, with $v$ the harmonic conjugate of $h$, and then argue that $f$ must be nonconstant and entire, and therefore by Picard, it would be impossible to omit the entire line Re$(f)=0$. That ok?
Trivial question: does Polyhedron need at least two contraints? I think no because 1 dim polyhedron is a trivial and there are also null polyhedrons --- used here terms polyhedron/polytope interchangeable.
@MJD a linear equality, inequality -- statement here such as $x=3$ or $x<0$.
Now $\bf A \bar x =\bf b$ is the polytope with more constraints. I am investigating the Bertsimas book: http://chat.stackexchange.com/transcript/message/5953866#5953866
I believe it is true: the argument that shows that the center of the matrix ring is the set of scalar matrices also works in proving that the center of the invertible matrices group is the set of (non null!) scalar matrices. thus, if a matrix commutes with all invertible matrices, it commutes with every matrix
bah, I don't think it is quite right as it is. What I wrote shows that if an invertible matrix commutes with all invertible matrices, then it is (non null and) scalar, and thus commutes with all matrices
Oh, man. I tried Google search for set theories with multiple empty sets and I found the website of a crank who has set up a web site specifically to promote his version of set theory in which there are multiple empty sets.
@FortuonPaendrag, maybe you can help me with something. I want to show that if $f$ is entire, then there is an entire function $g$ such that $e^g=f$. I'm looking at this link: math.rice.edu/~idu/Sp05/AnalysisV11.pdf, ctrl+V for "nowhere zero". I understand everything, except why the integral $h$ has parameters $z_0$ to $z$. Would you explain this, if you have a sec?
$x_1+x_2+x_3=1$ is replaced with $x_1+x_2+x_3\leq 1$ and $x_1+x_2+x_3\geq 1$.
(only one of the constrains is active --- hey do they abuse terminolgy?! It is not a logical AND used in the book?! It only makes sense if the word "and" is "or", otherwise not)
