click on 'show steps' ... i think you are right up to that point ... integrate it, change t's back to u's and u's back to x ... you will have your answer
Let $ N dx + Mdy = 0 $ be a non exact differential equation. To make it exact, we multiply it by factor R(x,y), the Integrating factor.
How to find this integrating factor for differential equation if it depends on both x and y. So far i have found if it depends on y or x only ...
http://www.p...
@experimentX regarding the Wolfram show steps - it works fine for me in any other browser, but not in Chrome. But I see you were using Chrome when taking the screenshot. What version are you using of Chrome?
@tb just kidding, I hate mixing. I better work a bit. There is a postdoc from Oxford coming on Monday, we write a paper together - I should read what he did in the last month
@MarkDominus I enjoyed his Abelian Categories and his early writing quite a bit, but that book was a bit much for me to take. I'm sure it's full of beautiful ideas but it is a wee bit impenetrable.
I am just not able to do a single problem on my own, what am I suppose to do? People say move on but the problems only get more difficult and I definitely can't do any of the others
@experimentX If I'm reading tentaclenorm's post correctly, most of it is background to help give intuition. If you don't know the relevant theory, you can probably skip over it, and start at "This is your equation (3)." Read through the OP's workings too.
well time for a break from math I guess, 7 hours of math and four problems done, I will probably be able to finish all 8 sections of homework by monday with no problem
@Ilya Well, I guess that's something for people in boundary theory or potential theory. Did you look at the book by Krengel (Ergodic theorems) or the work of Foguel?
The question is based off of this question (which is based on a gaming.SE contest going on now), and this google spreadsheet trying to figure out the odds of winning
@Rachel The worst odds calculation is correct. But I think Zhen is correct: to get the exact odds, you'd have to do a complete enumeration of the outcomes.
I don't see why the total # of players drops by 6 each time in the best-case calculation, though.
@AntonioVargas Isn't that what the spreadsheet does? Each of the 25 rows reduces the number of entries by 6 or 1, depending on what set of data you'relooking at
Users can have between 1 to 6 entries, and no user can win twice
@Rachel Then yes, those tables are correct. Best case: everyone has 6 tickets, so there are roughly 130 players, and since there are 25 guaranteed prizes, there is about a 25/130 (approx 19%) chance to win.
$im(\overline{v}) \subset ker(\overline{u})$ was easy: If $f \in im(\overline{v})$ then $f = g \circ v$ for some $g \in Hom_R(M^\prime, N)$. And then $\overline{u}(f) = f \circ u = g \circ v \circ u = 0$.
For $ker(\overline{u}) \subset im(\overline{v})$ I am trying something like this:
$v$ is surjective, so $M^\prime \cong M / ker(v) \cong M/im(u)$. Let $\varphi: M /im(u) \to M^\prime$ be an isomorphism. Let $\pi: M \to M/im(u)$ be the projection. define $g:= f \circ \pi^{-1} \circ \varphi^{-1}$.
Now I want to show that $\overline{v}(g) = f$.
Is this how it's done?
$\overline{v}(g) = g \circ v = (f \circ \pi^{-1} \circ \varphi^{-1}) \circ v$.
@robjohn: to be honest, last hour showed I'm not in a good shape to give advice. I like the picture very much, but it's hard for me to go into details. I've found it very clear for such a complex picture
How could one express $$\int\limits_0^{\pi /2} {{{\sin }^{2n}}\phi d} \phi \int\limits_0^{\pi /2} {{{\sin }^{2n}}\beta d} \beta \int\limits_0^{\pi /2} {{{\sin }^{2n}}\alpha d} \alpha $$ in one integral only?
@Ilya do you like the Brouwer fixed point theorem? I was shocked yesterday when I found one interesting identical formulation: constant and identity functions can't be linked by a curve in the space of continuous maps
@MattN I just added the (homological-algebra) tag. Yes, it would be nicer but it's purely out of aestheticism that I'd add that. The meaning remains unchanged.
@MattN I'd prefer that, actually. Left exact just means that $0 \to A \to B \to C$ is exact (no surjectivity on the right). It's mostly used in the context with left exact functors: the functors that transform short exact sequences into left exact sequences.
@tb I know, I looked it up : ) Would you like me to delete the "left" then? I guess it doesn't add any information to the diagram. OTOH, it doesn't do any harm either...
if I have $six^2 2x$ can that be $1-cos^2 2x$ or do I have to do some crazy subsitution?
@Peter That is fine if you don't answer my questions, your answers are always pointlessly complex with excessive notation and mathy terms that aren't needed
@Jordan I tried to help you many times Jordan, but you have to put some of your effort in too. I hope you can progress with your math, but now I'm a little put off.
@robjohn Fine, I had some fun researching some history for an answer and now I'm restraining myself not answering a fun question because the OP only posted homework without any own thoughts so far.
@MattN No, it's not a typo. From the categorical perspective, the kernel of a homomorphism $f\colon M \to N$ consists of two data: the kernel object $\operatorname{Ker}(f) = \{m \in M\,:\,f(m) = 0\}$ and the inclusion map $\operatorname{ker}(f): \operatorname{Ker}(f) \to M$. Here I mean the latter.
Isn't anyone here also impressed by the fact that $$\sum\limits_{n = 2}^\infty {{{\left( { - 1} \right)}^n}\frac{{\zeta \left( n \right)}}{n}} = \gamma $$
probably close to that, 70 questions in this section, 72 in the next, 44, 73, 84, 50
about 400 questions, each taking me an hour
That would take me 16 straight days to do, but I can skip most the questions, only do the odds so that makes it 200, and then only do a select few out of those so that would make it maybe 75 questions, so 75 hours
This is just so incredibly difficult, how do I know if I am doing the right thing? I started with a problem and it branches off into 3 and each of those three branch off into multiple problems so I have like four pieces of paper trying to work out this problem and I do not know where I made the mistakes at.
Let $ N dx + Mdy = 0 $ be a non exact differential equation. To make it exact, we multiply it by factor R(x,y), the Integrating factor.
How to find this integrating factor for differential equation if it depends on both x and y. So far i have found if it depends on y or x only ...
http://www.p...
