Recent content by threeder

simply - elegant. thank you!

I tried different approach, and I think this looks more promising, but I might be making a mistake somewhere. Could anybody give me some hints? What I tried is: \int (a^2 - x^2)^n \,dx = \begin{pmatrix} f(x)= a^2 - x^2 | f'(x)=-2x\\ g'(x)=(a^2-x^2)^{n-1} |g(x)=\int(a^2-x^2)^{n-1} \,dx...

Yes, indeed I made the mistake, but it was more of a typo, as I still get the final result using the metod I described which does not lead to the final expression of: \int (a^2 - x^2)^n \,dx = \frac{x (a^2 - x^2)^n}{2n + 1} + \frac{2 a^2 n}{2n+1} \int (a^2 - x^2)^{n-1} \,dx + C Now, using...

I suppose I could have framed it more clearly, sorry. What I have done is: ∫(a^2 - x^2)^n dx = [f(x) = (a^2 - x^2)^n -> f'(x)=-2nx(a^2 - x^2)^(n-1), g'(x)=1 ->g(x)=x.] (in order to integrate by parts, I have chosen particular f and g) After using integration...

integration by parts I'm working through Apostol's Calculus. I have attached the problem. I need to derive the formula integrating by parts. It is not a hard problem, but I can't seem to understand how on Earth the author came up with that expression. I take f(x) = (a^2 - x^2)^n, so...

Even though it is a very old topic, I am working through apostol's calculus and got stuck on the same problem. What I don't get is how do we do that I thought that if we cover Q with a union of T rectangles such that Q \subseteq T, the infimum of \tau_R will be bigger than c, because the...

Well I wanted somewhat more formal approach but oh well. Since it is not homework, but rather individual study, I will stick to more efficient way then. So taking this approach, the answer to me is obvious - 40 people can speak only english at most, because there has to be at least 60 people...

Homework Statement At an international conference of 100 people, 75 speak English, 60 speak Spanish, and 45 speak Swahili (and everyone present speaks at least one of the three languages).What is the maximum number of people who speak only English? The Attempt at a Solution The first...

Sorry guys, I mixed things a little bit. The correct thing should have looked like this: {\sum_{n\geq 2r\geq 0}} (-1)^r \binom{n}{2r} = \begin{cases}(-4)^k &if ~n=4k,\\(-4)^k &if ~n=4k+1,\\0 &if ~n=4k+2,\\ \frac{1}{2}(-4)^{k+1} &if ~n=4k+3,\end{cases} Though, after Dicks' simple hint I was...

OK, sorry for being so ignorant about it. Here was my initial line: If we know who pick what, we know that the first one has 52 combinations of picking first card, second has 51 and the thirds has 50 which leads to 132600. Thought, I thought, since we do not know who picks what, hence the first...

I rushed into things because I got the answer while creating this topic :) The difference is 3! because I simplified things a little bit too much, right?

Homework Statement 1. Three people each select a different card from a single pack of 52 distinct cards. How mnay choices are possible if we record who selected which card, and if we forget who selected which card? 2. Three people each select a main dish from a menu of five items. How many...

am... maybe I was not clear enough. I found this hint, and was able to prove it using it. Indeed it was a result due to Cauchy, and it is so elegantly beautiful :)

Homework Statement Let X= \{a,b,c\} and Y= \{d,e\}. Write down and explicit bijection N_{|X×Y|} → X×Y The Attempt at a Solution Well I came up with the easiest method, just giving one value to each member of N_{|X×Y|} so I was just wondering whether there is another way of doing it not by...

thank you for the help! proved it but only with few more hints that I found on the internet. it requires a little bit more of mathematical ingenuity than I currently posses :)