Recent content by b0it0i

this problem actually hard two parts one dealt with when A1 C A2 C A3 C ... the other was when A1 contains A2 contains A3 contains but the book only gave that single hint. It didn't say it strictly applied to one or the other, and I'm assuming it meant that it should work for both cases. And...

i see. however, i should have worded that part better, noting that A1 super set of A2 superset of A3... what exactly is B2=A2\A1... i mean everything in A2, is technically in A1, since A2 subset of A1. So an element in B2 is an element in A2, yet is not an element in A1, which then implies it...

Homework Statement Given A1 superset of A2 superset of A3 superset of A4 ... and so on how can i construct sets B1, B2, ... so that each Bi's are disjoint. The goal is to get the infinite intersection of Ai = the infinite union of BiHomework Equations De morgans law: (AUB)^c = (A^c N B^c)...

i see, my professor also gave the hint to list all possible outcomes, which i forgot to include. so i did so, and denoted each outcome as such 1234 (1 = 1st envelope, etc.) if 1 is the largest amount, you can take off all the combinations with 1 in the front, since you discard the first...

Homework Statement 4 envelopes contain 4 different amounts of money. you are allowed to open them one by one, each time deciding whether to keep the amount or discard it and open another envelope. once an amount is discarded, you are not allowed to go back and get it later. compute the...

well, f(x) and g(x) are indeed two different functions... but what you're trying to prove does not require using "two different epsilons" the problem only states to prove that the limit of f(x) as x approaches c = 0 write out the symbolic form of that... which is, for all epsilon greater than 0...

i don't believe your proof is correct "Therefore |g(x)-0|<ϵ_0= |g(x) |<ϵ_0. Thus δ_0= ϵ_0." i have no idea why this is necessary. if you're already on limits of functions, it must mean you past limits of sequences. The way i did this problem follows very similarly to the proof of the squeeze...

i think the problem you're having with this proof is understanding what the question is asking. there are many other proofs with the same structure the problem is saying show that (ABC)-1 = C-1B-1A-1 in other words, they want you to show that the inverse of ABC is actually (behaves like)...

im sure it was a typo and he/she meant (AB)T (AB) = BT AT (AB) use associativity in that line, and as others have pointed out not commutativity, to "regroup/rewrite" it and the fact that A AND B are orthogonal... and you get the identity matrix do the same for AB (AB)T

the solution to this is pretty simple focus on the definition you provided "If A is some nxn orthogonal matrix, than Click to see the LaTeX code for this image where I is the nxn identity matrix." now if you look at the question it tells you to show that AB is also orthogonal now apply...

can you use the theorem: every convergent sequence, is a cauchy sequence just prove the sequence converges, cite the theorem, and you're finished

since w is a function of x and y and x and y are functions of t the chain rule is dw/dt = dw/dx dx/dt + dw/dy dy/dt compute everything above and you're set an easy way to remember the chain rule is to draw a tree diagram w x y t...

I see, and you made that conclusion because of the Dimension theorem? a linear function is 1-1 iff it's nullity is 0 iff rank = dim (P^2(R))

Homework Statement the question i have is more of a conceptual question, i have no idea how to prove that a mapping would be an immersion. thus i have no clue how to start the assigned problem here's the specific problem: prove that (e~) is an immersion : note (e~ means phi tilda) Let F...

hey thanks a lot prove: x > 0 --> 1/x > 0 assume 0 < x and assume 1/x less than or equal to 0, x cannot equal 0. then there exists a unique 1/x s.t. x(1/x) = 1. since 1/x is unique, 1/x cannot equal zero. therefore 1/x < 0 since 0 < x and 1/x < 0 0. 1/x > x . 1/x 0 > 1 which...