Recent content by MattL

I think arcsinh is ok on all of \mathbb{R} With arccosh x has to be greater than or equal to one, but I can't remember the conditions for arctanh

thanks haven't done that since a-level and had forgotten it completely!

Just a quick question Can anyone give a method to derive arcsinh(x) from the definition of sinh(x)? Thanks

No problem. Think I should be able to give this question a fair go now.

I'm having trouble with evaluating [Triple Integral] |xyz| dx dy dz over the region (x/a)^2 + (y/b)^2 + (z/c)^2 <= 1 Do I need to use some sort of parametrisation for the region, and is there some way of dealing with the absolute value function without integrating over the eight...

yeah, oops, just realized it's about similar triangle and it doesn't matter where the origin is...

OK, I've got this question to do: Find complex numbers representing the vertices of a triangle ABC given that the midpoints of the sides BC, CA, AB are represented by complex numbers z_1, z_2, z_3 respectively. Thing is, I don't know where I'm taking the origin to be; if I took it at A...

how do I conjugate them? I've only recently started permutations and group theory (and yet I've been given this question :confused: ), so my knowledge is still quite basic ( :frown: ) EDIT: oh, I see...

Yep, there seems to be a pattern of (m+1 m+2) = (1 2 ... n)^(n-m+1) (1 2) (1 2 ... n)^(m-1) and after trying some in Maple, I think that it does hold for other n. Still, that doesn't give me a complete set of transpositions. (Another thing - would be ok to use the inverse of (1 2 ... n) in...

After giving it a go, I've managed to get, for n = 5 (2 3) = (1 2 3 4 5)^4 (1 2) (1 2 3 4 5) (3 4) = (1 2 3 4 5)^3 (1 2) (1 2 3 4 5)^2 (4 5) = (1 2 3 4 5)^2 (1 2) (1 2 3 4 5)^3 So I can get the transpositions (1 2), (2 3), (4 5), but I can't work out how to get all of the others that I...

Explain why the permutations (1 2) and (1 2 ... n) generate all of Sn, the symmetric group (the group of all permutations of the numbers {1,2,...,n}? Perhaps something to do with the fact that (1 2 ... n) = (1 2) (1 3) ... (1 n)? Other than that I haven't got a clue - help! (please!) Thanks