Recent content by ryou00730

Yea I realized this soon after posting, thanks for the help anyway.

Homework Statement Given that the minimal polynomial of a over rationals is x^4+x+8, find the minimal polynomial for 1/a over Q. Homework Equations I know there is a lot of work done out there for finding the min. polynomials of a^k for k>0, however I've never seen anything with a^k for...

Continously differentiable f: R^n --> R^m not 1-1? My course is over with now, but I never could figure out this question. It's pretty much been haunting me ever since, and the internet has not given me a proof that convinces me. My problem is determining why: A continuously di...

But does taking the principal branch of square root z not deal with that? Does the principal branch mean we only take the principal roots of z?

Homework Statement Does the principal branch square root of z have a Laurent series expansion in the domain C-{0}? The Attempt at a Solution Well I'm not really sure what a principal branch is? I believe that there is a Laurent series expansion for z^(1/2) in C-{0} because originally our...

hmmm, not entirely. I follow most of it, just the part where you get down to a 1 size subset. Why does when k=1 necessarily have to have c1 and ck+1 in the size one subset? I thought it was okay to have just c1 in the size 1 subset as proven in the base step?

Oh I didn't notice you posted a reply. So this is a false claim because P(1)= (c1), and we did not prove that it could also equal (c2)?

Mistake: The inductive step fails from n=1 to n=2. If we assume S=(c1, c2..., ck, ck+1) is a set of k+1 cpus, then when k=2, S=(c1, c2). If we take S-c1 then this is a set of k cpus and we get S-c1 = (c2) should be a set of k cpus (specifically k=1), but we already showed k=1 is P(1) = (c1). So...

I don't exactly follow what you want me to do. So you want me to verify in the proof for k=1? So you mean you want me to verify that it works for P(2) or? Now that I look at it, the proof looks correct, or is there something in there actually completely wrong?

I'm not really certain. I feel like he would need another base case as well.. like base of P(2), because just because one cpu is made by this "same manufacturer" doesn't mean that 2 are, or all are for that matter. I'm not really sure here what's wrong though, or what I'm suppose to be doing.

Homework Statement Consider the following induction proof. Discuss whether the proof is correct or not, and if not, explain precisely why not. Theorem: Every cpu is made by the same manufacturer. Proof: We prove this by induction. Let P(n) mean that for any set of n cpu’s , they are all...

No it should have said for n>6.

It's exactly the same method as last proof, except we use base case of 7 and 8, then set up the same way as last proof we did on the previous assignment, you get down to f(n)> (root2)(n-1) + (root2)(n-2) and we are guaranteed that (root2)(n-2) is always less than (root2)(n-1) for n>6. So make...

thank you:)

thank you :), so in general with strong induction, you prove a base, then your induction hypothesis is that it works for all numbers between your base up until some value n, and you have to prove using this, that it also works for the n? thank you for all your help!