Recent content by CrazyCalcGirl

I did get it to satisfy the Cauchy Riemann conditions. I think since f(z) is given to be analytic that we already know it has continuous first partials at U and V. The conjugate will obviously also have this as well.

Homework Statement If f(z) is analytic at a point Zo show that the Conjugate(f(z conjugate)) is also analytic there. (The bar is over the z and the entire thing as well.) The Attempt at a Solution I know if a function is analytic at Zo if it is differentiable in some neighborhood...

ok here ok now that you got rid of all those crazy a's here is what I got. Part 1. isomorphic to Z4 abcd bcda cdab dabc part 2. not isomorphic to Z4 abcd badc cdab dcba Anyone else get this?

:( All I can say is LOL you might want to work on the table with all a's in it. It has to follow group properties. I hope that's not the grade you want in this class.

oh god. I feel dumb now. I'm not even going to post back. I got it. Thank you.

Homework Statement Suppose that the function f is continuous on [a,b] and X1 and X2 are in [a,b]. Let K1 and K2 be positive real numbers. Prove that there exist c between X1 and X2 for which f(c) = (K1f(X1) + K2f(X2))/(K1+k2) Homework Equations The Attempt at a Solution I...

hmm both ways I did it I plugged in the endpoints.. so if the first way correct that I originally posted or is the second way correct where you set f(x)= 2^x-3x? Thanks : )

I know the sum of two continuous functions is continuous and same for a multiple, but I don't see how this applies to me using the IVT. The problems requires I use that theorem. If you think the way I did it is wrong how about this...? If I use your idea to let f(x)= 2^x-3x now I can just...

Homework Statement Use the Intermediate Value Theorem to show that the equation 2^x=3x has a solution c element of (0,1) Homework Equations The Attempt at a Solution Ok I know this theorem is usually very easy, but I've never done one where I couldn't easily solve for x and plug in...

I get that part of it, but my question is on the 2nd part of the squeeze theorem. It states abs(g(x))<=M for all x not zero. then lim f(x) x g(x)=0 as X--> a. Can someone show me the proof to this part and explain it. The book shows no further information and I'm confused as to what it means.

ok well .. If you want me to solve for L I moved it over and completed the square..then I get L=+/- (sq root(2)) +1 so the positive number matches the 2.14 I found before. So when you asked me to show sqrt(2x+1)>x you get the same thing with an inequality sign. The range that this is true for is...

hmm Well the problem I'm having is that my professor will not allow us to use anything at all if we have not done it in analysis yet. I used induction because it was in a prior section. I'm not sure I've seen an example of how to find the true upper bound. We've done problems where we were...

I've got the increasing part down. I started off with S(n+2)/S(n+1) >1 and basically used algebra to get it down to S(n+1) > Sn. That is where I stopped. I'm think that is enough to show its increasing for all n >=x. Now for the bounded above part. I am not sure this part works. I knew it was...

ok.. So I can see by plugging it in on my calculator that it is definitely increasing and I'm getting the limit to be about 2.41421. Is that what you got? Ok so if I can show it is bounded above (which it obviously is) and it is a monotone increasing sequence then it converges by the monotone...

Homework Statement The problems states let S1 = 2 and S(n+1) = (square root(2Sn+1)) show that Sn is convergent. Homework Equations The Attempt at a Solution I am not really sure how to go about this proof. Sometimes it is easier to prove something is cauchy than to show it is...