Recent content by Applejacks

ugh I forgot about that. Every prime except 2 in an odd number so that left my mind. Thanks to both of you

This is a stupid question but how do I determine it's cyclic? I don't remember covering this in our notes. I was looking at more examples, Z_97,98 and 99 and was wondering what approach I would need to take then. 97=prime=p 98=2 x 7^2=2p^k 99=3^2 x 11=/=2p^k or p^k 97 and 98 have PRs but 99...

What are the primitive roots of Z_32? \varphi(\varphi(32))=8 However you must first check that there is a primitive root. A PR exists if (a) n=2,4 (b) n=p^k (c)n=2p^k According to the solutions, Z_32 has no primitive roots. Is this correct? 32=2^5 which fulfills one of the conditions (b) so...

Ok that clears it up. I wasn't too sure if we could take a formal power series to n=infinity, which leads me to another question. If the n! was in the numerator instead, for the example you provided,would we get the same result (element of R[[X]] but not R[X])?

Not so much a question. Rather, I don't quite understand the concept. R[[X]] is the formal power series a0 + a1x + a2x^2... R[X] consists of all elements in R[[X]] which have only finitely many non-zero coefficients. Can someone give me an example? Would, say, two different elements of...

I ended up with a zero using Murphid's approach. I'm not sure why I get a different answer using LCKurtz method but I'll look into it either way. Thanks for the help guys.

Okay that makes sense. Inside the brackets I end up with the scalar xd()/dx + yd()/dy + zd()/dz. Then the right side vector acted on this left-side scalar and gave me x(1/r,0,0) + y(0,1/r,0) + z(0,0,1/r) = (x,y,z)/r. Lastly there was the 1/r from the start on the left side giving me (x,y,z)/r^2...

Well the question asks (r(hat) dot ∇)r(hat) I thought this equaled (r(hat))^2 dot ∇r(hat)= 1 dot ∇r(hat)=∇r(hat)

Homework Statement Let r=(x,y,z). Find ∇r(hat). Homework Equations r(hat)= (x,y,z)/sqrt (x^2+y^2+z^2) ∇f=df/dx x + df/dy y... The Attempt at a Solution Okay I'm having a complete brain freeze at the moment. I know the denominator is a magnitude but am I still supposed to use the...

I see now. However, my final answer should have 2pi instead of pi/2 right? I reworked it and that's what I got.

Well that's sort of my point. The first one is outside the unit circle since a>1, hence z becomes >1. Only the second one is inside the circle so how come I'm not using that one instead?

Sorry ignore my last post. I forgot to transform dt into dz. Here is my working out: http://imageshack.us/f/215/unled2lwl.png/ I just don't understand why I have to use the positive residue. What piece of information tells me that the negative part isn't inside the unit circle.

Homework Statement http://imageshack.us/photo/my-images/15/unledflsq.png/Homework Equations A simply connected domain D in the complex plane is an open and path connected set such that every simple closed path in D encloses only points of D.The Attempt at a Solution The answers are a,c and d.I...

Okay I found my mistake. It should have been \frac{2z}{(z-(a+\sqrt{a^2-1}))(z-(a-\sqrt{a^2-1})} Now the residues are: \frac{-a-\sqrt{a^2-1}}{\sqrt{a^2-1}} \frac{a-\sqrt{a^2-1}}{\sqrt{a^2-1}} So finally I have 2ipi*sum(res)=2ipi* -2=-4ipi Is this right?

Well I'm looking at the two singularities: a+\sqrt{a^2-1} a-\sqrt{a^2-1} For a>1, the second one tends to zero. So we should have 2ipi*-i/2 *sum (res)=pi* \frac{-1}{2\sqrt{a^2-1}}?