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Actually, I can't really see a problem. So...am I really just that rusty? What's the answer supposed to be?

It shouldn't be too terrible. The obvious thing to note is that (b - a) = -(a - b), for example.

It seems like you're going about the problem correctly except that obviously the splitting field won't be as described if the quadratic is reducible. Other than that, unless I'm forgetting something (which would hardly be surprising), what you're saying is completely true.

In general. Well, for nonnegative integer exponents anyway.

The definition of the second partials is just the partial derivatives of the first partials. Why couldn't you just use the same method as before?

Homework Statement Show that f is 2-pi periodic and analytic on the strip \vert Im(z) \vert < \eta, iff it has a Fourier expansion f(z) = \sum_{n = -\infty}^{\infty} a_{n}z^{n}, and that a_n = \frac{1}{2 \pi i} \int_{0}^{2\pi} e^{-inx}f(x) dx. Also, there's something about the lim sup of...

Well, it looks like you're trying to find the power series of ln(5 - x) by differentiating the series for 1/(5 - x) term by term. But ln(5 - x) is the integral of 1/(5 - x) (give or take a sign).

Oh, I meant for the intersection to be trivial. I'll think about what you said though.

I think changing the order of integration is the way to go. You'll get an x^3 term in the integral with respect to x. Then it's easy.

Homework Statement Show that G is isomorphic to the Galois group of an irreducible polynomial of degree d iff is has a subgroup H of index d such that \bigcap_{\sigma \in G} \sigma H \sigma^{-1} = {1} . Homework Equations The Attempt at a Solution I know that if G acts transitively as a...

Homework Statement I'm trying to come up with an example of a quartic polynomial over a field F which has a root in F, but whose splitting field isn't the same as its resolvent cubic. Homework Equations The Attempt at a Solution Well, I know the splitting field of the cubic...

I think you need to rethink your bounds on that one...

Without additional assumptions on m and n, the implications aren't true...

Bump before bed

Isn't your first question essentially, "Can you construct an integer multiple of a constructable angle?" Well...can you?

Homework Statement Let p(z) be a polynomial of degree n \geq 1. Show that \left\{z \in \mathbb{C} : \left|p(z)\right| > 1 \right\}[/tex] is connected with connectivity at most n+1. Homework Equations A region (connected, open set) considered as a set in the complex plane has finite...

Off the top of my head (that is, take this with a huge grain of salt), I think the approximating surface doesn't really matter in the answer, but the cone/cylinder might give the simplest (or maybe easiest to visualize) way to get to the answer.

1. First, the integers aren't even a group under multiplication. So you should be using addition as your operation. Under addition the integers are an (infinite) cyclic group. What does that tell you? 2. I'm not sure what you mean by mapping to an infinite number of items. But for the first...

I think I understand... The splitting field is the minimal field that contains all the roots of the minimal polynomial, and anything that's Galois over the rationals contains all the conjugates of \alpha (i.e., the roots of the minimal polynomial). So it contains the splitting field. Thanks...

I believe the splitting field is just the original extension adjoin i. So that's handy. But it seems like that really shouldn't be the Galois closure. Why would any Galois extension of the rationals that contains \mathbb{Q}(\alpha) have to contain i? It seems like it would only have to be a...

Homework Statement Find the Galois closure of the field \mathbb{Q}(\alpha) over \mathbb{Q}, where \alpha = \sqrt{1 + \sqrt{2}}. Homework Equations Um...the Galois closure of E over F, where E is a finite separable extension is a Galois extension of F containing E which is minimal...

Well, I believe that to use the Weierstrass M-test you have to find a bound that works for all x on whatever set you're testing convergence in, not just a given x in the set. That is, your bound a is allowed to be a function of n, but it cannot be a function of x.

Yeah, I'm pretty sure the Weierstrass M-test tests for uniform convergence by essentially testing for absolute uniform convergence...

Here's a tip that will get you through a good chunk of analysis problems: add 0 inside the absolute value sign in a "creative" way.

Homework Statement Okay, I'm trying to explicitly determine the Galois group of x^p - 2, for p a prime. Homework Equations The Attempt at a Solution Okay, so what I've come up with is that I'm going to have extensions \textbf{Q} \subset \textbf{Q}(\zeta) \subset...

No, that's not true. Just because |x - xo| is less than or equal to r doesn't mean that |x| + |xo| is too. Really, think more intuitively about what |x - xo| being less than r says.

Well, if you don't mind a cheat-ish answer, you could just look at the logarithm, then use the facts that logarithmic and exponential functions are continuous and that products of convergent sequences converge to the product of the limits

In general, a subring of R is a subset of R which is a ring with structure comparable to R. So you don't actually have to show all the axioms because multiplication being associative and distributive is inherited just by being a subset of R. Similarly, some of the additive group structure is...

What are the requirements for a subset of R to be a subring?

Yeah, I hate to keep doing this, but I still haven't figured it out. So I guess I'll try one more time. Sorry to be irritating.

You're completely right. It doesn't imply g^8 = e. You're going to have to do a bit more work than just finding an element of the preimage of g'. Unfortunately, I'm having trouble coming up with the exact proof off the top of my head, so I can't give you much of a kick in the right direction...

Ah, of course. I must have been thinking of something else. I've got to stop posting when I haven't slept.

I guess I'll give this a quick bump before I go to bed. Maybe I was wrong about missing something simple and I actually do have to use the Galois condition.

Er...aren't those fields are isomorphic? Take a map that's identity on the rationals, and root 2 mapped to root 3, right? I mean, strictly speaking I don't think that the degree being the same is enough to guarantee an isomorphism (if I had to guess, I'd say the automorphism groups of the...

Wow, I'm really slow today. Okay, g(z) is not entire because, as you said, it's not analytic at 0. Why does your argument fail? Because the integral \displaystyle\oint_{c} g(w) dw is not defined if c passes through 0, so it's not 0 for every closed curve in the complex plane. However, since...

I'm pretty sure that what you have is \displaystyle\oint_{c} g(w) dw = \displaystyle\oint_{c} \frac{f(w)}{w} dw = 2 \pi i * n(c, 0)*f(0) So for that to be zero for all closed curves, f(0) = 0. Dunno if that helps any. And, as always, I could be wrong.

Well, we can't really tell you if it said that you can't use unit vectors. Regardless, it's clearly true for any unit vector because, using your notation, the equation says ||v||^2 = v*v = ||v||. Edit: Yeah, I'm really out of it today.

Grad. Don't ask me how I managed it. I personally vote for black magic.

Homework Statement I need to show that, given F \subset E \subset K \subset L (K/F is Galois but I don't know how important that is for the part of the problem I'm having trouble with) and a homomorphism \phi:E \rightarrow L that's the identity on F, that \phi(E) \subset K. Edit: Yeah, if it...

By the way, my example map is completely wrong. That's what happens when you answer algebra questions while doing complex analysis I suppose.

Well, we haven't officially discussed path homotopy and its implications for contour integration yet (much to my chagrin), but I think I'm getting the idea. I think I could look at log(\frac{z+1}{z-1}), then maybe use that to define a square root in the region I want. Anyway thanks.

So prove that (n(n+1)/2)^2 is equal to the right hand side. Much easier to verify.

I don't quite understand what a branch point is apparently. I mean, as I understand it, yes, -1 and 1 are branch points, but I thought that any point where the multi-valued "function" actually started to be multi-valued was a branch point. Oh well, -1 and 1 is the answer you're looking for I...

Well, the second part isn't correct at all (\sum^{n}_{i=1} i^2 = \frac{n(n+1)(2n+1)}{6}), but that wasn't really my point. Look at the left hand side of your proposed equality.

I think it's just because t^2 = x doesn't necessarily imply that t = sqrt(x).

Well, log(z) has its branch points for z less than or equal to zero, so we get branch points for log(z^2 - 1) whenever z^2 - 1 \leq 0. So I guess it's -1 \leq z \leq 1?

What is \sum^{n}_{i=1} i?

You interpreted correctly. Anyway, I think the branch points all lie on the real axis less than or equal to 1, assuming we're using the principal branch of the log.

That's how I got the "antiderivative" I mentioned, but if that's right then I really have no idea how to choose the right branch of the logarithm. Do I have to see what z + \sqrt{z^2 - 1} does to my path?

Infinity is just \infty. Dunno about the other. Anyway, it looks correct to me.