Well, let me put up some more of what I have found - still no solution.
Since a,b \in Z(P) we know that P \subseteq C(a) and P \subseteq C(b). Since P \subseteq N(P) we also know that P \subseteq N(P) \cap C(a) and P \subseteq N(P) \cap C(b) .
Further, x^{-1} C(a) x = C(b)...
[SOLVED] Conjugates in the normalizer of a p-Sylow subgroup
Homework Statement
Let P be a p-Sylow subgroup of G and suppose that a,b lie in Z(P), the center of P, and that a, b are conjugate in G. Prove that they are conjugate in N(P), the normalizer of P (also called stablilizer in other...
Halls, the OP seemed to be the problem you mention but the original image that she uploaded was the other one, so I think it's ok. The poster just needs a bit more care in where parentheses go.
The main point is just that every polynomial is its own Taylor Series, irrespecitve of about what point it is taken. It will always come out to be itself.
The process wouldn't differ, i.e. if you wanted to take the Taylor Series of a polynomial about x = a you would evaluate all your derivatives at x = a instead of at x = 0. As an example, if I took the Taylor series of the general 3rd degree polynomial at x = 1 I'd have
f (1) = A + B + C + D...
1 + tan^2 = sec^2 is not equivalent to 1 - sec^2 = tan^2? (step 2) Looks like you missed a negative sign, pretty small error that apparently got magnified later on.
Here's an example:
Take the Maclaurin series of f(x) = Ax^3 + Bx^2 + Cx + D (degree 3 polynomial)
f(0) = D
f ' (x) = 3*Ax^2 + 2*Bx + C so f ' (0) = C
f '' (x) = 3*2*Ax + 2*1*B so f ''(0) = 2!B
f ''' (x) = 3*2*1*A so f ''' (0) = 3!A
f ''''(x) = 0 (and all higher derivatives as well are...
It only requires finding derivatives and evaluating them at x = 0. When the original function is a polynomial I am hopeful that you find taking a derivative to be a piece of cake!
It's pretty clearcut. Just follow the recipe for building a Maclaurin series for a polynomial of the general form, and lo and behold the answer will be the original polynomial.
How would you find the "slope of a line that is tangent to the graph"? Once you have the slope, what point would lie on the tangent line if it is tangent to the graph? Once you have the slope, and a point, how do you get the equation of the line?
I was looking at this, but I'm not sure that the method can work as nicely here (the author says there's no foolproof method)
http://www.asaurus.net/~buhr/academic/2004-1-ubc-math263/handouts/curve.pdf
Above you aren't using the spherical definition for z.
Edit: Hmmm, this is harder than I thought, hope I didn't bring you to a dead end. I'll keep working on it.
Edit 2: Somehow you need to combine the two equations together to get something like U^2 + V^2 = constant^2, at which point you...
Now the tricky part. You need to eliminate a parameter so that your curve of intersection expresses the set of points of the intersection with one parameter (this way it's a curve rather than a surface).
So, x_3 is the 3rd component of x correct? If that's the case (I don't see what else it could be) then I agree that it doesn't seem to be unique as we have
(1) x_1 = -15 + 10t
(2) x_2 = -3
(3) x_3 = t
Now t can be chosen to be any real number. Are you sure that you copied them correctly?
:approve:
A picture should help with the proof. Draw a circle with x and y at random points and x0 at the center. Then try to fit the triangle inequality in with your picture.
You are proving a statement about A U B, so you need to start with an arbitrary element in A U B. You can then invoke the known properties of A and B individually to prove the statement regarding the union.
You want to show H is a normal subgroup of G not the reverse. Showing that H is a subgroup follows from the fact that y has order 4. Do you see why? To show it's a normal subgroup of G consider the left (or right) cosets of H in G. How many are there?
This is good
\tau is a mapping and \tau^{-1}(g) is an element of the group on which the mapping is defined. Your question is not making sense. You have successfully shown that that the inner automorphisms are normal in the group of automorphisms. You can stop there.
Each basis vector in V that satisfies fi(ej) = \delta ij defines a system of 3 equations in 3 unknowns. Each of these systems gives you one of the basis vectors.
Well, yeah, thinking is always good, but he *thought* he should try the quadratic formula and I think he's right in the sense that it will give him the correct answer.