Recent content by Mathdope

You should have 0/0 not 6/0. In any case, try a trick similar to what you did with the radical in the denominator.

In both cases equivalent points lie on the same plane (sphere). What are the equations of these planes (spheres)?

OK solved, if anyone wants to see solution let me know.

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)...

*bump*

[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...

What's the remainder on division by 3 of the 3 factors you have exhibited?

One way would be to substitute for x and y, which would result in a function of t alone.

What kind of sadistic prof assigned this?

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.

I don't quite understand what you mean when you say X/B is not a union of X/A. Can you clarify please?

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...