Recent content by ThePerfectHacker

Use the following fact, $$ \oint_{\Gamma} \frac{dz}{z-a} = \left\{ \begin{array}{ccc} 2\pi i & \text{ if } & a \text{ is inside the contour } \\ 0 & \text{ if } & a \text{ is not inside }...

There is a simpler approach, but it does not use CR-equations. I was only doing it that way because I thought you wanted to see it by CR equation. The easier method is that if $f(z)$ is analytic on an open set $U$ then $e^{f(z)}$ is analytic on that same set. If $g(z)$ is analytic on $U$ and...

We showed that $u_n > \tfrac{A}{2n}$ and $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges, therefore $u_n$ diverges.

Here is an easier suggestion. Use partial fractions first.

Your inequality is not correct as stated. Take $u_n = \frac{1}{n}$ as counter-example. See it does not work. Instead here is an alternative one. Let $\varepsilon = \frac{A}{2} > 0$. There exists an $N>0$ large enough so that if $n>N$ then $|nu_n - A| < \frac{1}{2}A$. In particular...

This is wrong. You took an extra derivative. You can also write $\frac{e^{iz}}{z^3}$ as a power series and notice the only term with $z^{-1}$ in it has coefficient equal to $\frac{(iz)^2}{(2!)z^3} = -\frac{1}{2z}$ When you integrate $z^{-1}$ over that region you get $2\pi i$. Therefore, the...

Let $z = x+iy\not = 0$. Verify $\frac{1}{z} = \frac{x}{x^2+y^2} - \frac{y}{x^2+y^2}i$ and use $e^{a+bi} = e^a(\cos b + i\sin b)$. With that you can find the $u$ and $v$.

By assumption you start with a lin. ind. set, therefore no vector in it can be zero. (Or show that if you have a list of vectors, and a zero vector is among them, then that list is automatically lin. dep.)

I am not sure, but I would not be surprised if DG is not used anywhere outside of math and physics. Even something as basic calculus is not used by engineers on problems they work with. It is true that there are problems from engineering that need calculus to solve them, but engineers do not...

You have shown above that there is exactly one subgroup of order $p$. Let us call it $P$. Furthermore, this is a Sylow $p$-subgroup because $p$ is the largest power of $p$ which divides $|G|=pm$. Thus, by the third theorem all Sylow $p$-subgroups are conjugate. Consider $gPg^{-1}$. Show that...

The theorem says that if $\Omega$ is a set of ideals where $\leq$ simply means containment then all the maximal elements are prime ideals. What is a "maximal element"? It is an ideal $I \in \Omega$ with the property that $I$ is not contained in any other ideal (other than itself) from...

Compute the discriminant of the cubic polynomial (it is irreducible). Then check if the discriminant is a square or not, in the square field. If it is a square the group is $A_3$, if it is a non-square then it is $S_3$.

"Maximal element" with respect to the ordering $\leq$, is different from saying "maximal ideal". The confusion is that the word 'maximal', is being used in two different contexts.

Here is an easier proof that minimizes the use of $\varepsilon$'s . First define $X_n = \mathbb{Z}/p^n \mathbb{Z}$, and give these finite sets the discrete topology. The $p$-adic integers are the projective limit of the $X_n$. Now consider the product, $$ X = \prod_{n=1}^{\infty} X_n $$ This...

If two Borel measures agree on intervals of the form $[a,b)$ then they agree on anything in the $\sigma$-algebra which is generated by those intervals.