@DHMO no, exp(z) has one and only one value for every possible complex number z. perhaps you're thinking about exp(z) not being a one-to-one function (i.e. there are different complex numbers z,w for which exp(z)=exp(w)).
In mathematics, de Moivre's formula (also known as de Moivre's theorem and de Moivre's identity), named after Abraham de Moivre, states that for any complex number (and, in particular, for any real number) x and integer n it holds that
(
cos
(
x
)
+
i
sin
(
x
)
)
n
=
cos
(
n
...
@arctictern you are right. I was thinking about the multifunction $z^w$
@NaCl or you can just convert $\cos y+i\sin y$ to $e^{iy}$
@TedShifrin so an affine subspace of $\mathbb{R}^n$ is just a "linear" object, like a plane, line, point, cube, etc. etc. which need not contain the zero vector?
@meow-mix, I'd really enjoy that as well. One of my "dream jobs" is to teach high schoolers math the way I think it should be taught. There are all sorts of problems with intro math education.
Common core is good, but I'm fighting helpers in the library and parents who can only do the old algorithms and din't understand adding/subtracting by regrouping or compensating. But when the kids can't do basic arithmetic, it's all sorta pointless.
I'm surprised you elected for topology. What'd you focus on? IIRC, you aren't the biggest fan of all the separation axioms and such. Finding obscure counterexamples, etc.
A smallish fraction of Pete's class felt like re-writing Steen & Seebach :P
Not so ridiculous, no. They wanted topology cuz the class was canceled. And they were good enough that we went very fast. They did some Munkres exercises no student had ever done before with me.
Hello!! I want to calculate the limsup and liminf for the sequence $a_n=(-1)^{\frac{n(n+1)}{2}\sqrt[n]{1+\frac{1}{n}}$. I tried to find the lim a_{2k} and lim a_{2k-1} but I find anything... Could you give me a hint?
so the projective group $\mathrm{Proj}(1)$ is a set of equivalence classes over the general linear group $\mathrm{GL}(2,\mathbb{R})$ whose equivalence class is "can be transformed from one to the other via only scalar multiplication"?
@DHMO you can partition reals according to the asymptotics of the number of 0s and 1s in their binary expansions. since we're talking asymptotics, both parts will include a real with any given initial segment of binary digits.
R^2 consists of 2-tuples whereas R^4 consits of 4-tuples, so technically there is no element of R^4 that is an element of R^2. unless you identify R^2 with the set of 4-tuples which have last two coordinates 0, but then a 2-dimensional subspace of R^4 may not be related at all to this copy of R^2 existing within R^4
let's consider R^3 instead. you realize there is more than one plane existing in 3D right? they can't all be called R^2, only at most one of them can, in which case the other planes are not called R^2.
also, we say linear combination instead of expansion
Take meow's example. The span of {(0,1,0,0),(0,0,1,0)} is the set of all vectors of the form (0,a,b,0). But every element of R^2 is of the form (a,b). Or, if you identify R^2 with a subspace of R^4 in the usual way, R^2 refers to the vectors of the form (a,b,0,0). Therefore, the span of {(0,1,0,0),(0,0,1,0)} does not equal R^2, i.e. {(0,1,0,0),(0,0,1,0)} does not span R^2.
is $\zeta(x)-\frac{1}{1-x}$ holomorphic everywhere? I proved $\lim_{x\to 1^+}\left(\zeta(x)-\frac{1}{1-x}\right)=\gamma$, and if that is true for x as a complex variable then the singularity should be removable
