What exactly does it mean to say that $D_{2n}$ is the symmetry group of a regular $n$-gon? What does it mean to say that $\Bbb{Z}_n$ and $D_{2n}$ act on a regular $n$-gon? How does that work out? What's the set that these groups are acting on?
So, we view the $n$-gon as living in $\Bbb{R}^2$, and the set $D_{2n}$ acts on is the set of all ordered pairs making up the $n$-gon? So, from this perspective $D_{2n}$ is really a collection of $2 \times 2$ matrices; and when one does all the calculations, one can view it more abstractly as $D_{2n} = \langle r,s \mid ... \rangle$?
Hmm...that's a much more natural perspective---to me at least. It just seems weird when books claim that the abstract group $D_{2n} = \langle r,s \mid r^n = s^2 = (rs)^2 = 1 \rangle$ is the symmetry group and that acts on something in $\Bbb{R}^2$. It seems more natural to start with this matrix picture of acting on points in $\Bbb{R}^2$, and then move to the abstract picture.
Let $A$ be a $n\times n$ complex matrix. Assume that $A$ is self-adjoint and $B$ denote the inverse of $A+iI$, where $I$ is identity matrix of order $n\times n$. Then all eigenvalues of $(A-iI)B$ are
purely imaginary
real
of modulus one
of modulus less than one
My attempt: I discarded option...
Also no. We know that $\sum_m x^m=\frac{1}{1-x}$. If $x$ is irrational and $y=1-x$, then $x=1-y$. So $y$ cannot be rational unless $x$ is also rational. Now, assume $\frac{1}{y}=r$ where $r$ is rational. Then $yr=1$ which is impossible unless $y$ is rational. Thus $\frac{1}{1-x}$ is irrational.
Well, I'm tasked with showing that if $\mathfrak{M}=(M,x)\in R[[x]]$ with $M$ maximal in $R$, then $\mathfrak{M}$ is maximal in $R[[x]]$. However, by my understanding, $(M,x)\subsetneq (M,x,nx)$.
Today I think Google chrome was updated. So, bookmark bar only shows in new tab, and after opening a web page, bookmark bar disappears. So, in particular, I can't click on 'Start ChatJax'. What to do?
When Bakarka become uncommonly on, the discussion about algebraic topology dies out, only to be replaced by even more incomprehensible barrage of field extension questions
But then, my rambles are equally incomprehensible, so that serves as a good counterweight
I'm on mobile so latex is a challenge but I Don't know if I got this right because it's like an n choose k but I think it's like to the r power. I could upload the screenshot
I am thinking 3 choose k for i
And n choose k to the r power
For the second part
But it says unlimited supply so I'm also thinking 3^r choose k
For the first part and then n choose k to the r for the second part
hmm... I don't quite get 7.6, so you mean let spam, chicken and mahi be s,c,m. Then they are asking for a lunch with k plates, how many kinds of s,c,m combinations for each k (allowing repetitions of some s,c,m)?
so if it is a 5 plate lunch I have something like: sssss scmsc smmcc etc.?
hmm, in that case, let $r$ be some large number, then you have a total of $3^r$ plates and you want to order them in $k$ plates. Thus that will be $3^r$ choose $k$, and since it stresses it is unlimited, it means $r$ has to be taken an infinite limit somehow.
I think I need some information on 7.4, cause I am not that familar with generating functions in combinitorics rpoblems
$3^r$ choose $k$ make sense since $3^r$ is just some very large natural number
I think one good way to interpret "unlimited supply" is that given k plates, you have at least k of each item thus you can request as much item you want to fill in the plates
That way I can create my outline and reference sheet along with handouts
Oh s--- I think I got it
On 7.5 with the donuts the unlimited vegan donuts is $(1+x+x^2+...+x^r)$
So $(1+x+x^2+...+3^r)$?
Observation 12. Suppose there is 1 object of type A, 1 object of type B, 1 object of type C, and 3 of type D. Then he number of ways to pick k objects from this set is equal to the coefficient of $x^{k}$ in the polynomial $(1+x)(1+x)(1+x)(1+x+x^{2}+x^{3})$
This has to be related with the second part of 7.6
Ah I finally remember, the n choose k are the binomial coefficients, and hence their generating function is $(1+x)^n$. Thus in 7.6 there are $3^r$ objects in total, thus the generating function is $(1+x)^{3^r}$
7.7 is thus when there are unlimited supply of unlimited types of food, thus taking the limit $n$ of $(1+x)^{n}$ should give you the result as required for geometric series
I love how wiki contains "Johann Peter Gustav Lejeune Dirichlet", "Peter Gustav Lejeune Dirichlet", "Gustav Lejeune Dirichlet", "Lejeune Dirichlet", and "Dirichlet"
> His paternal grandfather had come to Düren from Richelette (or more likely Richelle), a small community 5 km north east of Liège in Belgium, from which his surname "Lejeune Dirichlet" ("le jeune de Richelette", French for "the youth from Richelette") was derived.
Since $\sqrt{-3}$ satisfies the equation $x^2+3=0$, we have $[\Bbb Q(\sqrt[3]2,\sqrt{-3}):\Bbb Q(\sqrt[3]2)]\le2$, hence degree must be $2$ since we observed above that $\Bbb Q(\sqrt[3]2)$ is not the splitting field.
I can't see why '$\Bbb Q(\sqrt[3]2)$ is not the splitting field' says that $x^2+3$ irreducible over $\Bbb Q(\sqrt[3]2)$. In fact, they showed earlier there that '$\Bbb Q(\sqrt[3]2)$ is not the splitting field for $x^3-2$'.
Sometimes I wish there is an easier way to determine how the roots of a n degree polynomial changes when we make an infintesimal change on one of its coefficients
Graphically, polynomials are pretty special because the elements $x^k$ for each $k$ forms a dense orthogonal basis
meaning that tweaking just one of these coefficients, and then compare that with the original, the other powers of x contributed to the graph does not change, which is why the space of polynomials behave like a vector space
but the motion of the roots are quite unpredictable with each change, at least I don't know of an analytical formula that describes them
Let $A$ be a $n\times n$ complex matrix. Assume that $A$ is self-adjoint and $B$ denote the inverse of $A+iI$, where $I$ is identity matrix of order $n\times n$. Then all eigenvalues of $(A-iI)B$ are
purely imaginary
real
of modulus one
of modulus less than one
My attempt: I discarded option...
