$|\Bbb N^{\Bbb N}|$ is not bigger than $|2^{\Bbb N}|$ for instance. we can inject the former into $\Bbb R$ via continued fractions. (so this example applies with both $|X^{\Bbb N}|$ and $|{\Bbb N}^X|$ interpretations).
@GPerez are you talking about ring isomorphism or K-isomorphism?
Is there a term for a function of the form $x' w' A w y$ where $x, y, w$ are variables and $A$ is some matrix? if $x = w= y$ then this would be some quartic form, but what aout this generalization?
@NicholasMancuso $w'Aw$ is a quadratic form, definitely. If $x = y$, there's something to be said about a specific type of distance metric, I forget the name.
@NicholasMancuso polynomials of all sorts have been investigated, don't see why this particular form would have a name out of the zoo of possible forms they can take
@NicholasMancuso multilinear form A(p,q,r,s) would be linear in p,q,r,s. in which case A(w,w,x,y) would be quadratic in w and linear in x,y, and A(z,z,z,z) would be quartic in z, etc.
I'll take a look, I've been using Boothby and I'm noticing a lot of things left to the reader that I'd have liked to have been told up-front instead of toyed with.
Mike, did I tell you that I got a tee-shirt for my retirement that had the subtitle ... Retirement: A Geometric Approach (there was a geometric series)
7
Oh, that's graduate stuff, @Axoren. You'll have to ask me directly.
On an aside, just last week, I saw the most devastatingly simple yet confounding linear algebra problem. It's due in two days as the first question on an otherwise simple refresher homework for CS 724.
Most of the class had trouble with it and I solved it this morning: Let $x$ and $y$ be orthogonal vectors and let $P$ be some idempotent matrix. True or false: $Px$ and $Py$ are orthogonal.
My solution was to simply normalize to $\hat x$ and $\hat y$, extend them to a basis of the whole space $\beta$ and select $P = [ \hat x, \dots \hat x]_\beta$.
@Axoren in 0 and 1 dimensions, there is no 90 rotation. in 2 dimensions, you need to specify which 90 rotation. in 3 or more dimensions, there's uncountably many 90 rotations (and then I'd have to explain what that even means).
The equation for the exponential function of a quaternion $q = a + b i + c j + dk$ is supposed to be $$e^{q} = e^a (\cos(\sqrt{b^2+c^2+d^2})+\frac{(b i + c j + dk)}{\sqrt{b^2+c^2+d^2}} \sin(\sqrt{b^2+c^2+d^2}))$$
I'm having a difficult time finding a derivation of this formula. I keep trying to...
assuming you define a^b to be exp(b ln(a)), you still need to define ln() for quaternions, which already has issues for complex numbers. but I mean we talk about i^i and stuff with the usual branch cut, so you can do that for i,j,k
does the null set come in pairs? by standard rules the fundamental theorem of algebra would be broken if we say no so i think its still yes in that case
The equation for the exponential function of a quaternion $q = a + b i + c j + dk$ is supposed to be $$e^{q} = e^a (\cos(\sqrt{b^2+c^2+d^2})+\frac{(b i + c j + dk)}{\sqrt{b^2+c^2+d^2}} \sin(\sqrt{b^2+c^2+d^2}))$$
I'm having a difficult time finding a derivation of this formula. I keep trying to...
