suppose $M$ a matrix, and it is the representation of foreign basis $B_A$ in terms of the standard basis $B_S$, so $M = B_S(B_A)$ . but... what is the meaning of $M^{-1}$?? $M^{-1} = B_S{B_A^{-1}}$?? does the inverse of a basis make sense?
see, this is what memorization does to undergrads. $M^{-1}M = I$, repeat until I pass
but: if you say it is "the inverse of the transformation taking one basis to another"
then... $M^{-1} = B_A(B_S)$?
ah
this makes sense: $B_S \mapsto B_A \mapsto B_S$
bless you, thanks
ok, I have entered a moment of semantic saturation, I thought that $I$ for 2x2 matrices $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ was fine up until this moment. it seems that we're mapping sets to sets, which are themselves maps of sets to sets. why would $I$ have any definite shape?
unless... that's only the identity element for $\mathbb{R}^2$
of course... $I$ is the identity mapping, and $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ is what happens to be the identity mapping to go from $\mathbb{R}^2$ back to $\mathbb{R}^2$ with the same input
i watched one of them. it was good. i don't know what i was expecting.
a guy came in the middle and asked about shares of partnership revenue. he dealt pretty well with that. didn't even throw chalk, which is what i would have done.
but yeah for each $\alpha$ you're asking for the probability of $x\le\alpha\le z$ six times (once for each permutation of $(x,y,z)$) and the $x\le\alpha$ and $\alpha\le z$ bits are independent
In "50 Challenging Problems in Probability", question #43 is the following:
"A bar is broken at random in two places. Find the average size of the smallest, of the middle-sized, and of the largest pieces."
The author gives what seems like a complicated geometric way of calculating the probabili...
i do like these problems. i thought a conjecture was true for about a year because the routine i used to test it was imposing a very highly non-uniform distribution on the unit ball of $\mathbb{C}^6$, weighted toward where the conjecture wasn't false.
when i discovered what i had done wrong i had to send some awkward email.
and so $s^{1/3}$ works because $P[r\le R]=P[s^{1/3}\le R]=P[s\le R^3]=R^3$
@leslietownes That sounds annoying
^Here's their code
Math.cbrt is the cube root
Math.random is uniform from 0 to 1
I appreciate that TWO_PI is apparently its own separate variable in memory for this programming language, presumably more efficient that 2*PI
And here's the 2D version
@leslietownes Here's a question. Say I want to get a random point on the surface of $S^n$ (so $n+1$ numbers satisfying $x_1^2+x_2^2+\dotsb+x_{n+1}^2=1$). How would I do that?
I know an elegant way to do this, but it involves first knowing how to sample from a normal distribution (given that we can sample uniform distributions), which isn't an obvious task
Which is, simply: sample $x_1,\dots,x_{n+1}$ from a standard normal distribution, then normalize the vector to obtain $(x_1,\dots,x_{n+1})/\sqrt{x_1^2+\dotsb+x_{n+1}^2}$
I suspect if you can't sample from normal distributions there should be some way to do induction on twos, that is, to sample from $S^n$ first sample from $S^{n-2}$ and do some stuff
Or no, sample from $B^{n-1}$
Does the projection $S^n\to B^{n-1}$ that forgets the two last coordinates ($(x_1,\dots,x_{n-1},x_n,x_{n+1})\mapsto(x_1,\dots,x_{n-1})$) preserve volume?
This is the higher-dimensional equivalent of the cylindrical projection
So the map $S^n\to B^{n-1}\times S^1$ is$$(x_1,\dots,x_{n-1},x_n,x_{n+1})\mapsto\left(x_1,\dots,x_{n-1},\dfrac{x_n}{\sqrt{x_n^2+x_{n+1^2}}},\dfrac{x_{n+1}}{\sqrt{x_n^2+x_{n+1^2}}}\right)$$
and uh I dunno we want the Jacobian of this or something?
if you write it as a difference of fractions you've got a sin^2 integral and a multiple of a two-higher sin^2 integral. there are recurrence relations for this stuff but maybe that isn't how you're supposed to approach it.
when i say sin^2 integral i mean integral of a positive integer power of sin^2.
maybe it's enough to know sin^(n+2) in terms of sin^n. hrm.
no matter. all of this went haywire when i discovered it was cosine down there. and this A^2 + B^2 stuff. what a goofy ride this problem is turning out to be.
my arm has finally stopped hurting from the covid vaccine.
my wife's uncle has a business that makes very accurate accelerometers. they have two uses, testing roller coasters, and something with testing helmets for the military. no other uses whatsoever.
apparently they're like orders of magnitude more accurate than any competitor, but there isn't a huge market for it, so they just have these really great graphs of acceleration if you ride a coaster or get RPG'ed.
the double pendulum ride would probably be rejected by an accelerometer test.
i loaned my book to someone who obviously figured out the value and has never returned it to me.
i hate maths. i see problems i have solved a million times over and suddenly have forgotten the trick. or a tiny modification to a nice problem turns intractable.
i do sympathize with forgetting tricks. i've forgotten so many tricks. i can usually remember that there is a trick, i just can't remember what the trick is.
Why can't there be a big North South East West marker on the canvas of a boxing ring, because then it would be clearer in your head if the radio announcer said something like "the Champ is circling Northeast."
Yep! My instructor is going to teach me about that in few days....But for now I wanted to stick to this algebraic method..But this algebraic method is yielding two equations..
The question is
$\int \frac{cosec^2x -2020}{cos^{2020}x}$=$\frac{Af(x)^B}{g(x)^{2020}}$+c where f($\frac{\pi}{6}$)=$\sqrt3$
Then, the value of $A^2+B^2+f(-\frac{\pi}{4}$)=....
I tried converting into tangent form , but it's not making sense further, nothing I am not getting any ideas I am now tr...
quick queston, if I have a parametric equation for a line, $x = l(a+1), y=lb$, how would I get an expression for the line in the form I am familiar with ie ($y=mx + b$?)
(i should probably have mentioned that $a, b$ in the parametric equation where not paramaters, that was kinda ambiguous
also evening @TedShifrin
I'm about to head to bed, and didn't get around to solvng the first excercise today
I took a misstep, wildly asserting that $t$ must be the projection of the vector from (-1,0) to (x,y) onto the y axis....this was naturally quite wrong, but I feel I learned something from my misadventure, and realise the question I need to ask myself is" given $t$ how do I get $x$ (or $y$)?"
sometimes people think of $\mathbb{R}^2$ as sitting inside $\mathbb{R}^3$ in that way. other times they do not. that type of identification is not automatic but it is often done.
in software you might get a type mismatch if you treated one as the other. or you might not. i can't think of a language that automatically pads lists with zeros to perform comparisons or arithmetic.
i thought she was playing with one of her toys, which i guess she was, in a sense. she really enjoyed it. i think she's got a spring in her step today because she's still on the high of the hunt.
we did keep her from killing a lizard once at the old house. very lucky lizard.
Can anybody please help me with cartesian product of a straight line with a circle? I understand (based on intuition) that it will be a cylinder with never ending axis but I don't understand how to deduce it using definition of cartesian product
If I take straight line L={((0,a): a \in \mathbb R }, C={(x,y): x^2+y^2=1 } and $L\times R={((0,a),(x,y)): a\in R and x^2+y^2=1 }$. I don't understand how to deduce that it's a cylinder from here.
not to be too formal here, but when you say 'is' a cylinder, you need a definition of cylinder. or to relax the meaning of 'is,' a map with sufficiently nice properties between that thing, and something that you agree is a cylinder.
i'd think of a as a (signed) "height" or something. one map would be from L x R to {(x,y,z) in R^3: x^2 + y^2 = 1} by sending (0,a),(x,y) to (x,y,a). the subset of R^3 is hopefully more clearly a cylinder.
and that map is a homeomorphism if everything is given the usual topologies
i would be as formal as is necessary for your context. whenever i say an object 'is' something, if it's not literally that thing and there is an implicit map in mind, i specify the map. but i'm very lazy with LaTeX. and it may not matter in your application.
in my first topology class i think a cylinder was just [something homeomorphic to] a cartesian product of a space curve and an interval or line. the curve might not even have to have been a circle.
A ruled surface (in $\Bbb R^3$) with rulings parallel lines. One usually refers to a directrix curve $\alpha(u)$ and then takes $x(u,v) = \alpha(u) + vA$ for $A\ne 0$ fixed.
the map i indicated earlier will be a continuous bijection with continuous inverse between your cartesian product (which unfortunately lives in R^4) and your favorite cylinder.
come to think of it, i don't know that there's one standard way to put a metric on a cartesian product of metric spaces. there are likely many choices, which are all equivalent in an appropriate sense.
this question is arguably missing a key piece of information, which is, what topology is put on the cartesian product (i.e. what does "open" mean for the cartesian product). if it is the usual "product topology," yes, more or less by definition.
if you have to route through metric spaces or something maybe the proof is more complicated.
yeah, it comes down to the definition. one definition is the topology generated by the sets U x V, with U and V open in the factors. there are other definitions.
I have to prove that:
$$Int(A\times B) = Int(A)\times Int(B)$$
Where $A\subset M$ and $B\subset N$, both $M$ and $N$ metric spaces.
The problem is that the exercise does not specify the metric, so I need to try to prove it using a generic metric.
If $x\in Int(A\times B)$, then an open ball ca...
I have to prove that:
$$Int(A\times B) = Int(A)\times Int(B)$$
Where $A\subset M$ and $B\subset N$, both $M$ and $N$ metric spaces.
The problem is that the exercise does not specify the metric, so I need to try to prove it using a generic metric.
If $x\in Int(A\times B)$, then an open ball ca...
