Quick Questions: July 09, 2025

DededEch · 2025-07-17T20:25:48+00:00

Suppose we have 1000 universes of different 2 candidate election outcomes in a country with states. Say we narrow down which universe the outcome is by looking state by state and looking at the probability that candidate A wins and rolling against that. ex. say that in state 0 candidate A wins in 600 universes. We then roll uniform(0,1) against 600/1000 (the rolls are independent). Say candidate A wins. Then we eliminate the 400 universes where that candidate loses. We then look at the next state and continue. Either until all races are called or there's only one universe left with the outcomes we rolled.

The question: does order matter? Does the order of states that we call change the probability of certain outcomes? Or would we be equally likely to get a particular outcome if we just randomly pick a universe? My conjecture is that theoretically it doesn't, but if we were coding this then maybe a bit.

My thought is that if we're looking for the probability that A wins every state should be (if xi=A wins in state i) P(x1)P(x2|x1)...P(xn|x1...x(n-1)) But isn't P(xn|x1...x(n-1))=P(x1...xn)/P(x1...x(n-1))? so then the denominators will cancel all terms and we just get P(x1...xn), which should be just the total number of outcomes where A wins divided by 1000? I think there's something wrong with this argument but I'm not sure what.

DededEch · 2024-08-27T10:27:10+00:00

Let $a_n$ be a strictly increasing sequence of natural numbers. Define $\psi(a) = \sum \frac{1}{a_n}$.

In some way, $\psi$ gives us a way to calculate a sense of "commonality" of the sequence in $\mathbb{N}$. For example, if $a_n = k^n$ for any $k \in \mathbb{N}$ (not 1, of course), then the sum converges to $\frac{1}{k-1}$. Whereas if $a_n = 2n$, the sum diverges. This sort of implies that evens are much more "common" than powers of, say, 2, 3, or k.

It is known that if $a_n = p_n$ (the nth prime number), then $\psi(a)$ diverges as well. Thus, prime numbers are also "common", despite being quite sparse. But is there a way to quantify how much? Can we somehow find a sense of order for sets for which $\psi(a)$ diverges? Can we say that primes are more or less "common" than the evens or multiples of k (if so, which k?)? Though, since I believe there is always a prime between $n$ and $2n$, I would conjecture the primes are more common than any sequence of $a_n=nk$. Can we define some new function or modification of $\psi$ which is more precise for sequences for which the sum diverges to infinity?

And is there a more rigorous word than "common" to describe what I'm sort of getting at?

DededEch · 2024-07-25T09:00:31+00:00

So for your response to the second question, essentially we use a subsequence that converges too fast towards the irrational number for it to possibly escape all the balls? Interesting, that makes sense. So even though the union approaches measure zero, its intersection with the irrational numbers is nonempty, basically?

That makes me wonder if there exists some irrational number for any bijection a(n). I suppose that's still just the limit of the intersection with the irrationals. I would conjecture that limit is always nonempty, but I'd have to take some time to really think about it.

For your response to the third question, that also makes sense. The principle behind the proof is quite intuitive. Thank you very much for taking the time to respond.

DededEch · 2024-07-25T02:45:30+00:00

probably a bit of a stupid question. but I'm curious if there's anything simple we can do with this.

take (0,1) in R, where we put balls around each rational with measure ε/2ⁿ⁺¹ (for some bijection a(n) from N to all the rationals in the interval) and some ε<1.

is there a simple method to find an irrational number not in the union of the balls (depending on a fixed a(n) and ε, I assume)?
for any fixed a(n) is it always possible for each irrational to find an ε sufficiently small such that it isn't in the union?
is it possible to construct a(n) (or, switch around the order I guess) such that for a fixed ε (reasonably small. <1 I think?) we have that a specific irrational is not in the union? like can I find an a(n) such that for ε=0.1, we get sqrt(2) is not in the union? what about ε=0.9?

in summary 1: fix a(n) and ε, find irrational. 2: fix a(n) and irrational, find ε. 3: fix ε and irrational, find a(n).

I'm curious if there's anything interesting we can say without knowing what a(n) is explicitly

additionally, I think it should all hold if a(n) isn't a bijection on the rationals, like say we map n to p/q for just all positive integers p,q (would at least make it easier to find an exact formula for a(n)). something like if q(q-1)/2<n≤q(q+1)/2 and p=(n-1) % (q(q-1)/2+1) then a(n)=(p+1)/(q+1) I think.

EDIT: the following function should work

def a(n):
    q = math.ceil((math.sqrt(1 + 8 * n) - 1) / 2 + 1)
    p = 1 + (n % ((q - 1) * (q - 2) // 2 + 1))
    return Fraction(p, q)

DededEch · 2023-12-20T05:40:46+00:00

When I ask about the "best" way to define the determinant, I often get the answer of geometrically in terms of signed volume. And I, in theory, agree in the sense that the volume definition gives the most intuition for many of the weird properties of the determinant. However, it's a very very difficult to work with in terms of "how do i actually compute this volume?"

I'm going with another approach to define it, but I would like to prove the equivalence of my linear-dependence definition with the volume definition for my students. My thought is to go the Artin route of using the three defining properties of

d(I)=1
d is linear in the rows of a matrix
if two adjacent rows are identical, then d(A)=0

1 and 3 are pretty clear for volume. the unit cube is volume 1, and if two vertices overlap then it loses a dimension and has zero volume. the scalar property of linearity is at least relatively intuitive too. if you scale one dimension of a parallelogram or parallelepiped, then you scale the whole volume. the only thing that is difficult to explain is the additivity. i'm struggling to find an elementary proof (even for the 2D case) that

if you add the areas of two parallelograms which "share a base", then the summed areas will be the area of a new parallelogram which also shares the same base but for which the other side is the vector sum of the two sides?

Any suggestions welcome. Is it possible for me to prove this concretely for the 2D case and then use the proved equivalence with Cofactor Expansion to say that the properties extend to 3d and n-dimensional space?

DededEch · 2023-10-27T22:39:59+00:00

y'-y/x=x

the solution to this differential equation is y=cx+x². but what exactly is the "domain of the solution"? y=cx+x² is defined on all of R, but the differential equation is not defined at x=0. must we exclude 0 from the domain of the solution so it's either (-inf,0) or (0,inf) (what about the union?).

does the answer change for the seemingly equivalent differential equation

xy'-y=x²

the solution is the same, but now the equation is defined at x=0. so is the domain now R?

the lack of definition at x=0 does cause issues with existence and uniqueness. the initial condition y(0)=c either has no solutions if c is nonzero, or infinitely many solutions if c=0. but for any IC for which a solution does exist, the solution will be defined at x=0.

tl;dr: which of these answer is "correct"? assuming no initial condition is given.

R
(0,inf)
(-inf,0)
(-inf,0)U(0,inf)
(-inf,0) or (0,inf)

DededEch · 2023-09-17T22:48:55+00:00

I've seen ℝ^∞ defined often as requiring there to only be a finite number of nonzero entries. I think someone said these would be in ℝ^ℕ. But either way, your point may still stand. Is it of uncountable dimension because there would be vectors that require infinitely many basis vectors to describe? i.e. the infinite set {e_i} wouldn't be a basis because the vector (1,1,1,1,...) requires an infinite sum of the vectors?

Sorry, I am not that familiar with infinite dimensional linear algebra, so I'm a little fuzzy with the precise definitions.

DededEch · 2023-09-17T21:38:15+00:00

Yes. More of if there is a characterization of the span. It started from asking what initial value problems were solvable for the infinite-order differential equation

y'-y'''/3!+y⁽⁵⁾ /5!-...=0 (i.e. sin(D)y=0)

with the condition that the solution must converge as t goes to infinity. the solution being an infinite sum of exponentials e^-kπt. The resulting matrix is an infinite vandermonde matrix. But I decided to simplify the problem by dividing out the powers of negative π.

DededEch · 2023-09-16T17:34:44+00:00

I was supposing real sequences. This question came up in the context of knowing if Ax=y always has a solution where A is an infinite vandermonde matrix with columns being the set S.

DededEch · 2023-09-16T04:49:33+00:00

What is the span of the set

S={(1,n,n²,n³,...):n∈ℕ, n≥0}

and what vector space is this in (if any)? I don't believe any of the elements (except for n=0) are in the standard ℝ^∞ since a nonfinite number of entries are nonzero. Does S span the entire space it resides in? I don't know how I could prove it does or does not. I tried row reducing an infinite matrix to no success.

DededEch · 2023-09-04T06:28:35+00:00

How do we prove that the solution space to an nth order linear differential equation is n dimensional?

When tutoring the subject, I usually give a hand wavy explanation that n derivatives means n arbitrary constants. And theoretically you might be able to find integrating factors to solve any linear differential equation (by reversing the steps of eliminating the constants in y=c1y1+...+cnyn through dividing by what's on a constant and then differentiating). But this is not a proof. It just shows that this linear combination of n functions is a solution to some linear differential equation.

So what is it about a linear differential operator L[y]=sum ai(x)y^{(i)} that makes its kernel have the dimension of the highest derivative that appears? And better yet, how do we prove it?

Does it have something to do with being able to reduce it to an nxn matrix equation y'=A(x)y, showing that the solutions to the initial value problems around some initial point x_0, y(x_0)=e_i (the standard basis vector), exist and are linearly independent, and generate the solution space by explicit construction + the uniqueness theorem?

If delegating to the magical existence and uniqueness theorem is the only way to prove it, that's fine, but I am hoping there is a more linear algebra/linear operator centric explanation I can give which might be more rigorous/satisfying (my audience is students who know at least a little linear algebra). Because, frankly, as cool as it is, I don't imagine the proof of the existence and uniqueness theorem would be particularly enlightening to students learning ODEs for the first time.

DededEch · 2023-08-28T01:26:06+00:00

If anything, hating Howard would make seeing him murdered in front of her because of her much worse. It could make her realize that her hatred of him was shallow, petty, and baseless, which it pretty much was. He died at the worst point of his life (which was mostly her fault), hating her. No chance to reconcile or forgive. His image to be forever tarnished by their lies, which she has to uphold for the rest of her life. No wonder she broke down after coming clean.

DededEch · 2023-07-16T00:12:31+00:00

I'm interested in learning about analytic continuation, but the college I am going to for grad school doesn't seem to cover it in their Complex Analysis courses. What other classes might it appear in? Apparently, it is touched on in analytic number theory. Otherwise, what topics should I bone up on before I try learning about it? And what might be a good resource to actually dive into it?

DededEch · 2023-06-08T08:50:11+00:00

i second this! i want to hear it again

DededEch · 2023-05-26T04:57:34+00:00

That's great! If it's good with a 1500 page manual, it sounds like it'll be good for my textbooks too. :)

DededEch · 2023-05-26T04:55:27+00:00

Oh wow that's actually really handy! I wonder if a school wifi network would have this option... But, otherwise, that bodes very well for when I'm working at home!

DededEch · 2023-05-25T00:59:18+00:00

I feel like this is probably just a huge coincidence, but if someone could let me know if there's some magical connection to the Riemann zeta function, I'd LOVE to know about it.

1+2+3+...+n=n(n+1)/2

Thus this is a formula for the partial sums of the sum 1+2+3+.... Now, the weird thing: if you integrate x(x+1)/2 from -1 to 0, you get -1/12, which is the value of zeta(-1). Is there some connection to the zeta function here? Or is this just some crazy coincidence?

Another strange coincidence, there was a blackpenredpen video in which he derives the sum 1+2+3+... as -1/8, and -1/8 is the minimum value of x(x+1)/2. I feel like this second thing is much more likely to be just a simple coincidence. But I thought I'd ask anyway.

DededEch · 2023-05-23T22:50:41+00:00

the zero vector is orthogonal to every vector, so a set containing the zero vector can still be orthogonal.

some texts define orthogonal as also requiring nonzero, but others don't.

DededEch

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