@orlp I wonder, though. I see the downvotes. Did you make those? (I wouldn't have if I were you; they're good faith efforts to answer your question. Unless they were definitely, glaringly wrong I wouldn't downvote.)
I only upvoted one (for a clever protocol), and I commented on it to that effect, because even that one I am not convinced the numerical answer is correct.
@Faust I am guessing that the complex calculus course teaches you a lot of computations without the proofs, and the complex analysis one teaches you the proofs.
@orlp I've made some further comments on the answers. In particular:
I think this can be further improved. For example, it can be agreed that the symbols $0$ and $1$ shall never be used to encode a wrong bit. This is possible because the other wrong symbol can always be used. So for example, if the next bit is a $1$ but Alice only has $0$ and $2$, she can send $2$. And then Bob can be certain of ALL $0$ and $1$ symbols, regardless of block size. Something more can surely be done with this info. — Wildcard6 mins ago
well thats what he claims math 301 covered and i got an A+ in it but i have no analysis pre-req hes offered to waive it im just trying to decide if i can get by
i can manage the definition and maybe even prove certain things are but only by leaning on theorems maybe i just watch it too many interesting topics not to go.
(i) ac-method if a quadratic can be factored with integers, (ii) quadratic formula if it can't, (iii) factoring by grouping if there's four terms and it's nice, otherwise for less nice but still elementary higher degree polynomials it's (iv) rational roots theorem + synthetic division. (v) plus there lots of special case formulas (which are just shortcuts), like difference of squares, difference of cubes, sum of cubes, perfect square trinomials
let's try 6x^2-x-2. here, ac=-12. list out all pairs that multiply to -12: 1,-12 2,-6 3,-4 4,-3 6,-2 12,-1
now find the pair that adds to the middle coefficient, -1. that pair is 3,-4. therefore we can split the middle term -1x into 3x and -4x, after which we factor by grouping
Let $a$ and $b$ be real numbers. The complex number $4-5i$ is a root of the quadratic $$z^2 + (a + 8i) z + (-39 + bi) = 0.$$ What is the other root?
How would I go about finding $a$ and $b$ or finding the other root? I'm totally lost on this problem. I'm currently a student in pre-calculus. T...
you can verify easily that (i) and (iii) are solutions, then use them to write down the equation of the secant line thru them, then verify if (ii) and (iv) are not on that line
the vertex of the parabola is inside the circle (something you can verify by hand) which guarantees there should be only two points of intersection
== Translingual ==
=== Symbol ===

(computing) The object replacement character, sometimes used to represent an embedded object in a document when it is converted to plain text....
"We have now thoroughly discussed non-principal ultrafilters, interpreting them as voting systems which can extract a consistent series of decisions out of a countable number of independent voters."
Is there a term/phrasing to describe a magma that satisfies a given set of axioms but does not require additional structure? Something like "at most" or "maximally" ___
I know it's imprecise but wondering if there's some informal jargon for this
@EricSilva So suppose $M$ is a complete Riemannian manifold and $p \in M$ is a point. The critical locus of $\exp_p$ is exactly the conjugate locus $C(p)$ of the exponential map - the intuition I have was this:
Take a "polar chart" on $T_p M$ with lines $t \cdot v$ coming out of the origin for $t \in [0, 1]$ where $v := v(s)$ is a curve in $T_pM$ with $v(0) = v$ and $v'(0) = w$ for $s \in [-\epsilon, \epsilon]$ and we exponentiate that in $M$ below to get a parametrized surface $f(t, s) = (d\exp_p) tv(s)$ below. $q = \exp_p v$ is conjugate to $p$ if $d/ds f(1, 0) = 0$
So usually one would expect that this literally means the geodesic $\gamma(t) = \exp_p tv(0)$ merges with a bunch of nearby geodesics at $q = \gamma(1)$, so that the "spread" of the geodesics near $q$ is 0.
I was looking for a scenario where that doesn't happen. I guess it suffices to look at examples where $f(1, s) = \exp_p v(s)$ looks like a $y = x^3$ curve in the base $M$?
So that we end up having a critical point at $s = 0$ anyway, but it doesn't have the "merging" behavior
@Balarka I had a problem about this at some point and I recall that my solution was a deformed 2 sphere that I got by multiplying the standard metric by some ad hoc function
@Daminark no, not particularly, but I wanted to work out by hand what happens to all the usual geometric quantities under Ricci flow and it took a while to bash out all the formulae
@EricSilva I think I see what the picture of the exponential map should be like in that case; I'm more or less content with that unless I happen to find an example in a manifold I know and adore
4th year I haven't thought about much. Algebra for sure, logic if I don't do it this year, and some kind of algebraic topology. Plus that computational/metric geometry class at TTIC
@skullpatrol It seems to be private rather than deleted
2
When you try chat.stackexchange.com/rooms/20352/math-mods-office you get: "The room you are attempting to access is private. If you believe that you should have access to this room, you can request access to it."
so far, I have both absolute sum $|x+y|$ and absolute difference $|x-y|$ commutative over $\Bbb R$, but not associative or admitting an identity element or invertibility
Hi, I was wondering if I could find some help here. I am trying to calculate the small axis of an ellipse where I know the coordinates of the left and right points, plus another points that belongs to the ellipse.
Well, you should be able to find them from this. They are the points with the same combined distance to each of the three points you have. They are also on the line between the two end points
@PVAL-inactive That sounds really great! I had a small taster of a Ledaig once, it was a nice smokey whisky that also had some nice flowery notes, sorta reminiscent of Highland Park. A lot better though :) Also it wasn't cask strength so I'll assume that's going to be even better!
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