A spoiler implemented with toggles that doesn't toggle back to the original text: $\require{action}\toggle{\text{You will never see this again.}}{\toggle{1+1=2}{1+1=2}\endtoggle}\endtoggle$
@karl is there a reason lee defines smooth emveddings as smooth immersions that are topological embeddings instead of diffeomorphisms onto their images?
nvm
ignore that
@Ted The idea is simple, the symbol-pushing is unpleasant
I created a community ad for the "Puzzling" proposal (since, as I mention in the comments, you already have the related puzzle and recreational-mathmatics tags here.
Note, that currently the link marks me as the "referral person"; if anyone has a problem with that, please, edit the link to remove me from it. I simply felt it would be a shame letting the rep go to waste if no one is listed as the referrer.
Hi, off topic: If $0\le x/y < u/v \le 1$ I do not see how it can ever be the case that $yv-ux=1$. For if we cross multiply the first inequality we obtain $yv-ux>1. Am I correct?
I need a function whom graph is the following (the Franck and Hertz experiment graph):
Currently I wrote the following function:
function comptuteValue (value) {
var period = 4.9
, low = 6
;
function addOrSubstract (x) {
if (x < low) {
if (x <= p...
I made a root function that uses newtons method, on the wikipedia for newtons method is written: "the convergence is at least quadratic (see rate of convergence) in a neighbourhood of the zero, which intuitively means that the number of correct digits roughly at least doubles in every step.", however my algorithm dosn't have quadratic convergence, infact it takes it 125 steps to get to $\sqrt{10}$ with initial guess of 5
ok, I really don't know what algorithm I was using, now it seems to converge in less then a second for $\sqrt{5000}$, instead of the 5 minutes it took before, thanks @Mike
If you move around finitely many things, then the $s_n$ are eventually the same as they were before - so the series converges to the same value, like your intuition says it should.
Because $\sum |a_n|$ doesn't converge, the absolute values of the individual terms don't decrease "fast enough". That ultimately means that the "tail end" of the series - which is where all the action really happens! - doesn't just go to 0 as fast as we want. Fiddle with it too hard and it's not going to play nice back.
Show that I can rearrange something to diverge to infinity * And do you not get why I'm so confused-ish? It's so weird that you can *reshape* something like that. I mean… it's just infinities?
Oh, no, I understand. But there's a difference between not understanding the rearrangement theorem which is weird as shit, and not understanding conditional convergence.
sqrt(1450640651560125604123100321213254625408235072380235417052340238052341085231420520371820423052707250528702402052740270201230273152702701207250270527012702530527023702) is calculated to 117 digits after the comma in 4 seconds
@Mike do you know any good ways of calculating initial guesses for n^(1/x)
this dosent seem right 13210^(1/7) with newtons method... @Mike 1886.85714285714285714286 1886.71428571428571428571 1886.57142857142857142857 1886.42857142857142857143
@Mike my convergence slows down the higher roots i try to calculate
@Mike Halley's method apparently has cubic convergence, wouldn't using this one be even faster?
I have a function of the form $f(x) = \frac{1}{x^n - a}$, and I need to programmatically find the n'th and (n-1)'th derivative of the function. Since the function has this specific form and that the power is tied to which derivative it is, I was wondering if there's an easier method than simply t...
Is there a proof of $\pi_n(x) \sim \frac{x}{\log(x)}\frac{(\log\log(x))^{n-1}}{(n-1)!}$ of approximately the same complexity as the prime number theorem, where $\pi_n(x)$ is the number of positive integers $\le x$ who have $n$ prime factors?
A fun topic of debate for the day: Should something be considered mistreatment if the being in question has never known any different? Alternatively, how do mistreated beings react to proper treatment after having known nothing else?
Related topics: human rights, animal rights, plant rights (is that a thing?), Aristotle, Plato's "Allegory of the Cave"
So far I have: $$ \begin{aligned} \int \dfrac{1}{x^2 + 1} \text{ d}x & = \dfrac{1}{2i} \int \dfrac{1}{x-i} \text{ d}x - \dfrac{1}{2i} \int \dfrac{1}{x+i} \text{ d}x \\ & = \dfrac{1}{2i} \log \left| \dfrac{x-i}{x+i} \right| + \mathcal{C} \end{aligned} $$ I know that the integral of a real valued function is a real valued function so how do I take the real part of this final result?
@Sawarnik @skullpatrol Yo! Have you guys got any idea of how to proceed with my above integral? Also, which text are you reading @Sawarnik that helped you grasp the rigorous definitions of limits? :)
@KhallilBenyattou My book is the worst for understanding the defintion of limit! I mostly the idea learned it from Khan Academy, then by thinking upon it. My book just states the defintion and gives one proof using it :(
@KhallilBenyattou I haven't done any calculus with complex numbers! :/
@N3buchadnezzar $$\left(\frac{z^2-1}{z^2+1}\right)^n = -i^n \iff \frac{z^2-1}{z^2+1} = \exp\left(\frac{(2k+1)\pi i}{n}\right)\cdot i =: \zeta_k$$ Then you get $$z^2 = \frac{1+\zeta_k}{1-\zeta_k}, \; 0 \leqslant k < n.$$
@GabrielR. Not officially in school, the British A-Level is literally just calculus, vectors and complex numbers (and a bit of differential equations).
@Alyosha How far have you gone in calculus ? Do A-Level classes (is it still considered part of high school btw ?) approach calculus with rigor (I mean epsilon-deltas, proofs...) ?
@GabrielR. No, ocr.org.uk/Images/67746-specification.pdf this is the specification. It's very routine. It's still considered high school, in America the equivalent of 12th grade.
@Alyosha I reckon just over 2 months should be enough to crack it! I've done a few questions here and there, but they were pretty nice integrals and differential equations.
@GabrielR. Pretty much everyone does them. There used to be S-Levels (of which the STEP mathematics paper is a remnant of), which were for the best people, but they were stopped a few decades ago.
@Alyosha Thanks for the info. Yeah I took a special math course for advanced students when I was in high school two years ago. Now I'm doing this en.wikipedia.org/wiki/…
I need a function whom graph is the following (the Franck and Hertz experiment graph):
Currently I wrote the following function:
function comptuteValue (value) {
var period = 4.9
, low = 6
;
function addOrSubstract (x) {
if (x < low) {
if (x <= p...
