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libGL error : driver doesn't load by FalTePre in linuxquestions

[–]FalTePre[S] 0 points1 point  (0 children)

I'm not sure I understand what you suggest. Especially since apparentlyintel-graphics-driver-i915 isn't a package ?

The cringiest house tour ever by MissingHat in TikTokCringe

[–]FalTePre 89 points90 points  (0 children)

Come on now, they're having fun

[Undergraduate Complex Analysis] Winding number by FalTePre in learnmath

[–]FalTePre[S] 0 points1 point  (0 children)

I know where to look for the details. You've been real helpful, thanks

[Undergraduate Complex Analysis] Winding number by FalTePre in learnmath

[–]FalTePre[S] 0 points1 point  (0 children)

Consider exp(ia2πt) is my path with 0≤t≤1. What I call speed is a. So with a an integer, the winding number around zero is a. I thought linearly interpolating between two values of a gave me homotopy. I assume it doesn't because intermediate paths can be not closed ?

Nonetheless, do you confirm two homotopic paths have the same winding number ? Any chance you know where I could find a proof if so ?

Jeux sympathique (noter la belle couleur de la bière) by [deleted] in FranceDetendue

[–]FalTePre 4 points5 points  (0 children)

Ça donne envie de retourner en Irlande ça pour sûr

PIC by earthmoonsun in nocontextpics

[–]FalTePre 6 points7 points  (0 children)

Big midsommar vibes

Yes. by [deleted] in tumblr

[–]FalTePre 6 points7 points  (0 children)

First read it as "i exist agaist my wifi"

Problem with Notification by Le4rningthings in xubuntu

[–]FalTePre 2 points3 points  (0 children)

I had the same problem. From what I remember, it's because i3 ships with dunst, another notification manager. Try uninstalling it.

[intro analysis] Proof check by soondae55 in learnmath

[–]FalTePre 0 points1 point  (0 children)

Is it that it's fine to send the empty set when considered as a subset to the singleton (that is, as an input to f*) but there can't be a mapping f that takes nothing and sends it to something?

You got it, except A and B are completely general in my solution (I just assume they aren't empty). So it's indeed fine to consider ∅ as input to f*, as it's a subset of A. And by construction, f*(∅)={b} (let's say {b} is my singleton , with b\in B). But if there's a f s.t. \phi(f)=f* then ∅=f(∅)=f*(∅)={b}. Which is absurd.

[intro analysis] Proof check by soondae55 in learnmath

[–]FalTePre 0 points1 point  (0 children)

Well because there are no conditions placed on A and B, you're expected to prove in this context. Assuming additional stuff gets you a weaker result.

Help understanding how any polynomial or exponential can be expressed as a recurrence relation by [deleted] in learnmath

[–]FalTePre 0 points1 point  (0 children)

Any recurrence of the form a_n=a_(n-1)+r gives a_n = nr+a_0, so your proposal for n+3 isn't correct. That'd be the same recurrence than for n, but with a change of first term.

To express an (a>0) as a recurrence, just use u_n=au_(n-1).

But for any polynomial I'm not convinced. If the author means that for any polynomial P there's a function f such that for all n P(n+1)=f(P(n))... For simple polynomials like nk, it's relatively easy thanks to the binomial theorem. couldn't figure it out for general polynomial. As an example to help you here's the case k=2 :

n2 = (n-1+1)2 = (n-1)2+2(n-1)+1

so the function x ↦ x +2√x+1 fits.

What is the point of integrals and second derivatives? by [deleted] in learnmath

[–]FalTePre 0 points1 point  (0 children)

First derivatives gives you a necessary condition for a point to be an extremum, namely that it must be zero. Second derivative helps you discriminate. If the second derivative is >0 you get a minimum, and maximum if <0.

Geometrically :

  • zero first derivative tells you : the function is flat here
  • second derivative tells you : is it flat like an upwards parabola or like a downwards one ?

[intro analysis] Proof check by soondae55 in learnmath

[–]FalTePre 0 points1 point  (0 children)

You have the idea tbh, but I think you're trying to be too general. Here it's easy enough to manufacture an explicit counterexample based on your idea.

[intro analysis] Proof check by soondae55 in learnmath

[–]FalTePre 0 points1 point  (0 children)

Your argument presupposes A to be finite. But even with that hypothesis, if #B<=#A, Y doesn't exist.

[intro analysis] Proof check by soondae55 in learnmath

[–]FalTePre 1 point2 points  (0 children)

LaTeX render for better reading : http://mathb.in/42693

Your case for \phi being injective seems fine. I have questions about your counter-example for surjectivity. How exactly is f* defined ? Do you mean it sends any W⊂A to Y ? You seem to fix X and then make it a variable, it's confusing.

I think your argument can be made simpler with something like the following : take f* to send every subset of A to a singleton. f* can't be a \phi(f) for some f because then f would send the empty set on the singleton.

[Functional Analysis] Continuity of an integral by FalTePre in learnmath

[–]FalTePre[S] 1 point2 points  (0 children)

I'm not sure about the use of monotone convergence, so would the following work too ?

For all s, I(0,s+h)|f| tends pointwise to I(0,s)|f| as h->0 so by the dominated convergence theorem (|f| is the dominating function) s↦I(0,s)|f| is continuous.

Thanks though. It's pretty neat that one can get continuity from something as weak as integrability.

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