I should really do something exercise-like, my neck has been bothering me quite a bit in the last few days. What's a sport that's not so "sporty"?
@MartinSleziak Unfortunately, I have that already I think. Because my finger slipped on one of my first days on SE. I removed the vote immediately but the badge stayed.
There's a slight difference (< vs. $\le$), not sure if that is important. If that matters much, then we should close the other question leaving this one open.
@tb Yes; I'm just being safe after seeing Andre's comment in the other question. I think there'll always be some disagreement unless the post is an exact duplicate.
Is there a way to tell Google to please not tell me about links from some websites by default? I'm having absolutely no use for mendeley.com or dict.cc or all of those wiki scraper sites, for example.
@tb Suppose you use the Firefox search bar, just change the setting of the Google query to automatically include the "not in site ..." as in the custom search.
@tb Yes. When you do a search you can hover over the hit you want to block. Then "Block xyz.com" appears and you click on that. It will be stored to your search settings.
@KorganRivera I didn't answer any of your questions, but it would be nice if you marked some the answers you got as accepted if they were helpful: see here how to do it.
Hmmm... 30 more questions on AC, I just need to add answers to 4 existing ones, then I might hit silver specialist together with the Bronze one. That would be impressive.
But if you don't want to have a Google account: Chrome is really good. Much faster than FF (I can't run FF on either of the laptops here) and more reliable (it restores your previous session if it crashes with 100% certainty).
And Chrome has a Personal Blocklist extension which does what you want.
: )
Of course you already have a Google account but Chrome is cool : )
Chrome is very good. I resisted it for a long time in favour of FF. But FF started stalling every 20 mins after an update and I switched. I don't regret it.
@Matt well, I don't do that on a daily basis and I have some rights there. And they don't ask me to share my entire life with them because it's oh, so convenient.
Guys, I feel a little defeated in trying to understand infinitesimals. I'm sure you all think this is hilarious. But if I can't understand this, then I'm yet again stalled. How did you guys come to terms with them, later in your studies?
do you know the history? Calculus was invented based on the notion of infinitesimals. There were serious logical difficulties found in it, and a new theory developed based on limits. In modern times using some quite deep ideas from logic a new rigorous theory of infinitesimals was created.
In terms of pragmatics, everyone uses the limits theory of calculus - and it's very good. The infinitesimal theory is slightly useful but it's quite esoteric.
@QED No. This is my question as best as I can put it: I understand that lim_{x->a} f(x) = f(a), but then to say that the gradient of the tangent curve is some value, is like saying that when x=a, then f(x) = f(a). The whole point of the limit, I thought, was to say, instead, that we don't know what f(a) is, but we can say that it approaches some value.
But saying the gradient of the tangent is a value, is different that saying it approaches some value. So, I'm having a problem with understanding that the limit is the value.
@KorganRivera, lim_{x->a} f(x) = f(a) is a theorem for continous f
@KorganRivera, the function you are taking a limit of, when you compute derivatives, is not continous
that function is not 'f', for example if f(x) = x, then to find the derivative of f at x=1... you need to compute lim_{h -> 0} (f(1+h)-f(1))/h.
so you're actually finding lim_{h -> 0}g(h) where g(h) = h/h. and that is not continous at h=0
@KorganRivera, Doesn't the theorem that if lim_{x->a} f(x) = A and lim_{x->a} f(x) = B then A = B answer that confusion about whether the thing really "is" or is not
@QED So if I'm working with function f, and f is continuous, my derivative dy/dx is by definition not continuous, since it is undefined at dx=0. Is that correct so far?
that way you can leave a message there, and even if nobody is around - they can see it when they join (rather than here, where it gets scrolled away by the other chat)
I have problem with showing that the limit of the following function
$$\frac{
\sqrt{\frac{3 \pi}{2n}} -
\int_0^{\sqrt 6}(
1-\frac{x^2}{6}
+\frac{x^4}{120})^ndx}{\frac{3}{20}\frac 1n \sqrt{\frac{3 \pi}{2n}}}$$
equal to $1$, with $n \to \infty$.
@QED When I said, "So if I'm working with function f, and f is continuous, my derivative dy/dx is by definition not continuous, since it is undefined at dx=0." I guess what I'm saying is that (f(x+h)-f(x))/h is not continuous since it's not defined at h=0.
@KorganRivera There are lots of things wrong with that: dx=0 is wrong. dy/dx - what/s y? "dy/dx is by definition not continuous" it's not a function how can you ask whether or not it's continous, ... etc.
In general this stuff with 'dy/dx' is supposed to help as some kind of memory aid, but since there's no rigorous mathematics behind it - all it's going to do is confuse people
in fact there was a big controversy about it since using it in obvious ways suggested by the notation leads to wrong results
@QED I'll work on trying to understand that the gradient of the tangent is the limit, rather than the gradient of the tangent approaches the limit. I'll read your proof. Thanks for your help. I think I just need some sleep. O_O
@KorganRivera, note that I never said anything about gradients or tangents.
@KorganRivera, you can do absolutely everything here without it: that stuff is just a geometric picture that's suppose to help people visualize what's going on
@NikhilBellarykar Either way, don't highlight everyone and ask them to check out some link. If you have a specific user which you think can say something in particular feel free to highlight them; you may also address "to all", but don't highlight several people like that.
@NikhilBellarykar No. I know what the link is. I have no idea why I am looking at it, what should I do about it, and frankly I have enough as it is. I use this chat to vent, not to exercise my better judgment.
Mmmm, so a number ending in 6 can only end in even digits, a number divisible by 7 can end in all digits, but a number divisible by 5 can only end in 0 and 5
I understand that. What I'm trying to say is that I've realised that if I have two functions f(x) and g(x) and g(x) = (f(x+h)-f(x))/h, then lim_{h->0}g(x) = L = f'(x). Is that also wrong?
@QED So now it makes sense to me that the derivative is the limit. What I think I was doing in my head was saying to myself that g(x) isn't continuous at x=h so how can I evaluate g(h)? But that's not what's happening. The derivative is the limit, not g(h).
while for continous functions f, you can simply evaluate lim_{x -> a} f(x) = f(a)... For discontinous functions, the limit may or may not exist and computing it will require more work
@KorganRivera, in that case you'll need to be proving $\forall \varepsilon > 0,\,\,\,\, \exists \delta,\,\,\,\, \forall x,\,\,\,\, 0 < |x - a| < \delta \implies |f(x) - L| < \varepsilon.$ by picking some correct L (somehow)
This is good. I've learned that a function that is differentiable at all points is continuous, but a continuous function is not necessarily differentiable.
Hey guys, I have a short question a friend of mine asked me which I cannot answer because I have not learnt about measure theory (or whatever is needed to answer the question) yet. He asks what is wrong with \int_0^{2 \pi} \frac{d}{dn} e^{inx} dx when he applies Lesbegue's dominated convergence theorem, because apparently, if he first integrates and then derives, the result is 0 but if he first derives and then integrates it's not 0. Does anyone know?
I appreciate you guys helping with my noob questions. I don't get to talk to anyone about this stuff, just study on my own from library books and online. So thanks :)
I have problem with showing that the limit of the following function
$$\frac{
\sqrt{\frac{3 \pi}{2n}} -
\int_0^{\sqrt 6}(
1-\frac{x^2}{6}
+\frac{x^4}{120})^ndx}{\frac{3}{20}\frac 1n \sqrt{\frac{3 \pi}{2n}}}$$
equal to $1$, with $n \to \infty$.
