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6:10 AM
Hi @JohnRennie
 
@cOnnectOrTR12 Hi :-)
 
I have a simple question. When we find derivatives we take limits of deltaf/deltax. Value of Deltaf/deltax never stops at a value and goes on decreasing as deltax decreases. Limit of this is the value which we never get from deltaf/deltax. So is it right to give a value to dy/dx which it can never have?
 
You'd have to ask a mathematician. I believe that exactly how derivatives are defined in maths is complicated. I am just a physicist, and all I can say is that derivatives work i.e. they do describe what happens when we apply them to real physical systems.
2
 
Good to know :)
It’s silly but Who should I ask by the way? Anyone particular?
 
You should be able to Google this, though be warned that it will probably get complicated as when mathematicians get stuck into something they usually make it complicated :-)
 
6:24 AM
@JohnRennie such a relief
 
:-)
 
7:19 AM
lol
 
glS
8:02 AM
@Semiclassical in this case, I just meant an ON basis of states $u_k$ such that $|\langle \psi,u_k\rangle|=1/\sqrt N$ with $N$ dimension of the space
@Semiclassical by the way, I think this could be a nice question to ask on the site. You might get interesting perspectives on how to look at it
 
8:50 AM
@cOnnectOrTR12 From en.wikipedia.org/wiki/Mean_value_theorem In mathematics, the mean value theorem states, roughly, that for a given planar arc between two endpoints, there is at least one point at which the tangent to the arc is parallel to the secant through its endpoints.
 
9:45 AM
0
Q: Answering weak questions in comments

Roger VadimI encounter quite often questions that are based on poor research, misunderstanding basic definitions, and likewise. In my opinion such questions should be closed rather than answered. Nevertheless, it is my policy to give people a hint or send them to a relevant text, so that they can improve th...

 
10:00 AM
@PM2Ring ok! So why do we take limit of the slope as dy/dx. It is that value which we can never achieve by simply taking the slope.
 
10:11 AM
@cOnnectOrTR12 If f is a nice smooth, continuous function, then the slope of the tangent at (x, f(x)) is the limit of the slope of the secant from (x-h, f(x-h)) to (x+h, f(x+h)). And there will always be some c in the interval x-h <= c <= x+h where the tangent at c is parallel to that secant.
So as h approaches 0, the slopes of the secants joining x-h to x+h approach the slope of the tangent at x. In other words, when we know m, the slope of the tangent at x, then we know that the slope of any small secant in the neighbourhood of x is approximately equal to m.
This is very useful. It means that if f and its 1st derivative are continuous in the neighbourhood of x, then small changes to x result in approximately linear changes to f(x). More formally, $$f(x+\Delta x) \approx f(x) + \frac{dy}{dx}\cdot\Delta x$$
 
10:34 AM
0
Q: Helping out in comments, when the question is too weak

Roger VadimI encounter quite often questions that are based on poor research, misunderstanding basic definitions, and likewise. In my opinion such questions should be closed rather than answered. Nevertheless, it is my policy to give people a hint or send them to a relevant text, so that they can improve th...

 
@PM2Ring The slope of the secant will be our rate. This slope can go on to infinity and never stop at a value. Taking its limit and so assigning dy/dx a value can be approximately right but never exact because there is no exact value that exist.
 
@cOnnectOrTR12 "The slope of the secant will be our rate." Sorry, I don't know what that means. If the slope is infinity then the function grows without bound at that point, eg for $f(x) = 1 / x$ as $x \to 0$
"there is no exact value that exist." But that's not true. As I said before, in the region from x-h to x+h there's always a c where the tangent at c is parallel to the secant. And as h approaches 0, c & x merge. So the secant is guaranteed to converge to the tangent.
 
10:52 AM
Can you verify when we find deltafx/deltax then that is a rough rate between some point to some point. And taking limit means it is exact rate at a point
 
@cOnnectOrTR12 That's right. As Delta x approaches 0, the rough slope gets closer & closer to the exact slope at x.
 
“So the secant is guaranteed to converge to the tangent.“ How can the slope of secant converge to slope of tangent when we know there are infinite numbers between any two numbers
 
Geometrically, as we "zoom in" towards the curve, the apparent curvature gets smaller & smaller, and that section of the curve looks more & more like a straight line. Of course, no finite segment of the curve is truly straight, but we can make it as straight as we want by zooming in sufficiently.
@cOnnectOrTR12 It doesn't matter that there are infinite numbers. As we zoom in, the variations in the slopes of the secants always get smaller.
 
@PM2Ring yes! But it will never stop
 
...what do you think "converging" means?
the whole point of a limit is that it is a value you can't reach in a finite amount of steps, but you can get arbitarily close to it
 
11:07 AM
@ACuriousMind so the value which we can never attain is finally assigned to dy/dx. Even though it’s a very small difference Is this mathematically right?
 
what PM2Ring said is still right - the limit is the slope of the tangent at x
and it is also a fact that this slope of the tangent is the limit of the slopes of the secants
while we can't attain the derivative as the slope of a secant, you shouldn't think of it as some sort of fiction or assignment - it really is the slope of the tangent, it's not some weird "unattainable" value in principle, it's just unattainable as the slope of a secant
 
By the way converging means “still converging”
Not converged!
Thanks @PM2Ring
 
Let's try a simple example, $f(x) = x^2$. The slope of the secant from $(x, f(x))$ to $(x+h, f(x+h))$ is $$m= \frac{x^2+2xh+h^2 - x^2}{h}=2x+h$$. So as $h\to 0$ then $m\to 2x$.
And if instead we look at the secant from $(x-h, f(x-h))$ to $(x+h, f(x+h))$ we get $$m= \frac{x^2+2xh+h^2 - (x^2-2hx+h^2)}{2h}=2x$$. That is, the secant is parallel to the tangent at $x$.
So the slope is not some tiny bit different to $2x$, it's exactly $2x$
 
 
2 hours later…
1:12 PM
@glS yeah, I’ll put something together. Do you think it’d be better here on PSE or over on the QC SE?
 
 
2 hours later…
glS
3:15 PM
@Semiclassical it's up to you really. It would be on-topic on both sites. The main difference is that you'd probably get a more "informational" point of view on qc.SE, and a more "standard quantum mechanics" one here. Then again, many people hang out in both, so the difference wouldn't that large
 
3:35 PM
Right. So a bit of a coin flip.
 
 
2 hours later…
5:17 PM
Now -this- is a proper grad student whiteboard:
Note the “please, do not erase!”
 
5:33 PM
the random arrows are what makes it authentic
 
that's what handwriting is good for
 
5:47 PM
i like the "how did they get it? this is weird" marginal comment
also "why is it there???"
 
Sometimes I hava such comments in my notes
 
yeah
it's all too human
 
6:05 PM
worldline FOMO
 
 
1 hour later…
7:34 PM
0
Q: How to bring an old question back to the users' attention?

Xfce4The question we want to ask might be posted by another user years ago. It is possible that the question did not get any answers or the answers were/are not satisfactory. What should we do for each case? How can we bring an old question back to the attention of the community, i.e. make it appear i...

 
8:25 PM
 
I don't understand what Atsumata Honda is doing on page 4 figure 2
 
@Slereah $\mho$: You're doing circuit problems and you're too cool to use $\Omega^{-1}$
 
Is he doing some analogue of Riemann surfaces (Lorentz surfaces) in $\Bbb L^3$ where the space like curve can travel on the surface to reach the different parts of the surface
confused about what happens to the metric at the singularity loop - (where the four surfaces intersect in fig. 2)
my guess is that it degenerates?
and then the metric "changes" about the singularity loop which makes sense that it's called a type-changing metric
 

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