The h Bar

user228700

6:03 AM

@JohnRennie I'm looking forward to it anyway :-)

Arguement version:
Given a function $f : A \rightarrow B$ the domain can always be partitioned into the following disjoint sets: $\emptyset$, $I=\{x\in I | f|_I \text{ is injective } \}$, $N=A-I$. Similarly the range $B$ can always be partitioned into the following disjoint sets: $\emptyset$, $J=\{y\in J | f^{\leftarrow}|_J \text{ is injective }\}$, $K=\{y \in K | f(A-I)=y\}$ and $S=B-K-J$

That is: $A=\emptyset \cup I \cup N$ and $B=\emptyset \cup J \cup K \cup S$

user228700

@MAFIA36790 I will, thanks :-) I've another question and I swear I won't keep bothering u all day, but um, "Consider a conical pendulum having a bob of mass $m$ suspended from a ceiling through a string of length $L$. The bob moves in a horizontal circle if radius $r$." What part of this implies that the bob is moving with constant speed?

Secret

Now for any $f$, $\emptyset$ always map to itself. If $g$ is the left inverse of $f$ then $g \circ f = \text{id_B}$ Now since $f$ preserve unions, $g \circ f (A)=g \circ f (\emptyset) \cup g \circ f(I) \cup g \circ f(N)$. The first term is empty as mentioned previously, For the second term, it is I. However for the third term in order for $g \circ f$ to be the identity in B, $g$ cannot be be a function as it will be mapping 1 to many. Therefore if $g$ is the left inverse, $N$ must be empty

Since $N$ is empty, f is injective

John Rennie

6:19 AM

> The bob moves in a horizontal circle of radius $r$

There's a clue there :-)

Secret

Now if h is the right inverse, then $f \circ h (B)=f \circ h (\emptyset) \cup f \circ h(J) \cup f \circ h(K) \cup f \circ h (S)$. The first term is again empty. For the second term, it is J. For the 3rd term, it will be some subset of $N$ which maps back to $K$ by $f$, thus it is ok. For the 4th term, the preimage of $S$ is empty, hence $f\circ f^{\leftarrow}(S)=\emptyset \neq S$. Therefore in order for $f \circ h = \text{id}_B$, $S$ has to be empty. Thus f is surjective

Typo: The 2nd previous message $\text{id}_B$ should be $\text{id}_A$, and "identity in B" should be "identity in A"

user228700

@JohnRennie Hm, OK...

John Rennie

@KaumudiHarikumar circular motion?

I suppose we also need to stipulate that the force is constant, but then the Earth's gravity usually is :-)

Secret

Disjoint sets can become intersecting after the mapping, that is why f does not preserve intersection and difference

ACM's example of $f: \{0,1\}\rightarrow 0$ illustrate that concretely

I have a analytic attitude about counterexamples: I don't just find a counterexample and showed it is indeed a counterexample, but also try to understand the underlying reason why it is a counterexample. For me, finding just one counterexample is sufficient but not perfect, I always aim for the set of all counterexamples of a given proposition and then choose the most pathological possible one out of it

2

A: Ideas of finding counterexamples?

My favorite approach in finding counterexamples whenever I am not familiar enough with the mathematical object $M$ is a top down approach. Start with the most general example of $M$, and the find the condition that all counterexamples of M has to violate, then pick one concrete counterexample fr...

While always works, it is insanely time consuming (depending on how many parameters you need to fiddle with in the mathematical object), thus we should stick to educated guess and trial and error unless absolutely stuck

user228700

6:57 AM

@JohnRennie Well, yeah, but there's tension as well and there's nothing to say that those two forces are equal in the vertical direction...

John Rennie

@KaumudiHarikumar I suspect you are over analysing this :-)

user228700

@JohnRennie Yeah, that's me :/

user228700

Unless I know that the velocity is constant. I'm finding it a little difficult to believe that simply because the radius is constant, the velocity is also constant. Why should that be? It can traverse the length of the same circle with a smaller(/bigger) time period, no?

John Rennie

Well is there any component of any force that acts tangentially to the circular motion?

Secret

It seems the left inverse has to be unique, but not necessary for right inverses. Might check this with Acuriousmind when I rewrote the diagrams into statements and set notations...

user228700

7:01 AM

@JohnRennie Yeah! Wait, oh no. No, no, there isn't. Well, there is, but not in the direction that will tend to increase/decrease its speed in the horizontal plane. I've got it. Thanks! :-)

Secret

7:18 AM

http://thepasqualian.com/?p=1113
Ok fine, you can map the non surjective part of the range back to anywhere in the domain and not affecting the identity map. I really have to spend more time thinking about "do nothing techniques" like this

DanielSank

Hi, everybody.

user228700

@DanielSank Hello! :-) Is it the weekend yet, for u?

DanielSank

7:38 AM

@KaumudiHarikumar Yes, officially.

It's past midnight on Friday evening.

Secret

7:59 AM

user228700

8:12 AM

@DanielSank Ah, I see :-)

Secret

Dictionary order means that which set corresponds to the first argument in the cartesian product matters on what order type is obtained

One of the challenges in visualising a matrix lies on having a reliable representation on $\mathbb{R}^2$ that reflect this ordering of the axes

Fence (mathematics)

In mathematics, a fence, also called a zigzag poset, is a partially ordered set in which the order relations form a path with alternating orientations: a < b > c < d > e < f > h < i ... or a > b < c > d < e > f < h > i ... A fence may be finite, or it may be formed by an infinite alternating sequence extending in both directions. The incidence posets of path graphs form examples of fences. A linear extension of a fence is called an alternating permutation; André's problem of counting the number of different linear extensions has been studied since the 19th century. The solutions to this counting...

Up and down we go...

user228700

8:41 AM

@JohnRennie: Are u there?

Secret

user116211

9:11 AM

@Secret Is it related to Hasse Diagram? I've not read that though.

Secret

No, its a tangent google search caused by reading a section on Munkres where it is said every set have at most one least and most element

user116211

I have studied ordered sets; well-ordered, total-ordered, partially-ordered; but didn't encounter that till.

Secret

My mathshacking personality means I have a habit of pretty much asking "what if" for any theorem or lemma that is made to break down

Often this will initiate a google search tangent to whatever I am reading, and then I learn new mathematical structures as a result, since it appears that even when a lot of theorems and lemma breaks down, the resulting mathematical object is often not absurd

The Fence poset above illustrate a set can have multiple largest and smallest elements

user116211

I have read so far the theorem that the greatest and the smallest elements of an ordered set are unique; they are quite easy to prove.

user228700

9:49 AM

@MAFIA36790: Hey :-) Are u busy at the moment?

John Rennie

Hi Kaumudi. I'm back if you wanted to ask something ...

user228700

@JohnRennie Oh, awesome! I was just about to lose hope and leave :-P

John Rennie

Good timing then! :-)

user228700

I'm having some trouble with friction again...

user228700

I'm trying to solve two-block problems involving friction.

user228700

9:52 AM

@JohnRennie Ah, yes! :-)

John Rennie

OK ...

user228700

I'll ask u a very basic question to see if that clears my doubt; suppose we have two blocks, with one block on top of the other. Some force $F$ is applied on the block in touch with the ground. I'm asked to find the acceleration of the two blocks, given that there is no friction b/w the ground and the bottom block and that the coefficient of static friction b/w the two blocks is $\mu_s$.

user228700

Let's forget about what I've been asked to find and consider the block on top.

John Rennie

OK ...

user228700

In the situation in which the two blocks move together, the accelerations are the same, so the net force acting on the two blocks are the same. In the case of the bottom block, we know that the only force in the horizontal direction is the force $F$. (Since there is no relative motion b/w the two blocks). Is all this correct, so far..?

John Rennie

10:00 AM

Yes, that's all fine so far

user116211

@JohnRennie @JohnRennie HALP!

user228700

OK. So this would mean that there is a force of magnitude $F$ acting on the upper block as well, but this force is the force of friction. My major problem with this is, how come friction pulls it in the direction of motion of the bottom block? How am I supposed to find its direction w/o knowing that it moves one way or another? I want to be able to say "Friction acts that way, so the body(depending on all the other forces that act on it as well) moves this way"...

John Rennie

@MAFIA36790 Your questions are usually too hard for me to help with, but I'm willing to try ...

@KaumudiHarikumar You know the acceleration of the upper block because the two blocks move together so they have the same acceleration. That acceleration is just $F/(M+m)$.

user228700

How am I supposed to do that? I'm having a hazard time wrapping my head around this...

user116211

I am reading Lagrange's undetermined $\lambda$ method for one kinematical condition; I'm not getting one thing....

John Rennie

10:06 AM

@KaumudiHarikumar oops sorry, that was a reply to Mafia

user228700

@JohnRennie Well, yeah, that I know but isn't there any way for me to ascertain the direction of static friction without knowing which way the body is moving..?

user228700

@JohnRennie I gathered that; my questions are never too hard for you :-)

user228700

And damn, that was supposed to be "a haaaard time", not " hazard time"! Long live autocorrect!

user116211

@JohnRennie You have helped me positively so far...

Secret

Example of overthinking: Solving problem 1 4 dimensionally. NB only works if the 4D object is a cartesian product of two objects in 2D

John Rennie

10:08 AM

@KaumudiHarikumar the frictional force is a force applied to the top block by the bottom block.

user228700

@JohnRennie OK, please go on...

John Rennie

So if you look at how the bottom block is moving that tells you what direction the force is being applied it.

user228700

@JohnRennie But, but...how?

John Rennie

It's easy to think of the frictional force as being some entity in its own right, but it's just a force applied by something to something else.

user228700

@JohnRennie OK...

John Rennie

10:10 AM

So the direction of acceleration of the bottom block will be the direction of the force it applies on the smaller block.

user228700

@JohnRennie This doesn't make that much sense to me :-(

user116211

Anyways, Lanczos is investigating the variation of the function $F= F(u_1,u_2,\ldots,u_n)$...

user116211

With one auxiliary condition $f= f(u_1,u_2,\ldots,u_n)\,.$

John Rennie

@KaumudiHarikumar suppose instead of a friction force there was a piece of string linking the two blocks. The acceleration of the bottom block would produce a tension in the string and that tension would pull on, and therefore accelerate, the smaller block. Does that make sense so far?

@MAFIA36790 does auxillary condition mean a constraint?

user116211

Now, $$\delta F= \displaystyle \sum_i \frac{\partial F}{\partial u_i}~\delta u_i = 0$$

user116211

10:14 AM

@JohnRennie yep.

user228700

@JohnRennie Yes, it does, but does this analogy work all the time?

John Rennie

Well if the blocks aren't sliding wrt each other how is the frictional force different to the force in the piece of string?

user116211

The differential vanishes at a stationary value, after all; that's why $\delta F= 0\,.$

user228700

@JohnRennie OK, what if they are sliding w.r.t each other?

John Rennie

@KaumudiHarikumar then replace the piece of string by an elastic band.

user116211

10:16 AM

Also, $$\delta f= \displaystyle \sum_i \frac{\partial f}{\partial u_i}~\delta u_i = 0$$

user228700

I'm fishing for some rule that I can apply to every case.

user228700

A rule that makes intuitive sense...

user116211

But $\frac{\partial f}{\partial u_i}$ can't be inferred to be zero for $u_i$s are not independent of each other.

John Rennie

@KaumudiHarikumar the intuitive rule is that the frictional force is applied in the same direction that the bottom block is accelerating.

user228700

@JohnRennie No, no, sir! Not just for this case. For every situation there can be.

John Rennie

10:19 AM

Whenever there is a frictional force that means something is exerting a force on something else. Just look at the direction of the relative motion of the two bodies. The frictional force must act along that line.

user228700

@JohnRennie Both static and kinetic..?

user116211

In order to eliminate $\delta u_n$ assuming that $\frac{\partial f}{\partial u_n}\ne 0$ at the concerned stationary point, we would consider the other $\delta u_k$ as free variations.

John Rennie

@KaumudiHarikumar Yes

user228700

@JohnRennie Hm, OK. I'm gonna read about this some more. I'll brb and ask u if I encounter some other problem if u're still here then...

user116211

It is permissible however, to multiply a undetermined factor $\lambda$ a function of $u_i$s with $\delta f$ and add it to $\delta F $ as the result would still be zero.

Secret

10:23 AM

Same as ans

user116211

This would mean $$\sum_{k~=~1}^n~\left(\frac{\partial F}{\partial u_k} + \lambda \frac{\partial f}{\partial u_k}\right)~\delta u_k = 0\,.$$

user116211

In order to eliminate $\delta u_n,$ it is sufficient that $\left(\dfrac{\partial F}{\partial u_n} + \lambda \dfrac{\partial f}{\partial u_n}\right)= 0\,.$

John Rennie

OK, I'm with you so far, just! :-)

user228700

@JohnRennie: I found this:

user116211

This implies then $$\sum_{k~=~1}^{n-1}~\left(\frac{\partial F}{\partial u_k} + \lambda \frac{\partial f}{\partial u_k}\right)~\delta u_k = 0\,.$$

user228700

10:26 AM

user116211

But, then....

user116211

And here comes the part that couldn't be comprehended after hours of pondering over...

John Rennie

@KaumudiHarikumar yes, and ...

user228700

If we go back to our two block system and apply those rules to the block on top, we see that it works.

user228700

The block's not moving w.r.t the bottom block and the bottom block is accelerating, so there's a pseudo force acting on the block that acts in the opposite direction to the motion of the bottom block. Then, we see that friction acts toward the same side...

user116211

10:30 AM

Lanczos writes:

user116211

> [...] since only those $\delta u_k$ remain which can be chosen arbitrarily, the conditions of a free variation problem are applicable. This requires that the coefficient of each $\delta u_k$ shall vanish: $$\left(\dfrac{\partial F}{\partial u_n} + \lambda \dfrac{\partial f}{\partial u_n}\right)= 0~,~~~(k = 1,2,\ldots,(n-1))$$

user228700

This makes sense, no? Is this a good way to figure it out..?

John Rennie

@KaumudiHarikumar it seems a complicated way to describe something that is blatantly obvious. If you have system, any system, and you apply a force $\mathbf F$ then and response is going to be in the same direction as $\mathbf F$.

user116211

@JohnRennie: I didn't the get the above reason why all coefficients vanishes; didn't get what Lanczos meant by that ;/

John Rennie

@KaumudiHarikumar Well, I suppose you could come up with some system of springs and pulleys where that wasn't try, but in a simple system it's always going to be true.

user228700

10:34 AM

@JohnRennie Well, yeah, but I have difficulty in finding out in which direction that force acts so these steps act as a good guideline.

John Rennie

@MAFIA36790 to be honest I don't see it either. I think I'd need to look at the book. What book is it? If you give me the details I'll see if I can find it online.

user228700

I'm just worried if, by blindly following these steps, I will lose touch with the basic concept behind these steps, which tbh, I don't fully understand...

John Rennie

@KaumudiHarikumar there's certainly nothing wrong with using those steps to analyse a problem if you're unsure what's going on. As for the basic concept, just look at how object $A$ is moving or accelerating relative to object $B$.

user228700

@JohnRennie Well, I mean, yeah, but I do want to be sure about what's actually going on...

John Rennie

If you put you hand on something and apply a force isn't it obvious that the something is going to move in the direction of the applied force.

user228700

10:38 AM

@JohnRennie But the thing is that the direction in which the frictional force acts is not obvious to me!

John Rennie

Can you come up with an example of a question where the direction isn't obvious to you?

user228700

And I could use your string analogy but Idk if that works all the time, even in complicated systems, because I do need to work with complicated systems...

John Rennie

@MAFIA36790 Is it "The Variational Principles of Mechanics"

user228700

@JohnRennie Well, in the problem that I described before, the direction was obvious to me only because I knew that for the blocks to move together, the frictional force would have to act the direction in which it does. It was more of "look at what's happening, and then come up with the answer", rather than "come up with the direction in which the frictional force acts and then decide how it moves". Do you understand what I'm trying to say..?

John Rennie

@KaumudiHarikumar I think the trouble that after 40 years it's always obvious to me what direction the force acts, so it's hard for me to put my finger on what's troubling you. If you came up with a specific example I might be able to help better.

"look at what's happening, and then come up with the answer" seems a pretty good approach to me.

user228700

10:45 AM

OK, never mind this for now. I've found those "rules" that I had been looking for (I'll think about them so that they start making sense to me) and that's enough for now. Do you have a little more time to spend..? I'll ask u just one more question regarding actually solving problems. Is that OK..? (Now that we've discussed this much, it wouldn't take too much time...)

John Rennie

Yes, carry on

user228700

OK. To solve problems involving two or more blocks that are connected, my book has provided the following steps: (Yeah, they're a huge fan of steps)

user228700

One second.

John Rennie

@MAFIA36790 I've found that bit of Lanczos, but it's subtly different to what you described to me.

Up to: $$\sum_{k~=~1}^n~\left(\frac{\partial F}{\partial u_k} + \lambda \frac{\partial f}{\partial u_k}\right)~\delta u_k = 0$$ is fine

The next step is to rewrite this as: $$\sum_{k~=~1}^{n-1}~\left(\frac{\partial F}{\partial u_k} + \lambda \frac{\partial f}{\partial u_k}\right)~\delta u_k + \left(\frac{\partial F}{\partial u_n} + \lambda \frac{\partial f}{\partial u_n}\right)~\delta u_n = 0$$

user228700

John Rennie

10:56 AM

@MAFIA36790 does this make sense so far?

user228700

Shall I wait for u to finish answering MAFIA's question then?

John Rennie

@KaumudiHarikumar no, he's gone. Just as I figured out the answer to his question! Oh well.

Anyhow ...

So you're now pushing the top block to the right, yes?

user228700

Yes.

John Rennie

So which way is the top block going to move?

user228700

10:59 AM

Can you help me to understand these steps that they've given?

John Rennie

Groan, you want to do it the hard way? OK then :: John gives a big sigh :: :-)

user228700

@JohnRennie Um, yessir! :P I want to understand it...

user228700

I'll ask u some other time if u're bored/tired...

user228700

And, uh, why is it the hard way, exactly..?

John Rennie

OK this is the easy way:
which way is block A going to move?

user228700

11:03 AM

To the right, for sure.

John Rennie

and which way is block B going to move?

Secret

Need some help in understand the proof here:
http://dbfin.com/topology/munkres/chapter-1/section-3-relations/problem-13-solution/

> Besides, for any other lower bound x′ of S , x′∈L , therefore, x′≤x .
@MAFIA36790 But aren't we already showed that $x\leq x'$ since x is the lowest upper bound of $L$ (which is also a lower bound of $S$), thus how can $x' \leq x$ ?

user228700

That we don't know yet. We haven't been given if it actually moves/stays put. The steps have been given to find that out...

John Rennie

If B moves at all which way will it move?

user228700

@Secret I kind of feel sorry that I ask everybody questions but I am not educated to provide any answers to your questions and help you out as well :-(

user228700

11:05 AM

@JohnRennie Definitely to the right, along with A.

Secret

@KaumudiHarikumar don't worry, I know the best person to ask what questions, so I will be fine

John Rennie

And the frictional force is the force applied by block A to block B - true or false?

user228700

@Secret OK, good for you :-)

user228700

@JohnRennie True.

John Rennie

So in which direction does the frictional force act?

user228700

11:08 AM

To the riiight.

John Rennie

Bingo! What could be easier than that?

user228700

No sir, I'm trying to find out if the block will move at all!

user228700

Those steps are for that and I wasn't able to get my head around it...

John Rennie

Block B will move if the net force on it is non-zero - true or false?

user228700

@JohnRennie True true!(Well, it'll accelerate)

John Rennie

11:10 AM

And how many different forces are acting on block B?

user228700

@JohnRennie One, but why am I learning about the case in which the frictional force is not enough to make it accelerate along with A, then? When does that happen..?

John Rennie

Suppose both blocks move together then the acceleration is $a = F/(m_a+m_b)$. Yes?

user228700

No no, I was confused! There are two forces acting on B!

John Rennie

@KaumudiHarikumar Two forces?

user228700

@JohnRennie Yes! It's being pulled to the right and there's also the frictional force to the left!

John Rennie

11:14 AM

What frictional force to the left?

user228700

Sigh. No, that's A.

user228700

I'm so sorry. Please go on...

John Rennie

Suppose both blocks move together then the acceleration is $a = F/(m_a+m_b)$. Yes?

user228700

Yes.

John Rennie

And looking just at block $B$ we know that the net force on $B$ is related to the acceleration of $B$ by: $$F_b = m_b a_b$$ Yes?

user228700

11:17 AM

OK, no, wait. What about the case in which the frictional is somehow not enough to produce this acceleration?

John Rennie

For now we are assuming the blocks accelerate together so $a_b = a_a = a$.

user228700

@JohnRennie OK. Yes.

John Rennie

So we can substitute for $a$ and get: $$F_b = F\frac{m_b}{m_a+m_b}$$ Yes?

user228700

Yes, definitely.

John Rennie

We also know that the maximum value of the frictional force is $\mu$ times the normal force, so in this case it's $F_{max} = \mu g m_a$. So far so food?

user228700

11:21 AM

@JohnRennie Yep.

John Rennie

Ok then. Take the $F_b$ I calculated above, and if $F_b \lt F_{max}$ then there will be no slipping and $A$ and $B$ will move together.

user228700

OK...

user228700

Otherwise, B will accelerate with whatever acceleration $F_max$ is capable of producing...yeah?

John Rennie

Exactly!

user228700

And it'll slip and everything. And this slipping will cause the force of kinetic friction to act on A and then the acceleration of A will decrease..?

John Rennie

11:26 AM

If there is slipping fhe net force on $A$ is $F_{net}=F - F_{max}$

user228700

@JohnRennie Yes, OK.

John Rennie

And the acceleration of $A$ is just $F_{net} = m_a\,a_a$

user116211

Really sorry @JohnRennie; had to go to buy some drugs....

user116211

I'm reading your messages now...

Slereah

Finally got the internet back

user116211

11:27 AM

@Slereah Me too!!

user228700

@JohnRennie Awesome! So we consider what A is doing first and then think about B.

John Rennie

@MAFIA36790 Good timing, I think I've finally convinced K to believe me :-)

user228700

@JohnRennie :P Yes sir, that u have.

user116211

@JohnRennie Indeed it is.

John Rennie

@MAFIA36790 are you happy with: $$\sum_{k~=~1}^{n-1}~\left(\frac{\partial F}{\partial u_k} + \lambda \frac{\partial f}{\partial u_k}\right)~\delta u_k + \left(\frac{\partial F}{\partial u_n} + \lambda \frac{\partial f}{\partial u_n}\right)~\delta u_n = 0$$

user116211

11:29 AM

@JohnRennie yes; totally; the second term has to be zero to get rid of $\delta u_n\,.$

user228700

@JohnRennie: OK great. I'm done now. Thank you so much sir! I'll try my very best to leave you alone for the rest of the weekend. No promises though :P

user228700

Have a nice day!

John Rennie

@MAFIA36790 yes, and since $\lambda$ is an arbitrary constant we simply choose it's value to be: $$\lambda=-\frac{\partial F}{\partial u_n} / \frac{\partial f}{\partial u_n}$$

user116211

@JohnRennie sure.

John Rennie

And that eliminates the $\delta u_n$

user116211

11:32 AM

@JohnRennie yep.

John Rennie

Easy peasy :-)

@KaumudiHarikumar and you. Now stop working, it's Saturday!!! :-)

user116211

@Secret; did you ping me?

Secret

yup

(see ping. It's something related to order theory somewhat)

user116211

@JohnRennie, what about the other case then; that was the main question.

user116211

@Secret checking...

user116211

11:35 AM

1 hour ago, by MAFIA36790

> [...] since only those $\delta u_k$ remain which can be chosen arbitrarily, the conditions of a free variation problem are applicable. This requires that the coefficient of each $\delta u_k$ shall vanish: $$\left(\dfrac{\partial F}{\partial u_n} + \lambda \dfrac{\partial f}{\partial u_n}\right)= 0~,~~~(k = 1,2,\ldots,(n-1))$$

user116211

@Secret What do you want to know? Could you jot down it more explicitly?

Secret

I don't understand how the final line of that proof does not contradict the fact that x is the lowest upper bound of $L$

> Besides, for any other lower bound x′ of S , x′∈L , therefore, x′≤x .

i.e. I cannot see how there can be lower bound x' that are greater than x, as x is already the lowest upper bound of L
Original link: http://dbfin.com/topology/munkres/chapter-1/section-3-relations/problem-13-solution/

user116211

@Secret Lowest upper bound is supremum, right?

Secret

yup

user116211

okay.

user116211

11:39 AM

@Secret Specifically which theorem are you reading? Which page of Munkres?

Secret

Q13 p.29 2nd Ed

user116211

@Secret checking.

Secret

It's an exercse but I am unable to sketch a proof thus I checked the ans, and I was still confused by the ans's proof

user116211

@JohnRennie, could you provide insight why the other coefficients have to be zero to get free variations?

user116211

@Secret I'm looking, wait a bit...

John Rennie

11:44 AM

@MAFIA36790 I think the point is that we have an $n$ dimensional configuration space where all the dimensions are orthogonal, so in the absence of a constraint the minimum requires all the $\partial F/du_k$ to vanish. Does that make sense so far?

user116211

@JohnRennie Somewhat making sense; I've to re-read.

user116211

Ah! seen the question @secret.

Slereah

ВеданЪ КолодЪ - Ведьма (Городище 2014) Russian Etno Folk/Medieval

►

John Rennie

@MAFIA36790 the constraint forces you to move on an $n-1$ dimensional subspace, so with $n$ degrees of freedom the motion is overconstrained and the $u_k$s cannot all be orthogonal. But eliminate one of the DOFs and you're back to $n-1$ DOFs in an $n-1$ D space.

I have to go now. Back in a few hours.

ACuriousMind

@Secret Once again, you need to read more carefully: The $x'$ in the two statements $x'\leq x$ and $x\leq x'$ are not the same.

user116211

11:55 AM

WTH ._.

Secret

@ACuriousMind The first x' are upper bounds of L, the second x' are lower bounds of S that are in L, but are those second x' s upper or lower bounds (or both) in L?

ACuriousMind

@Secret I don't understand the question. The second $x'$ are lower bounds of S, and you defined L to be the set of all lower bounds of S, so $x'\in L$.

@MAFIA36790 You're reading the statement wrong. It's not saying that to get a free variation, the coefficients need to vanish. It's saying that because you have free variations, the coefficients have to vanish.

Slereah

Is the mapping of hyperreals to real the contraction of the halo to a point

ACuriousMind

Does anyone except you even know the first thing about hyperreals here?

Slereah

Also can two halos ever intersect

I don't know

ACuriousMind

12:07 PM

Also, "halo"?

Secret

(I am going to relabel the x' to avoid confusion) Therefore, L has the lowest upper bound x such that for any upper bound x1 of L : x≤x1 . We show that x is the greatest lower bound of S . First, it is a lower bound of S , as any element z in S is an upper bound of L , implying x≤z for all z in S . Besides, for any other lower bound x2 of S , x2∈L , therefore, x2≤x .

Slereah

The halo is the set of all values that are related by an infinitesimal

Secret

is it impossible to have x≤x2 because the lower bound of S has to be less than the lowest upper bound of L?

Slereah

The halo of x is $\{y | y = x + \text{some infinitesimal}\}$

ACuriousMind

@Secret By definition, lower bounds $x_2$ of $S$ are elements of $L$. By definition, $x$ is the least upper bound of $L$, so $x_2\leq x$ by definition of what an upper bound is. I'm not sure what your problem here is.

@Slereah What kind of object is x allowed to be?

Slereah

12:11 PM

Hm

Any object technically I suppose

Although for mapping to reals I suppose it should be part of the finite subalgebra

ACuriousMind

Then it would appear that contracting halos does not merely produce the reals

John Doe

Hi all

Slereah

Hello

Secret

But x is the least upper bound of L, which means there are some $x2 \in L$ that are lower bound in S, but at the same time can be greater than x? (because x is the least upper bound of L, there are upper bounds of L that are greater than x). How do we rule these x2 out?

John Doe

Quick QM question, for elementary particles of $\frac{1}{2}$-spin ($s= \frac{1}{2}$), the eigenvectors are denoted by the symbols $| \uparrow \rangle = | + \frac{1}{2} \rangle$ and $| \downarrow \rangle = | -\frac{1}{2} \rangle$. Do you know what the symbols $| \uparrow \downarrow \rangle$ and $| \downarrow \uparrow \rangle$ denote?

ACuriousMind

12:14 PM

@JohnDoe Typically a state of two particles, one of which is spin up and the other spin down. That is, $\lvert \uparrow\downarrow\rangle = \lvert \uparrow\rangle\otimes\lvert \downarrow \rangle$.

Slereah

the problem of hyperreals is that it's hard to google without getting a lot of pop results

also all introductions try to do it in an intuitive way

ACuriousMind

@Secret And how could an element of L possibly be an upper bound of L?

Slereah

Give me real sequences plz

John Doe

@ACuriousMind Oh okay that makes sense, thanks.

Secret

It can if it is the maximum element of L(?)

ACuriousMind

12:17 PM

@Secret Yes, and then it would obviously be itself the least upper bound, i.e. it would be equal to x.

Slereah

"Fix a nonprincipal ultrafilter $F$. Let $\Bbb R^\Bbb N$ be the set of all sequences of real numbers, and say $(a_n) \cong (b_n)$ if $\{n : a_n = b_n\}$ is big. Then the hyperreals are the set of equivalence class $^* \Bbb R = \Bbb R^\Bbb N / \cong$"

That's the stuff

ACuriousMind

"big"

Slereah

A noble spirit embiggens the smallest man, @ACuriousMind

I think the word the word they grasp for is "cofinite"

ACuriousMind

Good ol' Jebediah

Slereah

cofinite is p. big

ACuriousMind

12:19 PM

So, they mean "two sequences are equal if they only differ in a finite number of terms"?

Slereah

Yes

ACuriousMind

If so, writing it that way with "big" seems almost obfuscatory

Slereah

It big

ACuriousMind

What if they only mean "infinite"?

Hm, that wouldn't be an equivalence relation

Slereah

"An ultrafilter on $\Bbb N$ is a collection $F$ of subsets of $\Bbb N$, called "big", such that 1. $\Bbb N$ is big and $\emptyset$ is not big. 2. If $A$ is big and $A \subset B$, then $B$ is big. 3. If $A$ and $B$ are big, then $A \cap B$ is big. 4. For any $A$, either $A$ or $\Bbb N \setminus A$ is big."

Bigness defined

Secret

12:24 PM

@ACuriousMind Ok so we have x2≤x≤x1≤z. Is the reason that the upper bounds x1 of L is not the greatest lower bound (unless x=x1, which is the case where x is the max of L) of S because it is not in L hence by definition not the lower bounds of S?

Slereah

IIRC it's equivalent to cofiniteness being big

Although...

If the set is the set of all even numbers

Is $A$ big or is it $\Bbb N \setminus A$

ACuriousMind

Yeah, the set of cofinite sets isn't an ultrafilter

Slereah

I guess that might depend on the filter?

Alex

@ACuriousMind In holomorphic functional calculus, do you maybe know how it follows that $e^B =
\begin{bmatrix}
e^2 & 0 \\
0 & e^3
\end{bmatrix}
$ where $B = \begin{bmatrix}
2 & 0 \\
0 & 3
\end{bmatrix}$?

Do you maybe use the Maclaurin series $e^B = I + B + \frac{B^2}{2!} + \frac{B^3}{3!} + ...?$

ACuriousMind

@Secret Yes. If $x_1$ is not in the set of lower bounds, it's not a lower bound! What about that are you uncertain about?

@Alex Sure, for diagonal matrices, the $n$-th power is just the matrix where each entry is raised to the n-th power

Has nothing to do with "holomorphic functional calculus" though, this is just the standard matrix exponential

Secret

12:36 PM

@ACuriousMind I think it should be clear now. Sorry but I might be not skillful enough at supremum and infrimum probelms as whenever I tried to do them abstractly/algebriacally, I always end up something that is e.g. greater that the lower bound and less than the upper bound, and then the loical flow of the workings get tangled up in the process, and lost track of which elements are bigger or smaller

typo: logical

Alex

@ACuriousMind Yeah agreed, I just started reading the wiki entry on Holomorphic Functional calculus where they start by motivating functions of matrices.

Slereah

"Nonprincipal ultra lters exist, but you need the Axiom of Choice to prove this."

Nooooooooooooooooo

If the axiom of choice is involved I suspect that actually constructing an ultrafilter isn't possible

You just have to assume it exists

Secret

You cannot construct anything that is proved by the AFC

John Rennie

Sigh. Why oh why did I take this question seriously? Oh well.

“The statement made in the answers above is that gravitational waves have a null wavevector. Null wavevectors are null in all coordinate systems so your statement that gravitational waves may travel at speeds different from light is simply untrue.” John Rennie. No, Mr. Rennie, you are wrong. See Appendix of: Crothers, S.J., A Critical Analysis of LIGO's Recent Detection of Gravitational Waves Caused by Merging Black Holes, Hadronic Journal, Vol. 39, 2016, vixra.org/pdf/1603.0127v4.pdf — Stephen J. Crothers 38 mins ago

And that's Dr. Rennie please, Mr. Crothers :-)

Slereah

lol vixra

Secret

12:41 PM

Currently on Ch.4 of Munkres. Once I get to the first topology chapter, I will make sure all questions I have generated so far will be milled down before proceeding

Slereah

Ah apparently the thing is that all cofinite sets are big

John Rennie

Yes, there was me thinking Crothers might actually be interested in an answer, but no he just wanted an excuse to flog his latest work of fiction.

Slereah

But not vice versa

Secret

* Read Ch. 4 And man... numbers... I am bad at numbers...*

Still much better that a EM problem unless the system like to go around in cirlces

ACuriousMind

@JohnRennie Crothers really rubs you the wrong way, hm?

John Rennie

12:44 PM

@ACuriousMind no I'm not annoyed. Sort of vaguely amused really.

It's the predictability of it. We get an apparently bona fide question, then a few days later a comment saying we're all stupid and we should read his latest paper (on Vixra).

ACuriousMind

@Slereah So do these hyperreal rely on the choice of a specific ultrafilter or not?

Slereah

@ACuriousMind I think so?

John Rennie

@ACuriousMind And of course our other resident non-conformist is still flogging his favourite dead horse and downvoting everyone else.

Slereah

They require a non-principal ultrafilter which seems to require the AOC

So I can probably just ignore its specific construction, I'm guessing

John Rennie

But on the flip side I learned about the Peres metric, which I didn't know about before, and there's a link to a really good answer on Math Overflow that I am attempting to read, learn and inwardly digest.

So it was not a complete waste of time.

Slereah

12:49 PM

I should check on the hooch

John Rennie

@Slereah what did you use to press it in the end?

Slereah

Grate

A bit long but efficient enough

Grate into a pulp and then press in a filter

John Rennie

So in a few weeks time we need to watch for your posts becoming suddenly incoherent for a few hours :-)

Slereah

Psh, when's the last time you saw me drunk here

(It was during the New Year)

John Rennie

I never drink and drive Chrome - I learned the hard way not to do that :-)

ACuriousMind

12:53 PM

@JohnRennie That's a nice answer! @0celo7 it seems the answer to this question of yours sat on MO all along.

John Duffield

@JohnRennie : only I'm with Einstein, so don't call me the non-conformist. And do note this: "The result stands or falls by the choice of coordinates and, so far as can be judged, the coordinates here used were purposely introduced in order to obtain the simplification which results from representing the propagation as occurring with the speed of light. The argument thus follows a vicious circle.” Eddington [38 §57]". Like I've said before, it's a tautology.

Secret

Don't worry, NOBODY can beat me in terms of incoherent posts, (PS I am not even drunk, lol)

(Having said that, my posts have become a lot more coherent compared to when I first joined this chat. I guess ACM and co. should give themselves a pat on their back that our communication training do work)

Slereah

It is fermenting

I hope the balloon will be enough to keep all the fermentation gas

I should get a vapor lock soon

John Duffield

@JohnRennie : that really good answer is totally wrong.

Secret

What is the sugar source for your alcohol?

Slereah

12:56 PM

Apples

ACuriousMind

Should it look like that?

Slereah

Seems alright

As time goes the solid bits will deposit at the bottom

Secret

Cider

Cider (/ˈsaɪdər/ SY-dər), known as hard cider in North America, is an alcoholic beverage made from the fermented juice of apples. The juice of any variety of apple can be used to make cider, but cider apples are best. The addition of sugar or extra fruit before a second fermentation increases the alcoholic content of the resulting beverage. Cider is popular in the United Kingdom, especially in the West Country, and widely available. The UK has the world's highest per capita consumption, as well as its largest cider-producing companies. Cider is also popular in other European countries including...

Slereah

and become less cloudy

ACuriousMind

Have you already chosen an acquaintance you dislike who will try it first?

Slereah

12:58 PM

I think the obvious choice is coworkers

it's all about presentation, really

get a fancy glass bottle

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