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12:49 AM
@TedShifrin When your own students refuse to come to office hours because you suck at teaching, you probably have a lot of time to browse stackexchange.
Office hours? Probably doesn’t even have those.
Maybe a sign on the door that says, "If you are here for office hours, the proof is trivial."
@TedShifrin thanks, the only example I know of in this vein is RP2.
Real Grassmannians of various dimensions among others.
It seems like with spaces that are 'twisted' rather than have holes of a certain dimension the forms fail to pick up on it. After thinking about it for a few hours my best guess at an explanation is that when you have a "hole" of dimension n, it fails to affect all the n-k forms and then suddenly there is an n form that fails to be exact, whereas the twist constrains forms regardless of their dimension allowing d to have full rank all the way through. Dunno if that's right or can be formalized.
1:19 AM
Oh, I see what you mean about full rank. You mean it on the boundary level. Yes, right. Or there might not be any nonzero closed forms at all. I guess the previous $d$ still has full rank.
1:32 AM
@TedShifrin What an immature, self absorbed, and condescending thing to say to somebody who is on the face of things putting in an effort to learn the discipline. It is well acknowledged that Spivak is or at least close to the most challenging introductory text for Calculus........some people forget that there was a time that they struggled with things too......... anyways no more bad energy on that.
there were pelicans at the beach
sholdn't there be pelicans at the beach?
sometimes you go and it's only gulls and other shore birds. depends on the time.
I’m late to the party re: that one post but
lots of containers out in the harbor, too.
1:42 AM
Read that they are finally starting to clear the back log in the port....
The idea that you can judge someone’s “talent” from one post is…rather breathtaking
but now we have Omaricon....so...
especially when talent is cultivated
I don't believe in it just coming form osmosis
Plus…not all math is Spivak Calculus
1:43 AM
even in child prodigies.....they just happened to be in an environment that cultivated it a lot earlier than most
@dc3rd huzzah for doom scrolling
@dc3rd I think there can be “talent” in the prodigy sense, but it’s by far the exception
yeah, should've not scorlled up to see al the previous convo.
And it won’t go far without a level of discipline
I agree
Was doom-scrolling Twitter re: omicron
And idfk
1:46 AM
ah.....I stay away from twitter....but the way most news is presented nowadays it is all very hyperbolic purposefully
so it doesn't matter the source you end up catching the "doomsday" commentary of the events
it's framed that way on the BBC and the Guardian as well
2:07 AM
Wikipedia's news section is probably enough news.
@TedShifrin Just wanted to say, I appreciate your comment below this question. I flagged two comments; one has been deleted after being handled by a mod; the first still remains, and my flag is pending. I just wanted to say thanks for your response there.
2:22 AM
Robjohn didn’t remove it because he wanted my comment to have context. Winning personality and effective teacher, that a**.
I suppose my semi-rude comments to people who post verbatim homework with zero effort might sound similar, but they’re not.
they don't sound similar to me. although i would probably make them under an alias. Shed Tifrin.
An uncanny alias.
2:39 AM
Not sure I have earned the credit you gave me there, @TedShifrin . I wasn't thinking about the "boundary" explicitly (although maybe I was implicitly without realizing it).

What I was thinking about what that the sphere is the universal cover of RP2. And I was trying to come up with 1-forms and 2-forms that would or wouldn't induce well-defined 1-forms or 2-forms on RP2. My reasoning was that if it were possible for forms to tell me something about the 'twist' in the space, then either (1) there needed to be some forms on RP2 that cannot work on the sphere OR (2) there need to be some form
2:54 AM
Hello World..
And I think I convinced myself that (1) is impossible. Every well-defined form on a space seems to induce a well-defined form on its universal cover.

(2) is possible, you just have to have a form that acts at a point and its antipode in an asymmetric way, but seems to restrict things in a way where if $\omega$ is an n-form on the universal cover that is not well-defined on the covered space, then a non-zero $d \omega$ is similarly ill-defined. So you don't run into the situation that would generate non-trivial cohomolgy where the twist disallows $\omega$ but preserves what was $ d \omega$,
What are the outcomes of torque? in sense of we can calculate torque any point within the body of outside the body it gives different results.
What are the benefits and disadvantages of this behavior of torque?
123 I don't understand the question. Are you asking about the fact that we can compute torque about any point, and we get different answers depending on the point we choose?
3:12 AM
@KevinDriscoll Yes
So if it the case we can not measure torque as single value as in force case.
What are pros and cons of this idea of torque at which we can have different values depend on point where we want to measure the torque ?
3:31 AM
@123 This reflects a fundamental difference in moving in a straight line (translating) vs. moving in a circle (rotating).

Suppose you move along some path in space. There is a unique answer to the question "how far away is the final point from the starting point?" It's some distance, and that distance doesn't depend on how I choose to overlay coordinates onto space.
@KevinDriscoll But if we measure force the result is always the same. It does not matter from where we calculate.
@123 However, if instead I ask "what angle does the final point make with the initial point?" Then that does NOT have a unique answer. For example, if I choose the origin of my coordinates to be exactly in the middle between where you started and where you ended, then the answer will be 180 degrees (assuming that you did not start and end in the same spot). However, suppose I choose my origin to be very far away from where you were moving around.
Torque does not have single answer like force. We can have different answers if we choose point of reference at different position.
In that case, teh angle that you moved through will be very small with respect to this very far away origin
The fact that torque gives a different answer depending on what point we calculate the torque with respect to follows exactly from this fact that the angle that you move through depends on where we choose our origin to be
@KevinDriscoll I think all inertial frame of references calculate same value of force irrespective where we choose to the origin close or far.
3:39 AM
That is correct. And this is a result of the fact that all inertial reference frames agree on distances and times (at least clasically)
@KevinDriscoll In force equation we don't have cross product or it does not depend on point where we calculate it. It is F = ma
But all inertial reference frames do NOT agree about angles
The angle that you move through depends on where you take the origin to be
@KevinDriscoll Pls share equation because F = ma i don't see use of angle. In M = r x F , i can see r distance can be change by me and cross product said it need angle between r and F
@123 Right
$\vec{F} = m \vec{a}$ no angles
$\tau = I \alpha$, the rotational version of Newton's 2nd Law, is all about angles
@KevinDriscoll Yes i agree with torque need angle .
3:46 AM
Indeed, so once you recognize that torque is all about angles, well then we better NOT get a unique answer. Because the angle that you move through depends on what point you take to be the origin.
@123 https://imgur.com/GazBhsJ

Consider the path shown here in blue. If you take the origin to be at the black point, then the angle that the path moves through is 180 degrees. However, if you take the red point to be the origin, the angle that the path moves through is 0 degrees (it starts and ends at the same angle with respect to the origin).
@KevinDriscoll Thank you dear for the help...
So when we ask questions like "how fast is the angle changing over time as you move along this path?" (AKA "what is the angular velocity?") and "how quickly is the angular velocity changing?" (AKA "what is the torque?"), you should expect to get different answers depending on which point you take to be the origin
@KevinDriscoll Aaah okay.. It means it is only for rotation.
torgue is all about the moment
Because rotation itself is always non-inertial frame of reference. It gives different answer depend from where we calculate. Is that mean?
@copper.hat Hello ...
4:00 AM
@123 hi!
I am reading principle of mechanics synge griffith. I am still confused with translation and rotation of a rigid body.
rotation is more complex in which 3 cases arises (1) when there is no fixed point (free body) (2) When fixed point inside the body (3) When fixed point outside the body.
Fixed point mean rotational axes..
Yes, although describing it as a "fixed point outside the body" is a little weird, because there might not be anything fixed at the point
Pls explain simple case. If we have isolated rigid body and we apply only single force on it. What conditions need to fulfill for a body have pure linear motion?
The force must act along a line that passes through the center of mass
Well, it depends on what you mean by "pure linear motion"
Usually we wouldn't think of an object moving in a straight line as "rotating"
@KevinDriscoll Thank you to strong me. This is what exactly i am thinking. Pure linear motion mean rectilinear motion.
4:08 AM
But if the object is moving along any line that does not connect the center of mass of the object to the origin, then formally speaking it is rotating about that origin
@KevinDriscoll What if multiple forces acting on rigid body. What should be the condition for a body have pure translation?
So to be as general as possible, if a single force is acting on a body, it must act along the line connecting the center of mass to the origin of the coordinate system in order for the motion to be purely linear
Multiple forces is slightly more complicated. If you add all the forces together to find the net force, then the net force must act along a line connecting the center of mass and the origin.

AND the sum of all the torques about the center of mass of the object must also sum to zero
@KevinDriscoll If this is the case for body purely linear motion under multiple forces. I can conclude all the forces must be concurrent and the point of concurrency is the body COM. Is it correct?
The first condition precludes the possibility that the object is rotating about the origin. And the second condition precludes the possibility that the object is rotating about its own center of mass.
What do you mean by origin here? I can understand the COM but pls explain origin in this case.
4:14 AM
Ummm, no I don't think the forces need to be concurrent.
The origin is the point in your coordinate system that you label (0,0,0). In order to describe the motion of the object using cartesian coordinates, you must first choose which point to call the origin of your coordinate system.
@KevinDriscoll Concurrent forces means line of action of force meet at one point.
Yup, I'm familiar with the definition. And that's not a requirement
Errr well, hold on let me think more carefully
@KevinDriscoll If all the forces line of action pass through body COM, is it possible body can have torque in such case?
Hold on, let me show you a counter-example
@KevinDriscoll pls
4:19 AM
Suppose in this image that the origin is the green point, the black square is some rigid body. Each of the red arrows represents a force of equal magnitude F. Suppose that the upward and downward areas are all an equal distance from the COM.
The forces here are not concurrent. But the motion will be purely linear. The box will move straight toward the green point without rotating.
Here is an example where the forces ARE concurrent, but the point of concurrency is NOT the body's COM
So the condition that you gave is sufficient. But it is not necessary.
@KevinDriscoll In this example all upward forces balanced downward forces. Yes upward and downward forces are not pass through COM, the only unbalanced force is at left side of the body which pass through the COM of the body which reside between the rectangular box. Yes also forces are not concurrent but is has pure transation.
@KevinDriscoll Great this example help me to better understand the topic. how to think this upward and downward force because they cancel each other.
The way to think about it is that the torque caused by these forces is equal and opposite
Is there any case at which only more than two forces their line of action pass through COM of body but point of concurrency is not COM.
@KevinDriscoll Ooookay... Thank you this discussion very helpful to me.
No, there is no such case. The geometry is impossible. If there are >2 forces, than either they all act along the same line, in which case the "point of concurrency" is the entire line of action OR they act along different lines, but in Euclidian geometry non-parallel lines always intersect at exactly 1 point. So if all the lines pass through the COM, then it is the unique point where they all intersect.
@KevinDriscoll One more important question. Is it possible for free rigid body can have torque nonzero if line of action of all the forces pass through COM. Is it possible? If it is not possible then it is the only condition for pure translation.
4:33 AM
Unless you don't mean that each force's line of action contains the COM but only that the line of action of the net force contains the COM'
Yes, it is possible if you calculate the torque about a point that is not the COM
Such an object cannot be rotating about its own COM. But it may be rotating about some external point.
(In fact if the net force is not zero, it must be rotating about an external point)
@KevinDriscoll Pls share diagram. How to possible object can have nonzero torque when all forces line of action is COM.
Here i am considering free body (no fixed point internally or externally ).
I don't really understand what an external fixed point means
external fixed point means as in wire attached to sphere.
Ah, okay
Here the force acting on the body is F (upward). The green point is the origin, a distance d away from the center of mass
@KevinDriscoll You said in this comment. If all the forces line of action pass through COM body can have nonzero torque and can rotate about some external point. What does it means pls share example.
4:42 AM
The torque about the green point is (0, F, 0) x (-d, 0 , 0) = (0, 0, -F d)
So this object will move upward in a straight line (assuming it is at rest to start), but that counts as "rotating" about the green point (the torque is non-zero)
@KevinDriscoll Yes.. But what if there no fixed point internally or externally. What happened in this case?
There is no wire attached here
the black line is just a ruler, not a string attaching the object to a point
@KevinDriscoll Okay. So how i understand this green point as external point?
It is just some arbitrary point that you have chosen to calculate the torque about
This is why I was asking before about what you consider to be the definition of "rotating"
@KevinDriscoll Okay .. Pls explain if object is moving in straight line how can we calculate torque from it about any point.
4:45 AM
Most people would not consider an object moving in a straight line as "rotating," but if you take "rotating" to mean there is a non-zero torque about SOME point, then such motion is rotating
Do you know the definition of torque?
@KevinDriscoll Yes that's why i confused.
@KevinDriscoll Yes M = r x F,
Indeed, and what does F mean here?
It require only moment arm perpendicular distance from any arbitrary point to the line of action of force.
@KevinDriscoll Thank you to clear me the topic. I was previously considered torque only for rotation not for translation.
Right, so F is the force. And r is the position of the point where the force acts on the body relative to the point that you're calculating the torque about
So to calculate the torque about any point, you just change what you take 'r' to be
@KevinDriscoll How this case is important to calculate torque in this case. What is benefit of calculating in such cases?
@KevinDriscoll Also thank you very much to clear my confusions. Now i can better understand mechanics book.
Why we relate torque only for rotation not for translation?
4:52 AM
I don't think there is a benefit in this case to using the torque. It gives you exactly the same information as using the forces and F = ma in this case. But it is important that we understand this case and that our system is consistent, ie we do not run into any contradictions by considering the torque in this unusual way.
Torque relates only to rotations by definition. By defining the torque to be r x F, it is automatically 0 when F acts along r and always non-zero when F does not act along r.
@KevinDriscoll Force is not confusing topic due to its single value, we can understand magnitude of force intuitively and its effect later for other bodies . But torque does not have single answer it change depend on reference point.
@KevinDriscoll I think in this your example if we fixed green point or green point is the force position object it shows pure circular motion.
provided that the force always acts at the same point and turns along with the object, yes, you'll get pure circular motion
@KevinDriscoll Great this discussion is very very helpful. Just last confusion about rotation
Torque does not have unique answer. What is the benefit of having this nature of torque and what are the disadvantages?
5:08 AM
The benefit is that it lets us correctly describe how a single object can be rotating at different rates about many different points at the same time. For example, it can be rotating about its center of mass and also its center of mass can be moving in a circle about some external point at the same time
Like how the Earth both revolves on its axis and rotates around the Sun at the same time
The disadvantage, I guess, is that this can be kinda confusing because people sometimes expect torque to have a unique answer like force. But such a construction would be impossible. There isn't a useful definition of torque that is independent of what point you are considering rotations around like there is with force.
@KevinDriscoll Do i consider this comment as, If i considered me as reference point, when i look rotating object at short distance it look slow rotating , then the same object i see from far it looks faster?
Yes, but be careful here about what you are holding fixed when you make this comparison
For example, given an object that is moving at a fixed speed, it rotates much faster about you if you are close to it than it does when you are far away
@KevinDriscoll If object is rotating at fixed speed , say spinning about it COM. I can calculate torque about me as reference point and i can distance from body.
5:23 AM
Sure, yes. Although if the COM is at rest and the object is spinning about its COM, you will get 0 calculating the torque about you (or any external point)
@KevinDriscoll Why 0 torque in this case?
Oh actually, sorry, you won't necessarily get zero. I was thinking of a different way of splitting up torque.
I think previously i said rotating at fixed speed that's why you said 0 torque.
yeah, if the angular speed is fixed you will certainly get 0
but if not, then you won't
@KevinDriscoll Rotation is more complex then translation.
It is hard to understand because of torque multiple values at different point.
@KevinDriscoll Thank you for your time and sharing knowledge.
5:32 AM
Yup, things can certainly get a bit complicated. No problem
@KevinDriscoll I need to discuss few more cases. I will be back in 30min. Thank you.
Alright, I am probably done for the night
6:26 AM
Fun fact I found : One known fact in general topology 'If $q:X\to Y$ is a quotient map and $Z$ is a locally compact T_2 space, then $q\times 1:X\times Z\to Y\times Z$ is a quotient map' is also called Whitehead theorem.
6:38 AM
6:52 AM
I saw an exercise where they were asking the extremas of $\displaystyle f(x,y)= \frac{xy+x\sin(y)}{(x^2+y^2)^{\alpha} }$
Not sure how the author pretends to compute the partials in y.
The max i can say is that for $(x,y)\neq 0$ we have $\dfrac{\partial f}{\partial x} = (y+\sin (y))\left(-\left((2 {\alpha}-1) x^{2}-y^{2}\right)\right)\left(x^{2}+y^{2}\right)^{-{\alpha}-1} = 0 \iff (2{\alpha}-1) x^{2}-y^{2} = 0$
Which yields $y = \pm \sqrt{2\alpha-1} x$ and $\alpha > \dfrac{1}{2}$
But $\dfrac{\partial f}{\partial y}$ is really ugly. Even for $\alpha = 1$
7:17 AM
Perhaps compute the extrema for a fixed $x^2+y^2 = r$?
@copper.hat Uh do you mean with that restriction or using polar coordinates?
@Odestheory12 i have not thought it through, it was just a suggestion. maybe something might pop out.
Uhm then $\frac{\partial}{\partial x}\left(\frac{x y+x \sin (y)}{r^{\alpha^{2}}}\right)=r^{-\alpha}(y+\sin (y))$
Which is never 0 since $y\neq 0$
That would justity the absence of extremas, right?
Uhm but $r=g(x,y)$ so that differentiation is not correct
7:49 AM
@copper.hat since $r$ was random that should mean $f_x$ is never zero right?
@Odestheory12 sorry, i don't have time at the moment to help you. it was a suggestion, note that if you extremise with the constraint $x^2+y^2 = r$ you need to do more than just differentiate with respect to $x$.
No worries, thanks.
5 hours later…
12:23 PM
The expression that I got for $Q(x)$ was $x^2-(b+c)x+b^2-bc+c^2$ through long division. Completing the square on $2Q(x)$ gives $2(x-\frac{b+c}{2})^2+\frac{3(b-c)^2}{2}$. The question states that $2Q(x)$ can be rewritten as the sum of three expressions, each of which is a perfect square. But as you can see there is only two perfect squares in my expression. Not sure what they want me to do
Fearsome general topology question : math.stackexchange.com/q/4317500/668308
2 hours later…
2:27 PM
let $g:[0,1]^2\to\mathbb{R}, g(x,y) = 0$ be a function.
prove that the graph of $g$ is compact.
i think it's noted $\mathcal{G}(g)$
effectively, it is $\{(x,y,z)\in\mathbb{R}^3|z=0,(x,y)\in[0,1]^2\}$
i've gone with trying to prove that it's both bounded and closed.
and i think i've proved that it's bounded.
i assert that the ball $B((0,0,0),2)$ contains it.
It is well known that $\int 1/(1+x^2) = arctan(x) $ but why when i do the following substitute i get rubbish results?
Let $ sinh(u) = x , cosh(u) * du = dx $ then $\int 1/(1+x^2) = \int \frac{cosh(u)*du}{1+sinh^2(u)} = \int \frac{du}{coshu}$ mathematica gives after resubstuting $ u = arcsinh(x) $ the final function to be $ gd (Sinh^-1(x)$
indeed, if $p=(p_1,p_2,0)\in\mathcal{G}(g)$, then $0<p_1<1$ and $0<p_2<1$, so $p_1^2+p_2^2<2$. so $p_1^2+p_2^2<4$. thus $p\in B((0,0,0),2)$.
but i don't have a clear path for closedness, how should i prove that $\mathcal{G}(g)$ is closed?
try maybe to show that complement is open
hmm, alright. so we will show that $\mathbb{R}^3-\mathcal{G}(g)$ is open.
I really have no idea about that subject i just threw a suggestion, sometimes i get inspired by monkeys doing wierd stuff so maybe this inspires you :D
so your function builds everything to zero?
2:42 PM
i see :) well, it's actually the basic way but i sense that it would get complicated.
@MadSpaces yes, exactly.
obviously the complement is open.
ithink the theory of integration helps here
the graph has a jordan volume of zero. this means there is a finite number of squares that covers at with their volume being lesser epsilon.
What can you make out of that?
i don't know about jordan volume, though. let me read the definition.
oh okay then do not
Suppose you have an element in the compliment with no epsilon such that it is completly in the compliment.
then it means it has to for all epsilon none trivially intersect with your graph
i don't think i get the sentences above.
2:49 PM
i take an element $a\in\mathbb{R}^3-\mathcal{G}(g)$?
Well ... as i said in the integration theory, your set has jordan volume of zero, it has thus finite number of squares covering it, you can take hte interior of these squares and you have a finite cover. thus your set is compact
is that what you mean in the beginning.
But you said you do not know about that, so how about you prove it with balls and epsilons.
alright, balls and epsilons are good.
2:51 PM
we will directly use the definition of being an open set, right?
I guess
I think there is a even a Lemma or something about R \any finite set being open
so, we will construct a ball around $a$ we took above.
It is obvious that you will find an $\epsilon$ small enough to be completly in R - Graph
Your problem is writing this mathematically ... your graph is literally a line
that will reside completely inside $\mathbb{R}^3-\mathcal{G}(g)$, let's call this $G'$.
yes, that's exactly it. i would cut the space in about 4 pieces, i suppose.
Okay here is my idea
2:54 PM
up, down, left and right of the film.
right, i'm listening.
You try it be contradiction. If some point is not in the interior of compliment, thus for all epsilon it must cut the graph none trivially. If such is the case, then you show the point must be in the graph because (draw it you will see what im ean) it only is the case for such point to itersect always with the graph because the graph is a line.. this is a contradiction to the point being frm the compliment & oyu sowed all points in compliment can be fitted in balls inside compiment thus its open
but i have to go now, good luck trying that) it should not be hard
thanks, i will think about this.
hmm, i think that would work really.
3:14 PM
As a tip remember that any point you take with radius of epsilon will have the coordinates $(a_x+\epsilon, a_y+\epsilon) $ remember oyur graph consists of points of the construction $(c_x,0) $ it is now easy to construct a contradiction
3:57 PM
Why is it when i integrate functions such as $ \sqrt(1-x^2) $ and $\frac{1}{\sqrt(1-x^2}$ using the substution $cos(u) = x $ i get the same result but with a minus sign difference as when i integrate with $sin(u) = x$?
The first minus sign results in from the derivation of $cos(u)$ but then i do not recieve a second one anywhere to cancel it out!
For example:
$cos(u) = x , -sin(u)*du=dx$ then we get $\int \sqrt (1-x^2) = -1*\int sin^2(u)du = - * (1/2u-1/2 sin(u)cos(u))$ resubstuting u yields the result... The answer is given to be as i written but both signs positiv... ?
hi, why is the following true? Let $f : \mathbb{C} \rightarrow \mathbb{C} \cup \{\infty\}$ be meromorphic, and suppose $\mathcal{F}$ is a family of meromorphic functions define on the complex plane, such that for every $g \in \mathcal{F}$, $g^{-1}(1) = f^{-1}(1) \neq \emptyset$ and $g^{-1}(2) = f^{-1}(2) \neq \emptyset$. Then for any $a \in f^{-1}(2)$, there is some $B_{\epsilon}(a)$ for which $g(B_{\epsilon}(a))$ does not contain $1$ for any $g$
we have no additional assumptions on the family $\mathcal{F}$ by the way
@MadSpaces Don't neglect the integration constants. $-\sin(u)$ is just a shifted $\cos(u)$.
Oh wow yea i would not have thought of that!! Thanks!
@MadSpaces Take a look at the properties of the Gudermannian function en.wikipedia.org/wiki/Gudermannian_function
The Gudermannian's a bit obscure these days, but kinda cute. And its inverse is useful if you want to make Mercator maps.
Yea i just heard of it today, i will check it out
4:17 PM
I suppose it's also relevant in Special Relativity because it maps rapidity, $v/c=tanh(w)$, to a sin parameterisation of speed.
oh wait, im dumb, just let $B_{\epsilon}(a)$ be a neighbourhood of $a$ not containing any points in $f^{-1}(1)$...
The Gudermannian function is a way of connecting the hyperbolic & circular functions for people who're allergic to complex numbers. ;) But it also shows you some stuff that may not be immediately obvious when you just think of the hyperbolics as the circulars with imaginary args.
@MadSpaces the graph resides in $\mathbb{R}^3$. $a$ would be something like $(a_x+\epsilon,a_y+\epsilon,a_z+\epsilon)$
and so far, i have the following in the proof: for all $\epsilon>0$, $B(a,\epsilon)\cap\mathcal{G}(g)\neq\emptyset$
sorry, i've mixed up things a bit.
let me clarify.
don't mind the first message. start with the second one.
though, we are working on $\mathbb{R}^3$, that stays.
5:04 PM
@love_sodam I don't think it's possible. I don't know how to prove it, but it wouldn't surprise me if there's a clever argument involving interlocking rings of 8 squares. It's easy enough to ensure all 7 numbers are in any 3×3 block, eg the following Latin square.
0 1 2 3 4 5 6
2 3 4 5 6 0 1
4 5 6 0 1 2 3
6 0 1 2 3 4 5
1 2 3 4 5 6 0
3 4 5 6 0 1 2
5 6 0 1 2 3 4
6:01 PM
any1 can help?? my calculus sucks...
@love_sodam Nice!
@Jam: You should not use the same notation for a double integral and a line integral. Write the double integral in polar coordinates and be careful.
P.S. Don't use photos. They're obnoxious. Type things out here in MathJax.
@MadSpaces i've been trying to make our strategy work, but i think it won't.
it boils down to proving the original set is closed. our reformulation doesn't give us anything useful.
or maybe i should say, it boils down to proving that the original set has an adherent point that is outside.
if $u(x_0)=\fint u(y)dy $ over the whole $B(r,x_0)$ then $u(x_0)=\fint u(y)dy $ over $ \partial B(r,x_0) $
and this is not particularly easier than proving the original claim.
6:11 PM
@Jam: So you're ignoring my first comment. Assume $n=2$ for starters. Do NOT use the same notation $dy$ for both the double integral and the line integral over the boundary. You will make mistakes like crazy that way.
im not ignoring it i wanted to state the problem first hehe
and i think we should use some lemmas to prove that the graph $\mathcal{G}(g)$ is closed. to recall, we are trying to prove that $\{(x,y,z)\in\mathbb{R}^3|z=0,(x,y)\in[0,1]^2\}$.
Well, it's in the statement, isn't it?
but i cant even write it haha suppose i translate the first integral as a double integral from 0 to r and an integral to the boundary then?
I don't think it's literally correct, by the way. There are going to be some constants.
When you differentiate with respect to $r$, two things happen. The region changes and the function changes. You are going to get two terms.
6:15 PM
@TedShifrin could you guide me on the problem i've mentioned above?
@Jam Do you also know that $u$ is harmonic?
no :/
@sevdaicmis You've written something silly. $g(x,y)=0$ is a function?
i was about to ask. that would suit the problem nicely.
@sevdaicmis The problem as stated is certainly not correct. You'd better assume $g$ is continuous.
6:17 PM
i guess if it harmonic you can show that the integral at the boundary is the value at the centre and then show it is also equal to the integral over the whole sphere
yes, that's the function. i've had written earlier, and forgot rewrite it.
Well, you also have a Green's identity to use for integrating $\partial u/\partial n$.
i believe that the hypotheses do imply that u is harmonic, but that seems harder than what you'd do if you didn't know that.
though, in the end, it's not necessarily relevant. the goal is to show that the set i've written above is closed.
but now starting with the integral over the sphere i need to somehow differentiate and get the integral at the boundary at least what my intutition says but literally i dont know what im allowed to do or how
6:18 PM
@sevdaicmis No, the $g(x,y)=0$ doesn't belong there. You're talking about the graph of a continuous function $g$ on $[0,1]^2$. You want to prove the graph is a compact subset of $\Bbb R^3$. Easiest to use sequences.
@sevdaicmis What set is closed? This is getting frustrating.
Write $\int_0^r\int_0^{2\pi} u(r,\theta)r\,d\theta\,dr$ and you want to differentiate with respect to $r$. As I said, constants need to be taken care of, as well. You need to use the Leibniz rule — differentiating with respect to $r$ will have a term from the Fundamental Theorem of Calculus and a term differentiating the integrand with respect to $r$.
well, yes. the question was to find a two variable function of which the graph is compact. i've chosen $g:[0,1]^2\to\mathbb{R},g(x,y)=0$. isn't $\{(x,y,z)\in\mathbb{R}^3|z=0,(x,y)\in[0,1]^2\}$ the graph of $g$?
So you want to show that the graph of the $0$ function is closed. This is totally easy. It's true for any continuous function. What is your issue?
i'm trying to prove this with the basic way, that the complement of it is open.
So you do not know that the preimage of a closed set under a continuous function is closed?
$$u(x_0)=\int_{B(x_0,r)} u(y)dy= \int_{0}^r\int_{\partial B} u(y)dS(y)dr$$
6:25 PM
Can you do it your way in $\Bbb R^2$, where you can easily draw pictures? The same proof will work in $\Bbb R^n$ for any $n$.
i know. but chronologically, i got this question before it is mentioned in the course.
@Jam Be careful. What does $dS(y)$ mean?
OK, @sevdaicmis. So show the complement is open. Where is there a difficulty?
i can't just take an arbitrary point in it and try to construct an appropriate ball around it, right? that's my concern.
In it? You mean in the complement? Yes, of course you can.
because, there will be different cases around the graph.
yes, i mean the complement.
hmm, alright.
6:28 PM
No, no different cases.
i see. i thought i have to divide the set about 4 parts and construct corresponding balls.
Four parts? Where does that come from?
like, the above, bottom, right and left of the graph.
@TedShifrin surface integral
I don't like writing $dS(y)$. But OK.
6:31 PM
because, i don't see a general way to select an $\epsilon$ for radius of the ball. that's why i divide the set into four parts, to more easily construct balls.
@sevdaicmis That makes absolutely no sense. What do right and left mean?
Oh, because of the square. There are going to be 16 pieces if you do that.
Hello, please a hint to prove that $\sqrt{|x|}-\sqrt{|y|}\leq \sqrt{|x-y|}$ for all real x,y.
yeah, it's ill-defined. let me explain.
Oh, maybe not quite 16.
You don't need to make it that complicated, @sevdaicmis.
Start in two dimensions so you can draw pictures easily.
yeah :)
6:33 PM
@TedShifrin thats how my prof writes it.. and now what im supposed to do with that equality differentiating would give me zero on the left side..
Yes, but if you read what I said above, you'll get two terms on the right.
oh nice so its something like leibniz rulle ? because evans says if u differentiate that you just get the boundary integral
I believe the result is wrong, by the way. I think you need to have average value, hence constants depending on $r$.
i was wondering if that's what the slash meant in the integrals.
ye the integrals are averages but i didnt know the latex symbol
6:35 PM
you mean, i start with that $\mathbb{R}^2 - [0,1]^2$ is open, right?
So average is not the integral. You have to divide by area/volume.
yes exactly
ohhh and evans's identity is for integrals not averages
@leslie I had no idea what the slashes were.
Someone have an idea please
i had this and was confused
calc is so hard haha
6:37 PM
All of math is hard if you don't do it carefully.
@sevdaicmis Are you now working on the two-dimensional problem or your original problem? I'm lost.
There's a MathJax question on the mother meta. meta.stackexchange.com/q/372313 Is there a directive for a diameter symbol ⌀ in MathJax?
That looks like a \varemptyset or something.
Diameter? Huh?
My best effort: $\bigcirc\hspace-12mu/\hspace8mu$
i don't change the problem. you said "start in two dimensions", i figured you mean the two dimensional case.
Hmmm. The size works differently in an Answer preview.
6:40 PM
So you're starting with the graph of the zero function on $[0,1]$ and looking at the graph in the plane.
@TedShifrin Yep. Engineers use it. It's a bit like the emptyset symbol, except it has to be circular, whereas the emptyset symbol often looks like a slashed zero.
now i'm lost. are you suggesting to work on that first?
So you're going to need to use \rlap or \llap and fiddle @PM2.
That was my suggestion, yes, @sevdaicmis.
There are basically only two cases, @sevdaicmis, if you say it right.
see if you can fit that into your problem perhaps via case analysis
@leslie Doesn't the multiply by the conjugate trick work nicely if you set $x-y=h$?
6:45 PM
@TedShifrin alright i'm following now. i would probably split it for $y>0$ and $y\leq0$.
No, why is $y<0$ any different from $y>0$?
ted: probably? one of the answers in the link kinda does that. i try to solve by google whenever possible.
@TedShifrin yeah, they can be unified. but we are left with something like $x>0$ and $x\leq0$, and this is not much different.
I never do that.
i can't think of anything else.
6:48 PM
No, @sevdaicmis. Only when $y=0$ is there an issue. And then you know that $|x|>1$.
My "I never do that" was to leslie, but anyhow ...
So you should have two cases: $y\ne 0$ and $y=0$; in the latter case you know that $|x|>1$.
Make sure you see what's going on in a picture.
$$0=d/dr \frac{ \int_{B}u(y)dy}{a(n)r^N} <=> \int_{\partial B} u(y)dS(y)=N/r \frac{ \int_{B}u(y)dy}{a(n)r^N} <=> \int_{\partial B} u(y)dS(y)=\frac{N}{r} u(x_0) $$
i'm too lazy for my own good.
Of course $N$ is $n$. The numerator depends on $r$, too. Your $B$ is $B(r)$.
And you need to relate the surface integral divided by the area of the sphere. So how is $a(n)r^n$ related to the surface area of the sphere?
ye n is N
Maybe you skipped too many steps for me.
But something's quite wrong.
6:55 PM
i just did a product rulle differentiating
So what is $d/dr \int_{B(r)} u(y)\,dy$?
the integral on the boundary
from evans
the left hand side on what i wrote
OK. And then the RHS is coming from differentiating $r^{-n}$? You did that wrong?
thats -nr^(-n-1)
@leslietownes Well, you should be lazy when some people expect us to do everything for them and could have googled themselves.
@Jam Which is $-nr^{-(n+1)}$?
6:58 PM
i kept the r^-n to make it an average
so im left with -n/r
which is + when i take it to the right hand side
and the average becomes the value on the centre from the hypothesis
Write out the product rule carefully. I think there's an error. And we want the average over the sphere, too, so we expect to divide the surface integral by $r^{n-1}$.
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