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9:00 PM
@AkivaWeinberger That was my thinking, so I was wondering if there was a more specific term.
 
Yeah in 3D perhaps a surface
Actually a surface is the graph of a n-variable function isn't it ?
So it is more specific than a manifold but no in the "how many dimension does it have" way
I might be completely wrong though
 
"Surface (topology), a two-dimensional manifold"
and others
 
Okay, I'm completely wrong :)
 
Thanks!
 
Hi guys
question
is there any particular relation between the cross product of two vectors u and v and the projection of u onto a plane to which v is the normal vector?
in general, is there a relation between cross product and projections?
 
9:17 PM
Let $f\colon\mathbb R^n\to\mathbb R$ be a function, such that for all $x\in\mathbb R^n$ the partial derivative $(D_jf)(x)$ exists. Show the following: if $(D_jf)(x)=0$ for each $x$ and $j$, then $f(x)$ is constant.

So I began as follows: Assume $f$ is not constant. That means there are distinct $x_1,x_2\in\mathbb R^n$, such that $f(x_1)\neq f(x_2)$. Now I would like to use the mean value theorem along with the following function: consider $a\in\mathbb R^n$, and let $g_i(x)=f(a_1,\dots,a_{i-1},x,a_{i+1},\dots,a_n)$. We know that $(D_if)(a)$ exists iff $g_i'(a_1)$ exists. However, I'm kinda
 
That's not how I would do it, I'm not sure there is a clean way to make your idea work
 
@nbro Not that I know of
 
but suppose we have another vector w
 
@TedShifrin by researchers, I presume you are implying PhD's, post-docs and professors in the field?
 
@nbro You know the formula for the length of a cross product is: $\vert a\times b\vert=\Vert a\Vert\cdot\Vert b\Vert\sin(\theta)$ ? where $\Vert a\Vert$ is the length of vector $a$, and $\Vert b\Vert$ is the length of vector $b$
@Astyx oh okay, any other ideas?
 
9:21 PM
and we calculate $w \times v$, $u \times v$, $proj_P(u)$ and $proj_P(w)$..
 
Consider the one variable function $t\mapsto f(ta)$
Where $a\in \Bbb R^n$
 
somehow the ratio of the cross-products and the projections must be related
I simply don't know how to explain it
but I know that $||w \times v ||$ is somehow equivalent to $||proj_P(w)||$, where $P$ is the plane to which $v$ is the normal
if we consider both $w$ and $u$ at the same time
in other words, if $|| proj_P(w) || = ||proj_P(u) ||$, then $||w \times v|| = ||u \times v||$
 
@nbro Hm, could be.
Maybe they're equal if $v$ is a unit vector
 
@Astyx yea but, that way all our coordinates are being shifted with the same rate?
 
Hi @Akiva
 
9:27 PM
If $x_1$ is not multiple of $x_2$, how is this one-variable function going to help us?
 
@ShaV It's simpler than that, what is the derivative of that function ?
 
@Astyx I can only write out the definition for now...
Say $g(t)=f(ta)$, then $\begin{align}g'(t)=\lim_{h\to0}\frac{f(ta+ha)-f(ta)}{h}\end{align}$
 
Oh I see ...
Then you can do it by induction on $n$ with a smilar method
 
okay, I'm going to try that
 
let me reason
we know that $||w \times v|| = ||w||*||v||*|\sin(\theta)|$, where $\theta$ is clearly the angle between $w$ and $v$
 
9:40 PM
@SoumyoB Yes. There are, of course, some of us professorial types and plenty of grad students on MSE, but the serious research-level questions are on MO.
@Sha: Unless your two points line up on a coordinate direction, you'll have to use the chain rule with that approach.
 
Hi @Ted
 
@nbro: Your projection idea isn't going to work.
You need a vector in that plane that is also orthogonal to $w$ ...
Plus there's a question of the length and orientation of $v$.
rehi Zach.
 
and the scalar projection of $w$ onto the plane to which $v$ is the normal vector, should be $||w|| \cos(\beta)$, where $\beta$ should be equal to $\pi - \theta$...
 
@Ted She can show that the function is constant on lines, thus remove the coordinates one by one and thus avoid chain rule
 
@Ted If I know how to do a problem / seen it before, should I still just jot down my answer?
 
9:43 PM
Sure, @Astyx. I didn't read the whole thing.
Lines parallel to the coordinate axes, you mean.
 
Yup
 
Zach: I'm more interested in your working on proofs now, but what are you referring to?
@nbro: To convince yourself that you're wrong, suppose $v$ and $w$ start off orthogonal. Then $w$ will be in that plane you want to project onto. Oops. Totally not the right vector.
 
Hey everyone!
 
@Astyx So if I consider these distinct $x_1,x_2$ for which $f(x_1)\neq f(x_2)$, then we could say that we can go from $x_1$ to $x_2$ by 'shifting' along the coordinate axes?
 
hides from evil Demonark
 
9:47 PM
Hi @Daminark !
 
cackles
 
Hi @Dami
 
Sure, @Sha, sorry for interrupting if that's what you and Astyx had in mind. Moving along lines parallel to the coordinate axes, yes.
 
yea sorry I actually meant to tag Astyx :P because it was his idea
 
How's it going @Sha and @Astyx? How's Ted been? I thought he'd be here right about now but I can't find him...
 
9:48 PM
Yup. You can do a more general argument later for any set that is called path-connected (you can get from any point to any other by a (piecewise) differentiable path). Then you can use chain rule and get a more general result.
 
Hi @Dami
 
Zach, you didn't tell me to what problem(s) you were referring. Any ideas about that infinite dimensional example?
 
@Daminark pretty good! busy doing math on a Friday evening :P you?
 
@ShaVuklia You don't have to suppose their image are different, you just need to take $x_1$ and $x_2$ and show $f(x_1) = f(x_2)$ (you could of course suppose it's not true and show that's absurd, but that's not needed here and could be quite confusing for the reader)
 
Ted's unasked-for opinion: Avoid unneeded proofs by contradiction.
3
 
9:49 PM
^ My point
 
What've you been up to @Meow?
 
@Astyx ah yea, I got the idea!
 
Doing linear algebra exercises.
 
And @Sha I've finally started getting productive
 
Some students just fall in love with contradiction, suppose not, give the direct proof and say they've contradicted the not.
 
9:50 PM
@Ted hahaha :P
 
And actually you only have to show $f(x) = f(0)$ for all $x$
 
It's because we usually hate it at first
 
@Ted to clarify, the perpendicular subspace of the perpendicular subspace of an infinite dimensional subspace isn't equal to itself, right?
 
@TedShifrin You're right, in that case the length of the projection would be 0 but not the length of the cross product
 
Or was I right. I don't know
 
9:51 PM
@astyx Ah, nice!
 
Zach: Orthogonal complements :) ... Needn't be, not isn't. :)
Oh, sorry, @nbro, I forgot you were talking just about lengths.
 
yeah
 
If $w$ is in that subspace, then it is itself the projection. I see ... my mistake.
 
@Ted Sorry, lol.
 
I'm interested in the lengths, not the directions of the vectors, which can of course be different
 
9:52 PM
@Ted That's the kind of thing I'd do to spite this one friend of mine who doesn't believe in the law of the excluded middle (and by extension, proof by contradiction).
 
I've always wondered : is there a difference between a proof by contradiciton and a proof by contrapositive ?
 
Yes, @Astyx, definitely.
Look at the logical structure.
Try to prove $\sqrt2$ is irrational by contrapositive.
I see, @nbro. Then you're right, as the angle between $w$ and its projection is the complement of the angle between $v$ and $w$. ...
 
Hey @s.harp and @semi!
 
@TedShifrin yeah, that's exactly what I thought
 
9:55 PM
@Ted Let me look for a counter example
 
To what, Zach?
Hi @Semiclassic
 
Well in some sense we have $\sqrt 2 \in \Bbb Q \implies \text{False}$, so by taking the contrapositive, $\text{True} \implies \sqrt{2} \notin \Bbb Q$
 
Good luck with that, @Astyx.
 
I mean, an infinite dimensional subspace whose orthogonal complement's orthogonal complement isn't itself
 
Hi @ted
Hi chat
 
9:57 PM
Look at the logical structure. Contradiction is to show that $P\land \lnot Q$ leads to a contradiction. Contrapositive is to show that $\lnot Q$ implies $\lnot P$.
Zach: Go back to my discussion of this. I gave you the $V$ to consider!
Hi @Alessandro
 
Hi @AlessandroCodenotti
 
Hi @Alessandro
 
Oh, the set of vectors with finite length?
 
Look carefully.
Every vector in there has finite length.
Oh, you mean different sort of length.
 
Well, $P\implies Q$ is the same as $\neg P \lor Q$, and the statement $\neg(P \land \neg Q)$ should be equivalent to that as well, no?
 

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