I was just reviewing a series on inner products, and it declared that Legendre Polynomials correspond to different pitches of musical notes. youtu.be/pYoGYQOXqTk?t=446 I can't find anything about this on the Wikipedia page, and everything I've learned about, i.e., sound synthesis, uses sines or cosines as a basis for sound. Is there an alternative framework where Legendre Polynomials can be used as a basis for sound, or is the idea just that you can theoretically approximate the sines
or cosines using Legendre Polynomials in a Maclaruin-Series-ish way?
but i'm listening to music and the music can't stop
when i say 'common analogy,' haha. as common as the good old discussions of legendre polynomials. but, once i've heard before. and maybe used.
for purposes of the analogy i've heard, which might not be his, they don't really have to be polynomials, i think you want them to be orthogonal, and also complete in the norm induced by the inner product, so you can decompose all of your 'sounds' or signals or whatever you want to call them into your 'notes' for the analogy.
I guess a decomposition into Legendre Polynomials would be more like a Maclaurin/Taylor series in that sounds would converge at a point rather than a region like in the Fourier case, which might be bad.
When you're free @TedShifrin , give me a ping. I wanted to tie up some loose ends on two problems of yours that I was working on and I feel I'm almost done. But I'm stuck at the last steps
actually I'm sure you could help me @leslietownes with one of them since it is specific to section of the book....actually you might be able to help with both....
when i was in college i knew a guy who tweaked around with stereo equipment and guitar amps and things. someone had told him somewhere that things would sound better if his speaker box was dimensioned like the 'golden ratio.' so he made one. i remember googling it and there was TONS of this stuff out there from people in acoustics and sound design about it. including lines of amps you can buy that do this.
they tend to cite the ancient greeks or just random stuff about intervals of sound so i assume it is all BS, or a classic case of, sometimes we don't want things too narrow or square either.
but i dunno.
dc3rd i'm certainly here, may be kind of mentally in and out but here.
I'll ask you the topological one first then. Specifically I'm showing $X \times Y = \{(\mathbf{x},\mathbf{y}) \in \mathbb{R}^n \times \mathbb{R}^p: \mathbf{x} \in X \ \text{and}\ \mathbf{y} \in Y\}$ is a $(k + l)$ dimensional manifold. Given the assumption that $X$ and $Y$ individually are $k$ and $l$ dimensional manifolds respectively.
there's a lot of stuff in sound where if you go too far in any one direction it sounds bad, but somewhere in the middle is just right. a lot of the golden ratio in visual art is basically (1) it being shoehorned into works that weren't based on it if you actually measure them, or (2) a trend beginning i think in the 1700s or 1800s (when it really took off) where artists thought it was what they were supposed to use.
Shoot, now I want to patch together a Legendre Polynomial software synth, and also visualize what happens when you take a Taylor approximation of a Fourier approximation of a Taylor approximation of a Fourier approximation, etc., of some function. The useless programming projects proliferate :/
by assumption $(x_{n-(k+1)}, \dots x_n) = \phi_X(x_1, \dots, x_k)$ and $(y_{p-(l+1)}, \dots y_p) = \phi_Y(y_1, \dots, y_l)$ are $\mathcal{C}^1$ graphs.
I defined $\phi_{X \times Y}(\phi_X, \phi_Y) = (x_{n-(k+1)}, \dots x_n, y_{p-(l+1)}, \dots y_p)$. This is a composition of $\mathcal{C}^1$ functions so it is itself $\mathcal{C}^1$
But is this enough to claim that $X \times Y$ is a $(k+l)$ manifold? since I have said graph of it?
@leslietownes he would......lol....that's why I'm here cowering....🤣
manifold is defined as the set $M = f^{-1}({c})$, where at each point $a \in M$ we can locally represent $M$ as a graph over one of the $n$ coordinate hyperplanes.
and more thoroughly: a subset $M$ is an $(n-m)$ manifold if each point has a nbhd that is a $\mathcal{C}^1$ graph over the $(n-m)$ dimensional coordinate plane.
I'm waiting for Ted to come in hot and just shoot down my idea with a quick witted remark......I've become Pavlonian to his reactions....
well whatever f is a function on and hypotheses are on f the key here is locally. like our friend the unit circle is a 1 manifold but isn't a graph of any function of one variable. you can do the stuff like sqrt(1-x^2) and -sqrt(1-x^2) or even the graphs of sqrt(1-y^2) and -sqrt(1 - y^2) but we maybe needed both coordinate axes to graph with because sqrt(1-x^2) for example isn't C^1 at -1 or 1 and nor is its flipped cousin.
but around any point on the circle there's one of these graphs, where it's on it and the function is C^1 around it.
i'm not so much formlaly trying to engage with ted's definition which i still haven't seen all of, just trying to use graph the way normal people use it in calc 1.
if you go back to that locally business, proofs for basic stuff stay local. so here for example, to show X x Y is a manifold by whatever definition, might pick a point (p,q) in XxY and try to show that, at least somewhere near there we've got what we need.
where we might use different things at different points.
so your idea is kind of sound as a vibe but all of this definitional stuff really matters.
So it seems it would require me to be more detailed in what I defined. Specifically near whichever of the trouble points I may encounter we would use one of the given two graphs to deal with the point
it'll be something like, we need to show X x Y is locally whatever. pick any (p,q) in X x Y. because p is in X and X is assumed to be a manifold, we have some cool googles we can put on so X looks like a graph around p.
and same with Y. and then (this was your vibe) somehow just show you can stitch that data together because all we're doing is cartesian producting here.
to get whatever we'd better have at (p,q) if X x Y is going to be a manifold.
i think i could get a C in ted's class just with what i wrote above. he would have to acknowledge my attempts.
C on this problem. partial credit. not a C in his class, that's out of reach.
it would be like 'please try harder, come to office hours, 5/10'
mah...I put that one to bed for now...This was the onew where I had to plug and chug:
have this level curve $f(x,y) = 4y^3 - 3y - x = 0 $
given this as one of the local graphs $\phi_2(x) = \cos(\frac{1}{3}\arccos(x)),\ x \in (-1,1)$, and verifying it satisfies $f$
So after a bunch of algebra and using trig identities I've arrived here (may need to start chatjax for this) : $\cos(\frac{1}{3}\arccos(x))[2 \cos(2(\frac{1}{3}\arccos(x))) - 1] - x $
I need to figure out some sort of way to be able to get my $\cos$ and $\arccos$ to come together so I can do some cancelling....that's what I posit...
So the crucial conceptual point in the other remarks in the implicit function theorem. As long as $\nabla f(a)\ne 0$, you can locally near $a$ write the level hypersurface as a graph over one or more of the coordinate hyperplanes. Hence that notion.
triple angle formula?.......I've never encountered this......in spanish there is a saying "no apto para cardiacos"....."this isn't apt for those who have cardiac problems"......this particular algebraic manipulation is that...
Yup, triple angle formula. Best way to get it is to use deMoivre's formula, but if you don't know $e^{i\theta}$ stuff, just work it out from $\cos(2\theta+\theta)$.
@leslie Update: The maintenance guy did snake my drain and got new parts to put it back together with no leaks. The bad news is that the snaking has flooded the people down below me — I'm sure to win a popularity contest for that.
I'm just thinking off the top of my head, is it possible that even though I may have the two individual manifolds could there be a point that once we cartesian product them that could mess things up....
one time i was living under a person who decided to garbage dispose like a pound of raw carrots for some brain-charred reason. about half of them made it about halfway down the pipe, right below us, then clogged. she kept disposing the carrots and carroty water was backing up into our kitchen, out of our sink and all over the floor.
Yup, a different maintenance person just came to tell me I was flooding my downstairs neighbor. I had to fill him in on the history and the snaking. And I did apologize.
@dc3rd No, I'm assuming everyone can do the addition formula.
i mentioned one time i ta'ed for a 16a instructor who spent far too much time and grader energy on making people write the domains of trig substitutions. you know, because those folks are gonna need that again.
i mean, i abstractly understand why he might want people to do that, but those folks are gonna make algebra errors and such anyway and really? every time?
unless you could highlight to them the ideas of periodicity and such....but coming from that sort of class originally before the math bug took over my soul. The majority aren't even looking at things with that sort of perspectve
@TedShifrin this is where I come in......in my future self...in some way
i've forgotten what those terms mean but i did know at one point. it's always something like, get the idea by drawing a big dumb lattice, or prelattice, or whatever, and then turn it upside down or put it in the mirror or something.
i once flagged my old answer as I thought there was a mistake in it but since it was accepted, it could not be deleted so I flagged it but the flag was denied.
the desire is to show that if g and h satisfy the condition g^2 > 4h then the next inequality hold, but that there are also values of g and h that do not satisfy this condition that will also make that next inequality hold.
well, it's connecting the rref form to the determinantal definition of rank. the structure helps you evaluate the minors.
for the column space definition it helps to know that row operations don't change its dimension. this is doable in a couple of different ways. then when you're in rref, it should be visually clear that the r columns with pivots in them will be linearly independent and any remaining nonzero columns are in the span of the ones containing ones.
can also be proved algebriacally although i've never found an algebraic formulation of what rref "is" very comfy to work with.
might help to do an example. i'm looking for a non-crappy picture of a matrix in rref. math.se is letting me down.
the pivot columns are linearly independent because if a combination of them is the zero vector, just look at the pivot entries in the columns and read off that the coefficients of the linear combination has to be zero. all of those zeros in the rref help you see that algebraically.
and the non-pivot column is in the span of the first two pivot columns and you can even just read off what coefficients to take to write it as such.
the missing ingredient here is that if the original matrix looked very different, it might not be initially clear that learning the column space of the RREF should have anything to do with the column space of the original matrix. and they can be different spaces, in fact. but they have the same dimension. proving that depends on how you like to think.
fans of minors will spot the column you can remove to get a 3x3 determinant that is easy to evaluate. and there are no 4x4 minors to check.
Yes, it do works, What I thought was The Pivot Columns forms the basis of vectorspace having each of the PCs as a part of basis( like ijk) and also the rank as its dimension
if the rref of M is denoted rref(M), the PCs form a basis for CS(rref(M)), which can be a different vector space from CS(M). but the two column spaces are guaranteed to have the same definition (if nothing went wrong in the row operations :)
consider e.g. M the nxn matrix of all 1s. column space is the set of multiples of the column (1,1,1,...,1). RREF zeros out all but the top row of M (which it leaves unchanged) and the column space of that is the set of all multiples of the column (1,0,0,...,0). different spaces but helpfully same "size".
one of my favourite texts is, surprisingly, Rockafellar's "Convex Analysis". however, he does something i find distasteful which is he sometimes takes the union of objects from different spaces. Not an issue from a set perspective, but somehow ugly to me.
For example, the union of a subset of $\mathbb{R}^n$ and a set of equivalence classes of subsets of $\mathbb{R}^n$.
i find it disturbing, i generally assume that sets are subsets of some base set that has some uniformity.
he takes the convex hull of a the union of a set of points and a set of directions. a direction is an equivalence class (by translation) of half lines.
it adds an extra step when untangling results and making sure i really am understanding correctly.
i think i've seen that a few times before but not in my own work. like if you mod out by a really ugly equivalence relation where the classes have little to do with one another, or maybe the relation's not even defined at some 'edge' of your space where you'd like it to be. i could see doing things like that. same way you might not want to distinguish between points and equivalence classes that consist of single points.
he does have a way of viewing the sets in $\mathbb{R}^{n+1}$ that unifies in some sense, but the convenience of his expression is not compensated by the increased vigilance needed while reading.
This can be solved using elementary algebra just solving a quadratic polynomial. Look at the following derivation:
Given line segment P defined by points P1 and P2 and line segment Q defined by points Q1 and Q2 we can define the ray P(t) as:
P(t) = P1 + t V
Where t is a positive scalar and V i...
@TedShifrin I added as much detail as I could. Please check out my final edit. The actual problem is related to attribute grammar and CFGs so I am trying to avoid talking about that.
i'm kinda wondering if "isomorphic to" in that q means something weaker (i.e. not order isomorphism but order monomorphism). and if there are other conditions, probably implicit in this attribute grammar cfg world about what would be relevant to it, but maybe not baked into just the math there.
There's actually nothing wrong with Einstein, and it makes a good check that things are in fact well-defined. But I always put in the summation symbols, regardless.
With complex geometry, there's the additional twist of barred (conjugate) indices, so that we get either ${}_\gamma {}^\gamma$ or ${}_\gamma {}_{\bar\gamma}$, etc.
@monoidal I don't know why that makes him talk about a $1$-form.
me too. He sais $\nabla_{i}\nabla_{j}f=\nabla_{i}(\nabla_{j}f)$ and $\nabla_{j}f$ is a one form and we have $\nabla_{\mu}\omega_{\nu}-\Gamma^{\rho}_{\mu , \nu}\omega_p$ for a one form $\omega$
i had difficulty of making sense of many QM-flavored treatments of operator theory for this reason. ok, bra and ket, weird looking but fine. even a lot of the impressionistic scary looking use is sensible once you have the right definitions, which often are provided. but then you turn the page and run into what is basically a homework problem called 'figure out what i mean by this.'
yeah, my prof is making DG very unappealing to the students that have no background in DG, he's expecting us to guess by what he means in more than one instance
So the point is that we're really applying the connection to the actual $1$-form (not its components) $\omega_\nu\,dx^\nu$. The second term in that formula comes from $\nabla_\mu dx^\nu$.
it's tough. particularly in DG when in so many places a few assumptions or extra structure will allow correspondences between formally different things.
To me it's a directional derivative. If he is calling it a $1$-form, it's because it's the coefficient function of a $1$-form, as I said, namely (unsummed) $\partial_\gamma f\,dx^\gamma$. I cannot continue to be in the middle on this. Go bother him.
I guess this shows me what gratitude is. Stays locked into his way of thinking (which I truly do not understand), and makes no effort to understand what I think is the right way to approach the problem.
Is calculating singular homology of $\mathbb{S}^p\times \mathbb{S}^q$ straightforward? Do we have a formula relating homology of $X\times Y$ with homology of $X$ and homology of $Y$?
Given $u,v \in \mathbb R^2$ and it is also given that $u$ is a scalar multiple of $(1,3)$, $v$ is orthogonal to (1,3) and (1,2)=u+v. Then, I want to find u and v. My question is: we must assume inner product on R^2 as "dot product" here or not?
might be worth thinking whether the given data is compatible with an arbitrary inner product (or at least some other ones), and if so, working out a more general solution. but that is a sideshow and not the exercise.
yeah, that was insane. i remember someone saying he had been really long lived and looking him up while in grad school. had he lasted just a few more years, i could have bothered him with my homework.
without dot product and considering general inner product, we can say that $||(1,2)||^2= ||u||^2+||v||^2$ and $\langle(1,2),(1,3)\rangle=\langleu,(1,3)\rangle$ and this actually determines $u$ as u is scalar multiple of (1,3).
I thought about projection so here $(1,2)=u+v$ is like projecting $(1,2)$ onto span$(1,3)$ @professor Ted.
In this case, perhaps you're being too geometric, @monoidal, for starters. If I fix one of the spheres, what should I do with the other one? (Yes, suitably interpreted, these things will be "cylinder-like.")
The problem is that to write macros for in here we have to introduce them each conversation. In my own books, I had something like \id#1 (for ideal) and \ip#1 (for inner product) which put the langle #1 rangle. :)
the social sciences even have a few latin phrases that some people lean on. it seems to be a very much, either you never use that stuff, or it's in everything.
This can be solved using elementary algebra just solving a quadratic polynomial. Look at the following derivation:
Given line segment P defined by points P1 and P2 and line segment Q defined by points Q1 and Q2 we can define the ray P(t) as:
P(t) = P1 + t V
Where t is a positive scalar and V i...
