It can be useful to have the ability to think of functions as vectors. valen.summerresearch24@gmail.com
Пікірлер: 134
@valenfeldmann645 ай бұрын
Technically I think its mathematically correct to say "a set of functions with certain properties form a vector space" (For example the set of continuous functions on the real line form a vector space). For the purpose of Fourier transform intuition, let "function" just refer to the subset of functions that are a continuous mapping from the reals to the reals (although the definition of function is more broad than that).
@rizalpurnawan37965 ай бұрын
Yes, that's the abstract algebraic definition of vector space, a set whose elements made of a field equipped with additive and scalar multiplicative operations satisfying some axioms related to linearity (or vector space axioms). What commonly pops out of our heads when we hear vectors are arrays of numbers. In fact, it is an example of vector space, known as finite dimensional vector space. Sequences also form a countably infinite dimensional vector space. And functions from complete domain forms an uncountably infinite dimensional vector space. This subject is further studied in functional analysis. I know abstract mathematics is truly fascinating.
@v12985 ай бұрын
Lp-space and "linear space", are those the same things as the "vector space" here? I'm assuming the vector way of seeing these functions is related to norms and such?
@rizalpurnawan37965 ай бұрын
@@v1298Lp-spaces are in fact normed vector spaces. Every L-space is a vector space. The key concept for Lp-space is that every element of Lp-space has a finite p-norm. If you're asking "the vector way of seeing these function is related to norms and such?", well yes. In fact you can prove that every element of Lp-space (every measurable function with finite p-norm) satisfies vector space axioms. Another example of functions which form a vector space are continuous functions. In fact, continuity is a stronger notion than measurability. Additive and scalar multiplicative operations preserve continuity of functions. This briefly describes how continuous functions also form a vector space.
@rizalpurnawan37965 ай бұрын
@@v1298 Every Lp-space is a normed vector space, meaning that is a vector space equipped with a norm. We can paraphrase an Lp-space as a family of measurable functions whose every element has a finite p-norm. You can in fact prove that an Lp-space is a vector space by showing that all vector space axioms hold for an Lp-space. About your question "I'm assuming the vector way of seeing these functions is related to norms and such?", my answer is "yes", since the way we identify whether a function is an element of an Lp-space is showing whether this function has a finite p-norm. About "linear space" that you mentioned, if what you mean is the family of linear maps between a pair of vector spaces, then such linear maps do form a vector space. Note that the underlying vectors spaces mapped by the linear map may be either finite dimensional vector spaces or an infinite dimensional vector spaces such as function spaces.
@v12985 ай бұрын
@@rizalpurnawan3796 Great explanation. Would you believe I passed applied mathematics (where we learned this) just a month ago and already forgot? I'm feeling like a lost cause but people like you help me back on track.
@OhPuree425 ай бұрын
I still remember the first time I understood this in a quantum mechanics class. It was mind blowing
@o0QuAdSh0t0o5 ай бұрын
What kind of Maths is required for Quantum Mechanics?
@sheepcommander_5 ай бұрын
@@o0QuAdSh0t0o calculus is one of them!
@OhPuree425 ай бұрын
@@o0QuAdSh0t0o mostly basic linear algebra but also Fourier and functionnal analysis
@rizalpurnawan37965 ай бұрын
I'd say functional analysis. I am not a physicist, but I happened to read a book about quantum mechanics. The treatments in quantum mechanics are quite similar to measure theoretic probability theory. I thought it was a direct implementation of formal probability theory. But, the pioneers of quantum mechanics gave birth the theory earlier than Andrey Kolmogorov formalized probability theory. I heard that David Hilbert, a famous mathematician, was among the pioneers of mathematical formulation of quantum mechanics. Perhaps, I am mistaken in this, am I not?
@OhPuree425 ай бұрын
@@rizalpurnawan3796 you might have read a book written by a mathematician, and of course you can want to have a very detailled mathematical formalism of quantum theory. However, as a physicist I clearly don't see how measure theory would help me in a quantum physics problem. I'm sure it is studied by some theoretical physicists but at this point they are just doing maths and it is not "basic quantum mechanics" anymore.
@Lavabug5 ай бұрын
Beautiful. I have a phd education in physics and never once thought of a function as a column vector with infinite components. The integral form of inner products makes a lot more intuitive sense now, I typically just thought of whatever the integrand was getting multiplied by as just as a "discrete" basis vector in some direction, but it was a column vector all along!
@Manisphesto5 ай бұрын
Vectors are functions.
@jacobweinstein91365 ай бұрын
Not true. They are "object functions" in the categorical sense, but that just means that they return only themselves. Matrices (linear transformations) are functions with far more outputs than just themselves.
@octopodes76195 ай бұрын
@@jacobweinstein9136Viewing finite dimensional vectors as functions from finite sets into a field (i.e. treating an arbitrary finite set as a basis) can be quite helpful
@suppositorylaxative31795 ай бұрын
@@jacobweinstein9136not ‘not true.’ They may not be morphisms in the categorical sense, but you can still think of vectors as functions. I.e you can think of n-tuples of a field as functions from a set of n elements to a field. You can think of the free vector space of a set A as consisting of k-linear combinations of set functions from A to a field k.
@enricgarrigasanchez2605 ай бұрын
@@suppositorylaxative3179I agree with you in that strictly speaking, a vector can represent (and in maths a lot of the time when we say "to represent" we consider it as good as "to be") functions. However I liked the insight @jacobweinstein9136 provided because matrices are functions in a much much deeper way than vectors are
@angeldude1015 ай бұрын
Any element of any group can be thought of as a function on said group. Since vectors can be added, any given vector could be considered a function that takes another vector and outputs the sum of the input and function vectors.
@davethesid89605 ай бұрын
I've never thought about Fourier transform like that. Thanks!
@dougiehwang91925 ай бұрын
Mind Blowing!! Each time I dig into Linear Algebra, I open my eyes. I believe there will be more eye-opening moments.😮
@Joao-uj9km5 ай бұрын
Great video. I had good intuition on functions as vectors, but not in the sense of infinite (and uncountable!) dimensional vectors. Thanks for the video!
@silas64465 ай бұрын
"take the dot product with itself and square root the result" just made me realize that thats exactly the same as Pythagoras theorem
@TheJara1235 ай бұрын
Wonderful video please keep up.
@pielover2675 ай бұрын
Oh man this is great. That's an incredibly beautiful connection I probably would not have noticed on my own, thanks!
@user-tj1wn5ve3s5 ай бұрын
It's interesting how the space of all linear transformations between two subspaces is a vector space which means all the linear transformations are 'vectors'. which is like functions being vectors as well.
@joelwillis20435 ай бұрын
pretty basic mathematical fact - like chapter 3 of an undergrad text in liner algebra
@justrandomthings81585 ай бұрын
@@joelwillis2043 hi can you delete your account please
@valenfeldmann645 ай бұрын
@@joelwillis2043This video is kinda meant for undergrads. As an undergrad myself, I thought I’d make a video that would have been helpful to me last year.
@James-mk8jp4 ай бұрын
@@joelwillis2043besides being a prick what is the aim of saying something like this
@graysoncroom5 ай бұрын
Great video with beautiful animations. Keep up the good work!
@erdemcanaz63943 ай бұрын
Amazed, Thanks. This is a fairly new intuition for me.
@MultiAndAnd5 ай бұрын
they are also points in the space C, L^2, in the schwartz class.
@aram91675 ай бұрын
Unfortunately, the transition from R^n to infinite-dimensional function spaces isn't exactly trivial
@ZACKARIASIS5 ай бұрын
Valen, this was fantastic - the dot product is the foundation for all signal processing (continuous or discrete - I'm following you to hopefully see more
@halikiidrisswouche65725 ай бұрын
I l❤ve it! It has helped me to see more clear why functions are vectors! Thank you to much!
@DrSimulate5 ай бұрын
Nice one! Looking forward to more videos! :)
@eduardoabreu784 ай бұрын
vectors are lists of numbers, functions are (infinite) lists of numbers. integrals are dot products.
@DistortedV125 ай бұрын
Linear algebra was a gift from God
@moehassan_4 ай бұрын
this is mindblowing bro, excellent video intuition makes math so much more beautiful
@bArda265 ай бұрын
Neat! Waiting for more videos!
@aaxa1015 ай бұрын
Fourier transform as a change of basis (or reference system). Is there any other way to see it?
@valenfeldmann645 ай бұрын
3b1b had a good visualization for it that wasn’t as linear algebra focused: kzfaq.info/get/bejne/qdaFgdOqq5ucco0.htmlsi=i9dEOFWAYmxrt-sU (I’m not sure if you’ll be able to click the link but if you search Fourier transform 3b1b it should be the first one to pop up)
@aaxa1015 ай бұрын
The link works. Thanks for your reply! I @@valenfeldmann64
@kaganozdemir43325 ай бұрын
regression on fourier coefficients. it is a directly statistical idea, in that each amplitude is the average overlap of the sinusiod with the signal.
@jamesdurtka27094 ай бұрын
I'd encountered the concept of "functions as vectors" before but not the "functions as infinitely long column vectors." That tickled my brain
@AngeliqueMalouin5 ай бұрын
Thank you, looking forward to more content!
@69erthx11385 ай бұрын
...and vectors are tensors of rank-1, so spinors are da freaks of the manifold.
@jaf79795 ай бұрын
Concise and well put, bravo
@melchiortod294 ай бұрын
Awww man i feel like you didn't do the fourrier transform justice. It's sooo mindblowing when you start thinking about a projection of a function onto rotational space. It would have been more mindblowing if you didn't spoiler the fourrier transform before actually projecting it. But anyways, happy to see that it's not being gatekept
@sam080904 ай бұрын
Great video 🎉
@MrSomethingred4 ай бұрын
This is something you learn in physics by intuition because no one annotates their maths properly. And then multiplies a function by a matrix without explaining it
@paperclips13065 ай бұрын
This is omg. You gave me a lot of happiness. So beautiful. Bro make more videos
@sebastianfranco15074 ай бұрын
This is very cool thx, in computer science we use this, we have a finite set of points of a given function say image or audio and apply a transformation like Furier and to do that we use linear algebra and we can use optimizations known for líneal algebra algorithms, like the FFT.
@wyboo20195 ай бұрын
the vector space of functions is probably my favorite abstract vector space. and then start doing geometric algebra on it and it becomes even more fun
@highgroundproductions85905 ай бұрын
Well then you'd have to define the Clifford algebra on this space of functions, which is not an obvious task - it may have infinite dimensions. Geometric algebra is useful when you have notions of area or volume, and in infinite dimensions, these concepts are usually either useless or, worse, meaningless (I mean not area under one function or something, but the (hyper)volume of a set of functions themselves).
@aimsmathmatrix4 ай бұрын
@@highgroundproductions8590Yeah, it can often be infinite. Path integrals and integrals over spaces of functions don't converge that easily. It becomes hard even trying to formulate good measures on such spaces in the first place. Not to mention, tensoring functions / maps from function spaces that are complete won't give a complete space necessarily, so we have to think of the projective & injective tensor product as well. It's a pain in the ass tbh. But cool as well!
@davidawakim54735 ай бұрын
very cool, thank you!
@javicerverafrias76755 ай бұрын
Its funny because everything in math is a f*ing vector if you add the correct additive and multiplicative composition rules
@cocccix5 ай бұрын
I have studied for a while about image/video compression and it relies a lot on DCT (Discrete Cossine Transform). I actually never understood it very well, and treated this as pure magic. This video just made everything click inside my head. Great job.
@ValidatingUsername4 ай бұрын
Integratable functions map every point in each dimension to a novel point in another and a jacobian can be found that maps the change 😊
@alexeykulik26254 ай бұрын
Mind blowing! Can you make a video that goes more in depth about the topic?
@philkaw4 ай бұрын
This is arguably the starting point for functional analysis…
@damianoventurini23815 ай бұрын
That's really cool. What's a good book you'd suggest to study this subject a bit more?
@valenfeldmann645 ай бұрын
I’m not sure to be honest. I think 3b1b has good videos on linear algebra and he also had a video about the Fourier transform but he looked at it from a perspective that wasn’t quite as focused on linear algebra.
@rubikashree77245 ай бұрын
this is an amazing video
@vancevontaine5 ай бұрын
Well done.
@insertoyouroemail5 ай бұрын
It's easier to visualize higher dimensions than you think. If you've ever used a tool like Blender, a 3D model is built up of vertices which are vectors. These vectors can have more than three dimensions. It's in fact common for vertices to have five dimensions; x, y, z, u, v. How do you work with this? Simple, each vertex is viewed from two separate perspectives, one 3D showing x, y and z and one 2D showing u and v.
@valenfeldmann644 ай бұрын
That’s interesting. Do you find that you can get a feel for higher dimensional shapes by seeing 2 separate perspectives?
@insertoyouroemail4 ай бұрын
@@valenfeldmann64 I couldn't intuitively rotate an object along an arbitrary 5D axis, for example. I think it's unfortunately only useful for demonstrating simple concepts in higher dimensions. Like collision and motion.
@tarvos_trigaranvs4 ай бұрын
I think you should have used Fourier series instead of Fourier transform, because it's not really correct mathematically to think about the Fourier transform as a dot product with some outher function in the same space.
@arararara23824 ай бұрын
Functions mapping to a vector space are vectors. An arbitrary function would not be a vector.
@user-ov4sc3fc6b5 ай бұрын
Really interesting!! Very clear and appealing visuals and a great explanation!! Please keep up this phenomenal work, you got a new loyal subscriber ❤
@honorcolling55564 күн бұрын
Wow
@rizalpurnawan37965 ай бұрын
Good video! Did you use manim for this video?
@valenfeldmann645 ай бұрын
Yeah
@zemoxian5 ай бұрын
I have a terrible problem. Any time I see a video about complex numbers I immediately wonder how it applies to quaternions and other hyper complex numbers. And anything dealing with any of the complex or hyper complex numbers or vectors makes me wonder how that translates to multi vectors and geometric algebra. So I understand how to use the geometric product to multiply vectors into multi vectors but I’ve only seen it applied to finite numbers of dimensions, (unless the most universal geometric algebra is invoke.) But I’ve never considered using functions with continuous domains as vectors. Are there vector spaces using function based vectors that can do something like Clifford algebras? Considering that Clifford Algebras can be boiled down to vectors with orthogonal basis vectors that square to either 1, -1, or (occasionally) 0, I’d be interested in how that works with continuous domain functions. (I don’t think the function itself necessarily has to be continuous unless it has to be manipulated with calculus.) I guess the weirdest part is that a n dimensional vector space used to create a geometric algebra creates a new vector space with 2^n dimensions. So would the function vectors create a multi vector space with the power set of the domain as its domain? The more I think about it the more unlikely it looks for a continuous domain function based geometric algebra. Maybe if I were a mathematician I’d have a clearer idea of whether or not they’d be possible or actually interesting. What does the power set of a real interval look like anyway? The only thing that I’m kinda sure about is that there are more elements in the power set of [0, 1] than in the numbers in [0, 1]. I’m not even sure how to build a function on such a domain. I guess I’ll stop my rambling here.
@MrCmon1134 ай бұрын
I think those are the kinds of questions you rather ask the PhDs in your seminar than the youtube comments.
@ismailchealaud35325 ай бұрын
Great job and beautiful Animation, can u tell me what the logiciel that u use to make this animation 🙏🏽
@valenfeldmann645 ай бұрын
Manim, iMovie, and Desmos
@ismailchealaud35325 ай бұрын
Do you have any tutorial videos? And thanks for ur answer ❤️
@valenfeldmann645 ай бұрын
@@ismailchealaud3532 I haven’t personally made any tutorial videos but there are certainly good tutorial videos out there for Manim and iMovie. Desmos is an online graphing calculator that I could screen record so that’s a pretty easy tool to get used to. I’m no expert on Manim but I’d suggest if you’re just starting out to copy in some examples and see if you can get them to run and maybe try modifying them. (I think tutorial videos will probably be more helpful than my comment tho😂)
@ismailchealaud35325 ай бұрын
Thank u , u give me the keyword 😂 so that will help me so much thank you ❤️🫡
@another_lazy_learner5 ай бұрын
Cooooool!
@BJ-sq1si5 ай бұрын
The pidgeon holing of everything through linear Algebra must be stopped! Still cool video for pointing out the identities
@adityakhanna1135 ай бұрын
Tbh everything is Linear Algebra
@vitorschroederdosanjos65394 ай бұрын
Functions are infinite dimension vectors
@MultiAndAnd5 ай бұрын
Functions with compact sypport are functionalas on radon measures
@alexkrochak71975 ай бұрын
Could you please elaborate on how the bottom integral at 3:46 is derived?
@valenfeldmann644 ай бұрын
Thats a very good question and I'm honestly not entirely sure. This is a very weird integral because we the bounds are infinity and we are integrating over something that doesn't converge. My understanding (which could be wrong) is that we can define a fourier series expansion over an arbitrary interval and this integral is what we get when we take the limit as the interval approaches the entire real number line. With this definition, we get destructive interference between "orthogonal" functions when w ≠ w' (the inner product evaluates to 0) and constructive interference when w=w' (the inner product evaluates to 2pi). (We could define the integral to evaluate to something different if we picked an interval where our basis states aren't orthogonal but that wouldn't be as useful for us). This is my understanding anyway and I'm not sure how helpful this is. This link might be more helpful: math.stackexchange.com/questions/2340094/why-frac12-pi-int-infty-inftyeiwt-xdw-is-the-dirac-delta-func
@thataialaperrera8105 ай бұрын
Everything seems to be a vector
@angeldude1015 ай бұрын
If it looks like a duck, and it quacks like a duck, it's a duck. If it adds like a vector, and it scales like a vector, it's a vector.
@sudoku10995 ай бұрын
Very interesting, but some technical details of this video could be improved: -some visuals go too fast, and we don't have the time to "digest" them. Take more time when you have a lot to read. -the voice over is difficult to understand due to a weak sound level in general. Some audio mastering would help. I know that the goal was to be "intuitive", but anyway, these details are to be considered.
@SamsungGalaxy-vw9gy5 ай бұрын
I didn't understand a thing
@mariovicente5 ай бұрын
I get your point however I believe the expression «functions are vectors» is not mathematically accurate. What you mean to show here is that «a set of functions can be a vector space» and a few of the consequences of that fact. Or alternatively, «a function is a vector IN a function vector space». Because for a «function to be a vector» you need a few assumptions beforehand, as you showed yourself here. So, the statement «functions are vectors» is false. Anyway thanks for the video.
@MrCmon1134 ай бұрын
2:09 There shouldn't be any t on the left side there, I think.
@valenfeldmann644 ай бұрын
I think I see what your saying. All that notation means is that a and b are functions of t. I can see that being confusing because later in the video I just wrote a single letter to represent a function.
@oliveratack55815 ай бұрын
Enjoyed the video, I understood the maths presented but I now work as a teacher. If you want to present information to even a fairly wide (first year uni student and beyond audience) you need to take slower steps. Explain why (not necessarily with proof) where all of the steps come from, do not jump from functions as a vector space to Fourier transforms. Realistically, if you are trying to attract a reasonable audience, getting to Fourier transforms would be the end of a several episode series of videos. It is worth reminding yourself that the average person watching these videos is not only below the level of the average first year maths student but probably struggles to solve linear simultaneous equations. This is absolutely not a hate comment, I hope you continue to do what you're doing.
@peasant82465 ай бұрын
The voice of the narrator is too quiet but otherwise its a good video.
@user-th5ui4ib3y5 ай бұрын
I guess your functions have to have some field as a codomain to view them as vectors, otherwise is see no way to define a dot product.
@alexatg18205 ай бұрын
Actually no, it’s just that the inner product will be defined differently, not just as an integral
@jmathg5 ай бұрын
No field needed, you can study modules over a ring, which are analogous to vector spaces over fields, but we don't require the scalars to have multiplicative inverses. This causes the study of modules to be (in my opinion...) much more complex and sometimes strange than the study of vector spaces!
@suppositorylaxative31795 ай бұрын
@@jmathgit is definitely more complex lol. Whenever you add restrictions to a space, the result is always a nicer space.
@ethanmorris995 ай бұрын
A vector space doesn't necessarily need to be an inner product space (you can have vectors without a dot product)
@adamotmar97135 ай бұрын
Well yeah vectors are a different form of a co-ordinate system
@chaossspy67234 ай бұрын
Numbers are sets
@-mwolf5 ай бұрын
hm nice topic but i feel like it woukd have been very helpful for understanding if the video was 10 minutes longer and if you went inti more details on the explanarions. too many question(mark)s touched/opened but not answered by this video. if the focus is to teaxh and not to entertain, that is. ( i think there are enough entertainers already).
@br42524 ай бұрын
Functions arnt just vectors. Functions are values.
@trogdorbu5 ай бұрын
Is it just me, or is there something wrong with the notation he's using? Taking it at face value, he's repeatedly showing various forms of X = integral of X from minus infinity to plus infinity with respect to t. This is like saying X = f(X) or f'(X) = f(X) which can be true in special cases, but not generalizable.
@jmathg5 ай бұрын
I agree he could have written the product in the integrand without the dot to make it clearer, but there's nothing wrong with his notation. A simple example comes from a first or second class of Linear Algebra, when we do row operations like "R3 = R3 + 2R1", which logically would mean 2R1 = 0 but we're writing it to say "R3 becomes R3 + 2R1". So he's writing the two sides to emphasize different ideas. On the left is the dot product between two infinite dimensional vectors, and on the right is the convolution of two functions. Fourier Transforms were first used by Fourier on heat equations, and then applied to many different areas of physics and analysis (e.g. solving PDEs). The equation he shows reveals a way to study these things from an algebraic viewpoint with infinite dimensional vector spaces (e.g. Hilbert Spaces). I always say that Linear Algebra isn't just an area of study, it's a common language that mathematicians from all different fields can communicate in. So now they have this language of vector/metric spaces, eigenvectors/values, inner products, orthogonal bases, and so on, and that can help reveal new insights in both directions.
@trogdorbu5 ай бұрын
@@jmathg thank you so much for the thoughtful answer! So while a(t) dot b(t) appears on both sides of the equation, you think he's actually referring to different operands and operators on either side, right? That's kinda what I was thinking.
@jmathg5 ай бұрын
@@trogdorbuExactly! He makes this comparison at 2:28 and 3:24
@valenfeldmann645 ай бұрын
@@trogdorbuyou are exactly right! I probably should have made that more clear with my notation😅
@dasten1235 ай бұрын
volume is very low, almost can't hear you
@empireempire35455 ай бұрын
voice wayyy to silent compared to the music - and too silent in general
@davidmurphy5634 ай бұрын
Hmm.. you give an example of the components of a vector being a function and show them, plotted together, making a sine wave or whatever it was. But fundamentally the dimensionality of components is an expression of their distinctness. Just because the fourier transform or DCT _can_ express any output as a frequency function doesn't make it meaningful. Let's make a vector. Component 1 is the coal output of Europe. 2 is the average toenail size. 3 is the wattage of a ZX81. So we have a 3d vector. And now we're going to treat it as a function and express it as a frequency. That's pointless.
@valenfeldmann644 ай бұрын
You’re right that is pointless. Change of basis only makes sense if the new basis vectors span the same space. Fourier transform can’t be applied to just any vector but it can be applied to any element of the vector space that is functions on the real number line because our basis vectors span all of “function space”. (Good clarifying question)
@davidmurphy5634 ай бұрын
@@valenfeldmann64 Ok yes, I see what you're saying. Is there a reason why you used a fourier instead of a discrete cosine transform or is that just the convention?
@valenfeldmann644 ай бұрын
No reason other than that I thought Fourier transform was more interesting.😂
@davidmurphy5634 ай бұрын
@@valenfeldmann64Ah, fair enough! Dunno, I always preferred DCTs - starting in the middle is kinda neater in my books and cosine is officially cooler than sine! ;) Nice chatting and keep the vids coming!
@user-gh4lv2ub2j5 ай бұрын
No, all functions are not vectors, this is not true. Go look up a vector space and figure out why.
@JohnElmore-sf1hj5 ай бұрын
I assume what you meant to say is “not all functions are vectors” and you’d be correct. It should be pretty clear from context of the video that he is talking about continuous mappings from the reals to the reals. (After all it’s a video for intuition)
@FredGibbons5 ай бұрын
Classical physics Bohr model Black Scholes Are all "wrong" but I'm glad to have learned them because they made me better at thinking.
@willthecat38615 ай бұрын
@@FredGibbons Physics is supposed to be about 'reality' which we have... and will have... an imperfect understanding of. Mathematics is not like that.. at all. IMO, learning the wrong = 'untrue' things about mathematics leads to later persist ant confusions... and a lot of tedious untangling of 'mental knots'
@FredGibbons5 ай бұрын
@@willthecat3861Black Scholes is not physical and we can make better models of it. Often you need to learn simple models that are wrong before you can learn better models.
@MrOvipare5 ай бұрын
Agreed. It feels like an overly signal-processing perspective, while vectors have a deeply analytical foundation.