I realized that derivatives are linear
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In another sense, derivatives themselves are linear: for a function f: U -> V of vector spaces, the derivative (at some point) is a linear map from U -> V, (i.e. the derivative of the functions is a function Df: U -> L(U,V)) and this extends the concept of derivative to multiple dimensions as f(x+h) = f(x) + (Df)(x)h + o(h).
This seems ok at first derivatives but can become unwieldy as they became tensors higher rank.
Another question one might ask on learning that differntiation is a linear operator is what it’s eigenvalues are. For differentiation these are functions of the form f(x) = exp(ax). But one can construct other linear operators and from this you get Sturm–Liouville theory which is fantastic.
One final note is that much of this multidimensional derivatives and tensor stuff becomes a lot easier if one learns suffix notation (aka Einstein notation, aka index notation, aka summation convention), as well as perhaps a few identities with the kronecker delta or Levi-Civita symbol. Notation can break down a bit with arbitrary rank tensors: $a_{i_1,...,i_k}$ becomes unwieldy but writing $a_{pq...r}$ is ok.
Of the mappings between vector spaces, the most well behaving are the bounded linear operators, and the derivative doesn't belong to these. But yes, it's linear.
Edit: Originally wrote f(x) = k·sin(k·x), but meant f(x) = k·sin(x/k).
Additionally, it only really makes sense to talk about bounded operators between topological vectors spaces (as you need to make sense of what it means to be bounded), of which the most commonly dealt with are Banach spaces.
Reading your comment, I wondered how can you define bounded sets in a topological vector space (where you don't have a norm). The definition is cute: a set X if bounded if any neighborhood of 0 can be inflated to include the whole of X.
(Also, what's the most natural norm on C^k? The sup norm? I haven't done any functional analysis in years)
|f| = ∑ᵢ₌₀ᵏ sup|f⁽ⁱ⁾|.
On open domains you can also use the topology that forces uniform convergence on all compact subsets. But this will only give you a metric space, no Banach space (but you’ll include unbouded functions). This is needed for studying Brownian motions with an unbounded time domain.
Regarding the other question: If you take Cᵏ⁻¹ as a the co-domain differentiation will be continuous. IIRC to get unbounded linear maps defined on the whole Banach space you need the axiom of choice, you won’t be able to write one down.
The problem with differential operators is that they are usually only defined on a dense subset of the domain, and there they are not bounded. E.g. in quantum mechanics the space of states is L² but all the interesting observables are differential operators. You can weasel out of this situation by defining them on a subset of “physical states” (e.g. smooth wave functions of rapid decay). But they aren’t continuous anymore (the spectrum is unbounded). But on Hilbert spaces everything mostly works out fine. Physicist usually ignore those technical problems and still don’t make mistakes.
First course in functional analysis.
Before that, people usually take linear algebra, all the calculus courses, some kind of theorem-proving course for calculus (either analysis or real analysis or really rigorous versions of the calculus courses) and measure theory, maybe also topology.
I think this post may still be too wedded to the idea of linear spaces and vectors being arrays of objects - specifically in insisting on decomposing functions like sin and cos to Taylor Series. In fact, you can have a vector space where, in addition to polynomial terms, there are also dimensions for sin(x), tan(x), sin(x - pi), e^x, etc. The fact that you can't enumerate these dimensions, or even describe the set of them until given a set of vectors you're trying to describe, doesn't keep this from being a vector space.
I always viewed real functions as infinite-dimensional vectors in the "canonical" basis, that is, shifted Dirac impulses. I guess it can be transformed into your representation with a change of basis with some handwaving.
Suppose that there’s a function f, that can be written as an infinite sum (integral) of shifted Dirac impulses, but cannot be written in your representation as a sum of those “base functions”. Then simply add a new dimension to your representation that will correspond to f, so that f will be represented as 1 at this new dimension, and zero everywhere else. (In other words: add f to the base functions)
Repeat until you have covered every function.
So yes, a linear approximation of a linear function is the function itself.
In any case, it is probably a good thing to get a good intuition of what differentiation and derivatives are in the vector space setting before digging into differential geometry.
(y-b) = f'(a)·(x-a)
It's not at all obvious to me that this means that the function d(f) = df/dx is linear. It is linear, but I don't see how the tangent curve demonstrates it.
From this you obtain that the derivative of any linear combination is the linear combination of the derivatives: differentiation is linear.
I feel a lot of comments are saying "well of course they are!", not realizing that this is not about a new discovery.
For example, for multi variable calculus, the results would be very different.
Let's take the example of W.X
d/dx (W.X) = X.d/dx(W) + W.d/dx(X)
since W is not dependent on x, the first term is zero and we get the answer the author got.
Before drawing conclusions from the post, please remember the assumptions the author has taken.
_analytic_ functions have a Taylor _series_, but it would be incorrect to say that "most" functions have a taylor series, and a taylor series is not a polynomial.
Ehh, then again, it doesn't really matter, people who don't understand it still won't with a nomenclature change.
Mention some mathematically advanced idea: out come the pitchforks about how you don't need that, all you need is code/market size/scalability/product fit/investment/execution.
Mention a banality that anyone who studied algebra knows: frontpage.
I think learning is really hard work, and so most people's first reaction to hard work is to say No, and then go and construct an a posteriori rationale for why actually they shouldn't do that hard work (it's not that useful, you're never gonna use it, you're an expert at something else, etc).
Similar story for why asking data structures in job interviews is a bad idea when you're an applicant (but the people who have been hired and are hiring, do think it's good to ask)
I like your conservation-of-mental-energy interpretation.
Thanks for that.
There's nothing wrong with the OP at all—someone sharing the excitement of discovering something for themselves.
What's Lbda though?
During school I never understood what the math was for, so my unconscious brain never saw the necessity to actually learn it. Now I want to learn - with hugely better results.
This mechanism should be utilized much more often instead of shoving seemingly unrelated knowledge into peoples ears without letting them feel the need for it first.
I can't tell you the number of 'trivial' math facts that I have (re)discovered because they were in the context of something I cared deeply about.
The point isn't to remember D_x is a linear operator--math isn't about memorization. It's about understanding the context where this is a useful fact and knowing how to figure it out.
Learn it in Calc I and you can half-heartedly reference it (...isn't differentiation linear? I feel like I remember that from senior year of high school...).
Figure it out on your own and you own it for life.
Post it to the internet and you get ridiculed and mocked for it so that you wish you could forget it.
Totally agree with this part.
> Post it to the internet and you get ridiculed and mocked for it so that you wish you could forget it.
Telling everybody you meet about a basic fact that you just learned is mildly cute when a 6-year-old does it.
It's great that this person learned something that was new to them, sure. But that doesn't mean that they need to shout about it to the tens of thousands of their closest friends who read the front page of HN.
Your comment doesn't make sense, considering that you acknowledge math is in fact a prior requirement. Just because you didn't made heads or tails out of math during school that doesn't mean math ceased to be a prerequisite to learn applied math.
His video on determinants (for example) teaches what they actually do, rather than emphasising a seemingly arbitrary algorithm for you to blindly follow (which is how I was taught determinants).
But being clear, the footwork is the fundamental and can't be skipped.