[oifjsidjf]

Honestly I forgot the details, but basical the ENTIRE field of DSP stands on this fact.

Basicaly there exist some functions into which you can feed in sound waves and the output is guaranteed to still be a sound wave.

If you'd feed in a sound wave and if the function would corrupt it you would not be able to do any digital signal processing, since the output must be a wave.

Sound(wave) in -> Sound(wave) out, guaranteed to always be true.

[lisper]

> there exist some functions into which you can feed in sound waves and the output is guaranteed to still be a sound wave.

That in and of itself does not seem like a particularly insightful observation. It's just obvious that such functions exist. I can think of three of them off the top of my head: time delay, wave addition, and multiplication by a scalar. There must be something more to it than that.

[brendanclune]

In math, the obvious things aren't always true and the true things are often not obvious.

Trivially, the identity f(x) = x satisfies the guarantee as well. What amounts to insightful observation is the definition and classification of these functions. In exploring their existence in various forms, we can begin to understand what properties these functions share.

So the interesting part is not that this class of function _exists_, because of course it does! Your intuition has led you to three possible candidates. But if we limit ourselves to only the functions that satisfy the condition _wave-in implies wave-out_ what do they look like as a whole? What do these guarantees buy us if we _know_ the result will be a wave? For example, f(g(x)) is also guaranteed to be _wave-in-wave-out_. Again, maybe obvious, but it's a building block we can use once we've proved it true.

[c-baby]

Ok, but what is being asked is what interesting properties does such a function have. You still haven’t answered that question.

[docfort]

Let me try. Let’s say you have some operator F = crazyweirdblackbox. Like, it exponentiates and does a bunch more complicated things. You can apply this operator to some input function and transform it into some output function. The programming analogy would be some piece of code that works on a lambda and spits out another lambda. The finite-dimensional linear algebra equivalent is a matrix. You would ideally want F to be invertible (or undoable) for all inputs. Just in case.

If g(x) is an eigenfunction of F, then h(x) = F(g(x)) is actually just a rescaled of g(x): h(x) = ag(x), for some constant a. No matter how complicated and hard-to-compute F is, it boils down to just one number a, when acting on some special function g(x).

So what? This only applies to special g(x), and not any choice for g. Let’s say that I have some special function y(x) that isn’t necessarily g(x). But I have a whole bunch of eigenfunctions of F called g_k(x). If I knew that F was linear (kind of a prerequisite for computing eigenfunctions anyway), then I can decompose some output function y(x) into a weighted sum of g_k: y(x) = sum([w_k g_k(x) for k in range(infinity)]).

Some abuse of Python list comprehension there.

So instead of evaluating F, which might be very hard computationally or numerically, we can instead do a for-loop over potentially easier functions g_k. And store some weights w_k that essentially describe how F transformed an input function into y. Easy-peasy.

And maybe I don’t want to evaluate the entire sum. So I could choose to evaluate only the “important” terms in the sum to approximate y(x).

The entirety of signal processing, much of quantum mechanics, much of electromagnetism, and many other partial differential equations can fit into this framework. And we use PDEs to describe F when we really have no idea how to even to write it down. But we can still compute eigenfunctions and therefore make progress in evaluating F since we know the effect of F on certain special functions.

Does that help?

[kridsdale2]

It's been 15 years since I was in my EE degree, and I learned all this stuff to the point of intuition. Since then I've been a C-monkey at megacorp and not doing any real math beyond simple geometry.

Thanks for tickling the old neurons.

[lisper]

Yes, exactly.

This comment was downvoted into oblivion (I vouched it back from the dead), but I have no idea why. You hit the nail on the head.