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I was blown away in my Digital Signal Processing (DSP) class that eigen "values" exist for certain systems in the form of "waves". Basicaly you put in a wave m

I was blown away in my Digital Signal Processing (DSP) class that eigen "values" exist for certain systems in the form of "waves". Basicaly you put in a wave made from multiple sine and/or cosine waves through some function f(x) and the output is STILL a wave, though its frequency, amplitude and phase might change. Technicaly if I remember correctly this applies to all complex exponentials, since those can be rewritten in the form of e^(ix) = cosx + i*sinx. This formula also beatifuly shows how rotations and the complex exponentials are connected. So basicaly you don't just have eigen values, eigen vectors: you also have eigen FUNCTIONS (sine and cosine above are the eigen functions of f(x)). DSP basicaly revolves arounds functions that don't "corrupt" wave-like inputs (wave in -> wave out).

sverona 4y ago
In fact functions are just infinite-dimensional vectors. Almost all of the theory goes through unchanged. This is the basic idea of functional analysis.
- OJFord 4y ago
  Or rather size-of-their-domain-dimensional?
  - sverona 4y ago
    The GP kind of implied functions R->R, but yes.
  - ravi-delia 4y ago
    If you want, but you can do better. I believe, for instance, that at least continuous functions on the reals have a countable basis. Might even be as strong as measurable, not sure about that. That's how, for instance, fourier transforms work.
    - constantcrying 4y ago
      The fourier Transform essentially means that every L^2 (the square of the function has a finite integral) is completely "described" by an l^2 series (a series of numbers whose sum of squares converges), which is about the greatest piece of magic in mathematics. One very important piece here is that the term "function" is somewhat of a lie (since the result couldn't be true if it weren't). The result for measurable functions (not almost functions) shouldn't be true (I think). I am not even sure it is true for L^1 almost functions.
- constantcrying 4y ago
  Notably, one very important part which does not go through is that for mappings between infinite dimensional spaces linearity does not imply continuity. (E.g. a series of functions bounded by a constant can have arbitrarily large derivatives) A large part of functional analysis is dealing with that fact and its implication for PDEs.
bobbylarrybobby 4y ago
If you're into eigenfunctions, pick up any textbook on quantum mechanics. The hamiltonian is a linear operator whose eigenfunctions are the stationary states of the system ("stationary" because an isolated system in a stationary state will never leave that state) and whose eigenvalues are the observable values of the energy of the system. In general, there is a correspondence between observable quantities and Hermitian operators on wavefunctions: "measurement" is the application of a Hermitian operator to the wavefunction, and the values you may observe are the eigenvalues of the operator. So, for instance, energy is quantized in some systems because their hamiltonian has discrete eigenvalues.
deleted 4y ago
[deleted]
c-baby 4y ago
Maybe I'm missing what's interesting about this, but a function like f(z) = 5z + 2 would output a wave with changed amplitude and phase when z = sin(x). That doesn't seem that interesting to me, so f(z) must have some other interesting properties?
- oifjsidjf 4y ago
  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 4y ago
    > 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 4y ago
      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.
mrfox321 4y ago
https://en.wikipedia.org/wiki/Spectrum_(functional_analysis) https://en.wikipedia.org/wiki/Spectrum_(functional_analysis)
WastingMyTime89 4y ago
I’m not sure I understand but it seems to me you are just talking about eigenvalues in C. That’s interesting but not particularly remarkable because eigenvalues are defined for linear transformations of any vector space over a field.
topaz0 4y ago
Small correction: eigenfunctions are analogous to eigenvectors, not eigenvalues. In fact they _are_ eigenvectors, in the sense that they are vectors in a vector space of functions (or some restricted set of functions, e.g. continuous functions or periodic functions).

sverona 4y ago

In fact functions are just infinite-dimensional vectors. Almost all of the theory goes through unchanged. This is the basic idea of functional analysis.

OJFord 4y ago
Or rather size-of-their-domain-dimensional?
- sverona 4y ago
  The GP kind of implied functions R->R, but yes.
- ravi-delia 4y ago
  If you want, but you can do better. I believe, for instance, that at least continuous functions on the reals have a countable basis. Might even be as strong as measurable, not sure about that. That's how, for instance, fourier transforms work.
  - constantcrying 4y ago
    The fourier Transform essentially means that every L^2 (the square of the function has a finite integral) is completely "described" by an l^2 series (a series of numbers whose sum of squares converges), which is about the greatest piece of magic in mathematics. One very important piece here is that the term "function" is somewhat of a lie (since the result couldn't be true if it weren't). The result for measurable functions (not almost functions) shouldn't be true (I think). I am not even sure it is true for L^1 almost functions.
constantcrying 4y ago
Notably, one very important part which does not go through is that for mappings between infinite dimensional spaces linearity does not imply continuity. (E.g. a series of functions bounded by a constant can have arbitrarily large derivatives) A large part of functional analysis is dealing with that fact and its implication for PDEs.

bobbylarrybobby 4y ago

If you're into eigenfunctions, pick up any textbook on quantum mechanics. The hamiltonian is a linear operator whose eigenfunctions are the stationary states of the system ("stationary" because an isolated system in a stationary state will never leave that state) and whose eigenvalues are the observable values of the energy of the system. In general, there is a correspondence between observable quantities and Hermitian operators on wavefunctions: "measurement" is the application of a Hermitian operator to the wavefunction, and the values you may observe are the eigenvalues of the operator. So, for instance, energy is quantized in some systems because their hamiltonian has discrete eigenvalues.

deleted 4y ago

[deleted]

c-baby 4y ago

Maybe I'm missing what's interesting about this, but a function like f(z) = 5z + 2 would output a wave with changed amplitude and phase when z = sin(x). That doesn't seem that interesting to me, so f(z) must have some other interesting properties?

oifjsidjf 4y ago
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 4y ago
  > 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 4y ago
    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.

mrfox321 4y ago

https://en.wikipedia.org/wiki/Spectrum_(functional_analysis) https://en.wikipedia.org/wiki/Spectrum_(functional_analysis)

WastingMyTime89 4y ago

I’m not sure I understand but it seems to me you are just talking about eigenvalues in C. That’s interesting but not particularly remarkable because eigenvalues are defined for linear transformations of any vector space over a field.

topaz0 4y ago

Small correction: eigenfunctions are analogous to eigenvectors, not eigenvalues. In fact they _are_ eigenvectors, in the sense that they are vectors in a vector space of functions (or some restricted set of functions, e.g. continuous functions or periodic functions).