Math proof of common knowledge?
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- KVRian
- 1215 posts since 17 Apr, 2004
Is there a mathematical proof of the statement that any wave can be considered as a superposition of a number of sine waves? I have seen a few graphical examples, but I'd just like to see it presented mathematically. Thanks. 
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- KVRAF
- 4339 posts since 20 Feb, 2004
This is exactly the subject of Fourier Analysis. There are a number of theorems therein which describe which functions can be expressed in this way (the ones which aren't usually wouldn't be encountered as a "sound wave").
A well-behaved signature.
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Rusty Shackleford Rusty Shackleford https://www.kvraudio.com/forum/memberlist.php?mode=viewprofile&u=13683
- KVRist
- 307 posts since 24 Feb, 2004
It's pretty advanced stuff, and requires a good knowledge of calculus and some introductory Real Analysis, as well as the fundamentals of linear algebra.
The basic idea is this: If C[0,L] denotes the set of piecewise-continuous(ie only a finite number of discontinuities) functions on the closed interval from 0 to L on the real line, then this set is a vector space under normal addition of functions and scalar multiplication by real numbers. The set {sin( n*2*Pi*x/L)| n=1...infinity} is then a subset of C[0,L] so it makes sense to look at the subspace spanned by this set. Call this subspace T.
The functions generating the subspace T are mutually orthogonal under the normal inner product of C[0,L] (or any function space), and next you can show that the subspace is actually *dense* in C[0,L] - that is, for any piecewise continuous function f on [0, L] there is a sequence of functions in T (that is, a sum of sine functions) that converges to f in the metric induced by the inner product on C[0, L]. What this means is that this sequence will converge to f pointwise "almost everywhere", which means except for on a set of measure 0.
This whole notion of measure 0 is a bit advanced and requires some understanding of Lebesgue measure, and if you understand what that is, I doubt you'd even be posting this question. The basic idea is that any finite or countably infinite set has measure zero. Since [0,L] is uncountable, this means that the function that your series converges to is reasonably well behaved.
This just outlines a very basic proof of the existence of Fourier sine expansions. There's ways of expanding these methods to obtain much stronger results than "almost anywhere", and in the general case, there's no mention of sines at all, or even the real line.
Sorry to geek out, but you asked.
Some helpful books:
Real Analysis: Modern Techniques and Their Applications, by Gerald Folland <-- also has a proof of the Sampling Theorem(left as an exercise!
)
Real and Complex Analysis(i think?), by Walter Rudin
Now, these books are pretty hardcore if you haven't ventured into higher maths, so be warned. Rudin in particular can be very hard to read but if you can follow, his proofs are usually very good and well thought-out.
There was also a pretty interesting approach to this proof in one of the books on Integral Equations that I read... can't remember the name, but I'll try to remember. It's been a while.
Hope this helps - need any more info, just ask.
The basic idea is this: If C[0,L] denotes the set of piecewise-continuous(ie only a finite number of discontinuities) functions on the closed interval from 0 to L on the real line, then this set is a vector space under normal addition of functions and scalar multiplication by real numbers. The set {sin( n*2*Pi*x/L)| n=1...infinity} is then a subset of C[0,L] so it makes sense to look at the subspace spanned by this set. Call this subspace T.
The functions generating the subspace T are mutually orthogonal under the normal inner product of C[0,L] (or any function space), and next you can show that the subspace is actually *dense* in C[0,L] - that is, for any piecewise continuous function f on [0, L] there is a sequence of functions in T (that is, a sum of sine functions) that converges to f in the metric induced by the inner product on C[0, L]. What this means is that this sequence will converge to f pointwise "almost everywhere", which means except for on a set of measure 0.
This whole notion of measure 0 is a bit advanced and requires some understanding of Lebesgue measure, and if you understand what that is, I doubt you'd even be posting this question. The basic idea is that any finite or countably infinite set has measure zero. Since [0,L] is uncountable, this means that the function that your series converges to is reasonably well behaved.
This just outlines a very basic proof of the existence of Fourier sine expansions. There's ways of expanding these methods to obtain much stronger results than "almost anywhere", and in the general case, there's no mention of sines at all, or even the real line.
Sorry to geek out, but you asked.
Some helpful books:
Real Analysis: Modern Techniques and Their Applications, by Gerald Folland <-- also has a proof of the Sampling Theorem(left as an exercise!
Real and Complex Analysis(i think?), by Walter Rudin
Now, these books are pretty hardcore if you haven't ventured into higher maths, so be warned. Rudin in particular can be very hard to read but if you can follow, his proofs are usually very good and well thought-out.
There was also a pretty interesting approach to this proof in one of the books on Integral Equations that I read... can't remember the name, but I'll try to remember. It's been a while.
Hope this helps - need any more info, just ask.
- Rad Grandad
- 38041 posts since 6 Sep, 2003 from Downeast Maine
I got a question....how can you remember all that and not the name book?...Rusty Shackleford wrote:It's pretty advanced stuff, and requires a good knowledge of calculus and some introductory Real Analysis, as well as the fundamentals of linear algebra.
The basic idea is this: If C[0,L] denotes the set of piecewise-continuous(ie only a finite number of discontinuities) functions on the closed interval from 0 to L on the real line, then this set is a vector space under normal addition of functions and scalar multiplication by real numbers. The set {sin( n*2*Pi*x/L)| n=1...infinity} is then a subset of C[0,L] so it makes sense to look at the subspace spanned by this set. Call this subspace T.
The functions generating the subspace T are mutually orthogonal under the normal inner product of C[0,L] (or any function space), and next you can show that the subspace is actually *dense* in C[0,L] - that is, for any piecewise continuous function f on [0, L] there is a sequence of functions in T (that is, a sum of sine functions) that converges to f in the metric induced by the inner product on C[0, L]. What this means is that this sequence will converge to f pointwise "almost everywhere", which means except for on a set of measure 0.
This whole notion of measure 0 is a bit advanced and requires some understanding of Lebesgue measure, and if you understand what that is, I doubt you'd even be posting this question. The basic idea is that any finite or countably infinite set has measure zero. Since [0,L] is uncountable, this means that the function that your series converges to is reasonably well behaved.
This just outlines a very basic proof of the existence of Fourier sine expansions. There's ways of expanding these methods to obtain much stronger results than "almost anywhere", and in the general case, there's no mention of sines at all, or even the real line.
Sorry to geek out, but you asked.
Some helpful books:
Real Analysis: Modern Techniques and Their Applications, by Gerald Folland <-- also has a proof of the Sampling Theorem(left as an exercise!)
Real and Complex Analysis(i think?), by Walter Rudin
Now, these books are pretty hardcore if you haven't ventured into higher maths, so be warned. Rudin in particular can be very hard to read but if you can follow, his proofs are usually very good and well thought-out.
There was also a pretty interesting approach to this proof in one of the books on Integral Equations that I read... can't remember the name, but I'll try to remember. It's been a while.
Hope this helps - need any more info, just ask.
The highest form of knowledge is empathy, for it requires us to suspend our egos and live in another's world. It requires profound, purpose‐larger‐than‐the‐self kind of understanding.
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Rusty Shackleford Rusty Shackleford https://www.kvraudio.com/forum/memberlist.php?mode=viewprofile&u=13683
- KVRist
- 307 posts since 24 Feb, 2004
4 years of university math and a whole lot of pot 
- Rad Grandad
- 38041 posts since 6 Sep, 2003 from Downeast Maine
works for meRusty Shackleford wrote:4 years of university math and a whole lot of pot
The highest form of knowledge is empathy, for it requires us to suspend our egos and live in another's world. It requires profound, purpose‐larger‐than‐the‐self kind of understanding.
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- KVRist
- 111 posts since 5 Jun, 2005
I thought I'd read somewhere (?in The CSound Book) that even non-periodic waveforms could be analyzed . . . is this right?Meffy wrote:One slight adjustment: any periodic wave.
IIRC, square waves ("Walsh functions"?) can be used as wellbirrbits wrote:...the periodic function does not have to be a sine...
"This sentence is true"
(Take that, Epimenides!)
(Take that, Epimenides!)
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Rusty Shackleford Rusty Shackleford https://www.kvraudio.com/forum/memberlist.php?mode=viewprofile&u=13683
- KVRist
- 307 posts since 24 Feb, 2004
Thats right. Actually it works for any set of functions that satisfy certain technical conditions(specifically, that they be mutually orthogonal and dense).
By the way, in that original set, i forgot to include the function f(x)=1. Otherwise, what i said would only work for functions that are 0 at x=0. But I'm sure no one cares either
By the way, in that original set, i forgot to include the function f(x)=1. Otherwise, what i said would only work for functions that are 0 at x=0. But I'm sure no one cares either
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- KVRist
- 111 posts since 5 Jun, 2005
I care too!Rusty Shackleford wrote:Thats right. Actually it works for any set of functions that satisfy certain technical conditions(specifically, that they be mutually orthogonal and dense).
What are the functions that need to be mutually orthogonal? Does this refer to the fact that sin(x) and cos(x) are out of phase by 90 degrees? (and thus orthogonal?)
"This sentence is true"
(Take that, Epimenides!)
(Take that, Epimenides!)
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- KVRAF
- 1743 posts since 3 Dec, 2004
again this is reaching back a few years for me, but I beleive the stipulation is that the waveform has to be stationary. Which goes into a set of definitions that I cant quite remember at this time....xcomp wrote:I thought I'd read somewhere (?in The CSound Book) that even non-periodic waveforms could be analyzed . . . is this right?Meffy wrote:One slight adjustment: any periodic wave.
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- KVRist
- 111 posts since 5 Jun, 2005
Can any of you suggest a good book that explains the Fourier transform & other music-related math-intensive issues (eg, sampling theorem), but written for musicians? Rusty Shackleford -- I suspect that the books you suggested may be too advanced for me right now . . . it's been a few years since calculus & linear algebra. . . . maybe I'll understand them in another year or so, after some brushing up. BTW, I've read the Computer Music Tutorial, but the discussions didn't go into enough depth for me.
"This sentence is true"
(Take that, Epimenides!)
(Take that, Epimenides!)
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Rusty Shackleford Rusty Shackleford https://www.kvraudio.com/forum/memberlist.php?mode=viewprofile&u=13683
- KVRist
- 307 posts since 24 Feb, 2004
hehe i was hoping you wouldn't ask...xcomp wrote:I care too!Rusty Shackleford wrote:Thats right. Actually it works for any set of functions that satisfy certain technical conditions(specifically, that they be mutually orthogonal and dense).(Still trying to get a handle on this stuff!)
What are the functions that need to be mutually orthogonal? Does this refer to the fact that sin(x) and cos(x) are out of phase by 90 degrees? (and thus orthogonal?)
sin and cos are out of phase by 90 degrees, but thats not why they're orthogonal.... they're orthogonal as functions in C[0, L]... what that means is:
If f,g are functions then define <f,g> as the integral of f(x)*g(x) from 0 to L.
Lets call S_k(x) = sin (k *2*Pi*x/L) for simplicity (the _k means subscript k)
Then <S_m(x), S_n> = 0 if m is not equal to n. This is what I mean by orthogonal. It can be verified by pretty basic calculus.
The reason orthogonality is important is because these Fourier Series and the functions they represent are *basically* seen as one and the same because they agree "almost everywhere", so we may think of any function on [0, L] as just its Fourier Series.
Now if C_n is the Fourier coefficient of S_n(x) in the Fourier series of h(x),
C_n = sqrt (L/(2*Pi)) * < S_n(x) , h(x)>,
where the <,> means the inner product i described above. This is because h(x) = [a sum of terms of the form C_k * S_k(x)] and this inner product is linear in each entry, so <a+b,c> = <a,c> + <b,c> and 3*<a,b> = <3*a,b> = <a, 3*b>, for example.
If you actually do the calculation and sustitute this Fourier series for h in <S_n(x) , h(x)> and use the linearity of the product, all but one of the terms will be 0.
Anyways what this all means is that this property of sines easily allows you to calculate the coefficients. If the functions in your basis aren't orthogonal, things get a lot more hairy... for one thing, in this sort of situation your "Fourier Series" may not be unique! There could be many different series representations of a function because the basis functions might not be what is called 'linearly independent'. I'm rambling again, and i've probably lost everyone.
But yeah, I'm sorry to say I don't know of any music-related math books that offer any sort of rigorous proof of anything. If anyone knows of any, please share.
Hope I didn't confuse you guys more :S But it feels good to have to use that part of my brain again