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What do you mean? In terms of Fourier series, the significance is that a function of period 1 breaks down into a sum of components whose derivatives are integer

by Chinjut 11y ago

What do you mean? In terms of Fourier series, the significance is that a function of period 1 breaks down into a sum of components whose derivatives are integer multiples of 2πi times themselves (or, if you wish to avoid descriptions which make use of complex numbers, whose second derivatives are square multiples of -(2π)^2 times themselves). When people describe sines and cosines by breaking a full revolution down into 2π radians, they don't do it for no reason. They do it because this manifests itself loudly in the differential properties of sines and cosines.

goldenkey 11y ago
I mean that sin and cos are tied to Pi in terms of their period. The antipode and catipode. So one might as well group all 3 quantities as Pi related. And if you merely take sin or cos into your formula because they are continuously differentiable periodic functions, then well, couldnt you have just taken something else and the Pi really loses its 'sanctity'? I'm on my phone but I'm starkly betting that there are Fourier transforms that use diff periodic functions instead of the trig ones. Maybe computable ones instead of symbolic, although I'd have to concede profundity if that is the case
- avian 11y ago
  You are correct and that point in the article caught my attention as well. Fourier transform works perfectly OK if you redefine sine and cosine functions to have different periods. That's basically equivalent to a variable substitution in the integral. As far as I can see, the transform itself doesn't favor any specific definition and pi has no special connection to periodic functions like the article implies. It's also true that sine and cosine functions are most conveniently defined in terms of radians, which is what I think Chinjut is trying to say. But that is more or less unrelated to the Fourier transform.
  - Chinjut 11y ago
    Fourier transform of arbitrary functions from R to C works fine with whatever units. Fourier series of functions from R/Z to C (i.e., functions of period 1; i.e., functions f from R to C such that f(x + 1) = f(x) for all x) uses specifically cos(2πNx) and sin(2πNx) for integer N; you can't replace 2π here by any other value (since then you don't get the set of cosines and sines of period 1). That's the special connection between 2π and periodic functions: every function of period 1 decomposes into the sum of a series of functions whose 2nd derivatives are square numbers * -(2π)^2 * themselves. [Or, more cleanly, into the sum of a series of functions whose 1st derivatives are integers * 2πi * themselves]. Again, you cannot replace 2π with any other value in the statements of this paragraph (well, except its negation...).
    - avian 11y ago
      > functions f from R to C such that f(x + 1) = f(x) for all x) uses specifically cos(2πNx) and sin(2πNx) for integer N Define cos'(x) = cos(2πx/A) and sin'(x) = sin(2πx/A) (such functions could be defined without referring to π in their definition) Now f can be expressed as a weighted sum of cos'(ANx) and sin'(ANx) for integer N. > every function of period 1 decomposes into the sum of a series of functions whose 2nd derivatives are square numbers * -(2π)^2 * themselves You are correct that the scaled sin and cos functions I mention above still have the 2nd derivative equal to -(2π)^2 * themselves. I found that a weak argument that pi is somehow inherent in periodic functions though. First, this property of sin and cos is not necessary for or directly connected to the Fourier transform as far as I can see. Second, there are infinitely many families of orthogonal functions that can be used to decompose periodic functions in the same way sin and cos can be used and do not fit this rule of the 2nd derivative. Consider a family of square wave functions for instance.
      - Chinjut 11y ago
        Sure, let's say cos'(x) is the cosine of the angle corresponding to x complete revolutions (thus, cos' is of period 1), and similarly for sin'. Then every function of period 1 is a unique linear combination of cos'(Nx) and sin'(Nx) for integer N. We can say this. This is essentially the same as what we were saying already. My point isn't that π is used in the words we say when defining cos' and sin', but that it is part of their properties, making its presence known once we consider derivatives: the derivative of cos' is -2π * sin', and the derivative of sin' is 2π * cos'. [Er, the primes here don't denote derivatives, but just our newly scaled trig functions, of course] π (2π, etc.) truly is special for the Fourier transform: it is inherent in decomposing a periodic function into a sum of exponentials (and cosine and sine are just even and odd components of exponential functions). An exponential function from R to C is of period 1 if and only if the natural logarithm of the base of the exponential is an integer multiple of 2πi (i.e., just in case the function's derivative is an integer multiple of 2πi times itself). It's true that there are other (non-exponential) families of orthogonal functions that can be used for decompositions, but the fact that the Fourier transform is specifically in terms of a family of exponential functions is of some significance (for example, it means multiplication on one side of the transform corresponds to convolution on the other side; it's also what makes the Fourier transform and inverse Fourier transform essentially the same operation). Tl;dr: Periodic functions can be represented in various ways, but the Fourier series decomposition is a particularly useful representation because it "diagonalizes" differentiation (which is to say, it "diagonalizes" translation by tiny (and consequently also by arbitrarily sized) amounts). And what we discover in diagonalizing the differentiation operator on functions of a fixed period is that its eigenvalues are precisely the integer multiples of 2πi divided by the period. This is a fundamental connection between π and the differential(/integral/etc.) structure of periodic functions.

by Chinjut 11y ago

goldenkey 11y ago

I mean that sin and cos are tied to Pi in terms of their period. The antipode and catipode. So one might as well group all 3 quantities as Pi related. And if you merely take sin or cos into your formula because they are continuously differentiable periodic functions, then well, couldnt you have just taken something else and the Pi really loses its 'sanctity'? I'm on my phone but I'm starkly betting that there are Fourier transforms that use diff periodic functions instead of the trig ones. Maybe computable ones instead of symbolic, although I'd have to concede profundity if that is the case

avian 11y ago
You are correct and that point in the article caught my attention as well. Fourier transform works perfectly OK if you redefine sine and cosine functions to have different periods. That's basically equivalent to a variable substitution in the integral. As far as I can see, the transform itself doesn't favor any specific definition and pi has no special connection to periodic functions like the article implies. It's also true that sine and cosine functions are most conveniently defined in terms of radians, which is what I think Chinjut is trying to say. But that is more or less unrelated to the Fourier transform.
- Chinjut 11y ago
  Fourier transform of arbitrary functions from R to C works fine with whatever units. Fourier series of functions from R/Z to C (i.e., functions of period 1; i.e., functions f from R to C such that f(x + 1) = f(x) for all x) uses specifically cos(2πNx) and sin(2πNx) for integer N; you can't replace 2π here by any other value (since then you don't get the set of cosines and sines of period 1). That's the special connection between 2π and periodic functions: every function of period 1 decomposes into the sum of a series of functions whose 2nd derivatives are square numbers * -(2π)^2 * themselves. [Or, more cleanly, into the sum of a series of functions whose 1st derivatives are integers * 2πi * themselves]. Again, you cannot replace 2π with any other value in the statements of this paragraph (well, except its negation...).
  - avian 11y ago
    > functions f from R to C such that f(x + 1) = f(x) for all x) uses specifically cos(2πNx) and sin(2πNx) for integer N Define cos'(x) = cos(2πx/A) and sin'(x) = sin(2πx/A) (such functions could be defined without referring to π in their definition) Now f can be expressed as a weighted sum of cos'(ANx) and sin'(ANx) for integer N. > every function of period 1 decomposes into the sum of a series of functions whose 2nd derivatives are square numbers * -(2π)^2 * themselves You are correct that the scaled sin and cos functions I mention above still have the 2nd derivative equal to -(2π)^2 * themselves. I found that a weak argument that pi is somehow inherent in periodic functions though. First, this property of sin and cos is not necessary for or directly connected to the Fourier transform as far as I can see. Second, there are infinitely many families of orthogonal functions that can be used to decompose periodic functions in the same way sin and cos can be used and do not fit this rule of the 2nd derivative. Consider a family of square wave functions for instance.
    - Chinjut 11y ago
      Sure, let's say cos'(x) is the cosine of the angle corresponding to x complete revolutions (thus, cos' is of period 1), and similarly for sin'. Then every function of period 1 is a unique linear combination of cos'(Nx) and sin'(Nx) for integer N. We can say this. This is essentially the same as what we were saying already. My point isn't that π is used in the words we say when defining cos' and sin', but that it is part of their properties, making its presence known once we consider derivatives: the derivative of cos' is -2π * sin', and the derivative of sin' is 2π * cos'. [Er, the primes here don't denote derivatives, but just our newly scaled trig functions, of course] π (2π, etc.) truly is special for the Fourier transform: it is inherent in decomposing a periodic function into a sum of exponentials (and cosine and sine are just even and odd components of exponential functions). An exponential function from R to C is of period 1 if and only if the natural logarithm of the base of the exponential is an integer multiple of 2πi (i.e., just in case the function's derivative is an integer multiple of 2πi times itself). It's true that there are other (non-exponential) families of orthogonal functions that can be used for decompositions, but the fact that the Fourier transform is specifically in terms of a family of exponential functions is of some significance (for example, it means multiplication on one side of the transform corresponds to convolution on the other side; it's also what makes the Fourier transform and inverse Fourier transform essentially the same operation). Tl;dr: Periodic functions can be represented in various ways, but the Fourier series decomposition is a particularly useful representation because it "diagonalizes" differentiation (which is to say, it "diagonalizes" translation by tiny (and consequently also by arbitrarily sized) amounts). And what we discover in diagonalizing the differentiation operator on functions of a fixed period is that its eigenvalues are precisely the integer multiples of 2πi divided by the period. This is a fundamental connection between π and the differential(/integral/etc.) structure of periodic functions.