Fourier Series Relation between "Conjugate" functions

Question

Assume we have two fourier series

$$f(x) = A_0 + \sum_{i=1}^{N}\alpha_i\cos(2ix)$$

$$g(x) = A_0 - \sum_{i=1}^{N}\alpha_i\cos(2ix)$$

Obviously there are many realtions between those functions (the same average, sum is constant, etc.). But in terms of Fourier analysis I was unable to find any definition or naming for those "conjugates".

Just trying to find a full set of relations and also if we know the $f(x)$ or $g(x)$ to calculate the other one, except the obvious:

$$f(x)=\frac{2}{\pi}\int_{0}^{\pi}g(x)dx - g(x)$$

Also for those functions there is an obvious property: $$f'(x)=-g'(x)$$

In order to describe motiviation of looking this, assume there is a representation of a function $f$ in terms of fourier series. I would like to construct the other one, without using fourier series, as it will not give me the full vision of properties.

Just an example:

$$f(x)=\sin(x)\sin(2x)\sin(3x)\sin(4x)=a_0+\sum_{i=1}^{N}\alpha_{2i}\cos(2xi)$$

So we know the fourier series of $f$ and $g$:

$$g(x)=?????=a_0-\sum_{i=1}^{N}\alpha_{2i}\cos(2xi)$$

The question is what is the explicit form of the $g(x)$?

Answer 1

This is not an answer, but something too long for being a comment, aswering the following comment from this question's author: "@Joako my point is simple, lets look at the provided example. Could you please provide the $g(x)$ without knowing the Fourier coefficients? Or in other words is there a way to calculate the $g(x)$ without using Fourier transform to calculate $A_0$ and $\alpha_{i}$-s?

@GevorgHmayakyan not being an expert (I am electrical engineer, not matematician), I could try at least these two:

1. Don't knowing $f(t)$ analytically beforehand as a function (but having a "dense enough" set of samples from an equispaced set of points), I will try to find the function $r(t)=f'(t)$, then I will try to find $s(t)=\int r(t)\ dt$ (here without integration constant, or equivalently, making the integration constant of the antiderivative $C=0$), and calculate $A_0(t)=f(t)-s(t)$, if it results to be a constant, then I have an estimate of $A_0$ and I am done since $g(x)=2A_0-f(x)$;

2. For the $\alpha_i$ you have already that $\alpha_i^{f(x)} = -\alpha_i^{g(x)}$ so knowing the expansion of $f(x)$ define also $g(x)$ (with exception for constant terms / frequency $= 0$). But finding a cosines expansion is harder since you are ruling out the use a Fourier Series tools, since you could find those $\alpha_k = \int_{-\infty}^{\infty} f(x)\cdot \cos(2kx)\ dx$, but as electrician I use the Fourier Transform since is quite known, with Fast algorithms contained in many softwares (almost all of them have the FFT), and this integral in the analogical world is just filtering each frequency with a RC pass-band filter.

I change the index label to "$k$" for avoid confusion with the imaginary unit, which is highly involved into Fourier expansions which are done with complex exponentials.

Now, without the use of Fourier series, if you allow to still use functional dot product expansions, I think you could try something like the Clenshaw–Curtis quadrature.

Maybe another idea for the $\alpha_k$ I think could work, is realizing that the Fourier series and the Taylor series are related: if you take the Taylor expansion and change the real-valued variable into a unitary magnitude complex one you have now the Fourier series expansion, so you could try to make a match among coefficients as is shown in this other answer but for your specific cosine expansion, and then find the $\alpha_k$ as relations with the Taylor series coefficients which could be found as derivatives of $f(x)$, but my intuition tells me you would be taking in a long way which are the coefficients of a Fourier Series anyway.

Maybe another way is noting that a perfect pass-band filter in the frequencies is a sinc function on the real domain variables so you could use also the Convolution Theorem to find the $\alpha_k$-s, but again, the Fourier Series integral is simpler.