Obtaining Fourier series of function without calculating the Fourier coefficients

Question

In this question in one of the answers it's shown how to get from $$f\left ( x \right )=\sum_{n=1}^{\infty}\frac{\sin\left ( nx \right )}{10^{n}}$$ to $$f\left ( x \right )=\frac{10 \sin x}{101-20\cos x}$$ I recently came across a function similar to $f$ and I needed to compute its Fourier coefficients. That can be done using contour integration. But then I wondered whether it is possible to go backwards and see that the function is an imaginary or real part of some complex valued function which can be expressed as an infinite sum. Basically the whole process in reverse. I hope that isn't just a trial and error process

EDIT: I might have not expressed myself clearly. I do know how to get from the geometric progression to the function, it is basic precalc. It would be very elegant to be able to go backwards, that is, given the analytic expression for the function, trace it back to a geometric, or some other series. That could save us from a lot of work involving integration. Functions such as the one I posted above seem to be very fitting for this problem.

Answer 1

Since $\sin(0x)=0$, we may start the index at $n=0$. $$\sum_{n=0}^\infty \frac{\sin(nx)}{10^n} = \frac1{2i} \sum_{n=0}^\infty \frac{e^{inx} - e^{-inx}}{10^n} = \frac{1}{2i} \left(\frac{1}{1-\frac{e^{ix}}{10}} - \frac{1}{1-\frac{e^{-ix}}{10}}\right)$$

$$=\frac{1}{2i} \left( \frac{10}{10-e^{ix}} - \frac{10}{10-e^{-ix}}\right)$$

$$=\frac{10}{2i} \left(\frac{10-e^{-ix} - 10 + e^{ix}}{(10-e^{ix})(10-e^{-ix})} \right)$$

$$=\frac{10}{2i} \left(\frac{e^{ix}-e^{-ix}}{100+1-10e^{ix}-10e^{-ix}} \right)$$

$$=\frac{10\sin(x)}{101 - 20 \cos(x)}$$

Answer 2

Responding to the clarified version of the question: for a holomorphic function $F$ on a neighborhood of the unit circle, the Laurent expansion of $F$ on an annulus containing the unit circle is a Fourier series when $z$ is replaced by $e^{i\theta}$ and $1/z$ is replaced by $e^{-i\theta}$. Is this the sort of thing you're looking for?

Answer 3

Yes, this is possible. The way to do is by analogy: With rational functions (that don't have a pole at $x=0$), one may always use a partial fraction's decomposition to rewrite a function in terms as a linear combination of the functions $\frac{1}{1-cx}$ (or powers thereof) and a polynomial. Then we use $\frac{1}{1-x}=1+x+x^2+x^3+\ldots$ (for $|x|<1$) to get a sum. So, for instance, one can go from a rational function back to a series with rational functions like this: $$\frac{1}{x^2+1}=\frac{1/2}{1-ix}+\frac{1/2}{1+ix}=\frac{1}2\sum_{n=0}^{\infty}(ix)^n + (-ix)^n=\sum_{n=0}^{\infty}(-1)^nx^{2n}.$$ This is heavily related to the idea of generating functions - in particular, the coefficient of $x^n$ in a rational function will always obey some homogenous linear recurrence relation after a point.

Now, let's say we've been given the ratio of two trigonometric polynomials - that is a rational function in $e^{ix}$ such as $$f(x)=\frac{10\sin(x)}{101-20\cos(x)}=\frac{5i(e^{-ix}-e^{ix})}{101-10(e^{ix}+e^{-ix})}=\frac{5i(1-e^{2ix})}{101e^{ix}-10(e^{2ix}+1)}.$$ We then continue rearranging into a convenient form, where the denominator has constant term $1$: $$f(x)=\frac{-\frac{1}2i(1-e^{2ix})}{e^{2ix}-\frac{101}{10}e^{ix}+1}$$ Then, to begin partial fraction decomposition, we factor the denominator into terms of the form $(1-ce^{ix})$ which gives $e^{2ix}-\frac{101}{10}e^{ix}+1=(1-10e^{ix})(1-\frac{1}{10}e^{ix})$. Then we want to find the constants so that $$f(x)=c_0+\frac{c_1}{1-\frac{1}{10}e^{ix}}+\frac{c_2}{1-10e^{ix}}$$ which can be done by putting everything over one numerator and equating the numerator of this with that of our previous expression (other techniques for finding the coefficients of a partial fraction expansion work here too, though). One finds that the form of $f(x)$ is $$f(x)=\frac{i}2+\frac{-i/2}{1-\frac{1}{10}e^{ix}}+\frac{-i/2}{1-10e^{ix}}.$$ Now, we proceed with a little care, since naively expanding these terms gives only terms of the form $e^{ix}$ (so we wouldn't recover real valued numbers in the sum) and also wants to use the series $1+10e^{ix}+100e^{2ix}+1000e^{3ix}+\ldots$ which obviously doesn't converge. To get around this, notice that we can actually expand $\frac{1}{1-cx}$ using two identities: $$\frac{1}{1-x}=1+x+x^2+x^3+\ldots$$ $$1- \frac{1}{1-x}=1+\frac{1}x+\frac{1}{x^2}+\frac{1}{x^3}+\ldots$$ where the latter is valid when $|x|>1$ and the former when $|x|<1$. Given that $|e^{ix}|=1$, we expand the term $\frac{-i/2}{1-\frac{1}{10}e^{ix}}$ using the first identity and the term $\frac{i}2-\frac{i/2}{1-10e^{ix}}$ using the latter to get: $$f(x)=\frac{-i}2\left(1+\frac{1}{10}e^{ix}+\frac{1}{100}e^{2ix}+\frac{1}{1000}e^{3ix}+\ldots\right)+\frac{i}2\left(1+\frac{1}{10}e^{-ix}+\frac{1}{100}e^{-2ix}+\frac{1}{1000}e^{-3ix}+\ldots\right)$$ and putting them together into one sum: $$f(x)=\sum_{n=0}^{\infty}\frac{-i}{2}\cdot \frac{e^{nix}}{10^n}+\frac{i}{2}\cdot \frac{e^{-nix}}{10^n}=\sum_{n=0}^{\infty}\frac{\sin(nx)}{10^n}.$$

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One might note that repeated factors need special handling since they can result in non-linear denominators in the partial fractions decomposition - but the identity like: $$\frac{1}{(1-x)^n}={n-1 \choose n-1}+{n \choose n-1}x+{n+1 \choose n-1}x^2+{n+2 \choose n-1}x^3+\ldots$$ where, for fixed $n$, the expression ${n+c-1\choose n-1}$ is simply a polynomial in $c$ of degree $n-1$ - so you might get sums like $\sum_{n=0}^{\infty}\frac{n\sin(n)}{10^n}$ at the end.

Another restriction is that $(1-ce^{ix})$ shouldn't be a factor of the denominator for $|c|=1$ since then we can use neither expansion to get a convergent power series. I don't see any easy patch for this, but it holds for any function with no poles.