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By numerical computation it seems like, if $a_0 < a_1$: $$ \begin{multline} \log({a_0}^2 + {a_1}^2 + 2 a_0 a_1 \cos(\omega t)) = \log({a_0}^2 + {a_1}^2) \\ + \frac{a_0}{a_1}\cos(\omega t) - \frac{1}{2}\frac{{a_0}^2}{{a_1}^2}\cos(2\omega t) + \frac{1}{3}\frac{{a_0}^3}{{a_1}^3}\cos(3\omega t) - \frac{1}{4}\frac{{a_0}^4}{{a_1}^4}\cos(4\omega t) \ldots \end{multline} $$ If $a_0 > a_1$, the two term must be exchanged.

I'm quite confident in this solution but I cannot find a way to prove it mathematically...

Would be nice to know the reason of this result!

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    $\begingroup$ Looks like the real part of $\log(1+e^{i\omega t})$, give or take some constants. $\endgroup$ – Wojowu May 9 at 18:58
  • $\begingroup$ You are totally right! This is the good approach. $\endgroup$ – Alister Trabattoni May 9 at 21:03
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Let's consider $$ f(x) = \log (1+q^2+2q\cos x) = \log |1+qe^{ix}|^2 , $$ which differs from your function only by the additive constant $2\log a_1$ if we take $q=a_0/a_1$. Since $|q|<1$, we can use the Taylor series of $\log(1+z)$ to write $$ \begin{align} f(x) = 2\,\textrm{Re}\; \log (1+qe^{ix}) = 2\,\textrm{Re}\sum_{n\ge 1} (-1)^{n-1}\frac{q^n}{n} e^{inx} \\ = 2\sum_{n\ge 1} (-1)^{n-1}\frac{q^n}{n} \cos nx . \end{align} $$

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  • $\begingroup$ Just the answer I was looking for. Perfect! Good idea to use the complex notation. $\endgroup$ – Alister Trabattoni May 9 at 20:49
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    $\begingroup$ Sorry to be a stickler, but I think you need $(-1)^{n-1}$ in the summand of the Maclaurin series expansion above, unless I missed something? $\endgroup$ – Procore May 15 at 16:49
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    $\begingroup$ @Procore: Yes, thanks, corrected now. $\endgroup$ – Christian Remling May 15 at 18:12
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    $\begingroup$ @ChristianRemling I found the way to apply your solution to the more general case. See the edit on my original post. $\endgroup$ – Alister Trabattoni May 25 at 9:52
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    $\begingroup$ You could also add it as an answer, that would probably be clearer. @AlisterTrabattoni $\endgroup$ – RP_ May 25 at 11:56
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To apply the solution proposed by @ChristianRemling to the more general case stated in the title you need to do the following:

The formula implies $a > 0$ and $|b| < a$. To solve the problem, we reformulate the formula as: $$ \begin{align} r|1 + qe^{ix}|^2 &= r(1 + qe^{ix})(1 + qe^{-ix}) \\ &= r(1 + 2q\cos x + q^2) \\ & = a + b\cos x \\ \end{align} $$

This requires to solve the following system: $$ \begin{align} a &= r(1 + q^2) \\ b &= 2rq \end{align} $$

From the two possible solutions we take the one where $|q|<1$ which is mandatory for the Taylor expansion. $$ \begin{align} q = \frac{a - \sqrt{a^{2} - b^{2}}}{b} \\ r = \frac{a + \sqrt{a^{2} - b^{2}}}{2} \end{align} $$

Finally, Taylor series expansion solve the problem: $$ \begin{align} \log (a + b\cos x) &= \log r + 2\log | 1 + qe^{ix} | \\ &= \log r + 2 \sum {(-1)}^{n-1} \frac{q^n}{n} e^{inx} \\ &= \log r + 2 \sum {(-1)}^{n-1} \frac{q^n}{n} \cos nx \end{align} $$

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