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I am reading this book on trigonometric series, "Тригонометрические ряды от Эйлера до Лебега" (Trigonometric series from Euler to Lebesgue) , it is in Russian, and my Russian is abysmal. But there is a very interesting formula, it is:

$$\dfrac{1-r\cos(x)}{1-2r\cos(x)+r^2}=1+r\cos(x)+r^2\cos(2x)+r^{3}\cos(3x)...$$

$$\dfrac{r\sin(x)}{1-2r\cos(x)+r^2}=r\sin(x)+r^2\sin(2x)+r^{3}\sin(3x)...$$

As $r=1$ or $r=-1$

(The title of the book is above, it is on page 12)

We have: $$-\dfrac{1}{2}=\cos(x)+\cos(x)+\cos(3x)+...$$

$$\dfrac{1}{2}=\cos(x)-\cos(x)+\cos(3x)+...$$

He integrated the second series and obtain the trigonometric series:

$$\dfrac{x}{2}=\sin(x)-\dfrac{1}{2}\sin(2x)+\dfrac{1}{3}\sin(3x)-...$$

The first and second series are divergent, so I think the result doesn't make sense, although the formal method of deriving it may seem correct.

The third series is correct on the interval $[\pi, -\pi]$, because the series expansion converges to that function (Correct me if I am wrong here, I haven't studied Fourier series).

I wish to ask how did he obtain the above formula? Are these formula incorrect because the issue of convergence? Also, which papers contain these results?

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  • $\begingroup$ For the reference, could you please specify what page is it mentioned on? $\endgroup$ – ain92 Jan 27 at 14:10
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    $\begingroup$ If “the above formula” is the first one: Write using Euler’s identity $\cos x = (e^{ix}-e^{-ix})/2$, factor the denominator, find the partial fractions decomposition, expand the parts as geometric series, and convert from exponential functions back to trigonometric functions (Euler’s identity again). Similarly for the second one. — Don’t you have typos, though? (read $r = 1$ instead of $x = 1$, $\cos(2x)$ instead of $2 \cos(x)$, etc.?) $\endgroup$ – Michael E2 Jan 27 at 14:44
  • $\begingroup$ Fixed, thank you! $\endgroup$ – James Warthington Jan 27 at 20:45
  • $\begingroup$ @ain92, it is on page 12, I have added this in my question. Are you a Russian native speaker? On that page, it is said that this formula appears in 1748. Then above that there is a mention of the book title "введения в анализ бесконечных", which is "Introduction to the analysis of the infinite". This book was published in 1748. So I suspect that it may come from that book. If so it is going to be harder to find a reference. $\endgroup$ – James Warthington Jan 27 at 21:04
  • $\begingroup$ @JamesWarthington Yes, I am! I downloaded the book and carefully read that page. The author also specifies that the series are divergent, just as you do. Later he notes that Euler doesn't provide any domain of applicability for the result. Unfortunately nowhere I've found any citation on any publication; your interpretation is quite possible but not the only one. $\endgroup$ – ain92 Jan 28 at 1:17

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