I was introduced to the so-called $C+iS$ method when I tried to prove these 2 formulae from the work of Euler. I don't know how to call this method of technique. Upon searching it on youtube, I've come across several videos made by Indian teachers who call it $C+iS$ method. $C$ stands for cosine series, and $S$ stands for sine series. Essentially, it makes use of the Euler formula $e^{ix}=\cos(x)+i\sin(x)$. The method was described by A-Level student, who helped me to prove those 2 formulae, "pretty powerful". While I don't deny its versatility, I think it can be dangerous to over-rely on this method because the issue of convergence of trigonometric series is highly complicated and has a lot of nuances (I don't even have the basic understanding of their conditions of convergence). In order to fully grasp all kinds of convergence of trigonometric and Fourier series, one must have enough prerequisites to read and understand reference works such as Antoni Zygmund's (a famed Polish mathematician) "Trigonometric Series" or Nina Karlovna Bary's "A Treatise of trigonometric series". I have read the more elementary, serving as an introduction to the subject, "Fourier series" by Georgi P. Tolstov, but many nuances do not make their way to my head. It is a highly difficult subject, and I don't have a background in measure theory to fully appreciate it. This issue is what I think dangerous if one use the $C+iS$ method all the times to derive the "sum" of a given trigonometric series, no matter how meaningful that is.
Below, I give a rough sketch and some exercises I have done with this method.
Suppose you are given a sine or cosine series of the form:
$$a_0\sin(x)+a_1\sin(x+y)+a_2\sin(x+2y)+a_3\sin(x+3y)+...$$
$$a_0\cos(x)+a_1\cos(x+y)+a_2\cos(x+2y)+a_3\cos(x+3y)+...$$
One may note that the arguments of these 2 series form an arithemtic progression.
One put:
$$C=a_0\cos(x)+a_1\cos(x+y)+a_2\cos(x+2y)+a_3\cos(x+3y)+...$$
$$S=a_0\sin(x)+a_1\sin(x+y)+a_2\sin(x+2y)+a_3\sin(x+3y)+...$$
Then:
$$C+iS=a_0[\cos(x)+i\sin(x)]+a_1[\cos(x+y)+i\sin(x+y)]+a_2[\cos(x+2y)+i\sin(x+2y)]+...$$
Apply the Euler formula:
$$e^{ix}=\cos(x)+i\sin(x)$$
$$C+iS=a_0e^{ix}+a_1e^{i(x+y)}+a_2e^{i(x+2y)}+a_3e^{i(x+3y)}...$$
According to Sydney Luxton Loney's "A Treatise of Plane Trigonometry, part 2: Analytical Trigonometry" page 114, the $C+iS$ series fall into 4 categories:
1) Those depending on the summation on a geometrical progression
2)Those depending on the binomial series
3)Those depending on the exponential theorems, including sub-cases, the sine and cosine series
4)Those depending on the logarithmic and, as subcase, Gregory's series (the series for $\arctan(x)$ and $artanh(x)$
One can utilize these series to sum up the $C+iS$ series above. Please consult this table of Maclaurin series for reference.
Then, after summing up and doing some more works to refine it, one can express it in the form of $A+iB$. Equating this with $C+iS$, one can easily point out $C=A$ and $S=B$, thus one obtains the "sum" of these two trigonometric series.
$C+iS$ series depending on the geometrical series:
Problem 1:
Find the sum of $S_1=\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}\sin(nx)$ and $C_1= \displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}\cos(nx)$
Notes: This series can be derived easily using this formula:
$$S_1 = \sin(x)-\sin(2x)+\sin(3x)-\sin(4x)...=\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}\sin(nx)$$
$$C_1 = \cos(x)-\cos(2x)+\cos(3x)-\cos(4x)...= \displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}\cos(nx)$$
$$-1+C_1 = -1+\cos(x)-\cos(2x)+\cos(3x)-\cos(4x)...$$
$$-1+C_1+iS_1=-1+[\cos(x)+\sin(x)]-[\cos(2x)+\sin(2x)+[\cos(3x)+\sin(3x)]...$$
$$-1+e^{ix}-(e^{ix})^2+(e^{ix})^3-(e^{ix})^4+...$$
This is the geometric series, and its "sum" is $-\dfrac{1}{1+e^{ix}}$
So, we have:
$$-\frac{1}{1+e^{ix}}=-\frac{1+e^{-ix}}{(1+e^{ix})(1+e^{-ix})}$$
$$=-\frac{1+\cos(x)-i\sin(x)}{(1+e^{ix})(1+e^{-ix})}=-\frac{1+\cos(x)-i\sin(x)}{2\cos(x)+2}$$
Equating the real and imaginary part with $C+iS$, we have:
$$\boxed{-1+C_1=-\frac{1+\cos(x)}{2+2\cos(x)}\implies C=\frac{1}{2}}$$
$$\boxed{S_1=\frac{\sin(x)}{2\cos(x)+2}}$$
Using the same precedure, one can also derive the series of:
$$\boxed {S_2 =\displaystyle\sum_{n=1}^{\infty} \sin(nx) = \sin(x)+\sin(2x)+\sin(3x)+\sin(4x)...=\dfrac{\sin(x)}{2\cos(x)-2}}$$
$$ \boxed {C_2 = \displaystyle\sum_{n=1}^{\infty}\cos(nx)=\cos(x)-\cos(2x)+\cos(3x)-\cos(4x)...= -\dfrac{1}{2}}$$
Remark:
For the two cosine series $C_1$ and $C_2$ above, the "sums" obtaining by this purely algebraical method doesn't make sense since they are divergent series for all x. For a proof, please refer to this post. The sum can only made sense not in the sense of convergence. But this lies outside of my domain of knowledge right now. These series appear in "Subsidium Calculi Sinuum" of Euler, the last part of this paper.
For the two sine series $S_1$ and $S_2$ above, it is easy to notice that at $x=0$ and $x=k\pi$, with $k$ being any integers, the two series attain a finite limit, that is $0$. At $x=\pi/2$, for example, the $S_2$ series turn into $1+0-1+0+1...$, which is divergent. Hence the series only converge for $x=0$ and $x=k\pi$.
$C+iS$ series depending on the logarithmic series:
Problem 2
$$S_1 = a\sin(x)-\dfrac{1}{2}a^2\sin(2x)+\dfrac{1}{3}a^3\sin(3x)-\dfrac{1}{4}a^4\sin(4x)...=\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}\dfrac{1}{n}a^n\sin(nx)$$
$$C_1 = a\cos(x)-\dfrac{1}{2}a^2\cos(2x)+\dfrac{1}{3}a^3\cos(3x)-\dfrac{1}{4}a^4\cos(4x)...=\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}\dfrac{1}{n}a^n\cos(nx)$$
$$C_1+iS_1=a[\cos(x)+i\sin(x)]-\dfrac{1}{2}a^2[\cos(2x)+i\sin(2x)]+\dfrac{1}{3}a^3[\cos(3x)+i\sin(3x)]-...$$
$$C_1+iS_1=ae^{ix}-\dfrac{1}{2}(ae^{ix})^2+\dfrac{1}{3}(ae^{ix})^3-\dfrac{1}{4}(ae^{ix})^4...$$
Let $t=ae^{ix}$, we have
$$C_1+iS_1=t-\dfrac{t^2}{2}+\dfrac{t^3}{3}-\dfrac{t^4}{4}...=\log(1+t)=\log(1+ae^{ix})$$
$$C_1+iS_1=\log[\color{green}{1+a\cos(x)}+i\color{blue}{a\sin(x)]}$$
We apply the formula for complex logarithm: $\log(\color{green}{a}+i\color{blue}{b})=\frac{1}{2}\log[a^2+b^2]+i\arctan(\frac{b}{a})$
$$\log[1+a\cos(x)+ia\sin(x)]=\dfrac{1}{2}\log[(1+a\cos(x)]^2+a\sin^2(x)]+i\arctan\left(\dfrac{a\sin(x)}{1+a\cos(x)}\right)$$
$$\boxed{C_1=\dfrac{1}{2}\log[(1+a\cos(x)]^2+a\sin^2(x)]}$$
$$\boxed{S_2=\arctan\left(\dfrac{a\sin(x)}{1+a\cos(x)}\right)}$$
Problem 3
$$S_2 = \sin(x)-\dfrac{1}{2}sin(2x)+\dfrac{1}{3}\sin(3x)-\dfrac{1}{4}\sin(4x)...=\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}\dfrac{1}{n}\sin(nx)$$
$$C_2 = \cos(x)-\dfrac{1}{2}\cos(2x)+\dfrac{1}{3}\cos(3x)-\dfrac{1}{4}\cos(4x)...=\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}\dfrac{1}{n}\cos(nx)$$
This is just a special case of the problem above, where $a=1$, skipping all these steps, one arrive at:
$$C_2=\dfrac{1}{2}\log[(1+\cos(x)]^2+\sin^2(x)]$$
$$S_2=\arctan\left(\dfrac{\sin(x)}{1+\cos(x)}\right)$$
Use the identity $\tan\left(\dfrac{x}{2}\right)=\dfrac{\sin(x)}{1+\cos(x)}$, we can rewrite
$$S_2=\arctan\left(\tan\left(\dfrac{x}{2}\right)\right)=\dfrac{x}{2}$$'
Remark:
Again, here, we don't know what is the interval where this "sum" $\dfrac{x}{2}$ is valid using this algebraical method. In Fourier analysis, one can rigorously derive this series by stating that $S_2 = \displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}\dfrac{1}{n}\sin(nx)$ is the Fourier series representation of the function $\dfrac{x}{2}$ on the interval $[\pi, \pi]$. Since the $\dfrac{x}{2}$ is an odd function, it only has the sine series as its representation.
We have the formula for the coefficients of the Fourier series:
$$a_0+\sum_{n=1}^{+\infty}(a_n\cos(nx)+b_n\sin(nx))$$
$$a_0=\dfrac{1}{2 \pi}\displaystyle\int_{-\pi}^{\pi} \dfrac{x}{2}dx=\dfrac{1}{2 \pi}\left[\dfrac{x^2}{4}\right]_{-\pi}^{\pi}=\dfrac{1}{2 \pi}\left(\dfrac{(\pi)^2}{2}-\dfrac{(-\pi)^2}{2}\right)=0$$
$$b_n=\dfrac{1}{\pi}\displaystyle\int_{-\pi}^{\pi} \dfrac{x}{2}\sin(nx)=\displaystyle\sum_{n=1}^{+\infty} (-1)^{n-1}\dfrac{\sin(nx)}{n}$$
Problem 4
Find the sum of the series $\displaystyle\sum_{n=1}^{+\infty}\dfrac{1}{n}\sin(nx)$
$$S_3 = \sin(x)+\dfrac{1}{2}sin(2x)+\dfrac{1}{3}\sin(3x)+\dfrac{1}{4}\sin(4x)...=\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n}\sin(nx)$$
$$C_3 = \cos(x)+\dfrac{1}{2}\cos(2x)+\dfrac{1}{3}\cos(3x)+\dfrac{1}{4}\cos(4x)...=\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n}\cos(nx)$$
$$C_3+iS_3=[\cos(x)+i\sin(x)]+\dfrac{1}{2}[\cos(2x)+i\sin(2x)]+\dfrac{1}{3}[\cos(3x)+i\sin(3x)]+...$$
$$C_3+iS_3=e^{ix}+\dfrac{1}{2}(e^{ix})^2+\dfrac{1}{3}(e^{ix})^3+\dfrac{1}{4}(e^{ix})^4...$$
Let $t=e^{ix}$, we have
$$C_3+iS_3=t+\dfrac{t^2}{2}+\dfrac{t^3}{3}+\dfrac{t^4}{4}...=-\log(1-t)=-\log(1-e^{ix})$$
$$C_3+iS_3=-\log[\color{green}{1+a\cos(x)}-i\color{blue}{a\sin(x)]}$$
We apply the formula for complex logarithm: $\log(\color{green}{a}+i\color{blue}{b})=\frac{1}{2}\log[a^2+b^2]+i\arctan(\frac{b}{a})$
$$-\log[1-\cos(x)-i\sin(x)]=-\dfrac{1}{2}\log[(1-\cos(x)]^2-\sin^2(x)]-i\arctan\left(-\dfrac{\sin(x)}{1-\cos(x)}\right)$$
$$\boxed{C_1=-\dfrac{1}{2}\log[(1-\cos(x)]^2-a\sin^2(x)]}$$
$$iS_3=-i\arctan\left(-\dfrac{\sin(x)}{1-\cos(x)}\right)$$
Using the elementary trigonometric identities $\dfrac{\sin(x)}{1-\cos(x)}=\cot\left(\dfrac{x}{2}\right)$ and $\tan\left(\dfrac{\pi-x}{2}\right)$
$$\boxed{S_3=-\arctan\left(-\tan\left(\dfrac{\pi-x}{2}\right)\right)=\arctan\left(\tan\left(\dfrac{\pi-x}{2}\right)\right)=\dfrac{\pi-x}{2}}$$
Remark
Like above, $C+iS$ doesn't indicate the interval in which this equation holds. It should be noted that since the series in problem 3 is valid for the interval $[-\pi, \pi]$, replacing $x$ by $\pi-x$ sends $[-\pi, pi]$ to $[0, 2\pi]$
Historical remark
This series could be found in Euler's "Institutiones calculi differentialis" (Foundations in differential calculus, published first time in 1755) (chapter 4 "Concerning the conversion into series" page 526, Ian Bruce's translation) (page 271, Russian edition entitled "дифференциальное исчислениие", 1949) (page 388, first Latin edition, 1755). He derived series after a long and complicated consideration for the series of $\arctan(x)$. He wrote the series as below and differentiate this sine series to obtain the series in problem 1:
$$S=\dfrac{\pi}{2}=\dfrac{u}{2}+\dfrac{1}{1}\sin(u)+\dfrac{1}{2}\sin(2u)+\dfrac{1}{3}\sin(3u)...$$
$$S'=C=0=\dfrac{1}{2}+\cos(u)+\cos(2u)+\cos(3u)+\cos(4u)+\cos(5u)+...$$
$C$ is obviously a divergent series for all x.
If we integrate this divergent $C$ series, we obtained a true identity (though the interval for validity must be indicated).
Mathematicians of Euler's time freely differentiated and integrated trigonometric series since they viewed them as purely algebraic formulae to obtain accurate numerical series. The accuracy of the numerical series was what that mattered. To Euler, Lagrange and other XVIII century mathematicians, algebraic rules that held for finite algebraic expression can be extended to infinite expressions. This is what Cauchy call the "generality of algebra". However, it is untrue, as claimed by many historians, that Euler and his contemporaries had virtually no understanding of convergence. Joseph Louis François Bertrand (1822-1900) wrote in his "Traité Calcul différentiel et intégral" (published in 1864), page 230, section 228, a scathing critic about mathematicians of XVIII century as followed "Les géomètres du XVIII siècle attache peu importance à la convergence des séries. Les plus illustres d'entre eux fait illusion sur le peu de rigeur des raisonments où elle interviennent". This line of attack has its own merits, because convergence of functional series were still poorly understood. Another author, Martin Ohm, older brother of the famed German physicist Georg Ohm, wrote in his tract "The Spirit of Mathematical Analysis: And Its Relation to a Logical System" that:
"The theory of infinite series rests at the present upon a very bad foundation. All operations are applied to them as if they were finite, but is this permissible? I think not. Where is it demonstrated that the differential of the infinite series is found by taking the differential of each term? Nothing is easier than to give examples where this rule is not correct; for example,
$$\dfrac{x}{2}=\sin(x)-\dfrac{1}{2}\sin(2x)+\dfrac{1}{3}\sin(3x)-\dfrac{1}{4}\sin(4x)+...$$
by differentiating we obtain
$$\dfrac{1}{2}=\cos(x)-\cos(2x)+\cos(3x)-\cos(4x)+$$
an utterly false result, for this series is divergent." (page 2)
Euler was very well aware of the divergence of series, but he still found divergent series useful in obtaining accurate results. Louis Antoine de Bougainville, an explorer but had been a mathematician in his youth, a student of d'Alembert, called the "sum" of divergent series as "faux" (false), and convergent series as "vrais"(true) (see his Traité du Calcul Intégral pour servir de suite à l'analyse des infiniments petits de M. le Marquis de l'Hopital, volume 1):
Les series convergents sont celles dont les termes vont diminuant à l'infini. Nous prouverons plus-bas que ces series sont les seuls vraies.
Les series divergents sont celles dont les termes vont en augmentant à l'infini. (page 302)
The criteria for determining the convergence is rather simple and far from being sophisticated as what is presented in a Calculus course. Of all illustrous mathematicians of Euler's generation, it seems that d'Alembert was the most careful in assessing the convergence of series. Perhaps, for this reason, the d'Alembert criteria, which is found in modern textbook, bears his name. In the early 19th century, several great mathematicians such as Abel and Cauchy spent great efforts in establishing criteria for convergence of numerical and functional series. It took mathematicians of the later generation, from the first half of the 19th century till the 20th century, to figure out the issue of convergence and other issues related to trigonometric and Fourier series.
Yet divergent series have made a come back to mathematics. Euler's view was somewhat vindicated by the emergence of the field of mathematics that study asymptotic expansions (or series) of functions. But I shall stop my comment here as I have no knowledge of this field of mathematics.
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