ca.classical-analysis-and-odessequences-and-series
Empirical evidence suggests that, for each positive integer $n$, the following equality holds:
\begin{equation*}
\prod_{s=1}^{2n}\sum_{k=1}^{2n}(-i)^k\sin\frac{sk\pi}{2n+1}=(-1)^n\frac{2n+1}{2^n},
\end{equation*}
where $i=\sqrt{-1}$.
Is it a known equality? If it is true, would you please give me some insights on how to derive this equality?
This is merely a variation of your own proof, Matt, but I believe it makes things clearer.
The first step is to define $c:=1-a-b$. Then, your identity takes the form
$\dfrac{\Gamma\left(b\right)\Gamma\left(c\right)}{\Gamma\left(b+c\right)}+\dfrac{\Gamma\left(c\right)\Gamma\left(a\right)}{\Gamma\left(c+a\right)}+\dfrac{\Gamma\left(a\right)\Gamma\left(b\right)}{\Gamma\left(a+b\right)}=\pi^{1/2}\dfrac{\Gamma\left(\dfrac{a}{2}\right)}{\Gamma\left(\dfrac{1-a}{2}\right)}\cdot\dfrac{\Gamma\left(\dfrac{b}{2}\right)}{\Gamma\left(\dfrac{1-b}{2}\right)}\cdot\dfrac{\Gamma\left(\dfrac{c}{2}\right)}{\Gamma\left(\dfrac{1-c}{2}\right)}$
for $a+b+c=1$. This is symmetric in $a$, $b$, $c$, which means we are way less likely to go insane during the following computations.
Now, using the formula
$\dfrac{\Gamma\left(\dfrac{s}{2}\right)}{\Gamma\left(\dfrac{1-s}{2}\right)}=\pi^{-1/2}2^{1-s}\cdot\cos\dfrac{\pi s}{2}\cdot\Gamma\left(s\right)$,
the right hand side simplifies to
$\pi^{-1}\cdot 4\cdot\cos\dfrac{\pi a}{2}\cdot\Gamma\left(a\right)\cdot\cos\dfrac{\pi b}{2}\cdot\Gamma\left(b\right)\cdot\cos\dfrac{\pi c}{2}\cdot\Gamma\left(c\right)$
(here we used $2^{3-a-b-c}=2^{3-1}=4$), so the identity in question becomes
$\dfrac{\Gamma\left(b\right)\Gamma\left(c\right)}{\Gamma\left(b+c\right)}+\dfrac{\Gamma\left(c\right)\Gamma\left(a\right)}{\Gamma\left(c+a\right)}+\dfrac{\Gamma\left(a\right)\Gamma\left(b\right)}{\Gamma\left(a+b\right)}$ $ = \pi^{-1}\cdot 4\cdot\cos\dfrac{\pi a}{2}\cdot\Gamma\left(a\right)\cdot\cos\dfrac{\pi b}{2}\cdot\Gamma\left(b\right)\cdot\cos\dfrac{\pi c}{2}\cdot\Gamma\left(c\right)$.
Dividing by $\Gamma\left(a\right)\Gamma\left(b\right)\Gamma\left(c\right)$ on both sides, we get
$\dfrac{1}{\Gamma\left(a\right)\Gamma\left(b+c\right)}+\dfrac{1}{\Gamma\left(b\right)\Gamma\left(c+a\right)}+\dfrac{1}{\Gamma\left(c\right)\Gamma\left(a+b\right)}=4\pi^{-1}\cdot\cos\dfrac{\pi a}{2}\cdot\cos\dfrac{\pi b}{2}\cdot\cos\dfrac{\pi c}{2}$.
Since $b+c=1-a$, $c+a=1-b$, $a+b=1-c$, this rewrites as
$\dfrac{1}{\Gamma\left(a\right)\Gamma\left(1-a\right)}+\dfrac{1}{\Gamma\left(b\right)\Gamma\left(1-b\right)}+\dfrac{1}{\Gamma\left(c\right)\Gamma\left(1-c\right)}=4\pi^{-1}\cdot\cos\dfrac{\pi a}{2}\cdot\cos\dfrac{\pi b}{2}\cdot\cos\dfrac{\pi c}{2}$.
Now, using the formula $\dfrac{1}{\Gamma\left(z\right)\Gamma\left(1-z\right)}=\pi^{-1}\sin{\pi z}$ on the left hand side, and dividing by $\pi^{-1}$, we can simplify this to
$\sin{\pi a}+\sin{\pi b}+\sin{\pi c}=4\cdot\cos\dfrac{\pi a}{2}\cdot\cos\dfrac{\pi b}{2}\cdot\cos\dfrac{\pi c}{2}$.
Since $a+b+c=1$, we can set $A=\pi a$, $B=\pi b$, $C=\pi c$ and then have $A+B+C=\pi$. Our goal is to show that
$\sin A+\sin B+\sin C=4\cdot\cos\dfrac{A}2\cdot\cos\dfrac{B}2\cdot\cos\dfrac{C}{2}$
for any three angles $A$, $B$, $C$ satisfying $A+B+C=\pi$.
Now this can be proven in different ways:
1) One is by writing $C=\pi-A-B$ and simplifying using trigonometric formulae; this is rather boring and it breaks the symmetry.
2) Another one is using complex numbers: let $\alpha=e^{iA/2}$, $\beta=e^{iB/2}$ and $\gamma=e^{iC/2}$. Then,
$\sin A+\sin B+\sin C=4\cdot\cos\dfrac{A}2\cdot\cos\dfrac{B}2\cdot\cos\dfrac{C}{2}$
becomes
(1) $\dfrac{\alpha^2-\alpha^{-2}}{2i}+\dfrac{\beta^2-\beta^{-2}}{2i}+\dfrac{\gamma^2-\gamma^{-2}}{2i} = 4\cdot\dfrac{\alpha+\alpha^{-1}}{2}\cdot\dfrac{\beta+\beta^{-1}}{2}\cdot\dfrac{\gamma+\gamma^{-1}}{2}$.
Oh, and $A+B+C=\pi$ becomes $\alpha\beta\gamma=2i$. Now proving (1) is just a matter of multiplying out the right hand side and looking at the $8$ terms (two of them, namely $\alpha\beta\gamma$ and $\alpha^{-1}\beta^{-1}\gamma^{-1}$, cancel out, being $i$ and $-i$, respectively).
3) Here is how I would have done it 8 years ago: We can WLOG assume that $A$, $B$, $C$ are the angles of a triangle (this means that $A$, $B$, $C$ lie in the interval $\left[0,\pi\right]$, additionally to satisfying $A+B+C=\pi$), because everything is analytic (or by casebash). We denote the sides of this triangle by $a$, $b$, $c$ (so we forget about the old $a$, $b$, $c$), its semiperimeter $\dfrac{a+b+c}{2}$ by $s$, its area by $\Delta$ and its circumradius by $R$. Then, $\sin A=\dfrac{a}{2R}$ (by the Extended Law of Sines) and similarly $\sin B=\dfrac{b}{2R}$ and $\sin C=\dfrac{c}{2R}$, so that $\sin A+\sin B+\sin C=\dfrac{a}{2R}+\dfrac{b}{2R}+\dfrac{c}{2R}=\dfrac{a+b+c}{2R}=\dfrac{s}{R}$. On the other hand, one of the half-angle formulas shows that $\cos\dfrac{A}2=\sqrt{\dfrac{s\left(s-a\right)}{bc}}$, and similar formulas hold for $\cos\dfrac{B}2$ and $\cos\dfrac{C}2$, so that
$4\cdot\cos\dfrac{A}2\cdot\cos\dfrac{B}2\cdot\cos\dfrac{C}{2}$
$=4\cdot\sqrt{\dfrac{s\left(s-a\right)}{bc}}\cdot\sqrt{\dfrac{s\left(s-b\right)}{ca}}\cdot\sqrt{\dfrac{s\left(s-c\right)}{ab}}$
$=\dfrac{4s}{abc}\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$.
Now, $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}=\Delta$ (by Heron's formula) and $\Delta=\dfrac{abc}{4R}$ (by another formula for the area of the triangle), so tis becomes
$4\cdot\cos\dfrac{A}2\cdot\cos\dfrac{B}2\cdot\cos\dfrac{C}{2}=\dfrac{4s}{abc}\cdot\dfrac{abc}{4R}=\dfrac{s}{R}$.
This is exactly what we got for the left hand side, qed.
EDIT, September 2014: I wrote to Prof. Ecalle, it turns out (as I had hoped) that the fractional iterates constructed by the recipe below really do come out $C^\infty,$ including a growth bound, in terms of $n,$ on the $n$-th derivatives at $0.$ The key word phrase is Gevrey Class. Also, I recently put a better exposition and example of the technique at https://math.stackexchange.com/questions/911818/how-to-get-fx-if-we-know-ffx-x2x/912324#912324
EDIT Feb. 2016: given that there is new discussion of this, i am pasting in the mathematical portion of Prof. Ecalle's reply, which includes the references
Yes, indeed, any $f(x)$ real analytic at $0$ and of the form
(*) $f(x)=x+ ax^{p+1} +o(x^{p+1})$ for $a \not= 0$
admits natural fractional iterates $g=f^{o w}$ (right or left of zero) that are not just $C^\infty$ at $0$, but of Gevrey class $1/p$, i.e. with bounds of type
(**) $| g^{(n)}(0)/n! |< c_0 \cdot c_1^n \cdot (n/p)!$
Here, $g$ may denote any iterate of rational or real order $w$. You may find details in my publication no 7 on my homepage http://www.math.u-psud.fr/~ecalle/publi.html or again in publication no 16 ("Six Lectures etc"; in English), pp 106-107 , Example 2 (with $\nu=1$).
Here, Gevrey smoothness at $0$ results from $g(x^{1/p})$ being the Laplace transform of an analytic function with (at worst) exponential growth at infinity.
The "Six Lectures" are in Schlomiuk editor, 1993, Bifurcations and periodic orbits of vector fields / edited by Dana Schlomiuk. The reference is currently number 19 on Ecalle's web page, it reads:
Six Lectures on Transseries, Analysable Functions and the Constructive Proof of Dulac's Conjecture . Bifurcations and Periodic Orbits of Vector Fields, D. Schlomiuk ed., p.75-184, 1993, Kluwer
ORIGINAL: The correct answer to this belongs to the peculiar world of complex dynamics. See John Milnor, Dynamics in One Complex Variable.
First, an example. Begin with $f(z) = \frac{z}{1 + z},$ which has derivative $1$ at $z=0$ but, along the positive real axis, is slightly less than $x$ when $x > 0.$ We want to find a Fatou coordinate, which Milnor (page 107) denotes $\alpha,$ that is infinite at $0$ and otherwise solves what is usually called the Abel functional equation, $$ \alpha(f(z)) = \alpha(z) + 1.$$ There is only one holomorphic Fatou coordinate up to an additive constant. We take $$ \alpha(z)= \frac{1}{ z}.$$ To get fractional iterates $f_s(z)$ of $f(z),$ with real $0 \leq s \leq 1,$ we take $$ f_s (z) = \alpha^{-1} \left( s + \alpha(z) \right) $$ and finally $$f_s(z) = \frac{z}{1 + s z}.$$ The desired semigroup homomorphism holds, $$ f_s(f_t(z)) = f_{s + t}(z), $$ with $f_0(z) = z$ and $f_1(z) = f(z).$
Alright, the case of $\sin z$ emphasizing the positive real axis is not terribly different, as long as we restrict to the interval $ 0 < x \leq \frac{\pi}{2}.$ For any such $x,$ define $x_0 = x, \; x_1 = \sin x, \; x_2 = \sin \sin x,$ and in general $ x_{n+1} = \sin x_n.$ This sequence approaches $0$, and in fact does so for any $z$ in a certain open set around the interval $ 0 < x \leq \frac{\pi}{2}$ that is called a petal.
Now, given a specific $x$ with $x_1 = \sin x$ and $ x_{n+1} = \sin x_n$ it is a result of Jean Ecalle at Orsay that we may take $$ \alpha(x) = \lim_{n \rightarrow \infty} \; \; \; \frac{3}{x_n^2} \; + \; \frac{6 \log x_n}{5} \; + \; \frac{79 x_n^2}{1050} \; + \; \frac{29 x_n^4}{2625} \; - \; n.$$
Note that $\alpha$ actually is defined on $ 0 < x < \pi$ with $\alpha(\pi - x) = \alpha(x),$ but the symmetry also means that the inverse function returns to the interval $ 0 < x \leq \frac{\pi}{2}.$
Before going on, the limit technique in the previous paragraph is given in pages 346-353 of Iterative Functional Equations by Marek Kuczma, Bogdan Choczewski, and Roman Ger. The solution is specifically Theorem 8.5.8 of subsection 8.5D, bottom of page 351 to top of page 353. Subsection 8.5A, pages 346-347, about Julia's equation, is part of the development.
As before, we define ( at least for $ 0 < x \leq \frac{\pi}{2}$) the parametrized interpolating functions, $$ f_s (x) = \alpha^{-1} \left( s + \alpha(x) \right) $$
In particular $$ f_{1/2} (x) = \alpha^{-1} \left( \frac{1}{2} + \alpha(x) \right) $$
I calculated all of this last night. First, by the kindness of Daniel Geisler, I have a pdf of the graph of this at:
http://zakuski.math.utsa.edu/~jagy/sine_half.pdf
Note that we use the evident symmetries $ f_{1/2} (-x) = - f_{1/2} (x)$ and $ f_{1/2} (\pi -x) = f_{1/2} (x)$
The result gives an interpolation of functions $f_s(x)$ ending at $ f_1(x)=\sin x$ but beginning at the continuous periodic sawtooth function, $x$ for $ -\frac{\pi}{2} \leq x \leq \frac{\pi}{2},$ then $\pi - x$ for $ \frac{\pi}{2} \leq x \leq \frac{3\pi}{2},$ continue with period $2 \pi.$ We do get $ f_s(f_t(z)) = f_{s + t}(z), $ plus the holomorphicity and symmetry of $\alpha$ show that $f_s(x)$ is analytic on the full open interval $ 0 < x < \pi.$
EDIT, TUTORIAL: Given some $z$ in the complex plane in the interior of the equilateral triangle with vertices at $0, \sqrt 3 + i, \sqrt 3 - i,$ take $z_0 = z, \; \; z_1 = \sin z, \; z_2 = \sin \sin z,$ in general $z_{n+1} = \sin z_n$ and $z_n = \sin^{[n]}(z).$ It does not take long to show that $z_n$ stays within the triangle, and that $z_n \rightarrow 0$ as $n \rightarrow \infty.$
Second, say $\alpha(z)$ is a true Fatou coordinate on the triangle, $\alpha(\sin z) = \alpha(z) + 1,$ although we do not know any specific value. Now, $\alpha(z_1) - 1 = \alpha(\sin z_0) - 1 = \alpha(z_0) + 1 - 1 = \alpha(z_0).$ Also $\alpha(z_2) - 2 = \alpha(\sin(z_1)) - 2 = \alpha(z_1) + 1 - 2 = \alpha(z_1) - 1 = \alpha(z_0).$ Induction, given $\alpha(z_n) - n = \alpha(z_0),$ we have $\alpha(z_{n+1}) - (n+1) = \alpha(\sin z_n) - n - 1 = \alpha(z_n) + 1 - n - 1 = \alpha(z_0).$
So, given $z_n = \sin^{[n]}(z),$ we have $\alpha(z_n) - n = \alpha(z).$
Third , let $L(z) = \frac{3}{z^2}+ \frac{6 \log z}{5} + \frac{79 z^2}{ 1050} + \frac{29 z^4}{2625}$. This is a sort of asymptotic expansion (at 0) for $\alpha(z),$ the error is $| L(z) - \alpha(z) | < c_6 |z|^6.$ It is unlikely that putting more terms on $L(z)$ leads to a convergent series, even in the triangle.
Fourth, given some $ z =z_0$ in the triangle. We know that $z_n \rightarrow 0$. So $| L(z_n) - \alpha(z_n) | < c_6 |z_n|^6.$ Or $| (L(z_n) - n ) - ( \alpha(z_n) - n) | < c_6 |z_n|^6 ,$ finally $$ | (L(z_n) - n ) - \alpha(z) | < c_6 |z_n|^6 .$$ Thus the limit being used is appropriate.
Fifth, there is a bootstrapping effect in use. We have no actual value for $\alpha(z),$ but we can write a formal power series for the solution of a Julia equation for $\lambda(z) = 1 / \alpha'(z),$ that is $\lambda(\sin z ) = \cos z \; \lambda(z).$ The formal power series for $\lambda(z)$ begins (KCG Theorem 8.5.1) with $- z^3 / 6,$ the first term in the power series of $\sin z$ after the initial $z.$ We write several more terms, $$\lambda(z) \asymp - \frac{z^3}{6} - \frac{z^5}{30} - \frac{41 z^7}{3780} - \frac{4 z^9}{945} \cdots.$$ We find the formal reciprocal, $$\frac{1}{\lambda(z)} = \alpha'(z) \asymp -\frac{6}{z^3} + \frac{6}{5 z} + \frac{79 z}{525} + \frac{116 z^3}{2625} + \frac{91543 z^5}{6063750}\cdots.$$ Finally we integrate term by term, $$\alpha(z) \asymp \frac{3}{z^2} + \frac{6 \log z }{5} + \frac{79 z^2}{1050} + \frac{29 z^4}{2625} + \frac{91543 z^6}{36382500}\cdots.$$ and truncate where we like, $$\alpha(z) = \frac{3}{z^2} + \frac{6 \log z }{5} + \frac{79 z^2}{1050} + \frac{29 z^4}{2625} + O(z^6)$$
Numerically, let me give some indication of what happens, in particular to emphasize $ f_{1/2} (\pi/2) = 1.140179\ldots.$
x alpha(x) f(x) f(f(x)) sin x f(f(x))- sin x
1.570796 2.089608 1.140179 1.000000 1.000000 1.80442e-11
1.560796 2.089837 1.140095 0.999950 0.999950 1.11629e-09
1.550796 2.090525 1.139841 0.999800 0.999800 1.42091e-10
1.540796 2.091672 1.139419 0.999550 0.999550 3.71042e-10
1.530796 2.093279 1.138828 0.999200 0.999200 1.97844e-10
1.520796 2.095349 1.138070 0.998750 0.998750 -2.82238e-10
1.510796 2.097883 1.137144 0.998201 0.998201 -7.31867e-10
1.500796 2.100884 1.136052 0.997551 0.997551 -1.29813e-09
1.490796 2.104355 1.134794 0.996802 0.996802 -1.14504e-09
1.480796 2.108299 1.133372 0.995953 0.995953 9.09416e-11
1.470796 2.112721 1.131787 0.995004 0.995004 1.57743e-09
1.460796 2.117625 1.130040 0.993956 0.993956 5.63618e-10
1.450796 2.123017 1.128133 0.992809 0.992809 -3.00337e-10
1.440796 2.128902 1.126066 0.991562 0.991562 1.19926e-09
1.430796 2.135285 1.123843 0.990216 0.990216 2.46512e-09
1.420796 2.142174 1.121465 0.988771 0.988771 -2.4357e-10
1.410796 2.149577 1.118932 0.987227 0.987227 -1.01798e-10
1.400796 2.157500 1.116249 0.985585 0.985585 -1.72108e-10
1.390796 2.165952 1.113415 0.983844 0.983844 -2.31266e-10
1.380796 2.174942 1.110434 0.982004 0.982004 -4.08812e-10
1.370796 2.184481 1.107308 0.980067 0.980067 1.02334e-09
1.360796 2.194576 1.104038 0.978031 0.978031 3.59356e-10
1.350796 2.205241 1.100627 0.975897 0.975897 2.36773e-09
1.340796 2.216486 1.097077 0.973666 0.973666 -1.56162e-10
1.330796 2.228323 1.093390 0.971338 0.971338 -5.29822e-11
1.320796 2.240766 1.089569 0.968912 0.968912 8.31102e-10
1.310796 2.253827 1.085616 0.966390 0.966390 -2.91373e-10
1.300796 2.267522 1.081532 0.963771 0.963771 -5.45974e-10
1.290796 2.281865 1.077322 0.961055 0.961055 -1.43066e-10
1.280796 2.296873 1.072986 0.958244 0.958244 -1.58642e-10
1.270796 2.312562 1.068526 0.955336 0.955336 -3.14188e-10
1.260796 2.328950 1.063947 0.952334 0.952334 3.20439e-10
1.250796 2.346055 1.059248 0.949235 0.949235 4.32107e-10
1.240796 2.363898 1.054434 0.946042 0.946042 1.49412e-10
1.230796 2.382498 1.049505 0.942755 0.942755 3.42659e-10
1.220796 2.401878 1.044464 0.939373 0.939373 4.62813e-10
1.210796 2.422059 1.039314 0.935897 0.935897 3.63659e-11
1.200796 2.443066 1.034056 0.932327 0.932327 3.08511e-09
1.190796 2.464924 1.028693 0.928665 0.928665 -8.44918e-10
1.180796 2.487659 1.023226 0.924909 0.924909 6.32892e-10
1.170796 2.511298 1.017658 0.921061 0.921061 -1.80822e-09
1.160796 2.535871 1.011990 0.917121 0.917121 3.02818e-10
1.150796 2.561407 1.006225 0.913089 0.913089 -3.52346e-10
1.140796 2.587938 1.000365 0.908966 0.908966 9.35707e-10
1.130796 2.615498 0.994410 0.904752 0.904752 -2.54345e-10
1.120796 2.644121 0.988364 0.900447 0.900447 -6.20484e-10
1.110796 2.673845 0.982228 0.896052 0.896052 -7.91102e-10
1.100796 2.704708 0.976004 0.891568 0.891568 -1.62699e-09
1.090796 2.736749 0.969693 0.886995 0.886995 -5.2244e-10
1.080796 2.770013 0.963297 0.882333 0.882333 -8.63283e-10
1.070796 2.804543 0.956818 0.877583 0.877583 -2.85301e-10
1.060796 2.840386 0.950258 0.872745 0.872745 -1.30496e-10
1.050796 2.877592 0.943618 0.867819 0.867819 -2.82645e-10
1.040796 2.916212 0.936899 0.862807 0.862807 8.81083e-10
1.030796 2.956300 0.930104 0.857709 0.857709 -7.70554e-10
1.020796 2.997914 0.923233 0.852525 0.852525 1.0091e-09
1.010796 3.041114 0.916288 0.847255 0.847255 -4.96194e-10
1.000796 3.085963 0.909270 0.841901 0.841901 6.71018e-10
0.990796 3.132529 0.902182 0.836463 0.836463 -9.28187e-10
0.980796 3.180880 0.895023 0.830941 0.830941 -1.45774e-10
0.970796 3.231092 0.887796 0.825336 0.825336 1.26379e-09
0.960796 3.283242 0.880502 0.819648 0.819648 -1.84287e-10
0.950796 3.337412 0.873142 0.813878 0.813878 5.84829e-10
0.940796 3.393689 0.865718 0.808028 0.808028 -2.81364e-10
0.930796 3.452165 0.858230 0.802096 0.802096 -1.54149e-10
0.920796 3.512937 0.850679 0.796084 0.796084 -8.29982e-10
0.910796 3.576106 0.843068 0.789992 0.789992 3.00744e-10
0.900796 3.641781 0.835396 0.783822 0.783822 8.10903e-10
0.890796 3.710076 0.827666 0.777573 0.777573 -1.23505e-10
0.880796 3.781111 0.819878 0.771246 0.771246 5.31326e-10
0.870796 3.855015 0.812033 0.764842 0.764842 2.26584e-10
0.860796 3.931924 0.804132 0.758362 0.758362 3.97021e-10
0.850796 4.011981 0.796177 0.751806 0.751806 -7.84946e-10
0.840796 4.095339 0.788168 0.745174 0.745174 -3.03503e-10
0.830796 4.182159 0.780107 0.738469 0.738469 2.63202e-10
0.820796 4.272614 0.771994 0.731689 0.731689 -7.36693e-11
0.810796 4.366886 0.763830 0.724836 0.724836 -1.84604e-10
0.800796 4.465171 0.755616 0.717911 0.717911 3.22084e-10
0.790796 4.567674 0.747354 0.710914 0.710914 -2.93204e-10
0.780796 4.674617 0.739043 0.703845 0.703845 1.58448e-11
0.770796 4.786234 0.730686 0.696707 0.696707 -8.89497e-10
0.760796 4.902777 0.722282 0.689498 0.689498 2.40592e-10
0.750796 5.024513 0.713833 0.682221 0.682221 -3.11017e-10
0.740796 5.151728 0.705339 0.674876 0.674876 7.32554e-10
0.730796 5.284728 0.696801 0.667463 0.667463 -1.73919e-10
0.720796 5.423842 0.688221 0.659983 0.659983 -1.66422e-10
0.710796 5.569419 0.679599 0.652437 0.652437 5.99509e-10
0.700796 5.721838 0.670935 0.644827 0.644827 -2.45424e-10
0.690796 5.881501 0.662231 0.637151 0.637151 -6.29884e-10
0.680796 6.048843 0.653487 0.629412 0.629412 1.86262e-10
0.670796 6.224333 0.644704 0.621610 0.621610 -5.04285e-10
0.660796 6.408471 0.635883 0.613746 0.613746 -6.94697e-12
0.650796 6.601802 0.627025 0.605820 0.605820 -3.81152e-10
0.640796 6.804910 0.618129 0.597834 0.597834 4.10222e-10
0.630796 7.018428 0.609198 0.589788 0.589788 -1.91816e-10
0.620796 7.243040 0.600231 0.581683 0.581683 -4.90592e-10
0.610796 7.479486 0.591230 0.573520 0.573520 4.29742e-10
0.600796 7.728570 0.582195 0.565300 0.565300 -1.38719e-10
0.590796 7.991165 0.573126 0.557023 0.557023 -4.05081e-10
0.580796 8.268218 0.564025 0.548690 0.548690 -5.76379e-10
0.570796 8.560763 0.554892 0.540302 0.540302 1.49155e-10
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x alpha(x) f(x) f(f(x)) sin x f(f(x))- sin x
Best Answer
We have $$ \sum_{k=0}^{2n}(-i)^k\sin\frac{sk\pi}{2n+1}=\frac{h(s)-h(-s)}{2i},\quad\text{where}\\ h(s)=\sum_{k=0}^{2n}e^{i(-\pi/2+\frac{\pi s}{2n+1})k}=\frac{1-e^{-i\pi(2n+1)/2+i\pi s}}{1-e^{i(-\pi/2+\frac{\pi s}{2n+1})}}=\frac{1+i(-1)^{n+s}}{1+ie^{i\frac{\pi s}{2n+1}}}. $$ The numerators for $s$ and $-s$ are the same, and $$ \frac1{1+ie^{i\theta}}- \frac1{1+ie^{-i\theta}}=\frac{2\sin\theta}{2i\cos \theta}=-i\tan\theta, $$ so the product reads as $$ 2^{-2n}\prod_{s=1}^{2n} (1+i(-1)^{n+s})\tan \frac{s\pi}{2n+1}. $$ The product of $(1+i(-1)^{n+s})$ equals $2^n$, since the product of two consecutive guys equals 2. It remains to prove that $$ \prod_{s=1}^{2n}\tan \frac{s\pi}{2n+1}=(-1)^n(2n+1). $$ This should be well known, and in any case it is standard: using the formula $$ i\tan \theta=\frac{e^{2i\theta}-1}{e^{2i\theta}+1} $$ we get $$ (-1)^n\prod_{s=1}^{2n}\tan \frac{s\pi}{2n+1}=\prod_{s=1}^{2n} \frac{\omega^s-1}{\omega^s+1},\quad\text{where}\, \omega=e^{2\pi i/(2n+1)}. $$ We have $\prod_{s=1}^{2n}(z-\omega^s)=1+z+\ldots+z^{2n}=:P(z)$, therefore $$\prod_{s=1}^{2n} \frac{\omega^s-1}{\omega^s+1}=\frac {P(1)}{P(-1)}=2n+1$$