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22/07/19 · $\begingroup$ @LLWZ Thanks for providing the extra info I'm sorry, but I realized I didn't have a correct proof when I checked what I did before However, I agree with you the answer at the book, ie, the sum of the function is monotonous on $x_k,x_{k1}$, is an obvious mistake. On note Cn = k=0 cos(kx), et Sn = k=0 sin(kx) Calculons plutˆot CniSn Laformule de Moivre et l’identit´e g´eom´etrique donnent M´ethode pour calculer une somme de fonctions trigo, on passe en complexes!. E W} A F % # G W ' $ 0 * O c H 3 # @ (.
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Comme par hypoth`ese x n’est pas congru `a z´ero modulo 2π, eix 6= 1, on peut donc appliquer l’identit´e g´eom´etrique Cn iSn = k=0 co s (kx) in = k=0. 3 8 9 8 7 6 5 4 3 2 c d d c b = a m e j l = a @ k j d j i f = e c d h = g a ?. F ' x % ( % " ' uh !.
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