By Fredon D., Bertrand F., Maumy-Bertrand M.
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Additional resources for Mathematiques Analyse en 30 fiches
Sample text
2). (3). 8 Solution π π , , est continue et strictement croissante. 2 2 4 5 π π ≈ 1,32 ∈ − , , l'équation (1) admet une solution, Comme arcsin + arcsin 5 13 2 2 et une seule. 1 1 π π / − , , l'équation (3) n'a pas de solution. Comme arccos + arccos ≈ 2,55 ∈ 3 4 2 2 La fonction arccos, de [−1,1 ] dans [ 0,π ] , est continue et strictement croissante. 1 1 Comme arccos + arccos ≈ 2,55 ∈ [ 0,π ], l'équation (2) admet une solution, et 3 4 une seule. • La fonction arcsin, de [−1,1 ] dans − • L'équation (1) implique : 5 4 = sin (arcsin x) = x .
A En particulier : b f (x) dx |b − a| sup | f (x)| x∈[a,b] a • Sommes de Riemann 1 n→+∞ n n−1 lim f i=0 i n 1 = f (x) dx 0 Plus généralement, si (x0 ,. . ,xn ) est une subdivision de [ a,b ] dont le pas tend vers 0 quand n tend vers l'infini, et ci un point quelconque de [ xi ,xi+1 ] (le plus souvent xi ou xi+1 ), on a alors : n−1 n→+∞ 64 Analyse en 30 fiches b (xi+1 − xi ) f (ci ) = lim i=0 a f (x) dx .
On en cherche une base par la résolution de l'équation caractéristique : ar 2 + br + c = 0 (E). – Cas a, b, c complexes / 0, (E) a deux racines distinctes r1 et r2 . Toute suite vérifiant (1) est alors Si = du type : 56 Analyse en 30 fiches 1 1 u n = K 1 r1n + K 2 r2n où K 1 et K 2 sont des constantes que l'on exprime ensuite en fonction de u 0 et u 1 . b Si = 0, (E) a une racine double r0 = − · Toute suite vérifiant (1) est alors du 2a type : u n = (K 1 + K 2 n)r0n . – Cas a, b, c réels Si > 0 ou = 0, la forme des solutions n'est pas modifiée.