1. ´ ´
ANALISIS MATEMATICO - PRIMER CURSO - UNIVERSIDAD DE ZARAGOZA
´
FORMULA DE TAYLOR (DESARROLLOS LIMITADOS)
f (a) f (n (a)
f (x) = f (a) + f (a)(x − a) + (x − a)2 + . . . + (x − a)n + Rn (x);
2! n!
f (n+1 (t)
Rn (x) = (x − a)n+1 = o((x − a)n ), x → a
(n + 1)!
EJEMPLOS:
1 1
1) = 1 + x + x2 + x3 + . . . + xn + xn+1 .
1−x (1 − t)n+2
1 2 1 1 et
2) ex = 1 + x + x + x3 + . . . + xn + xn+1 .
2! 3! n! (n + 1)!
1 1 1 (−1)n−1 n (−1)n
3) log(1 + x) = x − x2 + x3 − x4 + . . . + x + xn+1 .
2 3 4 n (n + 1)(1 + t)n+1
α 2 α n α
4) (1 + x)α = 1 + αx + x + ... + x + (1 + t)α−n−1 xn+1 .
2 n n+1
1 3 1 1 (−1)n 2n+1 (−1)n+1 cos t 2n+3
5) sen x = x − x + x5 − x7 + . . . + x + x .
3! 5! 7! (2n + 1)! (2n + 3)!
1 2 1 1 (−1)n 2n (−1)n+1 cos t 2n+2
6) cos x = 1 − x + x4 − x6 + . . . + x + x .
2! 4! 6! (2n)! (2n + 2)!
1 2 17 7
7) tg x = x + x3 + x5 + x + o(x8 ), cuando x → 0.
3 15 315
1 5 61 6
8) sec x = 1 + x2 + x4 + x + o(x7 ), cuando x → 0.
2 24 720
1 3 3 5 5 7 1 · 3 · 5 · . . . · (2n − 1) x2n+1
9) arc sen x = x + x + x + x +...+ · + o(x2n+2 ),
6 40 112 2 · 4 · 6 · . . . · (2n) 2n + 1
cuando x → 0.
1 3 1 5 1 7 (−1)n 2n+1
10) arc tg x = x − x + x − x + . . . + x + o(x2n+2 ), cuando x → 0.
3 5 7 2n + 1
1 1 1 1 cosh t 2n+3
11) senh x = x + x3 + x5 + x7 + . . . + x2n+1 + x .
3! 5! 7! (2n + 1)! (2n + 3)!
1 1 1 1 cosh t 2n+2
12) cosh x = 1 + x2 + x4 + x6 + . . . + x2n + x .
2! 4! 6! (2n)! (2n + 2)!
1 2 17 7
13) tgh x = x − x3 + x5 − x + o(x8 ), cuando x → 0.
3 15 315
1 3 1 · 3 · 5 · . . . · (2n − 1) x2n+1
14) arg senh x = x − x3 + x5 + . . . + (−1)n · + o(x2n+2 ),
6 40 2 · 4 · 6 · . . . · (2n) 2n + 1
cuando x → 0.
1 1 1 1
15) arg tgh x = x + x3 + x5 + x7 + . . . + x2n+1 + o(x2n+2 ), cuando x → 0.
3 5 7 2n + 1