x0 , x1 ,...,xn i = 0, 1,...,n
xi
f ( f (x)
f ( f (x) P n (x) = f ( f (x0 ) + (x x0 )f [ f [x0 , x1 ] + (x x0 )(x )(x x1 )f [ f [x0 , x1 , x2 ]+... ]+ ...+( +(x x x0 )(x )(x x1 )...( ...(x xn 1 )f [ f [x0 , x1 ,...,xn ] = X n
−
−
−
−
−
−
−
(x0 , y0 ), (x1 , y1 ), ........., ........., (xn, yn ) x 0 xn
f ( f (x1 ) x1
f (x ) f ( f (x) − f ( f (x ) − f ( = x−x −x 0
0
0
f 1 (x) = f ( f (x0 ) +
0
f ( f (x1) x1
f (x ) − f ( − x (x − x ) 0
0
0
f 1 (X )
f ( f (x1 ) x1
− f ( f (x ) −x 0
0
+ b1 (x f 2(x) = b 0 + b f 2(x) = b 0 + b + b1x
− b x + b + b x 1
0
2
2
+ b (x − x )(x )(x − x ) − x ) + b 0
+ b2 x0x1
2
0
1
− b xx − b xx 2
0
2
1
f 2(x) = a 0 + a + a1 x + a + a2 x2
a0 = b = b 0 a1 = b = b 1
− b x + b + b x x −b x −b x 1
0
2
0
2
0
2
1
1
a2 = b = b 2
b0
x = x0
b0 = f = f ((x0) x = x1
+ b1 (x1 f ( f (x1 ) = b 0 + b
−x )
f ( f (x1) = f ( f (x0) + b + b1 (x1
0
−x ) 0
b1 =
f ( f (x1) x1
− f ( f (x ) −x 0
0
x = x 2
b2 =
f (f (x ) − f (f (x ) 2
1
x2
f n (x) = b 0 + b + b1 (x
−x
1
−
f ( f (x1 ) x1
f (x ) − f ( /(x − x ) −x 0
2
0
0
+ ... + + b b (x − x )(x )(x − x )...( ...(x − x − x ) + ... n
0
0
1
n−1
)
b0 , b1,...,bn [x0, f ( f (x0 )], )], [x1 , f ( f (x1 )], )], .. ..., ., [xn , f ( f (xn )]
= f ((x0 ) b0 = f
b1 = f = f [[x1 , x0 ]
b2 = f = f [[x2 , x1 , x0 ]
b2 = f [ f [xn , xn 1 ,...,x1 , x0 ] −
P 1 (x), P 2 (x),...,P N N (x) P N N 1 (x) −
P N N (x)
P 1 (x) = b 0 + b + b1 (x
P 2 (x) = b 0 + b + b1 (x
P 3 (x) = b 0 + b + b1 (x
−x ) 0
+ b (x − x )(x )(x − x ) − x ) + b 0
2
0
1
+ b (x − x )(x )(x − x ) + b + b (x − x )(x )(x − x )(x )(x − x ) − x ) + b
+ b1 (x P N N (x) = b 0 + b
0
2
0
1
3
0
+ b (x − x )(x )(x − x ) + ... + ... + + b − x ) + b b 0
2
0
1
P N N (x)
2
)(x − x )(x )(x − x )...( − x )(x ...(x − x 0
1
2
P N N 1 (x) −
P N + bN (x N (x) = P N N 1 (x) + b −
)(x − x )(x )(x − x )...( ...(x − x − x )(x 0
1
2
P N N (x) centros centros x0 , x1 ,...,xN
polinomio de N ewton
aN (x
(x
N
1
0
1
P N N (x)
2
N −1
)
P N N (x)
−1
)(x − x )(x )(x − x )...( ...(x − x − x )(x
N −1
)
N −1
)
bk P 1 (x).... ....,, P N N (x) P k (x) x0 , x1 ,....,xk x0, x1,....,xk+1
f ( f (x)
P 1 (x)
b0
b1
P 1 (x0 ) = f ( f (x0 ) y P 1 (x1 ) = f ( f (x1 ) a0 f ( f (x0 ) = P 1 (x0 ) = b 0 + b + b1 (x0
− x ) = b 0
0
b0 = f = f ((x0 ) f ( f (x1 ) = P 1(x1 ) = b 0 + b + b1 (x1
− x ) = f ( f (x ) + b + b (x − x ) 0
0
1
1
0
b1 b1 = (x0 , f ( f (x0 ))
f ( f (x1 ) x1
b1 (x1 , f ( f (x1 )) b 0 b1
f (x ) − f ( −x 0
0
P 1(x)
f ( f (x2 ) = P 2 (x2) = b 0 + b + b1 (x2
P 2 (x) x2
+ b (x − x )(x )(x − x ) − x ) + b 0
2
2
b0
0
2
1
b1
f ( f (x2 ) b0 b1 (x2 x0) (x2 x0 )(x )(x2 x1 )
− − − − − f (f (x ) − f (f (x ) f (f (x ) − f (f (x ) b = /(x − x ) − x −x x −x b2 = 2
0
1
0
2
2
2
b2 =
0
f (f (x ) − f (f (x ) 2
x2
1
−x
1
1
−
f ( f (x1 ) x1
1
0
f (x ) − f ( /(x − x ) −x 0
2
0
0
f x
x0 ,...,xn
n
f [ f [x0; :::; :::; xn] xn]
f ( f (x) f [ f [xk ] = f ( f (xk ), f [ f [xk ] xk
− f [ f [x ] −x , f [ f [x , x ] − f [ f [x , x ] f [ f [x , x , x ] = , x −x f [ f [x , x , x ] − f [ f [x , x ,x ,x ,x ] = x −x f [ f [xk 1 , xk ] = −
k−1
k−2
k−1
k−1
k−1
k
k−2
k
f [ f [xk
k−2
−3
k−2
k−1
k−1
k−2
k
−
j +1
−
,...,xk ] =
f [ f [xk
bk f ( f (x j )
k −3
k−2
j +1
−
−
.
k−3
,...,xk ] f [ f [xk j ,...,xk 1 ] xk xk j
− −
−
−
−
P N N (x) j
bk f ( f (x) bk = f [ f [x0 , x1 ,...,xk ].
xi x0 x1 x2 x3
, xk 1 ]
k
k
f [ f [xk j , xk
k−1
k
f ( f (xi ) f ( f (x0 ) f ( f (x1 , x0 ) f ( f (x2 , x1 , x0 ) f ( f (x3 , x2, x1, x0) f ( f (x1 ) f ( f (x2 , x1 ) f ( f (x3 , x2 , x1 ) f ( f (x2 ) f ( f (x3 , x4 ) f ( f (x3 )
x0, x1,...,xN N + 1 P N N (x)
[a, b]. N
f ( f (x j ) = P N N (x j ) paraj = 0, 1,...,N.
P N + b1 (x N (x) = b 0 + b
+ ... + + b b − x ) + ... 0
bk f [ f [x0 , x1 ,...,xk ]
(x
)(x − x )...( ...(x − x − x )(x
N
0
N −1
1
)
k = 0, 1,...,N. pN (x) f ( f (x) f ( f (x) = P N + E N N (x) + E N (x)
N +1
f (a, b)
∈ ∈ C
[a, b]
x
E N N (x) =
(x
∈ [a, b]
)(x − x ) · · · (x − x − x )(x 0
1
c = c( c (x)
)f N +1(c)
N
(N + + 1)!
Observacion.
E N N (x)
g(t)
g(t) = f ( f (t)
( t − x )(t )(t − x ) − P (t) − E (x) (t (x − x )(x )(x − x ) 0
1
1
1
0
1
x, x0 , x1
t
g g (x) = f ( f (x)
)(x − x ) = f ( f (x) − P (x) − E (x) = 0 − P (x) − E (x) ((xx −− xx )(x )(x )(x − x ) 1
0
1
0
1
1
1
1
)(x − x ) = f ( f (x ) − P (x ) = 0 − P (x ) − E (x) (x(x −− xx )(x )(x )(x − x ) (x − x )(x )(x − x ) g(x ) = f ( f (x ) − P (x ) − E (x) = f ( f (x ) − P (x ) = 0 (x − x )(x )(x − x ) g(x0 ) = f ( f (x0 )
0
1
0
0
0
1
1
0
1
1
1
1
1
0
0
1
0
1
1
0
1
1
1
1
0
1
x
(x0 , x1 )
g (t) d0
[x0 , x1 ]
x0 < d0 < x
g (d0 ) = 0 g(t) d1
x < d1 < x1
[x; x1 ]
g (d1 ) = 0 g (t)
t = d 0 g (t)
t = d = d 1 [d0, d1 ]
c
∈ [a, b]
g (c) = 0 g (t)
g (t)
g (t) = f (t)
(t (t − x ) − P (t) − E (x) (t(x−−xx) +)(x )(x − x ) 0
1
1
1
0
()g ()g (t) = f (t)
1
2 − 0 − E (x) (x − x )(x )(x − x ) 1
0
1
P 1(t) N = 1 t = c = c
P 1 (t) = 0 2 − E (x) (x − x )(x )(x − x )
0 = f (c)
1
0
1
E 1 (x) E 1 (x) =
(x
2
)(x − x )f (c) − x )(x 0
1
2!
|E (x)|
h
N N
N +1
h
|E (x)|
hN +1
N N
M N N +1
h
h
h hN +1.
Aproximacion O( O (hn ) h 0 f ( f (h) C g (h)
→
|
|≤ |
|
f ( f (h) f ( f (h) = O( O (g (h)) h C
| |≤
2
O (h ) |E (x)| = O( 1
N f (x0 ) E N N (x) De Orden Orden de g (h) C c N O(h )
|
N +1
|
| |
x
∈ [x , x ] 0
1
O(h2 )
h2M 2 /8
h2 2
2
|E (x)| ≤ C h ≈ O(h ) 1
f ( f (x) [a, b]
|f
(x)
| |f
(x)
|h| < 1,
hN +1
N
y = f ( f (x) = cos( cos(x) M 2 M 3 M 4 [0, [0,0, 1,2]
|
[0, [0,0, 1,2] f (x)
|
2
cos(x)| ≤ | − cos(0 cos(0,,0)| = 1,000 = M = M , |f (x)| = | − cos( sen(x)| ≤ |sen(1 sen(1,,2)| = 0,932039 = M = M , |f (x)| = |sen( cos(x)| ≤ |cos(0 cos(0,,0)| = 1,000 = M = M , |f (x)| = |cos( 2
3
3
4
4
P 1 (x)
h = 1,2 2
2
(1,2) (1, (1,000) = 0,18 |E (x)| ≤ (h 8M ) ≤ (1, 8 2
1
P 2 (x)
h = (1, (1,2/2) = 0, 0,6 3
3
M ) (0, (0,6) (0, (0,932039) √ = 0,012915 |E (x)| ≤ (h9√ ≤ 3 9 3 3
2
P 3 (x)
h = (1, (1,2/3) = 0, 0,4
|E (x)| ≤ 3
(h4 M 4 ) 24
≤
(0, (0,4)4(1, (1,000) = 0,001067 24 f ( f (x) = Ln( Ln (x)
x0 = 1; f ( f (x0 ) = 0 x1 = 4; f ( f (x1 ) = 1, 3863 x2 = 6; f ( f (x2 ) = 1, 7917 x3 = 5; f ( f (x3 ) = 1, 6094
|
3
f ( f (x) = b 0 + b + b1 (x
+ b (x − x )(x )(x − x ) + b + b (x − x )(x )(x − x )(x )(x − x ) − x ) + b 0
2
0
1
3
0
1
2
b0 , b1 , b2 , b3 b0 = f = f ((x0) b1 = f [ f [x1, x0] b2 = f [ f [x2, x1, x0 ] b3 = f [ f [x3, x2, x1, x0 ] f (x ) 1, 3863 − 0 − f ( = = 0, 4621 ⇒ b 4−1 −x f ( f (x ) − f ( f (x ) 1, 7917 − 1, 3863 f [ f [x , x ] = = = 0, 2027 6−4 x −x 1, 6094 − 1, 7917 f ( f (x ) − f ( f (x ) f [ f [x , x ] = = = 0, 1823 x −x 5−6 f [ f [x1 , x0 ] =
f ( f (x1 ) x1
0
1
0
2
2
1
1
2
1
3
3
2
2
3
f [ f [x2 , x1 ] x2
2
− f [f [x , x ] = 0, 2027 − 0, 4621 = −0, 05188 ⇒ b 6−1 −x 0, 1823 − 0, 2027 f [ f [x , x ] − f [ f [x , x ] f [ f [x , x , x ] = = = −0, 0204 x −x 5−4
f [ f [x2, x1, x0] =
3
3
2
1
0
2
0
2
2
1
1
3
f [ f [x3 , x2 , x1 , x0 ] =
f ( f (x) = 0, 4621(x 4621(x
f [ f [x3 , x2 , x1 ] x3
1
− f [f [x , x , x ] = −0, 0204 + 0,0, 05188 = 0, 00787 ⇒ b 5−1 −x 2
1
0
0
05188(x − 1)(x 1)(x − 4) + 0, 0, 00787(x 00787(x − 1)(x 1)(x − 4)(x 4)(x − 6) − 1) − 0, 05188(x
f (2) f (2) = 0, 0, 4621(2
− 1) − 0, 05188(2 − 1)(2 − 4) + 0,0, 00787(2 − 1)(2 − 4)(2 − 6) f (2) f (2) = 0, 0, 4621(1) − 0, 05188(−2) + 0, 0, 00787(−8) 0, 62882 f (2) f (2) = 0, f ( f (x) = Ln( Ln (x) f (2) f (2) = Ln = Ln(2) (2) = 0, 0, 69314
− 0, 62882 | ∗ 100 = 9,9, 3 | 0, 69314 0, 69314
3
f ( f (x) = cos(2x cos(2x)
x0
E N N (x) =
(x
x1
x2
)(x − x ) · · · (x − x − x )(x 0
1
)f N +1(c)
N
(N + + 1)!
xk , y k f ( f (x) = cos(2x cos(2x) f 1 (x) =
sin(2x x) −2 sin(2 f (x) = −4 cos(2 cos(2x x) 2
f 3 (x) = 8sin(2x 8sin(2x)
8 sin(2 sin(2))
sin(2) 2) (1, (1,5−1,3)(1, 3)(1,5−1,2)(1, 2)(1,5−1,9)(1, 9)(1,5−2)| = 0,016 sin(2 sin(2)) = 0,0145487 | E | |= | 8 sin( 3! f (1 f (1,,5)
+ b1 (x f ( f (x) = b 0 + b
+ b (x − x )(x )(x − x ) − x ) + b 0
2
a0 , a1 , a2 , a3 b0 = f = f ((x0 ) =
−085
b1 = f [ f [x1 , x0 ]
0
1
b2 = f [ f [x2, x1, x0 ]
f (x ) −0, 73 + 0, 0, 85 − f ( = = −1,2 ⇒ b 1, 2 − 1, 3 −x f ( f (x ) − f ( f (x ) −0, 79 + 0, 0, 73 f [ f [x , x ] = = = −0, 085 x −x 1, 9 − 1, 2
f [ f [x1 , x0 ] =
f ( f (x1 ) x1
0
1
0
2
2
1
1
2
f [ f [x2 , x1 , x0 ] =
f ( f (x) =
f (1 f (1,, 5) =
f [ f [x2, x1] x2
1
− f [f [x , x ] = −0, 085 + 1,1, 2 = 1, 592 ⇒ b 1, 9 − 1, 2 −x 1
0
2
0
2(x − 1, 3) + 1, 1, 592(x 592(x − 1, 3)(x 3)(x − 1, 2) −0, 85 − 1, 2(x
2(1, 5 − 1, 3) + 1, 1, 592(1, 592(1, 5 − 1, 3)(1, 3)(1, 5 − 1, 2) = −0, 99448 −0, 85 − 1, 2(1,
f (1,, 5) − p(1 p(1,, 5) |= | − 0, 98999 + 0, 0, 99448| = 0,00449 | f (1
\
\
D(k, j ) =
D(k, j
− 1) − D(k − 1, j − 1) x −x k
k− j
(xk , yk ) = (xk , f ( f (xk ))
k = 0, 1,...,N
P ( P (x) = d 0,0 +d1,1 (x x0 )+d )+d2,2 (x x0 )(x )(x x1 )+... )+...+ +dN,N (x x0 )(x )(x x1 )
−
dk,0 = y k
dk,j =
dk,j
−1
xk
− − −d −x
−
k−1,j −1 k− j
x0 = 1 y0 = 0 x1 = 4 y 1 = 1,3863 x2 = 6 y 2 = 1,7917 x3 = 5 y 3 = 1,6094 0 0 1,3863 0,4621 1,7917 0,2027 1,6094 0,1823
0 0 0 0 0,0519 0 0,0204 0,0079
− −
3 2 787x 787x 2769x 2769x 24727x 24727x 1717/2000 − − − − − − − − − −1717/
10000 0000
2000 0000
250 25000
− · · · (x−x
N −1
)
x 0 = 1, 3 y 0 = 0,85 x 1 = 1, 2 y 1 = 0,73 x 2 = 1, 9 y 2 = 0,79
− − −
−0,8500 0 −0,7300 −1,2000 −0,7900 −0,0857 2 13x 13x
409x 409x
−−− −−− 7
0 0 1,8571
70
101/ 101/28
◦
◦
◦
