ÌÓÔĞ Ñ --------

 !
" #" $ % %
% % & '
! $ % % ()
! #" !& $ & $ * !
! +)
,)-$ -. %)
! -
−−−−−−−−
10 + 10
"# x0 = −1 x1 = 0 x2 = 1 f (x) = sin
πx
2
/$$ 0 P2 (x)1 $% 2% /$$ & 3
& (x − x1 )(x − x2 )
(x − 0)(x − 1)
x(x − 1)
1
=
=
= (x2 − x),
(x0 − x1 )(x0 − x2 )
(−1 − 0)(−1 − 1)
2
2
2
(x − x0 )(x − x2 )
(x + 1)(x − 1)
x −1
L1 (x) =
=
=
= 1 − x2 ,
(x1 − x0 )(x1 − x2 )
(0 + 1)(0 − 1)
−1
(x + 1)(x − 0)
(x + 1)x
1
(x − x0 )(x − x1 )
L2 (x) =
=
=
= (x2 + x).
(x2 − x0 )(x2 − x1 )
(1 + 1)(1 − 0)
2
2
L0 (x) =
*% f (x0 ) = f (−1) = sin −π
= −1 f (x1 ) = f (0) = sin 0 = 0 f (x2 ) = f (1) =
2
π
sin 2 = 1 " x0 = −1 x1 = 0 x2 = 1 f (x) = sin πx
2
& /$$ 0
P2 (x) = f (x0 )L0 (x) + f (x1 )L1 (x) + f (x2 )L2 (x)
1
1
= (−1) (x2 − x) + 0(1 − x2 ) + (1) (x2 + x)
2
2
1
1
= (−x2 + x + x2 + x) = (2x) = x
2
2
"# |f (x) − P2 (x)| " ! $% 2% $ 3
f (x) = sin
πx π
π2
π3
πx πx πx
, f (x) = cos
, f (x) = − sin
, f (x) = − cos
.
2
2
2
4
2
8
2
" z(x) −1 1 f (z(x))
(x − x0 )(x − x1 )(x − x2 )
|f (x) − P (x)| = 3!
3
− π cos πz(x)
8
2
=
(x + 1)(x − 0)(x − 1)
3!
π 3 π3 3
πz(x) 3
|x − x|
=
|x
−
x|
cos
8 · 3! 2 48
1
4 |x3 − x| & " g(x) =
x3 − x $. . - $.
3 √
1
1 1
1 2 3
2
⇒ g ±
g (x) = 3x −1 = 0 ⇒ x = ±
− ±
=
= ±
3
3 3
3 9
"
√
√
π3 3
π3 3
π3
π3 2 3
3
max |x − x| =
=
|f (x) − P (x)| |x − x| 48
48 x
48 9
216
5 + 10
−x
"# f (x) = e − sin x & . $% 5 f (x) & *)'" ")
" ! a < b " " a, b !& f (a)f (b) < 0
& $!% - a = 0 b = 1 $.
. f (a) = f (0) = e0 − sin 0 = 1 > 0, f (b) = f (1) = e−1 − sin 1 = −0.47359
" f (0)f (1) < 1 " - *)'" $. [0, 1]
" f (x) = e−x − sin x & . " $
"# 6 " . )-. , 7 $)
! 8" 9
$% 6 $. . ! " 3
n an
bn
pn
f (pn )
:
;
:
;:
−<;
:
;:
=:
−7<=
7 :
=:
:=:
=7
- $. . p ≈ p4 = 0.5625 15
y = x − 4x − 5 y = e − 4x − 5 " >? ,
p0 = 3 ε = 10−3 8" .9
$% - " (x, y) %" x3 − 4x − 5 = ex − 4x −
5 ⇒ x3 = ex " " x f (x) = x3 − ex
& . f (x) & f (x) = 3x2 − ex & p0 = 3 $! ! f (p0 ) =
f (3) = 3 · 32 − e3 = 6.9145 = 0 "" >? , - $.
n−1 )
pn = pn−1 − ff(p
! $ " $ 3
(pn−1 )
3
x
f (3)
6.9145
f (p0 )
=3− =3−
= 2 ⇒ f (p1 ) = f (2) = 0.61094 > ε
f (p0 )
f (3)
6.9145
f (2)
0.61094
f (p1 )
=2− =2−
= 1.8675 ⇒
p2 = p1 − f (p1 )
f (2)
4.6109
f (p2 ) = f (1.8675) = 0.40915 × 10−1 > ε
f (1.8675)
0.040915
f (p2 )
= 1.8675 − = 1.8675 −
= 1.8572 ⇒
p3 = p2 − f (p2 )
f (1.8675)
3.9906
f (p3 ) = f (1.8572) = 0.63619 × 10−4 < ε
p1 = p0 −
n
pn
=;:
:;
f (pn )
=< 7:
= <7
0.40915 × 10−1
0.63619 × 10−4
- $. " 10−3 " p3 = 1.8572 10 + 5
"# f (x) = cos x − (x + 1)2 & x = 0 %
0
8 $ 9
$% 5 %" !
" 3
f (x) = cos x − (x + 1)2 ,
f (x) = − sin x − 2(x + 1),
f (x) = − cos x − 2,
f (x) = sin x,
f (0) = cos 0 − (0 + 1)2 = 0,
f (0) = − sin 0 − 2(0 + 1) = −2,
f (0) = − cos 0 − 2 = −3,
f (0) = 0.
- $.
f (0)
f (0)
f (0)
(x − 0) +
(x − 0)2 +
(x − 0)3
1!
2!
3!
−2
−3 2 0 3
=0+
x+
x + x
1!
2!
3!
3 2
= −2x − x
2
P3 (x) = f (0) +
0 "# :)@ " 3
P3 (0.05) = −2(0.05) − (0.05)2 = −0.10375
2
- $. |f (0.05) − P3 (0.05)| = | cos(0.05) − (0.05 + 1)2 − (−0.10375)|
= 2.6039 × 10−7
" | cos(0.05) − (0.05 + 1)2 − (−0.10375)|
|f (0.05) − P3 (0.05)|
=
|f (0.05)|
| cos(0.05) − (0.05 + 1)2 |
= 2.5098 × 10−6
, " ! ! $! & " ! 7
15 + 5
"# sin x −
x
1.4
= 0 [1, π/2] " !. )> ,
! g(x) & 8g(x) &)
" " $.
$9
x
= 0 ⇒ x = 1.4 sin x " g(x) = 1.4 sin x &)
$% sin x − 1.4
[1, π/2] " )> "" $.
6 g(x) & [1, π/2] " "
! *% g (x) = 1.4 cos x & x ∈ [1, π/2] ! & "
g(x) & ' ")
π/21 " 11 g(x) & %" " g(π/2) g(1) " g(π/2) = 1.4 sin
π
π
= 1.4 < = 1.5708
2
2
g(1) = 1.4 sin 1 = 1.1781 > 1
$!" x ∈ [1, π/2] ! g(x) ∈ [1, π/2] " 6 g(x) &)
[1, π/2] " '" &
|g (x)| = |1.4 cos x| = 1.4| cos x| 1.4 max π | cos x|
1 x 2
" h(x) = cos x & x ∈ [1, π/2] ! h (x) = − sin x < 0
." "" " ! - $.
|h(1)| = | cos 1| = 0.54030 > |h(π/2)| = cos
π
=0
2
"
|g (x)| 1.4 max π | cos x| = 1.4 cos 1 = k = 0.75642 < 1
1 x 2
' g(x) & [1, π/2]
" "# 89 g(x) & p0 = 1.4 )
!. ε = 10−6 ! $ 8% $9
$% p0 = 1.4 p1 = g(p0 ) = 1.4 sin 1.4 = 1.37961 n kn
$. |pn − p| ≤ 1−k
|p1 − p0 | ≤ 10−6 "
|pn − p| 0.75642n
0.75642n
|1.3796 − 1.4| 10−6 ⇒
(0.0204) 10−6 ⇒
1 − 0.75642
0.24358
0.75642n 0.11940 × 10−4 ⇒ n log(0.75642) log(0.11940 × 10−4 ) ⇒
n(−0.12124) −4.9230 ⇒ n 40.605
n 41 :
15
p0 = 1 p1 = 1.2 % , ln x = cos x !. ε = 10−3 8" .9
$% f (x) = ln x−cos x - . " "
x " 4 % , $ ε = 10−3 p0 = 1 n−1 )(pn−1 −pn−2 )
p1 = 1.2 ! . pn = pn−1 − f (pf (p
"
n−1 )−f (pn−2 )
f (p1 )(p1 − p0 )
f (1.2)(1.2 − 1)
= 1.2 −
f (p1 ) − f (p0 )
f (1.2) − f (1)
(−0.18004)(0.2)
= 1.2 −
= 1.3
(−0.18004) − (−0.54030)
f (p2 ) = f (1.3) = −0.51346 × 10−2 ⇒
|f (p2 )| = |f (1.3)| = 0.51346 × 10−2 > ε = 10−3
f (1.3)(1.3 − 1.2)
f (p2 )(p2 − p1 )
= 1.3 −
p3 = p2 −
f (p2 ) − f (p1 )
f (1.3) − f (1.2)
−2
(−0.51346 × 10 )(1.3 − 1.2)
= 1.3 −
= 1.3029
(−0.51346 × 10−2) − (−0.18004)
f (p3 ) = f (1.3029) = −0.11084 × 10−3 ⇒
|f (p3 )| = |f (1.3029)| = 0.11084 × 10−3 ≤ ε = 10−3
p2 = p1 −
- $. . p ≈ p3 = 1.3029 n
pn
f (pn )
−0.54030
−0.18004
−0.51346 × 10−2
0.11084 × 10−3
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=