数值分析课后答案_华中科技大出版社

¨
I0 = (1 − I1 )/1 ≈ 0.632121. )
2
Iprev=I0; For[n=1,n≤9,n++, temp=1-n*Iprev; II=Append[II,temp]; Iprev=temp]; II output {0.6321,0.3679,0.2642,0.2074,0.1704,0.148,0.112,0.216,-0.728,7.552} I9=(0.3679*0.1+0.1)/2 output 0.068395 II={I9}; Iprev=I9; For[n=9,n≥1,n- -, temp=(1-Iprev)/n; II=Prepend[II,temp]; Iprev=temp]; II output {0.632121,0.367879,0.264241,0.207276,0.170895,0.145525,0.126848,0.112061,0.103512,0.068395} Approximating Real Values II=Table[NIntegrate[x n*Exp[x-1],{x,0,1}],{n,0,9}] output {0.632121,0.367879,0.264241,0.207277,0.170893,0.145533,0.126802,0.112384,0.100932,0.0916123}
3
0.6,
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1 x f (x) (f (x) = ln x): 0.4 0.5 0.6 −0.916291 −0.693147 −0.510826 f (0.55) 0.5, 0.6 0.7 −0.356675 0.55 Lagrange L1 (x) = y0 ∗ x − x1 x − x0 + y1 ∗ x0 − x 1 x1 − x 0 = −1.60475 + 1.82321x ln(0.55) ≈ L1 (0.55) = −0.601986. Lagrange x0 = 0.4, x1 = 0.5, x2 = 0.6
·
In ∈ [e−1 /(n + 1), 1/(n + 1)], I8 , I9 err(Ik ) = Ik − Ik err(I9 ) = err(I0 ) ≈ 0.5 ∗ 10−4 , I9 − I9 = (−9)err(I8 ) = (−9)(−8)err(I7 ) = · · · = −9!err(I0 ), |err(I9 )| ≈ 362880 ∗ 0.5 ∗ 10−4 = 18.144. Ik I9 ≈ (e−1 /10 + 1/10) ∗ (1/2) = 0.068395, In−1 = (1 − In )/n I8 , I7 , · · · , I0 , In
ª
Ex 3:
x > 0,x xA x
δ,
ln x
x
δ,
|x − xA | ≤δ |xA | x −1 ≤δ xA x 1−δ ≤ ≤1+δ xA ln(1 − δ ) ≤ ln x xA ≤ ln(1 + δ )
Ç
ε = max{− ln(1 − δ ), ln(1 + δ )} = − ln(1 − δ ), ( 1/(1 − δ ) ≥ 1 + δ ). | ln( xx )| ≤ ε, − ln(1 − δ ) A

f (x) f
|(x − a)(x − b)| = |(x − a)(b − x)| ≤ ((x − a) + (b − x))2 /4 = (b − a)2 /4 1 f (b) − f (a) (x − a)] ≤ (b − a)2 max |f (x)| a≤x≤b b−a 8 a = 0, b = 1, f (a) = 0, f (b) = 1,
ln(0.55) ≈ L2 (0.55) = −0.59465.
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{7b |7BD e7}7~7A7Ã7BD Äf t4{Á7|7ÅÂ7 CEÂ{|Æe ÇÈÉ 'Ê ²Ë B ÌÍ $ !¯° e 0 1 Î |ÏÎ q Î g Î Ð | Ð | Ð | Ð Î |
(I) L2 (x) = y0 ∗ (x − x1 )(x − x2 ) (x − x0 )(x − x2 ) (x − x0 )(x − x1 ) + y1 ∗ + y2 ∗ (x0 − x1 )(x0 − x2 ) (x1 − x0 )(x1 − x2 ) (x2 − x0 )(x2 − x1 ) = −2.2171 + 4.06848x − 2.04115x2
Lagrange
f0 = 2.71828, f1 = 3.2863, f2 = 3.52761,
2 g¾h
Lagrange
p2 (x) = f0 ∗
(x − x1 )(x − x2 ) (x − x0 )(x − x2 ) (x − x0 )(x − x1 ) + f1 ∗ + f2 ∗ (x0 − x1 )(x0 − x2 ) (x1 − x0 )(x1 − x2 ) (x2 − x0 )(x2 − x1 ) = 1.93566 − 9.2914x + 10.074x2
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(1 − δ 2 ) ≤ 1, ln x 1
« © ­ ¯ ¯ ¬ ¦§ Ú ÛØ°£Ù¤Ý¢¥ Ü« ¯ £ Þ ¯ Æ ¡
Ex 7: In = −1 e ≈ 0.3679,
1 0
xn ex−1 dx, n = 0, 1, · · · ., I0 , I1 , · · · , I9 (
1
I0 = 1 − e−1 ,In = 1 − nIn−1 . ). I0 , I1 , · · · , I9 .
ª
I0 =
1 0
0
−1 ex−1 dx = ex−1 |1 . 0 = 1−e 1 0
ª ¾
xk − 1 ≤ (1 + δ )k − 1 xk A
½Ê
ε = (1 + δ ) − 1,
¶ ¬ Ñ ·×Ë Í§±¢ Þ Ú¡ ͧ±Ë ¨ Þ¢ ª Ú¡¢ °£ « Ë ²ÅÒ¡
k
(1 + δ )k − 1 ≥ 1 − (1 − δ )k , ( (1 + δ )k − 1 xk
(1 + δ )k + (1 − δ )k ≥ 2).
½
Ï ± ²ÅÒ¢ Ì · « ´Æ ±Ò
I9 = 1 − 9I8 ≈ 7.5520.
ÐÄ Â Æ ¦ Ë Æ Í§¡ ¬ ´ Æ Î£
I8 = (1 − I9 )/9 ≈ 0.103512, I7 = (1 − I8 )/8 ≈ 0.112061, I6 = (1 − I7 )/7 ≈ 0.126848, I5 = (1 − I6 )/6 ≈ 0.145525, I4 = (1 − I5 )/5 ≈ 0.170895, I3 = (1 − I4 )/4 ≈ 0.207276, I2 = (1 − I3 )/3 ≈ 0.264241, I1 = (1 − I2 )/2 ≈ 0.367879, ( Mathematica I0=1-0.3679 ouput 0.6321 II={I0};
ß Ú¡ ¨ ͧ±Ë Þ Ú¡ª ¢ °£ « ËÞ ²ÅÒ¡
Ex 2: x xA x δ, x xk δ, |x − xA | ≤δ |xA | x −1 ≤δ xA x 1−δ ≤ ≤1+δ xA (1 − δ )k ≤ (1 − δ )k − 1 ≤ ε = max{(1 + δ )k − 1, 1 − (1 − δ )k }, −ε ≤ xk −1≤ε xk A xk ≤ (1 + δ )k xk A
2
lm 3 FyGyyg y y l y m y y u y b y y y o y y B y U V y o y y B W h 0 © we e cd B ¡ ¢ £ W ¤ ¥ u ¦ § ¨ Bvwe #$ ª «¬­ ® ' d ² · ¸ ¹ º » !¯° e ±² 0 ³´µ '¶ 1 ¼¾½¾¿ PQab ¾ ¾8¾9¾À¾o¾c¾d¾Q
max f (x) − [f (a) + f (x) f (x) − [f (a) +
f (b)−f (a) (x 来自百度文库−a
f (b) − f (a) 1 (x − a)] ≤ (b − a)2 max |f (x)| a≤x≤b b−a 8
x0 = a, x1 = b − a)],
L1 (x),
L1 (x) =
In =
1 0
xn ex−1 dx = In
e−1 ≈ 0.3679,
¥ Ë ²Å ÆÒ¡
In
xn dex−1 = xn ex−1 |1 0−
nxn−1 ex−1 dx = 1 − nIn−1 .
I0 = 1 − e−1 ≈ 1 − 0.3679 = 0.6321, I1 = 1 − I0 ≈ 0.3679, I2 = 1 − 2I1 ≈ 0.2642, I3 = 1 − 3I2 ≈ 0.2074, I4 = 1 − 4I3 ≈ 0.1704, I5 = 1 − 5I4 ≈ 0.1480, I6 = 1 − 6I5 ≈ 0.1120, I7 = 1 − 7I6 ≈ 0.2160, I8 = 1 − 8I7 ≈ −0.7280,
a≤x≤b
max f (x) − [f (a) +
/8 = 1
f (x) = 4x − 3x 4x2 − 3x − x = 4x2 − 4x = 1
4
f (x) = 3xex − 2ex , f p2 , p2 (1.03) − f (1.03)
x0 = 1, x1 = 1.05, x2 = 1.07
x ∈ [a, b],
x = a, b,
Ψ(t) = f (t) − L1 (t) − [a, b]
f (x) − L1 (x) (t − a)(t − b) (x − a)(x − b) Rolle
a, b, x
Ψ(t)
Ψ (t)
(a, b)
ξ,
|f (x) − L1 (x)| ≤
f |(x − a)(x − b)| 2 1
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x0 = 0.5, x1 =
CE fgh PQRB89ipqrsButwvUWyxy9eyPQy89yBUyWy9yeP gh PQcdX Q89 r YW ` 89 PQabBSTUV BR
CE rs89 CE
3 f ∈ C 2 [a, b],
a≤x≤b
ln(0.55) ≈ L2 (0.55) = −0.596884. (II) x0 = 0.5, x1 = 0.6, x2 = 0.7
B
L2 (x) = −2.0273 + 3.37256x − 1.4085x2
f (1.03) − p2 (1.03) = 0.00011176 h h = 0.05, f − p2 ≤ h3 f (3) 4 ∗ (2 + 1) 0.053 ∗ 29.766/12 = 0.000310063. 7 x f (x) 1 3 3/2 13/4 f 0 3 2 5/3 Newton