c1 :=
(
(x = = p, p, f f 4)) 4)) c1 := . .09787176213 09787176213
c2 :=
(
(x = 0, f f 4)) 4)) c2 := . .09259259259 09259259259
c3 :=
(
(x = 1, f f 4)) 4)) c3 := . .09472344463 09472344463 f (4) (x)
c1
:= c := c11/24 := . .004077990089 004077990089 x x sin x
≤x
| sin x| ≤ |x| f ((x) = x − sin x f
≥0
x = 0
f ((x) = x f
−1 ≤ cos x ≤ 1 | sin x| ≤ 1
x f ((x) f
|
≤
−π < x < 0 < 0 sin(−x) ≤ (−x)
||
x π
f
∈ C [a, b] ∈
≥0
x
≥ sin x
0
≤x≤π
≥ −x > 0 sin x | sin x| = − sin x ≤ −x = |x| | sin x| ≤ |x| x
x1
x2
ξ
[a, b] x1
f ((ξ ) = f c1
− cos x ≥ 0 |x| ≥ π
x
f (x) > f f ( f (0) (0) = 0 x > 0 sin x = sin x x = x
|
f ′ (x) = 1
− sin x
x2
f ((x1 ) + f f f ((x2 ) 1 1 = f f ((x1 ) + f f ((x2 ). 2 2 2
c2
ξ f ((ξ ) = f
x1
x2
c1 f f ((x1 ) + c2 f f ((x2 ) . c1 + c2 c1
c1 =
−c2 1 (f f ((x1 ) + f + f ((x2 )) 2
c2
f ((x1 ) f
f ((x2 ) f
f ξ
f ((ξ ) = f
m
x1
x2
1 1 1 (f f ((x1 ) + f f ((x2 )) = f ( f (x1 ) + f f ((x2 ). 2 2 2
m = min{f f ((x1 ), f f ((x2 )} M = M = max{f f ((x1 ), f f ((x2 )}. m ≤ f f ((x1 ) ≤ M ≤ f f ((x2 ) ≤ M, c1 m ≤ c1 f f ((x1 ) ≤ c1 M c2 m ≤ c2 f f ((x2 ) ≤ c2 M. (c1 + c2 )m m
≤ c1f f ((x1) + c2f f ((x2) ≤ (c1 + c2)M ≤ c1f f ((xc11) ++ cc22f f ((x2 ) ≤ M. x1
ξ
x1
x2
x2 f ((ξ ) = f
c1 f f ((x1 ) + c2 f f ((x2 ) . c1 + c2
f ((x) = x2 + 1 x1 = 0 x2 = 1 c1 = 2 f
c2 =
−1 f ((x) > 0 f c1 f f ((x1 ) + c + c2 f f ((x2 ) 2(1) − 1(2) = = 0. 0. c1 + + c c2 2−1 √ 2
p∗
− √ √ ≤ p p∗
2
2
10−4 ,
p p∗
−
√
10−4
≤ √ ×
2
x
2
10−4 ;
−√ 2 × 10 4 ≤ p − √ 2 ≤ √ 2 × 10 4. √ 2(0 √ 2(1..0001) . 2(0..9999) 9999),, 2(1 −
∗
−
p∗
13 14
− 76 , 2e − 5.4 13 14
13 14
− 67 = 0.0 .0720
2e
= 0. 0 .929
6 7
= 0.857
− 5.4 = 5.44 − 5.40 = 0.0.0400 0400..
e = 2.72 72..
13 14
− 76 = 0.0720 = 1.1 .80 80.. 2e − 5.4 0.0400 |1.80 − 1.954| = 0.0788 0788,,
|1.80 − 1.954| = 0.154
1.954
13 14
13 14
− 67 = 0.0 .0710
2e
= 0.928
6 7
= 0.857
e = 2.71 71..
− 5.4 = 5.42 − 5.40 = 0.0.0200 0200..
13 14
− 76 = 0.0710 = 3.3 .55 55.. 2e − 5.4 0.0200 |3.55 − 1.954| = 0.817 817,,
|3.55 − 1.954| = 1.60 60,, π = 4 arctan 21 + arctan 13
P ((x) = x P
1.954
− 31 x3 + 51 x5.
P
1 2
= 0.464583
1 1 π = 4 arctan + arctan 2 3
|π − 3.145576| ≈ 3.983 × 10
≈
P
3.145576 145576..
ex
−e
−x
x
.
limx→0 f f ((x) f (0 f (0..1) −x
−e ex − e lim
−x
x→0
x
=1
= 0.3218107 3218107,,
|π − 3.145576| ≈ 1.268 × 10 |π |
−3
f ((x) = f
limx→0 ex
1 3
−1 = 0
limx→0 x = 0
ex + e−x 1+1 = = 2. x→0 1 1
= lim
−3
.
e0.100 = 1.11 f (0 f (0..100) =
ex
f ((x) f
≈
1.11 0.905 0.205 = = 2.05 05.. 0.100 0.100
−
≈ 1 + x + 12 x2 + 16 x3
1 + x + 21 x2 + 61 x3
e−x
− − 1
x + 21 x2
x
f (0 f (0..100)
e−0.100 = 0. 0 .905
≈ 1 − x + 12 x2 − 16 x3,
− 61 x3
=
2x + 31 x3 1 = 2 + x2 . x 3
≈ 2 + 13 (0 (0..100)2 = 2 + (0. (0 .333)(0 333)(0..001) = 2. 2.00 + 0. 0.000333 = 2. 2.00 00..
0 0111111 01111111111 1111 010100 0101001100000 110000000000 0000000000 000000000 000000000 0000000000 000000000 000000000 0000000000 0000000 .
+2
1023−1023
1+
1 2
2
+
1 2
4
+
1 2
7
+
1 2
8
1 1 1 1 83 = 2 0 1 + + + + = 1 + = 1.32421875 32421875.. 4 16 128 256 256
0 0111111 01111111111 1111 010100 0101001100000 110000000000 0000000000 000000000 000000000 0000000000 000000000 000000000 0000000000 0000000 .
0 011111 01111111111 11111 0101001 0101001011111 01111111111 1111111111 1111111111 1111111111 1111111111 1111111111 1111111111 111111111 1111 =1.32421875 21023−1023 2−52 =1. =1..3242187499999997779553950749686919152736663818359375 =1 3242187499999997779553950749686919152736663818359375,,
−
0 011111 01111111111 11111 0101001 0101001100000 10000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 000000001 0001 =1.324 =1. 3242187 218755 + 2 1023−1023 2−52 =1..3242187500000002220446049250313080847263336181640625 =1 3242187500000002220446049250313080847263336181640625..
f ((x) = 1. f 1 .01 01ee4x
− 4.62 62ee3x − 3.11 11ee2x + 12. 12.2ex − 1.99 99..
f (1 f (1..53) f ((x) f enx = (e ( ex )
n
f ((x) = ((((1. f ((((1 .01) 01)eex
3(1..53) e3(1
− 4.62) ex − 3.11) ex + 12. 12.2) ex − 1.99 99..
e1.53 = 4. 4 .62 = (4 (4..62)2 (4 (4..62) = (21. (21.3)(4 3)(4..62) = 98. 98.4,
2(1..53) e2(1 = (4 (4..62)2 = 21 21..3 4(1..53) e4(1 = (98. (98.4)(4 4)(4..62) = 455
1.01(455) f (1..53) = 1. f (1 = 460 455
− 4.62(98 62(98..4) − 3.11(21 11(21..3)+ 12. 12.2(4 2(4..62) − 1.99 − − 66 66..2 + 56. 56.4 − 1.99 = 5.00 − 66 66..2 + 56. 56.4 − 1.99 = −61 61..2 + 56 56.4 − 1.99 = −4.80 − 1.99 = −6.79 79..
f (1 f (1..53) = (((1. (((1 .01)4 01)4..62
− 4.62)4 62)4..62 − 3.11)4 11)4..62 + 12 12.2)4 2)4..62 − 1.99 = (((4. (((4.67 − 4.62)4 62)4..62 − 3.11)4 11)4..62 + 12 12.2)4 2)4..62 − 1.99 = ((0. ((0.231 − 3.11)4 11)4..62 + 12 12.2)4 2)4..62 − 1.99 = ( −13 13..3 + 12. 12.2)4 2)4..62 − 1.99 = −7.07 07..
7.61 6.79 + 7. 7.61 = 0.82 0.0710 0710..
|−
| − 7.07 + 7.7.61| = 0.0 .54
|
f l(y )
0.108
k
y
− y
f l(y ) y
≤
0. 5
× 10
−k+1
.
dk+1
≤ 5
dk+1 > 5 5.. dk+1
≤ 5,
− y
f l (y ) 0.dk+1 . . . = 0.d1 . . . y
dk+1 > 5 5,,
− y
(1 f l (y ) = y
× 10n k ≤ 0.5 × 10 × 10n 0.1
− 0.dk+1 . . .) × 10n 0.d1 . . . × 10n
−
−k
<
(1
−k
= 0.5
− 0.5) × 10 0.1
× 10
−k+1
.
−k
= 0. 0 .5
× 10
−k+1
.
T
0.995
≤ P ≤ ≤ 1.005 005,, 0.082055 ≤ R ≤ 0.082065 082065,, 287..61 287 15◦
≤ V ≤ 0.1005 1005,, ≤ 0.004205 0.004195 ≤ N ≤ 004205..
≤ T ≤ ≤ 293 293..42 42..
288..16 288
P
1.99
≤ P ≤ ≤ 2.01
0.0497
286..61 286 19◦
0.0995
292.16 292. 290..15 290
V
≤ V ≤ 0.0503 0503,,
≤ T ≤ ≤ 293 293..72 72..
= 17◦
.
n ∞
arctan x = lim P n (x) = n→∞
x2i−1 (2ii 1) (2
−
( 1)i+1
−
i=1
|4P n(1) − π| < 10
−3
10−10
π
∞
π = 4 ar arct ctan1 an1 = 4
−
( 1)i+1
i=1
1 2i
−1
n 4 2(n 2( n + 1) n
− 1 < 10
−3
4000 < 4000 < 2 2n n + 1. 1.
≥ 2000 2000.. 4 2(n 2( n + 1)
− 1 < 10
−10
n > 20 20,,000 000,,000 000,,000 000..
π π 1 = 4 arc arctan tan 4 5
1 − arctan 239 . 10−3
π
∞
π = 4
−
( 1)
i+1
i=1
∞
1 52i−1 (2 (2ii
− 1)
− −
( 1)i+1
i=1
1 2392i−1 (2 (2ii
− 1) i
i+1 i := 1 :
4 51 (1)
10 =
4 , 5
i = 2 :
4 53 (3)
=
−3
4 375
n
i = 3 :
4 55 (5)
=
4 = 2.56 15625
× 10
i
ai bj ?
i=1 j=1
i j =1
i n
ai b j
i n
n(n + 1) i = 2 i=1
− i
i=1
n
n
(n + 2)(n 2)(n 2 n
i
n
ai b j =
i=1 j=1
n
− i
i=1
bj
+ 1)
n.
−n
1=
ai
i=1
n(n + 1) 2
+ 2)(n 2)(n
.
bj ,
j=1
i
i
−1
n
−1
−n ai
1 2 (n
− 1)
− 1)
−1 .
i
n
1 2 n(n
n(n + 1) 2
−1
n(n + 1) 2
bj
1=
i
−4
.
A B C x1 x2 A = 0 B = 0 x1 =
D = = B B 2 D = 0
− 4AC
x1 =
D < 0 b = a =
−B/ B/(2 (2A A)
−C/B
x1
x1
√ −D/ D/(2 (2A A) −B/ B/(2 (2A A)
x1 = = a a + bi x2 = = a a bi (x1 , x2 )
−
B
≥ 0
√
= B + D d = B + x1 = 2C/d (2A x2 = d/ d/(2 A)
− − √ d = −B + D x1 = d/ d/(2 (2A A) x2 = 2C/d (x1 , x2 ) x F 1 (x) = L1 + O (xα ) c1
c2 F ((x) = c 1 F 1 (x) + c2 F 2 (x) F γ = =
{α, β }
F ((x) = c 1 L1 + c2 L2 + O (xγ ) F
x
F 2 (x) = L 2 + O xβ .
G(x) = F 1 (c1 x) + F 2 (c2 x).
G(x) = L 1 + L2 + O (xγ ) x
|x|
k1
|F 1(x) − L1| ≤ K 1|x|α = max(K max(K 1 , K 2 ) c = max(|c1 |, |c2 |, 1) K =
k2
|F 2(x) − L2| ≤ K 2|x|β .
= max max (α, β ) δ =
|F F ((x) − c1 L1 − c2L2| =|c1 (F 1 (x) − L1) + c2(F 2 (x) − L2)| ≤|c1 |K 1|x|α + |c2|K 2|x|β ≤cK |x|α + |x|β ≤cK |x|γ 1 + |x|δ γ ≤ K |x|γ , |x| F ((x) = c 1 L1 + c2 L2 + O (xγ ) F
−
|G(x) − L1 − L2| =|F 1 (c1x) + F 2(c2x) − L1 − L2| ≤K 1|c1x|α + K 2|c2 x|β ≤Kc δ |x|α + |x|β ≤Kc δ |x|γ 1 + |x|δ γ ≤ K |x|γ , |x| G(x) = L 1 + L2 + O (xγ )
√
−
F 0 = 1 F 1 = 1 limn→∞ xn = = x x
xn = F n+1 /F n x = 1 + 5 /2
′′
F n+2 = = F F n+1 + F n
lim xn = lim xn+1 = = x x
n→∞
xn+1 = 1 +
n→∞
x = 1 +
1 , x
x2
≡
˜n = 1 F 5
√
− x − 1 = 0. √ 1 + 5 /2
√ − √ x =
F n
1 , xn
1+ 5 2
n
−
1
5
2
n
.
F 100 100 ˜100 F 100
˜100 F 100
n
≥0
n := 98; f := := 1; s := 1 n := 98 f := 1 s := 1 i
1
n
l := := f f + + s; f := := s s;; s := := l l;; od od : :
l :=2 l :=3
f := 1 s := 2 f := 2 s := 3
l :=218922995834555169026 f := := 135301852344706746049 s := 218922995834555169026 l :=354224848179261915075
F 100 F 100 :=
1 (5)
(1 +
− − √ √ √ − −
(5) 2
F 100 F 100 :=
100
1
(5)
100
2
1 5
1 1 + 5 2 2
100
(F F 100) 100) 0.3542248538
1 2
× 1021
1 5 2
100
p3
f ((x) = f
√ x − cos x a1 = 0
[0,, 1] [0
b1 = 1
f ((a1 ) = f
f ((b1 ) = 0. f 0 .45970 p1 = f ((a1 ) < f < 0 0
1 1 (a1 + + b b1 ) = 2 2
f (( p1 ) < f < 0 0
f ((a2 ) = f
a2 = = p p 1 = 0.5
−0.17048 17048 < < 0 0,,
f (( p2 ) = 0.13434 f 13434 > > 0 0
−0.17048 17048 < < 0 0.. b2 = b 1 = 1
f ((b2 ) = 0.45970 f 45970 > > 0 0,,
p2 =
1 (a2 + + b b2 ) = 0.75 75.. 2
a3 = 0.5 b3 = = p p 3 = 0.75 p3 =
f ((x) = 3(x f 3(x + 1) x
f (( p1 ) = f
−1
1 (a3 + + b b3 ) = 0.625 625.. 2
− 21 (x − 1)
[ 2, 1.5]
−
f ((x) = 3(x f 3(x + 1) x
−
1 2
(x
p3
− 1) 1),,
f ((x) f x 1 1 1 2
x 2 2 x 2 1
f ( x)
22 2 2 2 20 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 2 22 2 2 2 2 2 2 2 2 2
2 22 2 2 2 2 2 2 2 2 2 2 2 2 22 2 2 2
22
a1 = 2 a2 = f (( p2 ) < f < 0 0
−
0 1 1 1 1 11 1 1 1 1
0 1 1 1 1 11 0 2 01 1 1 1 11 1 1 1 1
1
0
21
f ((a1 ) < f < 0 0 b1 = 1.5 2 f ((a2 ) < f < 0 0 b2 = a3 = = p p 2 = 1.125
−
0 1 1 1 1 1 1 11 1 1 1 1
−
−
2 2
1 4
1
2
f ((b1 ) > f > 0 0 f ((b2 ) > 0 f b3 = 0.25
−
3
x
− 41 f (( p1 ) > f > 0 0 p2 = −1.125 p3 = −0.6875
p1 =
y = = x x
y = tan x 10−5
x
x = tan x y = = x x y = tan x x = 4.5 y
10
y = x
5 10 x y =
tan x
210
g (x) = x
− tan x g (4 (4..4)
g
≈ 1.303 303 > > 0 0
g (4 (4..6)
≈ −4.260 260 < < 0 0,,
[4.4, 4.6] [4. p16 = 4.4934143
f ((x) = (x + 2)(x f 2)(x + 1)x 1)x(x
10−5
− 1)3(x − 2)
f
[ 3, 2.5]
− [−1.75 75,, 1.5] f ((x) = (x + 2)(x f 2)(x + 1)x 1)x(x
− 1)3(x − 2) 2),,
f ((x) f x 1 2
2 2 2 2 0 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
x 1 1
22 2 22 2 2 2
( x
2
0 1 1 1 1 1 11 1 11 1 1 11 1 1 11
22 2 22 2 2 2 2 2 2 2
x 3 1)
x 2 2
f ( x)
0 1 1 1 1 11 1 1 11 1 1 11
2 2 2 22 2 2 2 2 2 2 2 2 22 2
0 1 1 1 1 1 1 1 1 11
22 2 22 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 22 2 20 1 11
23
22
0 2 2 2 0 11 1 0 2 2 2 0 1 1 1 1 1 1
0
21
2
1
[ 3, 2.5] f a1 = 3 f ((b1 ) > f > 0 0 p1 = ( 3 + 2. 2 .5) 5)/ /2 = 0.25 f ( p1 ) < f ( < 0 0 f ((a2 ) < f < 0 0 b2 = = b b 1 = 2.5 f ((b1) > 0 f
−
−
0 1 1 1 1 11
−
−
3
f ((a1 ) < f < 0 0
x
b1 = 2.5 a2 = p 1 = 0.25
−
p2 = ( 0.25 + 2. 2.5) 5)/ /2 = 1.125 f (( p2 ) < f < 0 0 b3 = 2.5 f (b3 ) > 0 f ( [1..125 [1 125,, 2.5]
a3 = 1.125
−
b1 = 1.5 a2 = = a a 1 =
f ((a3 ) < f < 0 0
[ 1.75, 1.5] 1 a1 = 1.75 f ((a1 ) > 0 f f ((b1 ) < f < 0 0 p1 = ( 1.75 + 1 .5) 5)/ /2 = 0.125 f (( p1 ) < f < 0 0 1.75 f ((a1 ) > f > 0 0 b2 = p 1 = 0.125 f ((b2 ) < f < 0 0 [ 1.75 75,, 0.125] 1 1
−
−
−
−
−
− −
−
−
10−4
√ 3 √ 3 ≈ p
−
−4= 0
−
√ 3
f ((x) = x2 3 f [1,, 2] [1
x3 + x
−
14 =
1.7320
10−3 .
[1,, 4] [1
10−3
n
| p p − pn| < b 2−n a = 4 2−n 1 < 0 0..001;
3 n
p12 = 1.3787
× 103 < 2n.
≥ 12
n
1 p =
{ pn}
n
k=1
{ pn} pn
− pn
1 =
−
lim ( pn
k
n→∞
1/n
limn→∞ ( pn
− pn
1) =
−
− pn
1) =
−
0
0.
pn
n
| p pn − pn 1| −
L = 10 r = 1
h
12..4 = 10 0.5π 12
− arcsin h − h 1 − h21/2
0.01
V = 12 1 2.4
3
[0,, 1] [0 f ((h) = 12. f 12 .4 h
− 10
≈ p13 = 0.1617
r
2 1/2
0.5π − arcsin h − h 1 − h − h ≈ 1 − 0.1617 = 0.0.8383
211/3 p0 = 1 20 p 20 pn−1 + 21/p 21 /p2n−1 21 p 3n−1 21 pn = p n−1 3 p2n−1
pn =
−
−
pn = = p p n−1
−
p 4n−1 21 pn−1 p2n−1 21
−
1/2
pn =
−
21 pn−1
pn = g ′ (x) =
20 21
20 p 20 pn−1 + 21/p 21 /p2n−1 , 21
− x23
g (x) =
g ′ 211/3 =
20 21
− 212 = 67 ≈ 0.857
3
pn = = p p n−1
− p n3 p12− 21 , −
g (x) = x
n−1
g ′ (x) =
2 3
− x73
g ′ 211/3 =
2 3
2
′
g (x) = =
−
p 4n−1 21 pn−1 , p2n−1 21
2
3
3
2
(x2 21) 2x5 + x4 + 84x 84x3 63 63x x2 (x2
−
−
4 21x x x3 − 21 21x x − x4 + 21x 21x x3 − x4 − xx2−−21 = = 21 x2 − 21 x2 − 21 4
x − 21 3x − 4x − x − x 2x −
3 − x 3−x221 = x − 13 x + x72 = 23 x + x72
− 13 = 13 = 0.0 .333
pn = = p pn−1
g (x) = x
20x 20 x + 21/x 21 /x2 20 1 = x + 2 , 21 21 x
− −
− 21)2
.
=
3x4
− 63 63x x2 − 4x5 + 84x 84x3 − 2x4 + 2x 2x5 2 (x2 − 21)
g ′ 211/3
≈ 5.706 706 > > 1 1 21
1/2
pn =
1/2
,
pn−1
√ 21 − g (x) =
g ′ 211/3 =
′
2x3/2
g (x) =
21
=
x
√ 21 x1/2
,
− 12 √ 3
10−4
f ((x) = x2 f
−3 0 = x2
x
−3
x =
3 , x
g (x) = 0. 0 .5 x +
√ 3 ≈ p = 1.73205
p0 = 1.0
4
3 x
g 2
3x
g (x) = e
x
−e
10−5
=0
1 x 3
[0,, 1] [0
p0 = 1
p12 = 0.910015
g (x) = ln ln 3x2
[3,, 4] [3
p0 = 4
p16 = 3.733090 10−4
x
x = tan x
[4,, 5] [4 g (x) = tan x
p0 = 4
p1 = g g(( p0 )
[4,, 5] [4
≈ 1.158
x = tan x
1 1 = x tan x
+ g (x) = x x +
1 tan x
− x1
p0 = 4 p1 p3
≈ 4.61369 61369,,
p2 = 4.49596 49596,,
p3 = 4.49341
p4 ′
x
≤ k < 1
g ′ (x)
|g (x)| ≤ k
∈ [a, b]
g ′(x)
p4 = 4.49341 49341..
′
x
∈ [a, b]
|g (x)| ≤ k
≤ k < 1
p [a, b]
p = q p
p = p =
1 2
ξ
q
(a, b)
′
− q = g = g(( p p)) − g (q ) = g (ξ )( )( p − q ) ≤ k ( p − q ) < p − q, g (x) = 1 p0 = 0.7
−1 + √ 5 .
− x2
[0,, 1] [0
1 1 xn−1 + 2 xn−1
xn = 0 x0 > > 0 0 < x0 <
√ 2
x1 >
g
√ 2 x0 > 0
g (x) = x/ x/22 + 1/x. 1 /x. x =0 g ′ (x) > > 0 0 g 2 = 2.
√
1/x2 < 1 1//2
x1
√ 2 < ξ < x . 0
− x1
√
2 = g g((x0 )
x0 >
−g
− √ 2 > > 0 0
g ′ (x) = 1/2
√
√ 2.
√ 2 = = g g (ξ ) x − √ 2 , ′
0
x1 >
√ 2.
x0 1 x0 1 x0 + x1 = + < + = 2 x0 2 2 2
√
√ 2
,
√ 2 < x < x . 1
0
√
2 < xm+1 < xm < .. . < x0 .
{xm}
p = limm→∞ xm . p = lim
m→∞
p = p =
±√ 2.
x
m−1
2
+
1 xm−1
p 1 p = p = + , 2 p
xm >
√ 2
=
p 1 + . 2 p p2 = 2,
m limm→∞ xm =
√ 2.
√ 2 √ 2 √ 0 < x0 − 2 = x20 − 2x0 2 + 2, 2, 0 < x0 <
√
2x0 2 < x20 + 2
√
2 <
x0 1 + = x1 . 2 x0
− 1/x2.
x>
√ 2
√ x0 > 2 √ x = 2 0
0 < x0 <
limm √ xm = 2 √ 2 < x m
→∞
√ 2
limm→∞ xm =
1
√
0 < x0 <
xm
x0 > 0 √ = 2
2 < xm+1 < xm < .. . < x1
lim xm =
m→∞
√ 2 g
p
|g ( p p))| > 1
g
′
pn = p
′
0 < x
| − p| < δ
′
′
′
′
′
′
|g (x) − g ( p p))| ≥ |g ( p p))| − |g (x)|,
0 < x
| − p| < δ |g (x)| ≥ |g ( p p))| − |g (x) − g ( p p))| > |g ( p p))| − (|g ( p p))| − 1) = 1.1. 0 < | p p − p0 | < δ | p p1 − p| = |g( p0) − g( p p))| = |g (ξ )|| p p0 − p|, p0 p. 0 < | p p − ξ | < δ | p p1 − p| = |g (ξ )|| p p0 − p| > | p p0 − p|. ′
′
′
′
′
′
′
′
p
p
p
f ((x) = x f = x 2
−6
p
p0 = 1
f ((x) = x2 f
p2
−6
f ′ (x) = 2x
pn = = p p n−1
2
− f f f ((( p pnn 11)) = pn 1 − p n2 p1n −1 6 . −
−
−
′
−
(a, b) p0
|g ( p p))| > 1 ǫ = |g ( p p))| − 1 |g (x) − g ( p p))| < ε = |g ( p p))| − 1, ′
ξ
2.
g′
′
p
δ > 0
p0
√
n g′
x
∈ C [a, b]
√ 2
−
p0 = 1 p1 = = p p0
2 − p 02 p−0 6 = 1 − 1 −2 6 = 1 + 2.5 = 3.5
p2 = = p p 1
2 .52 − 6 − p 12 p−1 6 = 3.5 − 3 3.2(3 = 2.60714 60714.. 2(3..5)
f ((x) = x f = x 2
−6
p0 = 3
p1 = 2
pn = p n−1
p3
f (( pn 1 )( pn 1 − pn 2 ) − f . f (( pn 1 ) − f f f (( pn 2 ) −
−
−
−
p0 = 3
p1 = 2 p2 = p1
f (( p0 ) = 9 f
−
−6 = 3
f (( p1 ) = 4 f
− 6 = −2
f (( p1 )( p1 − p0 ) ( −2)(2 − 3) 2 − f = 2− =2− −2 − 3 −5 = 2.4 f (( p1 ) − f f f (( p0 )
f (( p2 ) = 2.42 f
− 6 = −0.24 −0.096 = 2.45454 f (( p2 )( p2 − p1 ) f ( −0.24)(2 24)(2..4 − 2) p3 = = p p 2 − = 2. 2 .4 − = 2. 4 − 45454.. f (( p2 ) − f f f (( p1 ) (−0.24 − (−2) 1.76
p0 = 3 p1 = 2 f ( p0 ) = 3 f ( f ( p1 ) = 2 f ( p2 = 2.4 f ( p2 ) = 0.24 f ( f ( p1 ) < f ( < 0 0 f ( p2 ) < f ( < 0 0 p1 p1 p0 p1 = 3 f (( p2 ) = 0.24 f p3
−
−
f (( p1 ) = 3 f
p2 = 2.4
−
p3 = = p p 2
f (( p2 )( p2 − p1 ) ( −0.24)(2 24)(2..4 − 3) 0.144 − f = 2. 2 .4 − = 2.4 − 44444.. −0.24 − 3 −3.24 = 2.44444 f (( p2 ) − f f f (( p1 )
√ 6 ≈ 2.44949 x 10
4
−
f ((x) = x f
− cos x
− cos x = 0
[0,, π/ [0 π/2] 2]
f ′ (x) = 1 + sin x
pn = p n−1
cos p cos pn 1 − p n1 +1 −sin p . sin pn 1 −
−
−
p0 = 0
p1 = 1 p2 = 0.75036 p3 = 0.73911 x
10
4
−
pn = = p p n−1
p4 = 0.73909 73909..
− cos x = 0
[0,, π/ [0 π/2] 2]
− cos cos p pn 1 )( pn 1 − pn 2 ) − ( pn( p1n−1cos cos p pn 1 ) − ( pn 2 − cos cos p pn −
−
−
−
−
−
−
2)
−
.
p0
p1 n
pn
0 1 2 3 4 5 6
0 1.5707963 0.6110155 0.7232695 0.7395671 0.7390834 0.7390851
x 10
p0
− cos x = 0
4
−
[0,, π/ [0 π/2] 2]
p1 n
pn
0 1 2 3 4 5 6 7
0 1.5707963 0.6110155 0.7232695 0.7372659 0.7388778 0.7390615 0.7390825
10−4 y = = x x
2
x
(1,, 0) (1 2
(1,, 0) (1
x, x y = = x x d(x) = (x − 1) + (x (x − 0) = x + x − 2x + 1. 1. 2
2
2
4
2
2
d
f ((x) = [d(x)]2 = x4 + x2 f x x
0 = f ′ (x) = 4x3 + 2x 2x
y = x (0..589755 (0 589755,, 0.347811) 347811)..
2
− 2.
− 2x + 1,1,
d(x)
p0 = 1 (1,, 0) (1
f ((x) f p5 = 0.589755 589755..
0=
1 1 2 + x 2 4 p0 =
p19 = 1.895489
p0 = 10 10π π
pn =
π 2
−x
10 f ((x) f
f := := x x
−>
f p := := x x
(x)
−x
− > ( (DD)()(f f )()(xx)
p15 = 1.895488
f (( pn−1 ) pn−2 f f (( pn−1 ) f
pn−1
f ((x) = cos x f
− x sin x − 12 co coss 2x.
− f f (( pn − f f (( pn
2 ) pn−1
−
2)
p0 = 5π
.
−
pn−2
p0 = = π/ π/44
100
−
f ′ (x) f := := x x
f p := := x x
→ cos( cos(x x) − x → − sin( sin(x x) − 1
Digits := Digits := 100; p 100; p00 := P := P i/ i/44 := 100 1 p00 := π p 4 n p11 := p
1
7
( p p00
− f f (( p p0) 0)/f /f p( p p0)) 0)) err := err := abs abs(( p p11 − p p0) 0)
p00 := p p p11 od
p7 = . .73908513321516064165531208767387340401341175890075746496 73908513321516064165531208767387340401341175890075746496 56806357732846548835475945993761069317665319,, 56806357732846548835475945993761069317665319 10−100
f ((x) = ln x2 + 1 f
−e
0.4x
cos πx
10−6 10−6 n 10−6 e0.4x x
2
ln x + 1
f (x) f ( f ((x) f cos πx πx = = 0
x = n/ n/22
p0 =
x e0.4x
n
−0.5
p3 =
p0 = 0.5 p3 = 0.4506567 p0 = 1.5 p5 = 2.2383198 p0 = 3.5 p4 = 3.7090412 7090412..
−0.4341431 4341431..
p3 = 1.7447381 p0 = 2.5
n
n
p0 = 24 24..5 p2 = 24 24..4998870
− 0. 5
y
y
12
600
30000
8
400
20000
200
10000
y
4
12
4
24 22
6 4
2
x
16
8
= A f (i) = A f ( n
x
210000 220000
i A = $750, $750, 000 A =
−
P [(1 + i + i))n i
P = P = $1500
P [(1 + i + i))n i
− 1] = 750000 −
20
x
2200
24
16
− 1] .
1500 (1 + i/ + i/12) 12)(12)(20) (i/ i/12) 12)
i f
−1 .
t/3 3 c(t) = Ate = Ate −t/
t
A
′
0 = c (t) = A 1
−
t 3
t/3 3 e−t/ .
c(3) = 3Ae 3 Ae−1
t = 3 A = 31 e t 0.25 = c = c((t) =
1 3
t/3 3 e te−t/ .
t
t = 11 11..083
t 0.25 = c = c((t) =
f ((x) = 33x+1 f
1 3
t/3 t/3
−
e te
+ 0. 0.75
1 3
e (t
(t−11 11..083) 083)/ /3
−
− 11 11..083) 083)ee
− 7 · 52 x f
f ((x) f
f f
10−16
f ((x) = 0 f f := := x x
− > 33x+1 − 7 · 52x
f := := x x
→ 3(3(3xx+1) − 7 52x
(f f ((x) = 0, x) ln(3/ ln(3/7) − ln(27 ln(27/ /25) (f f ((x) = 0, 0 , x)
(3x x+1) (3(3
− 7 5(2(2xx) = 0,0 , x) x
f ((x) f
.
( f f ((x) , x = 9.5.. ..11 11..5)
{
}
15
y
3 x 10
15
2 x 10
15
1 x 10
10.5 12 x
11.5
11
f ′ (x) f p := := x x
− > ( (DD)()(f f )()(xx) f p := := x x
→ 3 3(3(3xx+1) ln(3) − 14 5(2(2xx) ln(5)
:= 18; p 18; p00 := 11 := 18 p00 := 11 p i p11 := p
1
5
( p p00
− f f (( p p0) 0)/f /f p( p p0)) 0)) err := err := abs abs(( p p11 − p p0) 0)
p00 := p p p11 od
i
pi
| p pi − pi 1|
1 2 3 4 5
11.009 11. 009738 7380401 0401552 552503 503 11..009 11 009438 4389359 9359662 662827 827 11..00 11 0094 9438 38644 64426 2684 8448 4888 11..00943864 11 64442681716 11..0094386442681716 11
0.0097380401552503 0.0002991041889676 0.291697 2916978339 8339 10−6 0.2772 10−12 0
−
33x+1 = 7 52x
·
(3x (3 x + 1) 1) ln3 = ln 7 + 2x ln 5. 3x ln 3
− 2x ln5 = lnln 7 − ln 3,
x =
x(3ln3
ln 7/3 ln 7/3 = = ln27 ln25 ln27/25 ln27/
−
− 2 ln5) = lnln 73 ,
ln 3/7 − ln27 . ln27/ /25
10−5 0
x2
≤x≤1 f ((x) = x2 f
pn = = p p n−1
x
−
− 2xe
+ e−2x
f ′ (x) = 2x
− 2e
2
x
− 2xe
+ 2xe 2xe−x
−
−
′
−
2x
−
,
+ e−2 pn−1 . − pn−1 2e−2 pn−1 −1 e
− n−1
−
− 2e
− n−1
−
−
+ e−2x = 0
p
e − f f f ((( p pnn 11)) = pn 1 − 2 pn 1 −pn2e1 − p 2 pn+12 p 2 pn −
x
−
−
p0 = 0.5 p1 = 0.5
− (0 (0..01134878) 01134878)/ /(−0.3422895) = 0. 0.5331555 5331555.. 10−5
p13 = 0.567135
f ((x) =x f = x2
− 2xe x + e 2x, f (x) =2 = 2x − 2e x + 2xe 2xe x − 2e −
′
−
f ′′ (x) =2 + 4e 4e−x
pn = = p p n−1 p0 = 0.5
−
−
−
+ 4e 4e−2x ,
−
−0.342289542 f ( p0 ) = 5.291109744 ′′
(0..01134878)(−0.342289542) (0 − (−0.342289542) = 0.5680137 5680137.. 2 − (0 (0..011348781)(5 011348781)(5..291109744) 10−5
p3 = 0.567143 pn = 1/n p < 5 10−2 1 lim =0 n→∞ n
| p pn − |
lim
n→∞
p = 0
×
| p pn+1 − p| = lim 1/(n + 1) = lilimm n = 1,1 , n | p pn − p| n 1/n n + 1 | p pn − p| < 5 × 10 2 1/n < 0 < 0..05 →∞
→∞
−
n > 20
,
f ( pn−1 )f ′ ( pn−1 ) f ( . [f ′ ( pn−1 )]2 f f (( pn−1 )f ′′ ( pn−1 )
f (( p0 ) = 0 .011348781 f ′ ( p0 ) = f
p1 = 0.5
x
−
− 2xe
2x
−
n
pn = 10−2
k
pn = 10−n
lim
n→∞
k > 1
n+1
n+1
10−2
n+1
10−2 10−2 = l im = lim li m = 1, n n+1 2 →∞ n→∞ 10−2·2 n→∞ 10−2 (10−2n )
| p pn+1 − 0| = lim | p pn − 0|2 n
k > 1 lim
n→∞
| p pn+1 − 0| = lim | p pn − 0|2 n
→∞
10−(n+1)
10 nk
−
k
k
k k 10−(n+1) = lim li m = lim 102n −(n+1) k 2 − 2 n n→∞ 10 n→∞
k
pn = 10−n
k > 1
g (x) = x g ′ ( p p)) = 0
p
f
mf ((x) − mf f (x) ′
m
f
m f ((x) = (x f
p
q
− p p))mq (x),
f ′ (x) = m m((x
− p p))m
lim q (x) = 0. 0.
x→ p
1
−
q (x) + (x (x
− p p))mq (x), ′
m
g (x) = x
mf ((x) m(x − p p)) q (x) − mf = x − , m 1 f (x) m(x − p p)) q (x) + (x ( x − p p))m q (x) ′
−
g (x) = x
′
p))q (x) p − mq (mx)(+x −(x . ( x − p p))q (x) ′
x = = p p
g ′ ( p p)) = 1
mq (( p p)[ )[mq mq ( p p)] )] − mq = 0. 0. [mq ( p p)] )]2
f ′′′ p f
m 0 = f f (( p p)) = f ′ ( p p)) =
f
· · · = f (m
1)
−
( p p)),
m f (m) ( p p)) = 0. 0.
p
f
m f ((x) = (x f
p,
− p p))mq (x),
f ′(x) = m m((x
f
x = p,
lim q (x) = 0.
− p p))m
1
−
x→ p
q (x) + (x (x
− p p))mq (x) ′
f ′ ( p p)) = 0.
− 1)( 1)(x x − p p))m
f ′′ (x) = m m((m
2
−
q (x) + 2m 2 m(x p p))m−1 q ′ (x) + (x (x
−
− p p))mq (x) ′′
f ′′ ( p p)) = 0. k
≤ m k
(k)
f
j
k d (x − p p)) (x) = j dx k
m
j
q (k−j ) (x)
j =0 k
=
j =0
0
j
m(m
− 1)· · ·(m − j j + + 1)(x 1)(x − p p))m
j (k−j )
−
q
(x).
f (k) ( p p)) = 0
≤ k ≤ m − 1
f (m) ( p p)) = m m!! lim q (x) = 0.
x→ p
(m
f (( p f p)) = f ′ ( p p)) = . . . = = f f (m−1) ( p p)) = 0 f p :
− 1)
(m−1)
f ((x) =f f = f (( p p)) + f + f ′ ( p p)( )(x x =(x =( x ξ (x)
p.
− p p))m q (x)
−
f (m) q (x) =
f ((x) = ( x f
−
ξ (x)) , m!
x
( p p)( )(x x p p))m−1 f (m) (ξ (x))( ))(x x + (m 1)! m!
f − p p)) + . + . . . +
(m) (
− p p))m f
f (m) ( p p)) = 0.
f (m) (ξ (x)) . m!
f (m) ( p p)) lim q (x) = = 0. x→ p m!
p
m
α en = = p pn
− p
α = 1 +
|en+1| = λ > 0 0,, n |en|α n |en+1 | ≈ λ|en |α . |en| ≈ λ|en 1|α |en 1| ≈ λ
√ 5 /2
lim
→∞
1/α
−
−
−
|en|1/α.
− p p))m
C
n
| p pn+1 − p| ≈ C | p pn − p| | p pn 1 − p| λ|en |α ≈ C |en |λ 1/α |en |1/α , |en| −
|en|α ≈ Cλ
−
1/α−1
−
√
1+ 5 α = . 2
α = 1 + 1/α 1/α
f ((x) = e 6x + 3(ln f 3(ln 2)2 e2x
− ln 8e4x − (ln2)3 = 0.0 . ∆2
| p pn+1 − pn| < 0 0..0002 p0 = 0
2
∆
pˆ6 =
g (x) = cos(x cos(x
p16 = 0.183387 183387..
−
(0)
− 1) (0)
{ pˆn} −0.182888
(1)
p0 = 2
g (x) = cos(x cos(x
1+1/α /α |en|1+1 .
p0 (0)
− 1)
p0 = 2
= cos(2 − 1) = cos cos 1 = 0. 0.5403023 = cos(0. = g p cos(0.5403023 − 1) = 0. 0.8961867 8961867.. (0)
p1 = g p0 (0)
p2
(0) 1
p
(0) 1
2
− p(0) 0 (1) (0) p0 = p0 − (0) (0) (0) (0) p2 − 2 p1 − 2 p1 + p0 (0..5403023 − 2)2 (0 =2− = 2 − 1.173573 = 0. 0.826427 826427.. 0.8961867 − 2(0 2(0..540 5403023 3023)) + 2 (0)
g (x)
(0)
p0 = 1
(0)
p2 = 3
p1
(0)
(0)
p1 = p 0
−
p
(0)
p2
(0) 1
−
(0) p0
2
(0) − 2 p(0) 1 + p0
.
(1)
p0 = 0.75
(0)
p0
(0)
(1)
p2
p0
0.75 = 1
1 (0)
p1 = 1.5
−
− 12 p(0) 1 = (0)
p
3
2
−1
(0) 1
− 2 p(0) 1 +1 2
p − 2 p (0) 1
,
0.25 =
(0) 1
p 4
(0) 1
2
−1
− 2 p(0) 1 2
− 1.5 p (0)
+ 1, 1,
0 = p1
.
(0) 1 ,
p1 = 0 10−5
x = 0.5(sin x + cos x)
g (x) = 0. 0 .5(sin x + cos x). g (x) = 0.5(sin x + cos x) (0)
(0)
p0 = 0, 0 , p1 = g g(0) (0) = 0. 0 . 5, (0)
p2 = g g(0 (0..5) = 0. 0.5(s 5(sin in 0.5 + cos cos 0.5) = 0. 0 .678504051 678504051,, (1)
(0)
p0 = p0 (1)
−
p
(0)
p2
(0) 1
−
(0) p0
2
(0) − 2 p(0) 1 + p0
= 0. 0 .777614774 777614774,,
= 0.707085363 707085363,, = g p = 0.704939584 704939584,, p − p
= 0. 0 .704872252 704872252,,
= g p = 0.704815431 704815431,, = g p = 0.704812197 704812197,, p − p
= 0.704812002 704812002,,
(1)
p1 = g p0 (1)
p2
(2)
(1) 1
(1)
p0 = p0 (2) p1 (2)
p2
(3)
−
(1) 1
(2) 0
2
(1) − 2 p(1) 1 + p0
(2) 1
(2) 1
(2)
p0 = p0 = (3)
(1)
p2
(1) 0
(2)
p2
(2) 0
2
(2) − 2 p(2) 1 + p0
= 0.704812002 704812002,, = 0.704812197 = g p 704812197.. (3)
p1 = g p0 (3)
p2 (3)
p2
(3)
p1
(3) 1
(3)
(3)
10−5
p0 10−5
{ pn} pn = α > 1
{ pn}
p2 = 0. 0 .704812197
p
α > 1
p
1 nn
α p
α > 1
λ = lim
n→∞
| p pn+1 − p| . | p pn − p|α
λ
p lim p
− p = lim | p pn+1 − p| · | p pn − p|α n | p pn − p|α n − p { pn}
n+1
n→∞
1
−
→∞
pn =
= λ 0 = 0
·
pn+1 p = 0. n→∞ pn p lim
− −
p
1 nn
1/(n + 1)(n+1) nn = lim li m n→∞ n→∞ (n + 1)(n+1) 1/nn n n 1 = lim n→∞ n + 1 n + 1 1 1 1 = lim = 0 = 0. n n→∞ (1 + 1/n 1/n)) n + 1 e lim
·
α > 1 1/(n + 1)(n+1) nαn = lim li m α n→∞ n→∞ (n + 1) (n+1) (1/n (1 /nn ) lim
n
(α−1) 1)n n
n n = lim n + 11 n + 1 n→∞
= lim
n→∞
(1 + 1/n 1/n))n
α P n (x)
1)n n n(α−1) 1 lim = n→∞ n + 1 e
· ∞ = ∞.
α > 1
f ((x) = e x f
n x
x0 = 0.
pn = P n (x) ∆2
x = 1
pˆ0 , p ˆ1 , . . . , ˆ , p ˆ8
∆2 n
pn = = P P n (x) =
1x , k
k=0
ξ
0
x
k!
− eξ x − − = P ( ) = pn p p = P n x e xn+1 , (n + 1)! 0 pn − p = n≥0 e n+2 pn+1 − p e(ξ ξ) x (n+2)! x = = , e n+1 pn − p n + 2 x (n+1)! −
ξ1
1−
− ξ
ξ 1
0
1 e(ξ1 −ξ) x = 0 < < 1 1.. n→∞ n + 2
λ = li lim m
5
×
n
0
1
2
3
4
5
6
pn pˆn
1 3
2 2.75
2.5 2.72
2.6 2.71875
2.7083 2.7183
2.716 2.71 7182 8287 8700
2.71805 2.7182823
n
7
8
9
10
pn pˆn
2.71 7182 8253 5399 2.71 7182 8281 8188
2.71 7182 8278 7877 2.7182818
2.71 7182 8281 8155
2.7182818
∆2 10−7
pˆ6 p10
10−5 P ((x) = x4 P
− 2x3 − 12 12x x2 + 16x 16x − 40 40.. p0 = 1
p7 =
−3.548233
p0 = 4
p5 = 4.381113 381113..
(x + 3. 3 .548233)( 548233)(x x
P ((x) P
− 4.381113) = x = x2 − 0.832880 832880x x − 15 15..54521 54521,,
16712xx + 2.2.57315 . ≈ x2 − 0.832880 832880x x − 15 15..54521 x2 − 1.16712
0.58356
± 1.49419 49419i. i. P ((x) = x4 P
f (0) f (0) = [0,, 2] [0 f (4) f (4) =
−40
− 2x3 − 12 12x x2 + 16x 16x − 40 40..
−40 f f (1) (1) = −37
f (5) f (5) = 115. 115.
f (2) f (2) =
−56
P ((x) P
f (( 3) = f
−
−61
−3
f (( 4) = 88. f 88.
−
p0
p1
p2
0 2 2
1 3 3
2 4 4
− − −
p7 = 0.583560 1.494188 494188ii p6 = 4.381113 p5 = 3.548233
−
0.583 583560 560 + 1. 1.494188 494188ii
−
f ((x) = x4 f
− 2x3 − 5x2 + 12x 12x − 5, f
limx→∞ f f ((x) = (
−∞, 0) (0 (0,, 1) (1 (1,, 2)
∞
(2,, (2
f f (0) f (0) < < 0 0 f f (1) (1) > > 0 0 limx→−∞ f f ((x) =
x = 1
∞
∞)
f ′(x) = 4x 4 x3
0.5798 1.521 2.332
− 6x2 − 10 10x x + 12. 12.
f ′(x) 2
−1.5 0 = f ′′ (x) = 12x 12x2
x = 0.5
± √ 39 39//6
− 12 12x x − 10
f 60
y
40 20
22 23
21 220
1
2
3
x
10−4 600x 600 x4
f (2) f (2) < < 0
− 550 550x x3 + 200x 200x2 − 20 20x x − 1 = 0, 0,
0.1
≤x≤1
−2.432
−4
p0 = 0.1
p1 = 1
p0 = 0.55
p14 = 0.23233
p6 = 0.23235
p0 = 0.1
p1 = 1
p8 = 0.23235
p0 = 0.1
p1 = 1
p0 = 0 p1 = 0.25
p88 = 0.23025
p2 = 1
p6 = 0.23235
3
0.25 0.25
V = V = 1000 = πr = πr 2 h, h = 1000/ 1000/ πr 2
π(r + 0. 0 .25)2
(2πr (2 πr + + 0. 0 .25) 25)h. h.
M ((r) M
M ((r) = 2π(r + 0. M 0 .25)2 + (2πr (2πr + + 0. 0 .25) 25)h h = 2π(r + 0. 0 .25)2 + 2000/r 2000/r + + 250/πr 250/πr 2 .
− 2000 2000/r /r2 − 500 500/ /(πr 3 ).
M ′ (r) = 4π 4 π(r + 0. 0 .25) M ′ (r) = 0
r
r
≈ 5.363858 363858..
M ((r) M
M (5 M (5..363858)
2
1 + 22
1 1 60
+7
60
≈ 573 573..649 3
+ 42
1 60
r 2
.
4
+ 33
5
1 1 60
+4
60
6
+ 40
1 60
x3 + 2x 2x2 + 10x 10x = 20 20..
1.3688081078532 3. 2
1.36880810782137 −11 .
× 10
10
16
−
