迭代法在数列不等式中的应用

−
xn+1
=
xn2
−
xn
−
c
+
√ c
=
(
1
√ −c
−
xn)
(
√ c
−
xn)
,
a
1` a
s √
1 − c − xn > 0
√ xn < 1 − c.
a 1
xn 0
n1
√
√√
c − xn+1 (1 − c)( c − xn).
b
(b
√ c − xn
(1
−
√ c
)n
−1(√c
−
x1)
<
(1
−
√c)n−1
$ √
_1/ Sn=k+2 } _ .
`a}
S .
2
n=1
|xn+1
−
xn|
=
x2
−
x1
=
1 6,
.
S @ n 2
0 < xn−1 < 1,
1
+
xn−1
<
2,
xn
=
1
+
1 xn−1
>
1 2.
@ (1 + xn)(1 + xn−1) =
1
+
1
+
1 xn−1
(1 + xn−1) = 2 + xn−1
5 2,
@ |xn+1 − xn| =
yn > xn 1(n ∈ N∗ ) .
1 xn+1 y ( 2 ) n+1
xn yn
(n
∈
N∗
)
yn+1 − xn+1 xn+1
yn − xn xn ,
d yn+1 − xn+1 yn − xn
xn+1 xn
=
λn−1(n
∈
N∗
)
.
x1 x2
− −
y1 y2
+
x2 x3
− −
y2 y3
+
+
xn xn+1
E√ xn < c
n1
.
d s / xn+1 = xn − xn2 + c > xn, {xn}
p.
1 `a
p U {xn}
c
/ 0,
1 4
.
' h · 2 . ( 2 0 1 0
· ·22
p - {an}
a1
=
1,
an+1
=
c
−
1 an
1
Bp yl c
=
5 2
,
bn
=
1 an −
2
,
{bn}
B 2
: p S U ( 2)
h(x) = x2 + x, x > 0
kn+1 = 3xn2 +1 + 2xn+1.
/ xn2 + xn,
xn2 + xn = 3xn2 +1 + 2xn+1
@ s xn 2xn+1
xn+1 xn
4xn2 +1 + 2xn+1 = (2xn+1)2 + 2xn+1,
d 1
2,
I an < an+1 < 3
< c
.
1
an+1
−
2
=
5 2
−
1 an
−
2
=
an − 2 2an
,
s 1
an+1 −
2
=
2an an − 2
=
4 an −
2
+
2,
bn+1 = 4bn + 2.
@ bn+1
+
2 3
=
4
bn
+
2 3
E a1 = 1,
b1
=
a1
1 −
2
=
−1,
@
bn
+
2 3
/
y
:
−
1 3
1
1 +
xn
−
1
+
1 xn
−1
=
(1
|xn − xn−1| + xn)(1 + xn−1)
2 5
|xn
−
xn−1|
2 5
2
|xn−1 − xn−2|
2 5
n−1
|x2
−
x1|
=
1 6
2 5
n−1
.
h· 5 . ( 2 0 0 7
·
·22
p- √ {an} , a1 = 2, an+1 = 2 − 1 (an + 2), n = 1,
(i) (ii)
2 < bn a4n−3, n = 1, 2, 3, .
) % · 6 . ( 2 0 0 6
· · 21
p v {xn} {yn} x1 = x2 = 1, y1 = y2 = 2
xn+1
: ^ xn
=
λ
xn xn−1
,
yn+1 yn
λ
yn yn−1
(
λ
I p B p ( 1 ) x1 x3 x5
sL
B S .
n ∈ N∗
y x1 =
( 1 ) xn2 + xn = 3xn2 +1 + 2xn+1
( 2)
1 n−1 2
xn
1 n−2
2
.
: ( 1 )
f ′(x) = 3x2 + 2x,
@
( y = f (x) (xn+1, f (xn+1))
: $ (0, 0) (xn, f (xn))
@ xn2 + xn = 3xn2 +1 + 2xn+1.
1 4
×
1
n−1
+
1
n
+
+
1 2n−2
17
17
17
=
1 4
·
1 17n−1
1
1 −
− 1
1 17n
<
1 64
·
1 17n−2
(n
2).
17
d |b2n
−
bn|
<
1 64
·
1 17n−2
(n
∈
N∗).
U· 4 . ( 2 0 0 9
·
· 22
p { xn}
x1
=
1 2
,
xn+1
=
1
1 + xn
,
(n
2, 3, .
B yl ( 1 ) {an}
p - ( 2)
√
{bn}
b1
=
2,
bn+1
=
3bn 2bn
+ +
4 3
,
n
=
1,
2,
3,
,
2 < bn a4n−3, n = 1, 2, 3, .
1 1
√ an+1 = 2 − 1 (an + 2)
√
√
√
√
= 2 − 1 an − 2 + 2 − 1 2 + 2
@
a
−
an+1
=
1 ana
(a
−
an)
<
1 3
(a
−
an)
<
32(a
1 − an−1)
<
1 33
(a
−
an−2)
<
<
1 3n−1
(a
−
a2)
<
1 3n
(a
−
1).
@ Sn
>
log3
a a
− −
1 3
a − an+1 < a − 3,
@ an+1 > 3,
@ & c
>
10 3
B.
< Ec
/
2,
10 3
xn + c < xn,
' p / { xn}
1 x2 < x1 c < 0.
2
`G / p 1 { xn}
x1 = 0 x2 = c, x3 = −c2 + 2c.
1 x1 < x2 < x3 0 < c < 1.
1 xn < xn+1 = −xn2 + xn + c
n1
√ xn < c
`
0√ c
>) (p I - (
' -f o
· 1 . ( 2 0 1 2
1
· ·21
/ { xn}
p {xn} pE
x1 = 0, xn+1 = −xn2 + xn + c(n ∈ N∗ ) .
a /c<0
B 2
c
1
<