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Problems
Contests
National and Regional Contests
France Contests
French Mathematical Olympiad
1986 French Mathematical Olympiad
Problem 5
Problem 5
Part of
1986 French Mathematical Olympiad
Problems
(1)
iterating function f(x)=sqrt4(1-x)
Source: France 1986 P5
5/19/2021
The functions
f
,
g
:
[
0
,
1
]
→
R
f,g:[0,1]\to\mathbb R
f
,
g
:
[
0
,
1
]
→
R
are given with the formulas
f
(
x
)
=
1
−
x
4
,
g
(
x
)
=
f
(
f
(
x
)
)
,
f(x)=\sqrt[4]{1-x},\enspace g(x)=f(f(x)),
f
(
x
)
=
4
1
−
x
,
g
(
x
)
=
f
(
f
(
x
))
,
and
c
c
c
denotes any solution of
x
=
f
(
x
)
x=f(x)
x
=
f
(
x
)
.(a) i. Analyze the function
f
(
x
)
f(x)
f
(
x
)
and draw its graph. Prove that the equation
f
(
x
)
=
x
f(x)=x
f
(
x
)
=
x
has the unique root
c
c
c
satisfying
c
∈
[
0.72
,
0.73
]
c\in[0.72,0.73]
c
∈
[
0.72
,
0.73
]
. ii. Analyze the function
f
′
(
x
)
f'(x)
f
′
(
x
)
. Let
M
1
M_1
M
1
and
M
2
M_2
M
2
be the points of the graph of
f
(
x
)
f(x)
f
(
x
)
with different
x
x
x
coordinates. What is the position of the arc
M
1
M
2
M_1M_2
M
1
M
2
of the graph with respect to the segment
M
1
M
2
M_1M_2
M
1
M
2
? iii. Analyze the function
g
(
x
)
g(x)
g
(
x
)
and draw its graph. What is the position of that graph with respect to the line
y
=
x
y=x
y
=
x
? Find the tangents to the graph at points with
x
x
x
coordinates
0
0
0
and
1
1
1
. iv. Prove that every sequence
{
a
n
}
\{a_n\}
{
a
n
}
with the conditions
a
1
∈
(
0
,
1
)
a_1\in(0,1)
a
1
∈
(
0
,
1
)
and
a
n
+
1
=
f
(
a
n
)
a_{n+1}=f(a_n)
a
n
+
1
=
f
(
a
n
)
for
n
∈
N
n\in\mathbb N
n
∈
N
converges. [hide=Official Hint]Consider the sequences
{
a
2
n
−
1
}
,
{
a
2
n
}
(
n
∈
N
)
\{a_{2n-1}\},\{a_{2n}\}~(n\in\mathbb N)
{
a
2
n
−
1
}
,
{
a
2
n
}
(
n
∈
N
)
and the function
g
(
x
)
g(x)
g
(
x
)
associated with the graph. (b) On the graph of the function
f
(
x
)
f(x)
f
(
x
)
consider the points
M
M
M
and
M
′
M'
M
′
with
x
x
x
coordinates
x
x
x
and
f
(
x
)
f(x)
f
(
x
)
, where
x
≠
c
x\ne c
x
=
c
. i. Prove that the line
M
M
′
MM'
M
M
′
intersects with the line
y
=
x
y=x
y
=
x
at the point with
x
x
x
coordinate
h
(
x
)
=
x
−
(
f
(
x
)
−
x
)
2
g
(
x
)
+
x
−
2
f
(
x
)
.
h(x)=x-\frac{(f(x)-x)^2}{g(x)+x-2f(x)}.
h
(
x
)
=
x
−
g
(
x
)
+
x
−
2
f
(
x
)
(
f
(
x
)
−
x
)
2
.
ii. Prove that if
x
∈
(
0
,
c
)
x\in(0,c)
x
∈
(
0
,
c
)
then
h
(
x
)
∈
(
x
,
c
)
h(x)\in(x,c)
h
(
x
)
∈
(
x
,
c
)
. iii. Analyze whether the sequence
{
a
n
}
\{a_n\}
{
a
n
}
satisfying
a
1
∈
(
0
,
c
)
,
a
n
+
1
=
h
(
a
n
)
a_1\in(0,c),a_{n+1}=h(a_n)
a
1
∈
(
0
,
c
)
,
a
n
+
1
=
h
(
a
n
)
for
n
∈
N
n\in\mathbb N
n
∈
N
converges. Prove that the sequence
{
a
n
+
1
−
c
a
n
−
c
}
\{\tfrac{a_{n+1}-c}{a_n-c}\}
{
a
n
−
c
a
n
+
1
−
c
}
converges and find its limit. (c) Assume that the calculator approximates every number
b
∈
[
−
2
,
2
]
b\in[-2,2]
b
∈
[
−
2
,
2
]
by number
b
‾
\overline b
b
having
p
p
p
decimal digits after the decimal point. We are performing the following sequence of operations on that calculator:1) Set
a
=
0.72
a=0.72
a
=
0.72
; 2) Calculate
δ
(
a
)
=
f
(
a
)
‾
−
a
\delta(a)=\overline{f(a)}-a
δ
(
a
)
=
f
(
a
)
−
a
; 3) If
∣
δ
(
a
)
∣
>
0.5
⋅
1
0
−
p
|\delta(a)|>0.5\cdot10^{-p}
∣
δ
(
a
)
∣
>
0.5
⋅
1
0
−
p
, then calculate
h
(
a
)
‾
\overline{h(a)}
h
(
a
)
and go to the operation
2
)
2)
2
)
using
h
(
a
)
‾
\overline{h(a)}
h
(
a
)
instead of
a
a
a
; 4) If
∣
δ
(
a
)
∣
≤
0.5
⋅
1
0
−
p
|\delta(a)|\le0.5\cdot10^{-p}
∣
δ
(
a
)
∣
≤
0.5
⋅
1
0
−
p
, finish the calculation.Let
c
↔
\overleftrightarrow c
c
be the last of calculated values for
h
(
a
)
‾
\overline{h(a)}
h
(
a
)
. Assuming that for each
x
∈
[
0.72
,
0.73
]
x\in[0.72,0.73]
x
∈
[
0.72
,
0.73
]
we have
∣
f
(
x
)
‾
−
f
(
x
)
∣
<
ϵ
\left|\overline{f(x)}-f(x)\right|<\epsilon
f
(
x
)
−
f
(
x
)
<
ϵ
, determine
δ
(
c
↔
)
\delta(\overleftrightarrow c)
δ
(
c
)
, the accuracy (depending on
ϵ
\epsilon
ϵ
) of the approximation of
c
c
c
with
c
↔
\overleftrightarrow c
c
. (d) Assume that the sequence
{
a
n
}
\{a_n\}
{
a
n
}
satisfies
a
1
=
0.72
a_1=0.72
a
1
=
0.72
and
a
n
+
1
=
f
(
a
n
)
a_{n+1}=f(a_n)
a
n
+
1
=
f
(
a
n
)
for
n
∈
N
n\in\mathbb N
n
∈
N
. Find the smallest
n
0
∈
N
n_0\in\mathbb N
n
0
∈
N
, such that for every
n
≥
n
0
n\ge n_0
n
≥
n
0
we have
∣
a
n
−
c
∣
<
1
0
−
6
|a_n-c|<10^{-6}
∣
a
n
−
c
∣
<
1
0
−
6
.
function
algebra